mm-2639 == > http://www.iit.edu/~krawczyk/phdsza/02conc.htm Well, I ain't arguing because it is defined that way, but that is one lousy piece of terminology, concurrent has a strong implication of simultaneous, not coincident in space, to your average English speaker. Would it have hurt so much to use coincident, way back when? Greg Locock === Subject: Re: Concurrent forces have no turning effect?? > .... > Well, I ain't arguing because it is defined that way, but that is one lousy > piece of terminology, concurrent has a strong implication of simultaneous, > not coincident in space, to your average English speaker. > Would it have hurt so much to use coincident, way back when? .... This is a slight digression from the OP's question; but concurrent (from Latin running together) is a long-established term in geometry. Three or more lines are concurrent iff they all pass through a common point. Ken Pledger. === Subject: Re: Concurrent forces have no turning effect?? >My textbook says that concurrent forces have no turning effect, but I >don't quite understand this. >What are the conditions under which this applies? >If I hold a pencil horizontally and then drop it, the only considerable >force acting on it is the force of gravity. If, while it's falling, I >then hit it at one end, obliquely downwards, it spins. And yet the >direction in which I applied the force and the direction of gravity are >concurrent. >What am I missing??? (probably something very simple, but I will be >grateful for any help). >Chris Gravity acts as though it is applied at the center of mass. If you apply your force at the center of mass to be concurrent, it won't twist. --Lynn === Subject: football pool probability I'm a teacher at a high school (not math!) and I participate in a football pool with about 20 others (teachers and admin.) I was thinking about what the probability of winning the pool any given week would be. I realize there's some skill involved in the picking of teams (at least for some participants), but to simplify my thinking I assume that everyone picks at random. Assuming there are 14 games per week, intuitively I know the probability of picking 7 winners and 7 losers would be highest, 8 winners or 6 winners next highest, 9 or 5 next highest, and so on. The point is, I can see how you could determine the probability of picking any given number of winners, but that's not what I'm trying to determine. I'm trying to determine the probability winning the pool (i.e. picking more winners than any other pooler). I asked one of the math teachers (who's also a close friend of mine), and he said it's impossible to determine. Somehow, it seems to me like it should be possible. Is my friend right? If not, how could it be done? mike === Subject: Re: football pool probability > I'm a teacher at a high school (not math!) and I participate in a > football pool with about 20 others (teachers and admin.) I was > thinking about what the probability of winning the pool any given week > would be. I realize there's some skill involved in the picking of > teams (at least for some participants), but to simplify my thinking I > assume that everyone picks at random. Assuming there are 14 games per > week, intuitively I know the probability of picking 7 winners and 7 > losers would be highest, 8 winners or 6 winners next highest, 9 or 5 > next highest, and so on. The point is, I can see how you could > determine the probability of picking any given number of winners, but > that's not what I'm trying to determine. I'm trying to determine the > probability winning the pool (i.e. picking more winners than any other > pooler). I asked one of the math teachers (who's also a close friend > of mine), and he said it's impossible to determine. Somehow, it seems > to me like it should be possible. Is my friend right? If not, how > could it be done? With your simplifying assumption that everyone picks at random, on the face of it, everyone has an equal chance of winning. Then the probability of winning is1/n where n is the number of players. === Subject: Re: football pool probability >With your simplifying assumption that everyone picks at random, >on the face of it, everyone has an equal chance of winning. Then >the probability of winning is1/n where n is the number of players. I don't think it is that simple. I thought he was asking about the probability of beating everyone else, not perhaps tying for the win. --Lynn === Subject: Re: football pool probability >>With your simplifying assumption that everyone picks at random, >>on the face of it, everyone has an equal chance of winning. Then >>the probability of winning is1/n where n is the number of players. >I don't think it is that simple. I thought he was asking about the >probability of beating everyone else, not perhaps tying for the win. >--Lynn For 14 teams and 20 bettors, only 60% of the time will there be a unique winner. To be more exact, the probability of a unique winner subject to the given assumptions is exactly equal to the fraction a/b where a is equal to: 9152584839936420856383319612998238465103919271913413271172123613207837056843 646385 and b is equal to: 1517710072051350836655829614705874145814380343009484000977978445108518972816 5691392 As a decimal, the probability is approximately: 0.6030522567 thus, approximately 60% quasi === Subject: Re: football pool probability >>With your simplifying assumption that everyone picks at random, >>on the face of it, everyone has an equal chance of winning. Then >>the probability of winning is1/n where n is the number of players. > I don't think it is that simple. I thought he was asking about the > probability of beating everyone else, not perhaps tying for the win. The result is close enough. It's not as if I didn't consider the possibility. === Subject: Re: football pool probability > With your simplifying assumption that everyone picks at random, > on the face of it, everyone has an equal chance of winning. Then > the probability of winning is1/n where n is the number of players. Yeah, that makes sense to me. OK, now what if we factored in skill so that certain poolers had a better than 50-50 chance of picking winners on some games. Then could you still calculate your chances of winning? === Subject: Re: football pool probability >> With your simplifying assumption that everyone picks at random, >> on the face of it, everyone has an equal chance of winning. Then >> the probability of winning is1/n where n is the number of players. > Yeah, that makes sense to me. OK, now what if we factored in skill so > that certain poolers had a better than 50-50 chance of picking winners > on some games. Then could you still calculate your chances of winning? Maybe you want to compute the expected payoff? If there are 20 players of equal skill, they each should break even in the long run (minus the fees). If skills are unequal, then it depends a lot on how much better the skilled players are/and or how much dead money there is. The problem is that players tend to get better as they become more involved or interested and worse otherwise. It's not a fixed measure. Also, there is a schooling effect (as in schools of fish). One-on-one, a good handicapper wins almost all the time. The more fish, the less the edge of a somewhat better player, since it is more likely that one of their cards can beat him. On the other hand a highly skilled player can win much more than his share against a lot of weak opponents. Anyway, a major part of the value of a pool at work, as you know, is that it builds a sense of community, heightens rooting interest, gives people something to talk about, etc. === Subject: Re: football pool probability > With your simplifying assumption that everyone picks at random, > on the face of it, everyone has an equal chance of winning. Then > the probability of winning is1/n where n is the number of players. >> Yeah, that makes sense to me. OK, now what if we factored in skill so >> that certain poolers had a better than 50-50 chance of picking winners >> on some games. Then could you still calculate your chances of winning? >Maybe you want to compute the expected payoff? >If there are 20 players of equal skill, they each should >break even in the long run (minus the fees). If skills are >unequal, then it depends a lot on how much better the >skilled players are/and or how much dead money there is. Exactly. The more people playing at random, the higher the expected value for the skilled player. >Also, there is a schooling effect (as in schools of fish). >One-on-one, a good handicapper wins almost all the >time. The more fish, the less the edge of a somewhat >better player, since it is more likely that one of their cards >can beat him. On the other hand a highly skilled player can >win much more than his share against a lot of weak opponents. Right. As the number of players increases, all playing randomly except for one skilled player, the probability that the skilled player wins decreases, but the expected return for the skilled player per dollar bet increases. However, the skill factor is irrelevant unless the games are biased. If we assume that what is meant by a skilled player is a player who can correctly judge which team is more likely to win a given game, the greater the bias, the better it is for the skilled player. As an example, consider the following scenario: 14 games, 20 players. Of the 20 players, 1 is skilled, while the others play at random. The skilled player bets on the team most likely to win, whereas the other 19 players have no understanding of football, so they choose teams at random, 50-50. Assume that each game is strongly biased, with one team twice as likely to win as the other. In other words, the probabilities for each game are 2/3 to 1/3. Assume no fees to run the pool. Everyone bets a dollar and the pool is split equally among all players with the most number of wins. With these assumptions, the expected return for the skilled player is about $5.75 for each dollar bet -- a huge advantage, whereas for all the other players, the expected return for each dollar bet is only about 75 cents. So in effect, on average, all the other players are losing 25 cents of each dollar they bet to the skilled player. quasi === Subject: Re: football pool probability On 19 Nov 2005 10:51:44 -0800, arolotom@hotmail.com >> With your simplifying assumption that everyone picks at random, >> on the face of it, everyone has an equal chance of winning. Then >> the probability of winning is1/n where n is the number of players. >Yeah, that makes sense to me. OK, now what if we factored in skill so >that certain poolers had a better than 50-50 chance of picking winners >on some games. Then could you still calculate your chances of winning? I haven't looked at this kind of stuff for a very long time, so I could be wrong, but here goes: Lets say player number i has a given probability p_i of guessing right. Then if X_i is the number of teams the ith person chooses correctly, X_i has a binomail distribution, 14 trials with probability of success p_i, or B(14, p_i). You could consider X1, X2, ... X20 as independent {B(14, p_i)} random variables whose discrete probability function would be the product of the individual binomial functions. If your variable is X1, the probability that you win is: P( X1 > max{X2, X3, ... X20}) = P(X1 > X2 and X1 > X3 and ... X1 > X20) = product[i = 2 .. 20] P(X1 > Xi) =product[i=2..20] sum[k = j + 1..14, j= 2..14] {C(14, j)*(p_1)^j*(1 - p1)^(14-j) *C(14,k)*(p_i)^k*(1 - p_i)^(14-k)} where C(n,k) is the binomial coefficient. Sorry about the line breaks. The point is I believe it is possible to calculate although how well it models the real world problem is another question. --Lynn === Subject: Re: football pool probability >=product[i=2..20] sum[k = j + 1..14, j= 2..14] >{C(14, j)*(p_1)^j*(1 - p1)^(14-j) >*C(14,k)*(p_i)^k*(1 - p_i)^(14-k)} Typo alert, the j index isn't correct, should be: sum[k = j + 1..14, j= 0..14] --Lynn === Subject: translation of a set? How to prove measure(A+c/A)+ measure(A/A+c)-->0, when c-->0 A+c={a+c: a in A} A is a Lebesgue measurable set === Subject: Re: translation of a set? <10084543.1132425526923.JavaMail.jakarta@nitrogen.mathforum.org>, > How to prove measure(A+c/A)+ measure(A/A+c)-->0, > when c-->0 > A+c={a+c: a in A} > A is a Lebesgue measurable set Can you prove it whan A is an interval (a,b) ? Do you know something about approximating measurable A using intervals? === Subject: Re: integral of sin(x)/x On Thu, 17 Nov 2005 08:05:02 -0600, David C. Ullrich >On 16 Nov 2005 11:31:53 -0800, steiner@math.bgsu.edu >>The easiest way to do this integral(IMHO) is to use Laplace transforms. >>Any good book on this subject will have the following theorem: >The following theorem does not suffice. >The argument you propose is really the same as what >your argument, with the proof of the theorem inserted >in place of the theorem. He acknowledged that there >was a gap in his presentation - there's also a gap >in yours: >>Let F(t) be of class T with a < 0. Then >>[*] Int(F(t)/t, 0 ..infty) = Int(f(s) ds, 0.. infty), >>where f(s) is the Laplace transform of F(t). >The hypothesis a < 0 is important here. >(Whether it's important or not, it is a >hypothsis...) >>So the given integral equals the value of >>Int( ds/ (s^2 + 1), 0 .. infty ) = pi/2. >>So, why is sin t of class T and what does this mean? >>Def: A function F(t) is of class T if for some constant a it is of >>exponential order exp(at) and sectionally continuous. >>Here sin t is certainly continuous everywhere and >>is of exponential order because >>| exp(-t) sin t) | < 1 if t > 1. >So that sin(t) is of class T with a = 1. But >1 is not less than 0. In fact sin(t) is class T >with a = 0, but 0 is not less than 0 either. >(And the theorem above is simply not true in >general for a = 0. Or if it is I'm very surprised; >if a = 0 it's easy to give an example where both >sides of the equation (*) are +infinity; if it's >not possible to give an example where the left >side converges (as an improper Riemann integral) >but the right side does not I'd be surprised. >In any case a = 0 is certainly not allowed in >the theorem as you stated it.) >>Ray Steiner >************************ >David C. Ullrich In his post, Steiner stated a theorem, and gave a definition involving exponential order exp(at). He goes on to say that ... sin t is certainly continuous everywhere and is of exponential order because | exp(-t) sin t) | < 1 if t > 1. I read that to say he is using a = -1. In your post, your discussion uses a=1 and a=0. I must be missing something here, so can you say a little more about this? === Subject: Re: integral of sin(x)/x >On Thu, 17 Nov 2005 08:05:02 -0600, David C. Ullrich >>On 16 Nov 2005 11:31:53 -0800, steiner@math.bgsu.edu >The easiest way to do this integral(IMHO) is to use Laplace transforms. >Any good book on this subject will have the following theorem: >>The following theorem does not suffice. >>The argument you propose is really the same as what >>your argument, with the proof of the theorem inserted >>in place of the theorem. He acknowledged that there >>was a gap in his presentation - there's also a gap >>in yours: >Let F(t) be of class T with a < 0. Then >[*] Int(F(t)/t, 0 ..infty) = Int(f(s) ds, 0.. infty), >where f(s) is the Laplace transform of F(t). >>The hypothesis a < 0 is important here. >>(Whether it's important or not, it _is_ a >>hypothsis...) >So the given integral equals the value of >Int( ds/ (s^2 + 1), 0 .. infty ) = pi/2. >So, why is sin t of class T and what does this mean? >Def: A function F(t) is of class T if for some constant a it is of >exponential order exp(at) and sectionally continuous. >Here sin t is certainly continuous everywhere and >is of exponential order because >| exp(-t) sin t) | < 1 if t > 1. >>So that sin(t) is of class T with a = 1. But >>1 is not less than 0. In fact sin(t) is class T >>with a = 0, but 0 is not less than 0 either. >>(And the theorem above is simply not true in >>general for a = 0. Or if it is I'm very surprised; >>if a = 0 it's easy to give an example where both >>sides of the equation (*) are +infinity; if it's >>not possible to give an example where the left >>side converges (as an improper Riemann integral) >>but the right side does not I'd be surprised. >>In any case a = 0 is certainly not allowed in >>the theorem as you stated it.) >Ray Steiner >>************************ >>David C. Ullrich >In his post, Steiner stated a theorem, and gave a definition involving >exponential order exp(at). But he did not define the notion of exponential order exp(at) itsself. The definition (or if not _the_ definition, then in any case a definition that makes the theorem he cited correct) is that f has exponential order a if |f(t)| <= c e^(at) for some constant c. With that definition the function sin(t) is exponential order with a = 0 but not with any a < 0. It is certainly true that |e^(-t) sin(t)| <= 1. That does not say sin is exponential order with a = -1. (We _could_ give a different definition so that it would say that. But that definition would be very non-standard, and also irrelevant, because with that definition the cited theorem is simply false.) >He goes on to say that ... sin t is certainly continuous everywhere >and is of exponential order because | exp(-t) sin t) | < 1 if t > 1. >I read that to say he is using a = -1. The value of a might change depending on what side of the inequality it's on... >In your post, your discussion uses a=1 and a=0. I must be missing >something here, so can you say a little more about this? ************************ David C. Ullrich === Subject: Re: integral of sin(x)/x On Sun, 20 Nov 2005 11:06:15 -0600, David C. Ullrich >>In his post, Steiner stated a theorem, and gave a definition involving >>exponential order exp(at). >But he did not define the notion of exponential order exp(at) >itsself. The definition (or if not the definition, then >in any case a definition that makes the theorem he cited >correct) is that f has exponential order a if > |f(t)| <= c e^(at) >for some constant c. With that definition the function sin(t) >is exponential order with a = 0 but not with any a < 0. >It is certainly true that > |e^(-t) sin(t)| <= 1. >That does not say sin is exponential order with a = -1. >(We could give a different definition so that it would >say that. But that definition would be very non-standard, >and also irrelevant, because with that definition >the cited theorem is simply false.) I smell your cookin'. If one opens the door with a definition (using the well established notion of 'exponential order'), and invokes a well known theorem, best that it all be spot on. I suppose my mind-set was to just let that stuff slide. === Subject: Re: integral of sin(x)/x >On Sun, 20 Nov 2005 11:06:15 -0600, David C. Ullrich >In his post, Steiner stated a theorem, and gave a definition involving >exponential order exp(at). >>But he did not define the notion of exponential order exp(at) >>itsself. The definition (or if not _the_ definition, then >>in any case a definition that makes the theorem he cited >>correct) is that f has exponential order a if >> |f(t)| <= c e^(at) >>for some constant c. With that definition the function sin(t) >>is exponential order with a = 0 but not with any a < 0. >>It is certainly true that >> |e^(-t) sin(t)| <= 1. >>That does not say sin is exponential order with a = -1. >>(We _could_ give a different definition so that it would >>say that. But that definition would be very non-standard, >>and also irrelevant, because with that definition >>the cited theorem is simply false.) >I smell your cookin'. >If one opens the door with a definition (using the well established >notion of 'exponential order'), and invokes a well known theorem, best >that it all be spot on. >I suppose my mind-set was to just let that stuff slide. Which of course doesn't work if we're talking about an actual _proof_. The theorem is very easy to prove if a > 0. For a < 0 it _obviously_ simply false, in fact it's incredibly easy to give examples of functions that satisfy the hypothesis for a < 0 such that the relevant Laplace transform does not even exist for all s > 0. Now in the present case we have a = 0, which implies that the LT does exist for all s > 0, and hence that the result might be true. But the incredibly easy proof of the theorem for a > 0 doesn't work; if we're going to prove the result for sin(x) we need to give some justification for a certain crucial step in the proof of the theorem. Which is to say the theorem is certainly not irrelevant, it gives a good indication of why the result might be true, and gives an indication of one way one might try to prove it. But if we try to just slide through by simply _citing_ the theorem then there's a big gap in the argument (as I pointed out, exactly the same as the gap that Robert Israel acknowledged existed in his version of the proof.) ************************ David C. Ullrich === Subject: Re: integral of sin(x)/x On Mon, 21 Nov 2005 06:22:15 -0600, David C. Ullrich >>If one opens the door with a definition (using the well established >>notion of 'exponential order'), and invokes a well known theorem, best >>that it all be spot on. >>I suppose my mind-set was to just let that stuff slide. >Which of course doesn't work if we're talking about an >actual proof . >The theorem is very easy to prove if a > 0. For a < 0 >it obviously simply false, in fact it's incredibly >easy to give examples of functions that satisfy the >hypothesis for a < 0 such that the relevant Laplace >transform does not even exist for all s > 0. >Now in the present case we have a = 0, which implies >that the LT does exist for all s > 0, and hence that >the result might be true. But the incredibly easy >proof of the theorem for a > 0 doesn't work; if >we're going to prove the result for sin(x) we >need to give some justification for a certain >Which is to say the theorem is certainly not irrelevant, >it gives a good indication of why the result might be >true, and gives an indication of one way one might >try to prove it. But if we try to just slide through >by simply citing the theorem then there's a big gap >in the argument (as I pointed out, exactly the same >as the gap that Robert Israel acknowledged existed >in his version of the proof.) >>crucial step in the proof of the theorem. I know you are not saying it's illegitimate to cite known theorems as part of a proof, or in solving a particular problem. When that is done, it's necessary to demonstrate (when it's not 'obvious') that the cited theorem is logically applicable, and that all conditions imposed by the theorem are satisfied. In the case of Robert Isreal's solution, would it not have been sufficient to just say that The interchange of the order of integration in this step is justified by the such-and-such theorem? Same thing as using a theorem from Laplace transform theory, although in the problem under discussion here I don't see that there was any need to use the theorem in the first place in order to solve the problem using a Laplace transform approach. Well, anyway, I hope I haven't garbled the above too badly, and that we're at least pretty much on the same wavelength. === Subject: Re: integral of sin(x)/x Daniel Grubb says... >> f'(x) = 2/pi integral from -infinity to +infinity >> of sin(k)/k exp(ikx) dk >In your situation, you have the advantage that f(x) is also >in L1, so g(k) is continuous. As I already pointed out, this >shows that f(x)=f'(x) almost everywhere. Thus, if *both* >f and f' are known to be continuous in a neighborhood of >x_0, we get the result. So do you know that your f' is >continuous in a neighborhood of 0? Well, we can rewrite sin(k)/k exp(ikx) as follows: sin(k)/k cos(kx) + i sin(k)/k sin(kx) The second term integrates to 0 (being an odd function of k), so we know: f'(x) = 2/pi integral from -infinity to +infinity of sin(k)/k cos(kx) dk We can write sin(k)cos(kx) as 1/2 (sin(k+kx) + sin(k-kx)), so we have f'(x) = 1/pi integral from -infinity to +infinity of (sin(k+kx) + sin(k-kx))/k dk = 1/pi integral from -infinity to +infinity of sin(k+kx) dk + 1/pi integral from -infinity to +infinity of sin(k-kx)/k dk Assuming 1+x and 1-x are both positive (which they will be, if -1 < x < 1), we can perform a change of variables in the first integral to k' = k(1+x), and a change of variables in the second integral to k' = k(1-x). Thus we get f'(x) = 2/pi integral from -infinity to +infinity of sin(k)/k dk = f'(0) So f'(x) = f'(0) in the entire region -1 < x < 1. -- Daryl McCullough Ithaca, NY === Subject: A-algebra homomorphism sends 1 to 1? If B and C are A-algebras and f : B ---> C is an A-algebra homomorphism, does f necessarily send 1 to 1? If so, why? If it doesn't, I have another question. New notation. Let A be a k-algebra where k is a field. Let S be a multiplicative subset of A. Let A_S be the localization of A at S. Let g : A ----> A_S be the natural k-algebra homomorphism. Suppose h : A_S -----> B is a k-algebra homomorphism. Does h send 1 to 1? Does it send units to units? If it doesn't, finally let A, A_S, k be as above. Now let B be a k-algebra and N an ideal of B such that N^2 = 0. Let h : A_S ------> B / N be a k-algebra homomorphism. Does 1 go to 1? Do units go to units? Does h o g send S to the units of B / N? (where g was as above) THank you for your help, James === Subject: Re: A-algebra homomorphism sends 1 to 1? > If B and C are A-algebras and f : B ---> C is an A-algebra homomorphism, does f necessarily send 1 to 1? If so, why? Yes, by defintion. J. === Subject: Re: A-algebra homomorphism sends 1 to 1? >>If B and C are A-algebras and f : B ---> C is an A-algebra homomorphism, does f necessarily send 1 to 1? If so, why? > Yes, by defintion. I think it depends on which definition you use. I think the answer is no. Consider the mapping from R^2->R^2 (pointwise multiplication) given by (x,y)->(x,0). I cannot answer the other questions because I don't know what a localization is. === Subject: Re: A-algebra homomorphism sends 1 to 1? > If B and C are A-algebras and f : B ---> C is an A-algebra > homomorphism, does f necessarily send 1 to 1? If so, why? >> Yes, by defintion. > I think it depends on which definition you use. I think the answer is > no. Consider the mapping from R^2->R^2 (pointwise multiplication) > given by (x,y)->(x,0). > I cannot answer the other questions because I don't know what a > localization is. Stephen, in general, you are right. But James embeds the problem into the world of commutative algebra where localisations are typically talked about. These are generalisations of quotient rings; if S is a multiplicative subset of a ring A (i.e. 1 e S (!) & s,t e S => st e S), then S^{-1}A is the 'smallest' ring A' together with a natural ringhomomorphism i: A -> A' such that -roughly speaking- all elements of S become invertible in A'. Note that 'smallest' does not necessarily mean that i is injective. Localisations are sometimes called 'rings of fractions'. Bearing this in mind it is clear that James is speaking of rings *with* unity. So, your example does not apply in this case. And to make use of the (universal) properties of localisations, it is needed to have ringhomomorphism preserving the unity elements. Otherwise there would be no value of taking localisations. This is a bit more detailed as I have previously posted in this thread. I think this is all I can say about it up to this point. J. === Subject: Re: A-algebra homomorphism sends 1 to 1? >> If B and C are A-algebras and f : B ---> C is an A-algebra homomorphism, does f necessarily send 1 to 1? If so, why? >Yes, by defintion. Well, that would depend on whose definition. (It would also depend on the definition of A-algebra including the existence of 1, I suppose; but James has already implicitly assumed that B and C, at least, have 1s.) Herstein's textbook _Algebra_ doesn't insist on restricting the definition of algebra homomorphism that way. And if the intended applications are to endomorphism algebras of vectorspaces, then insisting that a homomorphism preserve 1 seems counterproductive. Lee Rudolph === Subject: Re: A-algebra homomorphism sends 1 to 1? >If B and C are A-algebras and f : B ---> C is an A-algebra homomorphism, does f necessarily send 1 to 1? If so, why? >>Yes, by defintion. > Well, that would depend on whose definition. (It would also > depend on the definition of A-algebra including the existence > of 1, I suppose; but James has already implicitly assumed that > B and C, at least, have 1s.) Herstein's textbook _Algebra_ > doesn't insist on restricting the definition of algebra > homomorphism that way. And if the intended applications > are to endomorphism algebras of vectorspaces, then insisting > that a homomorphism preserve 1 seems counterproductive. > Lee Rudolph I agree, but as James refers to constructions used only in commutative algebra, it is very likely to apply the definitions used in this discipline. There, in general, rings are commutative and unitary and ring homomorphisms preserve the unity element. J. === Subject: Re: The complete general solutions for eq. ax^2 +y^2 =z^2 ? Hi to All, Following: > My literature sources used to describe: > for eq. ax^2 +y^2 =z^2 > =z^2 ................................(1) > once x even number so solutions are: > y = am^2 -1 > z = am^2 +1 > and x = 2m > once x odd number so solutions are: > y =(am^2 -1)/2 > z =(am^2 +1)/2 > and x = m but also a;m odd numbers. > These solutions could be generalized into: > y = am^2 -k^2 > z = am^2 +k^2 > x = 2mk for m;k optional natural numbers > of gcd=1 > once x even number > y =(am^2 -k^2)/2 > z =(am^2 +k^2)/2 > x = mk for m;k optional odd numbers > of gcd=1 > once x odd number but also a odd > number > *) Then additional should be recognized fall for a > even > once a=4a1 so let: > 4a1x^2 +y^2 = z^2 > y = a1m^2 -k^2 > z = a1m^2 +k^2 > then 4a1x^2 = 4a1m^2 k^2 and x = mk > once m;k optional natural numbers of gcd=1 > but m even and k odd or inverse once a1 odd > then finally x=mk even, > then for a1 even always should be k odd > and once m optional natural number and of gcd=1 to > o k > so according to m even x also even or for m odd x > x odd. > Then for a=2a1 there are not solutions for x > odd > and should be taken solutions for x even > numbers... > Does this describes all possible solutions to > eq.(1) ? Does *) developments are covered with > first fall of x even number ? > With Compliments > Ro-bin > So far I can see *) developments are not covered > with developments for x even and there more not > with > developments for x odd. > *) there x=mk; y=a/4 m^2 -k^2; z=a/4 m^2 +k^2 > x even: > x=2mk; y=(am^2 -k^2)/2; z=(am^2 +k^2)/2 > where for sure y;z sentences could not be equal > after division by 2 to *)y;z sentences: > y'=a/4 m^2 -k^2/4; z'=a/4 m^2 +k^2/4 > x odd: x=mk; y=am^2 -k^2; z=am^2 +k^2 > and the differences are because of *) a/4 to a > values. > Does somebody could express more general opinion > to natural solutions to eq.1 ? > > Compliments > Ro-bin I can see, that there is to consider one more special case: let in eq. ax^2 + y^2 = z^2 a=a1^2 then for a1^2 x^2 +y^2 = z^2 or x1^2 +y^2 = z^2 we'll have for a1 or x even: 1): a1x = 2mn; y = m^2 -n^2; z = m^2 +n^2 once for x even we'll have also 2):x = 2mk; y = a1^2 m^2 -k^2; z = a1^2 m^2 +k^2 How it works ? See example: 1):let a1=m=3; x=2n=2; y=8; z=10; a1x=x1=6 2): x=2=2mk; m=k=1; y = 9 -1 = 8; z = 9+1 =10 *** or for a1 and x odd: 3): a1x =m^2 -n^2; y =2mn; z = m^2 +n^2 4): x= mk; y=(a1^2 m^2 -k^2)/2; z=(a1^2 m^2 +k^2)/2 What also works as example: 3): let a1x = 3^2 -2^2 = 5; y=12; z=13 4): so for x=1=mk; m=1;k=1 y=(5^2-1)/2 = 12; z=(5^2+1)/2 = 13 *** Therefore, we can be sure of unity of these forms of general solutions even expressed in some different forms and parameters. Again with Compliments Ro-bin === Subject: counterexample that demonstrates 4-Color Mapping is a fake Recently had a email from South-America wanting to know what my counterexample was for the 4-color mapping since it is a ill-defined problem. I gave my counterexamples I believe circa 1999, but perhaps I should re-do that work in a more clear and concise exposition. Counterexample that 4-Color Mapping is ill-defined and a fakery: Whenever we ignore essential elements of a proof domain such as the borderlines in color mapping we no longer have a problem of mathematics but rather something of sensory perception or psychology or sense feelings. The 2-Color Mapping proves via Jordan Curve theorem that 2 colors are necessary and sufficient to color all mapps. The counterexamples I gave circa 1999 was that of information theory that you never need 4 things to communicate any piece of information, you need only 2 things. All of language, all of pictures, paintings, drawings, all verbal communication, all visual communication is built of 2 things and never is 4 things needed to make any of these communications. Now when we look at maps, those are a subset of all verbal or visual communication and thus being a tiny subset of all visual communication and since visual communication is all built from 2 things, not 4 things then 2 Colors suffice to color all maps. CounterExample Proof that the 4-Color Mapping is a fakery. I will use the Computer to be ironic and funny and a slap in the face to those who have been mislead and those that are misleading of young people about Color Mapping. These people are not mathematicians but crooks of mathematics and logic. A Computer runs on 2 things-- gate open or gate closed. Binary operation. Gate open is represented by 1 and gate closed is represented by 0. Now a computer can display any and every map. Now suppose the 4 Color Mapping was well defined and true as per Appel and Haken claims. That would imply that there is a mapp for which computers are unable to ever display because this map requires not just two gates of either open or closed but 4 gates. Further yet, we are going to be even more universal by saying that writing on a computer and that every screen of a computer whether it be pictures or writing or text or anything is a mapping. And in this ocean of universality we are going to say that the gate open or gate closed of binary of 0 or 1 are 2 colors. So, now, if the 4 Color Mapping of Appel and Haken and many others who believe in this nonsense is a true and valid work of mathematics, then there must be a thought or idea or graphic or map or text or diagram that cannot be displayed by a computer because its 2 gates of open or closed are insufficient. So what the Appel and Haken nonsense implies is that a computer has to be built that runs on 4 things and not 2, in order to represent that item. And to further show how ridiculous and stupid the Appel and Haken 4 Color Mapping is if true, is that it treads even into the subject of physics and would say that positive and negative charge are not adequate to cover all matter and that there are 4 types of charges to cover all matter. Moving on further into physics, if the Appel and Haken 4 Color Mapping nonsense was true then Complimentarity in physics is also false because somethings need 4 things and not just 2 and that Mathematicians rarely have the brains to see when they bring a dude into the world that their dude wrecks havoc all over the place. They spend most of their time trying to explain their fake method to others because it is hidden and arcane and abstract, not because it is difficult but because it is a fake from the start. The Appel and Haken computer alleged proof is hundreds of pages long in computer program and the Wiles's FLT is likewise hundreds and thousands of pages long because a dude and fake do not clarify the world, they hide from the world of clear thought and view. So Appel and Haken and Wiles spend most of their time explaining their fakery to the world that is duped. But what the world should do for these fake proofs is ask the opposite-- suppose they are true. Suppose the 4-Color Mapping and Wiles FLT is true then how does that destroy common everyday truths about the world. For 4-Color Mapping it denies that computers can run on just 2 gates of open and closed, denies that physics is built on 2 things-- goes to infinity. Wiles's FLT in a sentence says that infinity ends, when the truth is that infinity goes on forever where 1,2,3,.... becomes ....9999 Ironic and funny that Appel and Haken have hid and hidden themselves behind a computer alleged proof and above I am offering a computer argument that shows their 4 Color Mapping is a utter and total dude fakery. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake The 4-color problem does not solely arise when you consider coloring regions; it is also a vertex-coloring problem. The fact you can't creatively think up new problems doesn't mean they don't exist. A vertex-coloring problem looks like: What is the minimum number of colors you need to color the numbers 1, 2, 3, ..., N subject to a lot of conditions of the form: a(i) and b(i) have to be colored differently. > Recently had a email from South-America wanting to know what my > counterexample was for the 4-color mapping since it is a ill-defined > problem. Your definitions are ill-defined. If S is a statement, a counterexample is DEFINED TO BE an object x such that S is false for x. A counterexample to the 4-color theorem is a map that requires more than 4 colors to completely color it. Your example only requires 2, and since 2 is less than or equal to 4, it's not a counterexample. What you have is a _re-interpretation_ of the coloring problem, and a dull one at that, since any competant mathematician can prove it in 5 minutes. Therefore no one is interested in it, therefore you found it not mentioned anywhere else. The proof of your result (which does not even need the Jordan Curve Theorem) is as trivial as 1 + 1 = 2. Color the regions one color, color the boundary a different color. You never have boundary next to itself, and you never have a region next to another region. QED. Next problem. > [...] > Whenever we ignore essential elements of a proof domain such as the > borderlines in color mapping we no longer have a problem of mathematics It still is a mathematics problem. If we replace the Parallel Postulate in the definition of Euclidean geometry with the statement Every pair of lines intersects, we still have something in mathematics (Riemann geometry). It's a DIFFERENT mathematics situation, but it's still a MATHEMATICS situation. It may not fit in with that narrow construct called reality, but it's still worth pondering. In this case, we get a more interesting question. In fact, we can think about coloring vertices, not regions, and the same problem will come up. You can also talk about coloring the EDGES of a network, and this also leads to the standard (non-AP) interpretation of the 4CT. You need to take off your blinders. > A Computer runs on 2 things-- gate open or gate closed. Binary > operation. Gate open is represented by 1 and gate closed is represented > by 0. Now a computer can display any and every map. Now suppose the 4 > Color Mapping was well defined and true as per Appel and Haken claims. > That would imply that there is a mapp for which computers are unable to > ever display You're confusing DISPLAYING a map with COLORING it. Those are two different things. You can display more than 2 colors by using 2 gates. 2 x 2 = 4 possibilities. You can get more using all possible combination of 3 gates. 2 x 2 x 2 = 8. You can get an arbitrarily large number of combinations by adding more gates. Learn how a computer display works before you pontificate about it. After all, you were the one who also said, Language cannot be reduced to 0s and 1s, despite the computer in front of you actually doing it. (This makes me wonder whether your computer display is black and white, since if it was color, you clearly made the same blunder.) > And to further show how ridiculous and stupid the Appel and Haken 4 > Color Mapping is if true, is that it treads even into the subject of > physics and would say that positive and negative charge are not > adequate to cover all matter and that there are 4 types of charges to > cover all matter. And hence everything in the universe has a charge of +1 (proton) or -1 (electron) and you cannot get an object with a larger charge. You're not even getting your physics right here, Ludwig von Ludwig. > [...] For Wiles's fake FLT denies that endless adding of 1 > goes to infinity. Wiles's FLT in a sentence says that infinity ends, No one would have taken a second look at his proof if that's what he claimed. > Ironic and funny that Appel and Haken have hid and hidden themselves > behind a computer alleged proof and above I am offering a computer > argument that shows their 4 Color Mapping is a utter and total dude > fakery. Of course your computer argument is false because it suggests that you know nothing about computers, physics, or mathematics. --- Christopher Heckman === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake as you say, 2 colors is no counterexample!... it seems that he is merely reifiying the Jordan curve theorem, without adding anything to it (otherwise, say, he could prove the curve theorem by assuming that 4CC is true, and finding a contradiction; this would be so, because inductive proofs are isomorphic to deductive proofs -- hoo-ray !-) I'm not sure what you mean by vertex-coloring, although the problem has historically been presented as a coloring of vertices of connected graphs -- the dual problem -- undoubtedly to save hand-coloring. (herein, AP's format would have all-hollow vertices with no coloring allowed.) > The 4-color problem does not solely arise when you consider coloring > regions; it is also a vertex-coloring problem. The fact you can't > creatively think up new problems doesn't mean they don't exist. > A counterexample to the 4-color theorem is a map that requires more > than 4 colors to completely color it. Your example only requires 2, and > since 2 is less than or equal to 4, it's not a counterexample. --ils ducs chez Kyoto! http://tarpley.net/bush8.htm http://larouchepub.com/other/2002/2903_chapter_11.html http://clowder.net/zubek/zubek.html http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake incidentally, the proof that four colors are *neccesary* is just, The Tetrahedron. sufficiency is another matter, that also does not seem to follow from AP's English (as an almost-second-language). thus: maybe we hate having our nation bankrupted for the sake of the Peak Oil doctrine -- you know, how much oil it takes to run a God-am war, not to mention getting the whole islamic jihad up against us with bogus invasion number two. anyway, the carbon emmissions-trading scheme of the Protocols of the Elder George of Kyoto (sik), was started in the USA, on Feb.12, not in any of the countries that signed it (the practice market was also here); why doesn' the Urinal say what effect this has had, if any? not that Gore might not have been worse, as implausibly undeniable as that might seem -- just read LaRouche to find out, why. --ils ducs chez Kyoto! http://tarpley.net/bush8.htm http://larouchepub.com/other/2002/2903_chapter_11.html http://clowder.net/zubek/zubek.html http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake The 4-color problem does not solely arise when you consider coloring regions; it is also a vertex-coloring problem. The fact you can't creatively think up new problems doesn't mean they don't exist. mathematics. Chris is deluded. Can he put the equivalency of the Jordan Curve theorem into vertex coloring? No, of course not. And what is ignored in the vertex coloring when the Jordan Curve is transfered to vertex coloring. Again, the onus and burden of proof that Appel and Haken have to demonstrate is how you can have a legitimate mathematical problem if you separate elements of a proof set and ignore them in the statement. Show us how that is logically acceptable and show us another case example in mathematics where elements are ignored yet a statement remains that is provable. What the 4-Color Mapping teaches us is that many things may look like mathematics but are in fact pieces of psychology, such as those optical illusions. Appel and Haken and Heckman are mistaking optical illusions as pieces of mathematics when they are just that-- optical illusions. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake > What the 4-Color Mapping teaches us is that many things may look like > mathematics but are in fact pieces of psychology, Actually, this is what sci.math cranks teach us. === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake > The 4-color problem does not solely arise when you consider coloring > regions; it is also a vertex-coloring problem. The fact you can't > creatively think up new problems doesn't mean they don't exist. > mathematics. You couldn't be more wrong! There are quite a few in graph theory, and several outside of the area of graph theory. One of the more bizarre involving cross-products in 3-dimensional space. Consider the following proposition: PROPOSITION: Start with the badly-expressed cross product: a(1) x a(2) x ... x a(n), where a(N) are 3-dimensional vectors in the set {i,j,k} = {(1,0,0), (0,1,0), (0,0,1)}; it is badly-expressed because the cross product operation is not associative. So put in parentheses in the expression until it is well-defined, and take two such parenthesized expressions. Then it is possible to assign i, j, or k, to each a(N) such that the two expressions are the same. For instance, (A x B) x (C x D) = ((A x B) x C) x D becomes true with the proper choice of A, B, C, and D, where you can choose A, B, C, and D from the set {i,j,k}. This proposition is equivalent to the Four Color Theorem. That is, if the 4CT (as defined by the mathematical world) is true, so is this proposition, and vice versa. There's another result relating to whether 7 divides into a product of expressions. > Chris is deluded. Can he put the equivalency of the Jordan > Curve theorem into vertex coloring? No, of course not. And what is > ignored in the vertex coloring when the Jordan Curve is transfered to > vertex coloring. Just because one proof of a result requires the JCT doesn't mean they all have to. For instance, Chebyshev's Theorem (For every positive integer N, there is a prime number between N and 2N) was first proved using complex numbers. So AP would have said that the complex numbers were an essential proof element. Several proofs followed, all using complex numbers. AP is right so far, but then Paul Erdos comes along and does it without complex numbers. So here, an essential proof element isn't necessary after all. AP is just being stubborn. He is, after all, in his 50s, and I suspect his cranial bones have closed over. --- Christopher Heckman === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake > The 4-color problem does not solely arise when you consider coloring > regions; it is also a vertex-coloring problem. The fact you can't > creatively think up new problems doesn't mean they don't exist. > mathematics. You couldn't be more wrong! There are quite a few in graph theory, and several outside of the area of graph theory. One of the more bizarre involving cross-products in 3-dimensional space. Consider the following proposition: PROPOSITION: Start with the badly-expressed cross product: a(1) x a(2) x ... x a(n), where a(N) are 3-dimensional vectors in the set {i,j,k} = {(1,0,0), (0,1,0), (0,0,1)}; it is badly-expressed because the cross product operation is not associative. So put in parentheses in the expression until it is well-defined, and take two such parenthesized expressions. Then it is possible to assign i, j, or k, to each a(N) such that the two expressions are the same. For instance, (A x B) x (C x D) = ((A x B) x C) x D becomes true with the proper choice of A, B, C, and D, where you can choose A, B, C, and D from the set {i,j,k}. This proposition is equivalent to the Four Color Theorem. That is, if the 4CT (as defined by the mathematical world) is true, so is this proposition, and vice versa. There's another result relating to whether 7 divides into a product of expressions. > Chris is deluded. Can he put the equivalency of the Jordan > Curve theorem into vertex coloring? No, of course not. And what is > ignored in the vertex coloring when the Jordan Curve is transfered to > vertex coloring. Just because one proof of a result requires the JCT doesn't mean they all have to. For instance, Chebyshev's Theorem (For every positive integer N, there is a prime number between N and 2N) was first proved using complex numbers. So AP would have said that the complex numbers were an essential proof element. Several proofs followed, all using complex numbers. AP is right so far, but then Paul Erdos comes along and does it without complex numbers. So here, an essential proof element isn't necessary after all. Chris Heckman's above example of where you ignore essential elements of the proving domain is not an example at all. It is a shifting of domain elements. I cannot give you an example in mathematics that is like the 4-Color Mapping because it is the only example in mathematics that abuses logic like this of ignoring domain elements. 4-Color Mapping is unique because it abuses logic. An example in physics would be if the engineers and physicists of the builders of the Mars rovers were to ignore everything of the force of electricity and magnetism when building the Rovers. Obviously the mission would have never been successfull, nor even begun. Or, let us say those physicists and engineers ignored everything of the StrongNuclear force, would the Rovers to Mars have been successful? Or if they took into account every force of gravity, EM and StrongNuclear but ignored everything of the force of Weak Nuclear would there have been a Mars Rover mission. It is an example such as this that compares to the 4-Color Mapping in mathematics. As for all of those alleged equivalencies that Chris believes in. Let me put him to a test that will shake his confidence. Logic dictates that if the 4 Color Mapping is a true statement of mathematics then its equivalences are also true. Let us call the 4 Color Mapping as A, and let us call Chris's graph equivalency as B So logic gives us A => B, and B => A Now comes the fun part, the part that reveals the phony baloney of 4 Color Mapping and the phony baloney equivalents. Let us call the Jordan Curve theorem application to 2 Color Mapping that 2 colors suffice to color all maps when we do not ignore borderlines and we color the borderline in blue ink and the interiors remain white. Call this J. Now the 4 Color Mapping and B ignore borderlines. J includes borderlines. Logic dictates that if A => B, and B => A, then J => X and X => J. What is X? It is the equivalent statement and proof in Chris's graph theory equivalency. Logic dictates that the graph theory alleged equivalency that Chris espouses must contain the 2-Color Mapping of the Jordan Curve theorem within that graph theory. If it does not contain it, then it was all phony baloney from the start. Now Chris says there is another equivalency in Cross Products to 4-Color Mapping. Let us call this one P. So A => P and P=> A. And again if this is really true then Logic would demand and dictate that J => P and P => J. So, what I am saying is that no-one disputes , not even Dik Winter, that when you ignore borderlines you have a 4 Color Mapping and when you do not ignore borderlines you have a 2 Color Mapping proven by JCT. So, now, can Chris show us how the 2 Color Mapping is existing in his alleged Graph theory equivalency and if not why not. Can he show how the Cross Product equivalency has the 2 Color Mapping equivalency and if not, why not. If the 2 Color Mapping with the Jordan Curve theorem is never translatable into any of the alleged 4 Color Mapping equivalencies then there is a good explanation why. The explanation is that 4 Color Mapping is all phony baloney from the start and that there never was an equivalency. Logic has this: A => B, and B => A J => X and X => J and A => P and P=> A J => P and P => J If Chris and his allies of Appel, Haken and others cannot show J => X and X => J and J => P and P => J, then the 4-Color Mapping was all a phony baloney from the very start, way back into the 19th century. The moral of the story is that when you *ignore* key elements of the domain of a proof of a statement, you abandon mathematics and am into a subject of psychology or sensation or optical illusion but not precision that is the subject of mathematics. So, Chris, show us how graph theory yields the 2 color mapping of the Jordan Curve theorem as proof. If none of those graph theory equivalents or the cross-product equivalent can yield the 2 color mapping with the JCT, then all of yours was phony baloney from the start. And someone should comment on the fact that suppose the 4 Color Mapping were true, then there exists some graphs or maps for which our present day computers can never display because our present day computers all operate on 2 parameters of gate open or gate closed. If the 4 Color Mapping were true then a computer exists for which there are 4 parameters of gate open, gate closed, and two more. Obviously this is absurd. And so the 4 Color Mapping is absurd. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake Now I feel it is best for me to explain this in a variety of ways to my opponents and critics because they are so brainwashed with their belief that telling them their mistake with one picture is not as good as telling them of their mistake in 2 pictures or 3 pictures. The more pictures I show them of their mistaken position, the easier it is for them to see their flaw. So I am going to show them a 3rd picture. Heckman, Appel, Haken and others believe the 4 Color Mapping even though it ignores borderlines. They also accept the 2 color mapping because it does not ignore borderlines and is proven by the Jordan Curve theorem. So they accept 4-Color Mapping and accept 2-Color Mapping. I, on the other hand, accept only the 2 Color Mapping because one cannot define or imagine curves without borderlines, so I see mapps without borderlines as not even mathematics but some sorcery witchcraft voodoo mysticism. So, now, the 4-Color Mappers have alleged to have found equivalent statements in graph theory, in cross product theory etc etc. So here is the test. The 3rd test. Combine together both the 4 Color Mapping and the 2 Color Mapping and union them together into one big statement. That if we ignore borderlines we have 4 Coloring and if we do not ignore borderlines we have 2 Coloring. Now go to each of those alleged equivalent statements and see if the combined colorings of 4 color and then 2 color come out of those equivalent statements. Chris Heckman claims that graph theory contains the 4 Color Mapping. Well, if that is true, then that same graph theory would also contain the 2-Color Mapping includind the Jordan Curve theorem look alike. You see, Chris accepts 4 Color Mapping and also accepts 2 Color Mapping when borderlines are not ignored and he accepts equivalent statements in graph theory to 4 Color Mapping. So be Logic, Chris should be able to pull out of Graph theory the 2 Color Mapping with its attendent Jordan Curve theorem. And do the same for all other alleged equivalent statements to the 4 Color Mapping. If Chris is unable to do that, means that the 4 Color Mapping was a fake all along. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake > Now I feel it is best for me to explain this in a variety of ways to my > opponents and critics because they are so brainwashed with their belief > that telling them their mistake with one picture is not as good as > telling them of their mistake in 2 pictures or 3 pictures. The more > pictures I show them of their mistaken position, the easier it is for > them to see their flaw. Can you tell me what the colours are of the boundaries in chess and checker boards? And why two colours are used for the fields? Is that not ignoring boundaries? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake > There's another result relating to whether 7 divides into a product of > expressions. Can you give a reference of this? === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake Deluded people tend to fantasize their delusion to other areas. The fact is that 2 colors is able to encompass all visual data and all visual language. Only buffoons of mathematics would then try to foist a statement that it takes 4 colors to do all maps. When maps are a tiny subset of visual data. Chris has never been able to show another piece of mathematics where key essential proving elements of the domain are ignored. 4 Color Mapping totally ignores borderlines. That is one of the reasons computers were brought in to cover the cases. Because when ignore borderlines, the mind can no longer handle weird and strange curves and intersections. Because, well, because when borderlines are ignored you no longer have a precise statement of what you are doing. What Chris needs to do is show us another example in all of mathematics where a key element of the proving domain is ignored and yet some mathematical statement remains that makes any sense. Scientists in physics, in chemistry, in biology in geology etc etc try hard not to ignore anything that influences their domain of experiment. Yet mathematicians blithely and stupidly hang onto the 4 Color Mapping. I think it is because computers are promoted and sales figures rise because of this computer link, when it is not even a mathematical problem in the first place but a fakery, a con-artist job. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake > Deluded people tend to fantasize their delusion to other areas. > The fact is that 2 colors is able to encompass all visual data and all > visual language. Only buffoons of mathematics would then try to foist a > statement that it takes 4 colors to do all maps. When maps are a tiny > subset of visual data. This is of course not true since it takes 3 colors to represent all colors - red, green, and blue for emitted light (CRT), and yellow, red, and blue for absorbed light (e.g. paintings). On the other hand it's entirely beside the point. You could express the 4CT as a problem of assigning distinct primes to territories of a map, for example, if you want. > Chris has never been able to show another piece of mathematics where > key essential proving elements of the domain are ignored. 4 Color > Mapping totally ignores borderlines. That is one of the reasons > computers were brought in to cover the cases. Because when ignore > borderlines, the mind can no longer handle weird and strange curves and > intersections. Because, well, because when borderlines are ignored you > no longer have a precise statement of what you are doing. You can define territories and adjacency without requiring borderlines be included in territories at all, viz: Two territories on a map are adjacent if there is a set of points, not belonging to either territory, which is a line and has length > 0, such that each territory comes arbitrarily close to each point on this line. Further, there are no two territories on the map which have only a set of points with length = 0 such that each territory comes arbitrarily close to each point in this set. (this to exclude the dart board example). So we have a definition of adjacency which requires borderlines, yet in which no borderline belongs to any particular territory. And with this definition of adjacency we can express the 4CT. And with this definition of adjacency, MORE than 2 colors are required for obvious and easy examples. Of course real mathematicians are too careful and precise to settle for such a loosey-goosey definition of adjacency as I just gave. Oh well. What I need to do is go see how the current mathematics expresses adjacency and the 4CT. Mathworld seems a good starting point. I'm going to go do that right now. Care to join me? Ken > What Chris needs to do is show us another example in all of mathematics > where a key element of the proving domain is ignored and yet some > mathematical statement remains that makes any sense. > Scientists in physics, in chemistry, in biology in geology etc etc try > hard not to ignore anything that influences their domain of experiment. > Yet mathematicians blithely and stupidly hang onto the 4 Color Mapping. > I think it is because computers are promoted and sales figures rise > because of this computer link, when it is not even a mathematical > problem in the first place but a fakery, a con-artist job. > Archimedes Plutonium > www.iw.net/~a_plutonium > whole entire Universe is just one big atom > where dots of the electron-dot-cloud are galaxies === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake > Deluded people tend to fantasize their delusion to other areas. You're the one claiming the world should bow down to you for finding the right interpretation, not me. > Chris has never been able to show another piece of mathematics where > key essential proving elements of the domain are ignored. In another post, I've given the example of Chebyshev's Theorem (between N and 2N there is a prime number, for all positive integers N >= 2). It was first proven with complex numbers, so AP would have claimed that complex numbers were an essential proof element. Erdos later proved the result without recourse to the complex numbers. So much for essential proof methods. > 4 Color Mapping totally ignores borderlines. Which is what makes it interesting. > That is one of the reasons > computers were brought in to cover the cases. Because when ignore > borderlines, the mind can no longer handle weird and strange curves and > intersections. Not true. There simply are too many cases for a human being. > What Chris needs to do is show us another example in all of mathematics > where a key element of the proving domain is ignored and yet some > mathematical statement remains that makes any sense. Done above. > Scientists in physics, in chemistry, in biology in geology etc etc try > hard not to ignore anything that influences their domain of experiment. This is because of unknown variables. There are no unknown variables in a mathematics result. --- Christopher Heckman === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake On 19 Nov 2005 21:22:39 -0800, Proginoskes > snip.... >You're not even getting your physics right here, Ludwig von Ludwig. > --- Christopher Heckman Back many years ago he was Ludwig Plutonium, working as a dishwasher at Dartmouth in Hanover VT. Then he was carrying on about Fermat's Last Theorem being false, and he generated similarly long threads. I posted a satire about him which you might find amusing. Amazingly enough, someone has preserved it at http://harlie.idsfa.net/~john/humor/fermat.chicken You might get a chuckle out of it. --Lynn === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake <1w6AQ0VW8aguTMPq54mlxSiR0Ca8@4ax.com> Back many years ago he was Ludwig Plutonium, working as a dishwasher at Dartmouth in Hanover VT. Then he was carrying on about Fermat's Last Theorem being false, and he generated similarly long threads. I posted a satire about him which you might find amusing. Amazingly enough, someone has preserved it at Since I first appeared on the Internet in 1993, there have been a ocean full of people in hatred of me and what I write and represent. These people are hate mongers and demonizers. They spend the majority of their time in demonizing others and never a ability to read what the person has to say or debate or discuss. Alan Schwartz is an example of this negative behaviour run amok and Lynn Kurtz is another example. People like Schwartz or Kurtz should have never been in the teaching profession, because teachers that demonize other people are never teachers. Last week I posted about NOVA TV show on Newton and commented on what a jerk that Hooke was, where Hooke demonized Newton and even tried to steal Newton's work. Now there is no law against demonizing people and doing it on the Internet, but I think there should be a ethical code in the teaching profession of colleges and universities that when they have a member of the campus demonizing people on the Internet that those persons should be put on an exit path from that college or university. Schwartz removed from that California University and Kurtz from that Arizona University as people who cannot discuss ideas and spend their lifetime in demonizing other people. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake <1w6AQ0VW8aguTMPq54mlxSiR0Ca8@4ax.com> > [...] > Back many years ago he was Ludwig Plutonium, working as a dishwasher > at Dartmouth in Hanover VT. Then he was carrying on about Fermat's > Last Theorem being false, and he generated similarly long threads. I > posted a satire about him which you might find amusing. Amazingly > enough, someone has preserved it at > http://harlie.idsfa.net/~john/humor/fermat.chicken > You might get a chuckle out of it. Yes, exactly. You could also use an anecdote like the following: the blackboard and called on Ludwig, one of the boys, to multiply it out. Ludwig said, There is no practical applicaton of this problem to the real world, so I won't do it. The problem you really want me to work out is 1 + 1, which does have applications. The answer is 2. Ludwig got an F for that day and never could figure out why. --- Christopher Heckman === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake <1w6AQ0VW8aguTMPq54mlxSiR0Ca8@4ax.com> this was pre-googolplex, as I recall, but I recommend the sci.math topic, Heaven's TV by Kurt Russel (?... I mean, Russel Myers, maybe the famous director). --ils ducs chez Kyoto! http://tarpley.net/bush8.htm http://larouchepub.com/other/2002/2903_chapter_11.html http://clowder.net/zubek/zubek.html http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake > On 19 Nov 2005 21:22:39 -0800, Proginoskes >>snip.... >>You're not even getting your physics right here, Ludwig von Ludwig. >> --- Christopher Heckman > Back many years ago he was Ludwig Plutonium, working as a dishwasher > at Dartmouth in Hanover VT. Actually, Hanover is in New Hampshire, not Vermont, though you can cross the Connecticut river from Hanover and be in Vermont in a minute or so. Rick === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake On Sun, 20 Nov 2005 14:27:01 -0500, Rick Decker >> Back many years ago he was Ludwig Plutonium, working as a dishwasher >> at Dartmouth in Hanover VT. >Actually, Hanover is in New Hampshire, not Vermont, though you >can cross the Connecticut river from Hanover and be in Vermont >in a minute or so. Yes, of course you're right. One of my high school classmates teaches there and lives cross the river and I have been there for a visit. Just a normal brain lapse. --Lynn === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake <1w6AQ0VW8aguTMPq54mlxSiR0Ca8@4ax.com> > On 19 Nov 2005 21:22:39 -0800, Proginoskes >You're not even getting your physics right here, Ludwig von Ludwig. > Back many years ago he was Ludwig Plutonium, working as a dishwasher > at Dartmouth in Hanover VT. Then he was carrying on about Fermat's > Last Theorem being false, and he generated similarly long threads. I > posted a satire about him which you might find amusing. Amazingly > enough, someone has preserved it at > http://harlie.idsfa.net/~john/humor/fermat.chicken > You might get a chuckle out of it. I'm well aware of AP's past history as a crank. I'm trying not to let this affect my judgment of his ideas, and barely making it at the moment. He does have a couple of problems, though: First, he doesn't have the mathematics background to claim to have proved anything. Second, and more importantly, he doesn't recognize his first problem. He may have some good ideas, but good ideas are not enough for a proof. (In fact a lot of great ideas lead nowhere.) Third, AP can't seem to let go of his mistakes and views attacks on them as attacks on him. BTW, are you the Lynn Kurtz at Arizona State? --- Christopher Heckman === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake On 20 Nov 2005 01:13:49 -0800, Proginoskes >BTW, are you the Lynn Kurtz at Arizona State? > --- Christopher Heckman --Lynn === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake <1w6AQ0VW8aguTMPq54mlxSiR0Ca8@4ax.com> > On 20 Nov 2005 01:13:49 -0800, Proginoskes >BTW, are you the Lynn Kurtz at Arizona State? I'm the Christopher Heckman who teaches there. --- Christopher Heckman === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake On 20 Nov 2005 16:06:16 -0800, Proginoskes >> On 20 Nov 2005 01:13:49 -0800, Proginoskes >>BTW, are you the Lynn Kurtz at Arizona State? >> >I'm the Christopher Heckman who teaches there. > --- Christopher Heckman Yes, I had that figured out some time ago. I'm sure our paths have crossed but I don't recall ever meeting you. --Lynn === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake > He does have a couple of problems, though: First, he doesn't have the > mathematics background to claim to have proved anything. Second, and > more importantly, he doesn't recognize his first problem. He may have > some good ideas, but good ideas are not enough for a proof. (In fact a > lot of great ideas lead nowhere.) Third, AP can't seem to let go of his > mistakes and views attacks on them as attacks on him. Those three problems are common to most cranks. Stephen === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake > Recently had a email from South-America wanting to know what my > counterexample was for the 4-color mapping since it is a ill-defined > problem. I gave my counterexamples I believe circa 1999, but perhaps I > should re-do that work in a more clear and concise exposition. > Counterexample that 4-Color Mapping is ill-defined and a fakery: > Whenever we ignore essential elements of a proof domain such as the > borderlines in color mapping we no longer have a problem of mathematics > but rather something of sensory perception or psychology or sense > feelings. The 2-Color Mapping proves via Jordan Curve theorem that 2 > colors are necessary and sufficient to color all mapps. > The counterexamples I gave circa 1999 was that of information theory > that you never need 4 things to communicate any piece of information, > you need only 2 things. All of language, all of pictures, paintings, > drawings, all verbal communication, all visual communication is built > of 2 things and never is 4 things needed to make any of these > communications. Now when we look at maps, those are a subset of all > verbal or visual communication and thus being a tiny subset of all > visual communication and since visual communication is all built from 2 > things, not 4 things then 2 Colors suffice to color all maps. > CounterExample Proof that the 4-Color Mapping is a fakery. I will use > the Computer to be ironic and funny and a slap in the face to those who > have been mislead and those that are misleading of young people about > Color Mapping. These people are not mathematicians but crooks of > mathematics and logic. > A Computer runs on 2 things-- gate open or gate closed. Binary > operation. Gate open is represented by 1 and gate closed is represented > by 0. Now a computer can display any and every map. Now suppose the 4 > Color Mapping was well defined and true as per Appel and Haken claims. > That would imply that there is a mapp for which computers are unable to > ever display because this map requires not just two gates of either > open or closed but 4 gates. > Further yet, we are going to be even more universal by saying that > writing on a computer and that every screen of a computer whether it be > pictures or writing or text or anything is a mapping. And in this ocean > of universality we are going to say that the gate open or gate closed > of binary of 0 or 1 are 2 colors. > So, now, if the 4 Color Mapping of Appel and Haken and many others who > believe in this nonsense is a true and valid work of mathematics, then > there must be a thought or idea or graphic or map or text or diagram > that cannot be displayed by a computer because its 2 gates of open or > closed are insufficient. So what the Appel and Haken nonsense implies > is that a computer has to be built that runs on 4 things and not 2, in > order to represent that item. Jeez, AP. Try displaying a square. It has 4 sides. Not 2. 4. Which is easily represented in binary by 100. Right? Ken > And to further show how ridiculous and stupid the Appel and Haken 4 > Color Mapping is if true, is that it treads even into the subject of > physics and would say that positive and negative charge are not > adequate to cover all matter and that there are 4 types of charges to > cover all matter. Moving on further into physics, if the Appel and > Haken 4 Color Mapping nonsense was true then Complimentarity in physics > is also false because somethings need 4 things and not just 2 and that > Mathematicians rarely have the brains to see when they bring a dude > into the world that their dude wrecks havoc all over the place. They > spend most of their time trying to explain their fake method to others > because it is hidden and arcane and abstract, not because it is > difficult but because it is a fake from the start. The Appel and Haken > computer alleged proof is hundreds of pages long in computer program > and the Wiles's FLT is likewise hundreds and thousands of pages long > because a dude and fake do not clarify the world, they hide from the > world of clear thought and view. So Appel and Haken and Wiles spend > most of their time explaining their fakery to the world that is duped. > But what the world should do for these fake proofs is ask the > opposite-- suppose they are true. Suppose the 4-Color Mapping and Wiles > FLT is true then how does that destroy common everyday truths about the > world. For 4-Color Mapping it denies that computers can run on just 2 > gates of open and closed, denies that physics is built on 2 things-- > goes to infinity. Wiles's FLT in a sentence says that infinity ends, > when the truth is that infinity goes on forever where 1,2,3,.... > becomes ....9999 > Ironic and funny that Appel and Haken have hid and hidden themselves > behind a computer alleged proof and above I am offering a computer > argument that shows their 4 Color Mapping is a utter and total dude > fakery. > Archimedes Plutonium > www.iw.net/~a_plutonium > whole entire Universe is just one big atom > where dots of the electron-dot-cloud are galaxies === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake Jeez, AP. Try displaying a square. It has 4 sides. Not 2. 4. Which is easily represented in binary by 100. Right? If it takes human civilization only 2 parameters to construct all visual languages-- black ink on white paper; takes only 2 parameters to construct all visual information -- black ink on white paper. Then only some buffoons in the mathematics departments would come along and say that a subset of all information needs 4 parameters. Tell me Ken, when was the last time you saw physicists making themselves into a buffoon by ignoring critical elements of a proving domain ( as in i.e. mathematicians ignoring borderlines in color mapping). I mean, say, the physicists building the Mars rovers last year, did they ignore say a force of gravity or EM forces etc etc. How about in sports, where we ignore say a force in the activity. Ignore gravity in diving at the Olympics, or say ignore EM in the marathon races. And have the judges at the Olympics ignore some forces. You see, only in mathematics can you have a academic environment where buffoons can foist a world class fake onto the general public. 4 Color Mapping is not mathematics. It is visual sense perception of psychology. 4 Color Mapping is like saying Beauty is high up and a troop of mathematicians thinking that this is a statement of mathematics when it is sense perception, not mathematics. There is no example in math or physics except the 4 Color Mapping where key elements of the proof domain are ignored. That is why it takes a computer program to hide this mockery of mathematics. It is not mathematics because no statement of mathematics can ignore elements of a proof domain. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: counterexample that demonstrates 4-Color Mapping is a fake your arguments become increasingly abstruse, as you throw the kitchen sink into it & it shatters into a billion porcelain pieces. anyway, with scissors, you only need one color, called paper. everyone but you knows how it's defined, although the computerized simulacrum of the proof is certainly up in the air (see what the author of _Four Colors Suffice_ actually says; supposedly, though, others have improved it, but I don't care; there has to be simpler methods, probably not involving the Jordan curve or n-adics .-) > feelings. The 2-Color Mapping proves via Jordan Curve theorem that 2 > colors are necessary and sufficient to color all mapps. --ils ducs chez Kyoto! http://tarpley.net/bush8.htm http://larouchepub.com/other/2002/2903_chapter_11.html http://clowder.net/zubek/zubek.html http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Quick proof that int_0^infty sin x /x dx = pi/2 by parametric integration > can someone give me a quick explanation why > integral from 0 to infinity of (sin(x)/x) dx > is pi/2? Why? Because that is the way it is. Dirichlet did it this way. Start from the elementary int_0^infty exp(-ax) cos bx dx = a/(a^2+b^2) integrate wrt b int_0^infty exp(-ax) (sin bx)/x dx = arctan (b/a) (a>0) Uniform convergence in [0,a] results from Abel's criterion (known by Dirichlet); make a->0 and you have int_0^infty (sin bx)/x dx = pi/2 Integrating again in b one gets the pretty result int_0^infty (1- cos bx)/x^2 dx = (pi/2) |b| and for b=2 int_0^infty (sin x/x)^2 dx = pi/2 === Subject: Re: Quick proof that int_0^infty sin x /x dx = pi/2 by parametric integration Sorry I forgot to qualify int 0^infty (sin bx)/x Êdx = pi/2 (b>0) === Subject: Re: Differentiation > For quaternions, yes. You get two derivatives, called left quaternion > derivative and right quaternion derivative of a quaternion differentiable > function, respectively.= > And what do you mean by that? > lim_{h->0} (f(x+h)-f(x))*h^{-1} > with limit in the quaternions seldom exists. For example, > it does not exist when f(x) = x^2 ... _differentiable_ function -- Ioannis --- Eventually, _everything_ is understandable === Subject: bounded variation Have read various (all quite similar definitions) of bounded variation (eg http://en.wikipedia.org/wiki/Bounded_variation) im slightly confused by what it is do i do the sum first then take the sup? (but isnt that just going to be the same as if i didnt take the sup?) if i take the sup of each term? that doesnt seem to make sense to me either... anyone shed some light on this.. === Subject: Re: bounded variation > Have read various (all quite similar definitions) of bounded variation > (eg http://en.wikipedia.org/wiki/Bounded_variation) > im slightly confused by what it is > do i do the sum first then take the sup? (but isnt that just going to be the > same as if i didnt take the sup?) Take the sup over all possible sums. If a function f is differentiable, then the total variation on an interval (a,b) could be defined as integral (t=a..b) |f'(t)| dt. (i.e. the integral of the norm of the derivative). The function has bounded variation if that is finite. === Subject: Re: bounded variation > If a function f is differentiable, then the total variation on an > interval (a,b) could be defined as integral (t=a..b) |f'(t)| dt. > (i.e. the integral of the norm of the derivative). The function has > bounded variation if that is finite. I have a question about the integral you mentioned. Is this a Riemann integral or a Lebesgue integral? Does it matter? What I mean is, for the equation V(f) = integral( | f ' | ) to hold, is it neccesary that | f ' | exist AND be Riemann integrable? Example: Let f (x) = x * Sqrt[x] sin(1 / x) when x=/=0, and f (0) = 0. We want to know the variation of f on the interval [0,1]. It can easily be checked that f ' exists on the interval [0,1], and that | f ' (x) | = | 3 / 2 Sqrt[x] sin (1 / x) - cos(1 / x) / Sqrt[x] | when x=/=0, and f ' (0) = 0. Thus, f ' is unbounded on [0,1], so it is not Riemann integrable. However, the Lebesgue integral of f ' on [0,1] exists and is finite ( | f ' | is bounded above by 1 / Sqrt[x], whose improper integral and Lebesgue integral converge on [0,1]). Does this imply that f is of bounded variation on [0,1]? === Subject: Re: bounded variation > If a function f is differentiable, then the total variation on an > interval (a,b) could be defined as integral (t=a..b) |f'(t)| dt. > (i.e. the integral of the norm of the derivative). The function has > bounded variation if that is finite. > I have a question about the integral you mentioned. Is this a Riemann > integral or a Lebesgue integral? Does it matter? What I mean is, for > the equation V(f) = integral( | f ' | ) to hold, is it neccesary that | > f ' | exist AND be Riemann integrable? Example: > Let f (x) = x * Sqrt[x] sin(1 / x) when x=/=0, and f (0) = 0. We want > to know the variation of f on the interval [0,1]. It can easily be > checked that f ' exists on the interval [0,1], and that > | f ' (x) | = | 3 / 2 Sqrt[x] sin (1 / x) - cos(1 / x) / Sqrt[x] | when > x=/=0, and f ' (0) = 0. Thus, f ' is unbounded on [0,1], so it is not > Riemann integrable. However, the Lebesgue integral of f ' on [0,1] > exists and is finite ( | f ' | is bounded above by 1 / Sqrt[x], whose > improper integral and Lebesgue integral converge on [0,1]). Does this > imply that f is of bounded variation on [0,1]? suppose I have some kind of obligation to respond. However, to answer your question, I would have to actually think, which I don't do so well anymore :-) But you seem to have sufficient grasp of the material to figure out the answer yourself, so consider it a homework problem! === Subject: Re: bounded variation > suppose I have some kind of obligation to respond. However, to answer > your question, I would have to actually think, which I don't do so > well anymore :-) But you seem to have sufficient grasp of the material > to figure out the answer yourself, so consider it a homework problem! :) I probably should have spent more time thinking about that question myself before posting it. === Subject: Re: bounded variation thank you for all your help. === Subject: Re: bounded variation > Have read various (all quite similar definitions) of bounded > variation (eg http://en.wikipedia.org/wiki/Bounded_variation) > im slightly confused by what it is > do i do the sum first then take the sup? (but isnt that > just going to be the same as if i didnt take the sup?) > if i take the sup of each term? that doesnt seem to make > sense to me either... > anyone shed some light on this.. I only saw one definition of bounded variation at the web page you gave, and it seems fine to me. As for which you do first, note that the definition says the supremum is taken over all partitions of [a,b]. Since the sum doesn't make sense until a partition is chosen, it has to be that the supremum comes last. It also doesn't make sense to take the supremum before the sum because then you wouldn't be taking the supremum of a collection of real numbers, and even if you somehow managed to do this, there would be no point to the sum since there would then only be one number involved (the number that is the supremum). You probably need to see some specific examples worked out to clear things up. Some of these might help: In particular, try: p. 210 in A First Course in Analysis by George Pedrick p. 252 in Mathematical Analysis 2edmetallury 2edse p. 245 in Advanced Mathematical Methods by Adam Ostaszewski Dave L. Renfro === Subject: Re: bounded variation >Have read various (all quite similar definitions) of bounded variation >(eg http://en.wikipedia.org/wiki/Bounded_variation) >im slightly confused by what it is >do i do the sum first then take the sup? (but isnt that just going to be the >same as if i didnt take the sup?) >if i take the sup of each term? that doesnt seem to make sense to me >either... >anyone shed some light on this.. The idea of total variation is to capture all the ups and downs of a function. Consider f(x) = sin(x) on [0, 2pi]. If you took the partition P: {0, pi, 2pi}. Then sum_P |sin(x_i) - sin(x_(i-1) | is just: | sin(pi) - sin(0) | + sin(2pi) - sin(pi) | = 0. This is a poor choice because we didn't pick up any variation in the function. Contrast that with the partition {0, pi/2, 3pi/2, 2pi}. Now you get the sum: |sin(pi/2) - sin(0)| + | sin(3pi/2) - sin(pi/2) | + | sin(2pi) - sin(3pi/2) | = 1 + 2 + 1 = 4. In this case we have in fact picked up all the ups and downs of the function. Add some more points to this last example and you will see that it never gets greater than 4. Pick a completely different partition and you will get a number between 0 and 4 in this example. So the total variation in a function is the supremum of all values over all partitions, and a function is of bounded variation if this number is finite. Hope that helps. --Lynn === Subject: Re: bounded variation > Have read various (all quite similar definitions) of bounded variation > (eg http://en.wikipedia.org/wiki/Bounded_variation) > im slightly confused by what it is > do i do the sum first then take the sup? (but isnt that just going to be the > same as if i didnt take the sup?) > if i take the sup of each term? that doesnt seem to make sense to me > either... The number of terms can be as large as you like, so probably no single sum achieves the supremum. If your function is piecewise monotone (a finite number of local maxima and minima), then achieve the supremum by choosing those maxima and minima for your sum. But usually that will not be the case. > anyone shed some light on this.. === Subject: Special Matrix Names????? Hello everyone, I am a newbie so please bear with me. I have three seperate matrices which I am sure are special cases and have names for them. Does anyone know what the names are? They only seem to depend on the i and j elements. #1 if i = j A(i,j) = i + j Otherwise A(i,j) = i*j This one has the very cool property of having each successive number within a row the sum of the previous number plus the row number (except diagonal). Does it have a name??? #2 if i = j A(i,j) = 2 if |i-j| = 1 A(i,j) = -1 Otherwise A(i,j) = 0 I believe this one is simply tridiagonal only? #3 A is symmetric with A(i,j) = i/j for j>= i Are there any special properties of these matrices and do they have names??? OxF1 === Subject: Re: Special Matrix Names????? >I have three seperate matrices which I am sure are special cases and >have names for them. Does anyone know what the names are? They only >seem to depend on the i and j elements. >A is symmetric with A(i,j) = i/j for j>= i >Are there any special properties of these matrices and do they have >names??? I don't know if there are special names, but #3 is interesting: its inverse is tridiagonal. In fact A = L L^T where L is lower triangular with L_{ij} = sqrt(2j-1)/i for i >= j; L^(-1) is bidiagonal with (L^(-1))_{ii} = i/sqrt{2i-1}, (L^(-1))_{i,i-1} = -(i-1)/sqrt(2i-1}. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Special Matrix Names????? >I have three seperate matrices which I am sure are special cases and >have names for them. Does anyone know what the names are? They only >seem to depend on the i and j elements. >#3 >A is symmetric with A(i,j) = i/j for j>= i >Are there any special properties of these matrices and do they have >names??? > I don't know if there are special names, but #3 is interesting: > its inverse is tridiagonal. In fact A = L L^T where L is lower > triangular with L_{ij} = sqrt(2j-1)/i for i >= j; L^(-1) is bidiagonal > with (L^(-1))_{ii} = i/sqrt{2i-1}, (L^(-1))_{i,i-1} = -(i-1)/sqrt(2i-1}. Slightly more generally: For any sequence {g(j): j=1..n} of distinct nonzero numbers, consider the matrix A with A_{i,j} = A_{j,i} = g(i)/g(j) for j >= i. Then A^(-1) is tridiagonal; A = L L^T where L is lower triangular, and L^(-1) is bidiagonal with (L^(-1))_{i,i} = g(i)/sqrt(g(i)^2 - g(i-1)^2) (L^(-1))_{i,i-1} = - g(i-1)/sqrt((g(i)^2-g(i-1)^2) (where g(0) = 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: Special Matrix Names????? I think some psychologists would say that the (correlation) matrix A is a simplex, or that it has simplex structure. >I have three seperate matrices which I am sure are special cases and >have names for them. Does anyone know what the names are? They only >seem to depend on the i and j elements. >#3 > >A is symmetric with A(i,j) = i/j for j>= i >Are there any special properties of these matrices and do they have >names??? > I don't know if there are special names, but #3 is interesting: > its inverse is tridiagonal. In fact A = L L^T where L is lower > triangular with L_{ij} = sqrt(2j-1)/i for i >= j; L^(-1) is bidiagonal > with (L^(-1))_{ii} = i/sqrt{2i-1}, (L^(-1))_{i,i-1} = -(i-1)/sqrt(2i-1}. > Slightly more generally: > For any sequence {g(j): j=1..n} of distinct nonzero numbers, > consider the matrix A with A_{i,j} = A_{j,i} = g(i)/g(j) for j >= i. > Then A^(-1) is tridiagonal; A = L L^T where L is lower triangular, and > L^(-1) is bidiagonal with > (L^(-1))_{i,i} = g(i)/sqrt(g(i)^2 - g(i-1)^2) > (L^(-1))_{i,i-1} = - g(i-1)/sqrt((g(i)^2-g(i-1)^2) > (where g(0) = 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: Special Matrix Names????? I've never seen #1 before. #2 is a very important engineering matrix and it turns up all the time. It is the finite elements form of the second derivative operator. You could call it the finite elements second derivative matrix. :) I've seen something like matrix #3 A(i,j) = 1 / (i + j) It occurs in the standard example of a system of lineal equations that is very difficult to solve numerically because of the way small errors get magnified. In essence, it is a harmless looking matrix whose inverse contains really huge numbers. However, your matrix may be better behaved. === Subject: Re: Special Matrix Names?????- >I've never seen #1 before. >#2 is a very important engineering matrix >and it turns up all the time. >It is the finite elements form of the second derivative operator. >You could call it the finite elements second derivative matrix. :) >I've seen something like matrix #3 >A(i,j) = 1 / (i + j) >It occurs in the standard example of a system of lineal equations >that is very difficult to solve numerically because of >the way small errors get magnified. >In essence, it is a harmless looking matrix whose inverse >contains really huge numbers. Not an answer to any of your questions, but interesting enough to deserve mention: NxN matrices with Aij = 1/(i+j-1) for i, j = 1...N are known as Hilbert matrices. They are notorious for their ill-conditionedness, and so are favorites in performance tests of linear algebra software systems. Inversion is a key test; the inverses of Hilbert matrices consist of integers. The condition number, i.e. the ratio of the largest eigenvalue to the smallest, grows exponentially with the order N of the matrix. Take a look at http://mathworld.wolfram.com/HilbertMatrix.html >However, your matrix may be better behaved. Happy computing: Johan E. Mebius === Subject: Re: Rational points on schemes > I tried that earlier since it's the definition of f(x). I haven't done > classical algebraic geometry either, so maybe I'm missing some > correspondences. Try to boil the situation down by choosing an affine neighborhood U of x and translating things in ring theoretic terms. J. === Subject: Re: Rational points on schemes <437D8D9F.8020108@web.de> <437FA24E.6070608@web.de> > I tried that earlier since it's the definition of f(x). I haven't done > classical algebraic geometry either, so maybe I'm missing some > correspondences. > Try to boil the situation down by choosing an affine neighborhood U of x > and translating things in ring theoretic terms. > J. So it boils down to showing that for an algebra map k[x_1,...,x_n] -> R there is a map Spec R -> A^n that sends f(x) to (f_1(x), ..., f_n(x)), and that in some terms prime ideals correspond to rational points. I'll try to figure this out... === Subject: Re: [1/3OT]Counting points on elliptic curves over F_p p prime removed a.l.a from f/u. > there is someone that know how Counting points on elliptic curves over > F_p p prime? (an easy way to compute #E(F_p)=order of group over > elliptic curve E, mod p p prime) > Can i see something on this? Crandall & Pomerance - Prime Numbers, a Computational Perspective The technique you use depends on the size of p. I believe if you grab a copy of the Miracl library, there's a SEA implementation as an example. For moderate sizes, it's a perfectly acceptible algorithm and implementation. Phil -- If a religion is defined to be a system of ideas that contains unprovable statements, then Godel taught us that mathematics is not only a religion, it is the only religion that can prove itself to be one. -- John Barrow === Subject: Re: Power series of elliptic curve differentials > If you look at J. Silverman's book The Arithmetic of Elliptic Curves, > Chapter IV you will find an algorithm for developing power series of > the Weierstrass coordinates of an elliptic curve around infinity. > Two pages further the invariant differential is displayed as a power > series in the new coordinates, but I don't see how to derive that > expansion. > Any help is greatly appreciated. > Oswald. Hi Oswald, In the middle of page 113 are the first few terms for the power series expansions of x(z) and y(z). In order to get the power series for the differential omega(z), use the formula omega(z)/dz = (dx(z)/dz)/(2y(z)+a1*x(z)+a3). In other words, fomally differentiate the series for x(z) as the numerator, and use the series for x(z) and y(z) to express the denominator as a series in z. Both numerator and denominator will then be Laurant series that start -2/z^3, so you can divide (formally) to get a power series in z. It appears, a priori, that the terms in that power series might not be integer polynomials in a1,...,a6, but might have 2's in the denominator. However, as explained at the bottom of page 113, there's another way to do the computation that shows that any denominators would be powers of 3. So in fact only integers are needed. I hope this answers your question, but if not, you can send me an email. Joe Silverman === Subject: Microsoft software for 80 % 0ff by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with SMTP id jAK036902379 for ; Sat, 19 Nov 2005 19:03:09 -0500 --------------------------------------------------------------------- === Subject: Defining the Fibonacci numbers with only one starting value The standard definition of the Fibonacci numbers (FNs) is: F(0) = 0; F(1) = 1; N > 1 => F(N) = F(N-2) + F(N-1). However, the FNs can also be defined, and generated, by: F(0) = 0; F(1) = 1; F(2) = 1; N > 1 => F(2*N-1) = F(N)^2 + F(N-1)^2 and F(2*N) = F(N+1)^2 - F(N-1)^2. (The inelegance of defining three starting numbers instead of two arises from the ambiguity of F(2) = F(2)^2 - F(0)^2). But, thinking this over, I've concluded that a modification of this alternate definition of the FNs actually allows using only _one_ starting number, as in F(3) = 2; N > 0 => F(2*N-1) = F(N)^2 + F(N-1)^2 and F(2*N) = F(N+1)^2 - F(N-1)^2. For N = 2, this yields F(3) = 2 = F(2)^2 + F(1)^2, which can be true only if F(2) = F(1) = 1. And then for N = 1, it yields F(2) = F(2)^2 - F(0)^2, which, for F(2) = 1, can be true only F(0) = 0 (or, F(1) = F(1)^2 + F(0)^2, again true for F(1) = 1 only if F(0) = 0). Interesting, defining a series in which each member except one depends on two others, starting with just one member. === Subject: Re: Defining the Fibonacci numbers with only one starting value > But, thinking this over, I've concluded that a modification > of this alternate definition of the FNs actually allows > using only _one_ starting number, as in > F(3) = 2; > N > 0 => F(2*N-1) = F(N)^2 + F(N-1)^2 and > F(2*N) = F(N+1)^2 - F(N-1)^2. > For N = 2, this yields F(3) = 2 = F(2)^2 + F(1)^2, > which can be true only if F(2) = F(1) = 1. Well, you'll need the assumption that F(n)'s are non-negative integers. > And then for N = 1, it yields F(2) = F(2)^2 - F(0)^2, > which, for F(2) = 1, can be true only F(0) = 0 > (or, F(1) = F(1)^2 + F(0)^2, again true for > F(1) = 1 only if F(0) = 0). Actually few people define F_n to be started from 0. Most of the definition starts from 1. > Interesting, defining a series in which each member except > one depends on two others, starting with just one member. With an extra assumption. === Subject: Re: Defining the Fibonacci numbers with only one starting value <437fddbc$0$51964$892e7fe2@authen.yellow.readfreenews.net> > [...] > Actually few people define F_n to be started from 0. Most of the definition > starts from 1. Most PROFESSIONALS define F_0 to be 0. This leads to the formula F_n = (1/ sqrt(5))[((1 + sqrt(5))/2)^n - ((1 - sqrt(5))/2)^n], as well as the result GCD (F_m, F_n) = F_GCD (m,n). --- Christopher Heckman === Subject: Re: Defining the Fibonacci numbers with only one starting value > Most PROFESSIONALS define F_0 to be 0. This leads to the formula > F_n = (1/ sqrt(5))[((1 + sqrt(5))/2)^n - ((1 - sqrt(5))/2)^n], > as well as the result > GCD (F_m, F_n) = F_GCD (m,n). > --- Christopher Heckman Of course, if there's F_0, then obviously it will be zero since you want all the result to be the same. There's even Fibonacci number with negative index, and I suppose you must know that. But does that important? I didn't say there's no F_0, I just said most people doesn't care about that. Whenever people're talking about Fibonacci numbers, what they have in their mind is 1,1,2,3,5... not 0,1,1,2,3..., that's all. === Subject: Re: Defining the Fibonacci numbers with only one starting value <437fddbc$0$51964$892e7fe2@authen.yellow.readfreenews.net> <43803151$0$41973$892e7fe2@authen.yellow.readfreenews.net> One can also define the Fibonacci numbers with one starting value as follows: F[0] = 0 F[n+1] = ( F[n] + sqrt(5F[n]^2+4(-1)^n) ) / 2 === Subject: Re: Defining the Fibonacci numbers with only one starting value > One can also define the Fibonacci numbers with > one starting value as follows: > F[0] = 0 > F[n+1] = ( F[n] + sqrt(5F[n]^2+4(-1)^n) ) / 2 One can define the Fibonacci numbers without any starting values: F[n] = (a^n - b^n) / (a - b) where a = (1 + sqrt(5)) / 2 and b = (1 - sqrt(5)) / 2. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: matlab help I was wondering if anyone knew how to define (in Matlab) the following nx1 vector x where x equals the transpose of: [x1, x2, x3, ..., xn]. The kind of vector you would use for problems like Ax=b, where you solve for x. === Subject: Re: matlab help You should post this in Matlab's newsgroup. Anyways, the transpose of X is X' === Subject: Re: matlab help > You should post this in Matlab's newsgroup. Anyways, the transpose of X > is X' The Matlab newsgroup is comp.soft-sys.matlab. All Matlab users should make use of it, the Matlab developers are regular participants. A column vector can also be input directly using the semicolon syntax, which means vertical concatenation: X = [x1; x2; x3; x4]; - Randy === Subject: Re: golbach conjeture called _Four Colors Suffice_, shows that they actually didn't finish it. whether or not others have improved it might be of interest ... to others. I'm quite sure that there is at least one elementary proof, but I wouldn't plug it unless I'd gone through it. --ils ducs chez Kyoto! http://tarpley.net/bush8.htm http://larouchepub.com/other/2002/2903_chapter_11.html http://clowder.net/zubek/zubek.html http://www.rwgrayprojects.com/synergetics/plates/plates.html === Subject: Re: golbach conjeture > called _Four Colors Suffice_, shows that they actually didn't finish > it. > whether or not others have improved it might be of interest ... > to others. It also had errors in it. That's why Robertson, Sanders, Seymour, and Thomas started from scratch for THEIR proof of the 4CT. > I'm quite sure that there is at least one elementary proof, > but I wouldn't plug it unless I'd gone through it. After 150 years of studying this problem, no one has come up with an elementary proof. That suggests to me that there ISN'T one, just like the more extreme Fermat's Last Theorem. --- Christopher Heckman === Subject: divergence of improper integral implies divergence of series Let f: [0, oo)-> [0, oo) and [sum](a_k) be given. Assume that for all sufficiently large k and all x in [k, k+1), f(x)<=(a_k). Prove that divergence of the improper integral (from 0 to oo) of f(x) implies divergence of [sum](a_k). (Do I use the comparison test here? If not, I'm clueless...) === Subject: Re: divergence of improper integral implies divergence of series > Let f: [0, oo)-> [0, oo) and [sum](a_k) be given. Assume > that for all sufficiently large k and all x in [k, k+1), > f(x)<=(a_k). Prove that divergence of the improper > integral (from 0 to oo) of f(x) implies divergence of > [sum](a_k). (Do I use the comparison test here? Well...what do you think? Yes, or no? Try giving an answer first, then maybe if you have made an error, people will be willing to help. One small hint: draw a picture. R,G. Vickson > If not, I'm clueless...) === Subject: Re: divergence of improper integral implies divergence of series > Let f: [0, oo)-> [0, oo) and [sum](a_k) be given. Assume > that for all sufficiently large k and all x in [k, k+1), > f(x)<=(a_k). Prove that divergence of the improper > integral (from 0 to oo) of f(x) implies divergence of > [sum](a_k). (Do I use the comparison test here? Well...what do you think? Yes, or no? Try giving an answer first, then maybe if you have made an error, people will be willing to help. One small hint: draw a picture. R,G. Vickson Ok, so I think I need to compare (a_k) with the integral (k to k+1) f(x)dx. What does the inequality f(x) <= a_k on [k,k+1) tell me about this integral? I think it says that: If f(x) <= a_k on [k,k+1), then integral(k to k+1)f(x)dx <= integral(k to k+1)(a_k)dx. Then the integral(1 to oo)f(x)dx= [sum](1 to oo)integral (1 to k+1) f(x)dx. I got this far... but I don't see how I can prove that f(x) diverges and thus that this proves that [sum](a_k) diverges.... === Subject: Re: divergence of improper integral implies divergence of series <24853406.1132456671165.JavaMail.jakarta@nitrogen.mathforum.org> > Let f: [0, oo)-> [0, oo) and [sum](a_k) be given. Assume > that for all sufficiently large k and all x in [k, k+1), > f(x)<=(a_k). Prove that divergence of the improper > integral (from 0 to oo) of f(x) implies divergence of > [sum](a_k). (Do I use the comparison test here? > Well...what do you think? Yes, or no? Try giving an answer first, then > maybe if you have made an error, people will be willing to help. One > small hint: draw a picture. > R,G. Vickson > Ok, so I think I need to compare (a_k) with the > integral (k to k+1) f(x)dx. > What does the inequality f(x) <= a_k on [k,k+1) tell me > about this integral? I think it says that: > If f(x) <= a_k on [k,k+1), then > integral(k to k+1)f(x)dx <= integral(k to k+1)(a_k)dx. Right. And what is the integral of a_k from k to k+1? Hint: it's very easy. By the way: did you take my previous advice to draw a picture? RGV > Then the integral(1 to oo)f(x)dx= > [sum](1 to oo)integral (1 to k+1) f(x)dx. > I got this far... but I don't see how I can prove that > f(x) diverges and thus that this proves that [sum](a_k) > diverges.... === Subject: Re: divergence of improper integral implies divergence of series > Let f: [0, oo)-> [0, oo) and [sum](a_k) be given. Assume > that for all sufficiently large k and all x in [k, k+1), > f(x)<=(a_k). Prove that divergence of the improper > integral (from 0 to oo) of f(x) implies divergence of > [sum](a_k). (Do I use the comparison test here? > Well...what do you think? Yes, or no? Try giving an answer first, then > maybe if you have made an error, people will be willing to help. One > small hint: draw a picture. > R,G. Vickson > Ok, so I think I need to compare (a_k) with the > integral (k to k+1) f(x)dx. > What does the inequality f(x) <= a_k on [k,k+1) tell me > about this integral? I think it says that: > If f(x) <= a_k on [k,k+1), then > integral(k to k+1)f(x)dx <= integral(k to k+1)(a_k)dx. Right. And what is the integral of a_k from k to k+1? Hint: it's very easy. The integral of (a_k) from k to k+1 is the [sum](a_k)? I really have no idea... and how would I represent this by a picture? > Then the integral(1 to oo)f(x)dx= > [sum](1 to oo)integral (1 to k+1) f(x)dx. > I got this far... but I don't see how I can prove that > f(x) diverges and thus that this proves that [sum](a_k) > diverges.... === Subject: algebra (?) number theory I am preparing for GRE math, and I dont understand how this problem is solved (the book im preparing with does not have solutions). Let x and y be positive integers such that 3x + 7y is divisible by 11. Which of the following must also be divisible by 11? (a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 The answer in the book is (d), but dunno how they got it? (I was looking in my number theory book and read about the linear Diophantine equation, but im not sure if I should apply it in here!?) === Subject: Re: algebra (?) number theory > I am preparing for GRE math, and I dont understand how this problem is > solved (the book im preparing with does not have solutions). > Let x and y be positive integers such that 3x + 7y is divisible by 11. Which > of the following must also be divisible by 11? > (a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 The mathematics has been pretty well explored in this thread, so here's something of a historical note. The Eotvos contests for Hungarian students in their last year of high school began in 1894. The first problem on that first exam was, Prove that the expressions 2x + 3y and 9x + 5y are divisible by 17 for the same set of integral values of x and y. See Hungarian Problem Book 1, which was Volume 11 in the New Mathematical Library, published in 1963. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: algebra (?) number theory Thx FOR all The answers. It was really like another review.. (using all ur answers) The way I would solve it next time, is Since 3x + 7y is divisible by 11, then 3x + 7y = 11z; Now x = (11z - 7y)/3 Note that I can multiply all answers by 3 since (3,11) = 1; Plugging x in each of the equations multiplied by 3, I get (A) 44 z - 28y + 18 y = 44 z - 10y (B) 11z - 7y + 3y + 15 = 11z - 4y + 15 (C) 99z - 63z + 12y = 99z - 51y (D) 44z -28y -27y = 44z - 55y <<<< (I would stop here and answer D!) === Subject: Re: algebra (?) number theory BTW, this was mainly Peter Schorn Answer. > Thx FOR all The answers. It was really like another review.. > (using all ur answers) > The way I would solve it next time, is > Since 3x + 7y is divisible by 11, > then 3x + 7y = 11z; > Now x = (11z - 7y)/3 > Note that I can multiply all answers by 3 since (3,11) = 1; > Plugging x in each of the equations multiplied by 3, I get > (A) 44 z - 28y + 18 y = 44 z - 10y > (B) 11z - 7y + 3y + 15 = 11z - 4y + 15 > (C) 99z - 63z + 12y = 99z - 51y > (D) 44z -28y -27y = 44z - 55y <<<< (I would stop here and answer D!) === Subject: Re: algebra (?) number theory dnblqZ4MPPR3enZ2dnUVZ_sKdnZ2d@comcast.com: > BTW, this was mainly Peter Schorn Answer. >> Thx FOR all The answers. It was really like another review.. >> (using all ur answers) >> The way I would solve it next time, is >> Since 3x + 7y is divisible by 11, >> then 3x + 7y = 11z; >> Now x = (11z - 7y)/3 >> Note that I can multiply all answers by 3 since (3,11) = 1; >> Plugging x in each of the equations multiplied by 3, I get >> (A) 44 z - 28y + 18 y = 44 z - 10y >> (B) 11z - 7y + 3y + 15 = 11z - 4y + 15 >> (C) 99z - 63z + 12y = 99z - 51y >> (D) 44z -28y -27y = 44z - 55y <<<< (I would stop here and answer D!) Just a couple of comments. (1) If the GRE is a timed multiple-guess exam, and you need to scribble that much algebra on paper to do the above, it may well be too slow to finish the whole exam. This is a few seconds, in your head problem. You might use another responder's idea (note that (3,-7) is trivially true, and adjust to (3,4) if you want the positive constraint), and then just substitute these specific numeric values. The multiplicative inverse idea is also very fast mental arithmetic (3*2 isn't, 3*3 isn't 3*4=12==> 1 is, so change 3x+7y into x+6y first). My hit-and-miss approach to find a sum that was a multiple of 11 was very fast, and will work on the exam since the numbers will be very small integers. [learning the exam more than learning the material] (2) Be sure that the question which... really means which -one- ..., and not which -one or more- .... If the latter, you would need to consider (e) also, and not stop after finding one answer. [more learning the exam] If the former, consider plugging into the potential answers in reverse order. [(e) is trivially not, so you only need to evaluate (d)] Should not reduce the amount of work, with an exam with randomly scrambled answers, but a might help, can't hurt approach. Lynn Killingbeck === Subject: Re: algebra (?) number theory > I am preparing for GRE math, and I dont understand how this problem is > solved (the book im preparing with does not have solutions). > Let x and y be positive integers such that 3x + 7y is divisible by 11. Which > of the following must also be divisible by 11? > (a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 > The answer in the book is (d), but dunno how they got it? > (I was looking in my number theory book and read about the linear > Diophantine equation, but im not sure if I should apply it in here!?) Apart from the more convoluted approaches you can solve problems of this kind easily using modular arithmetic: 3x + 7y is divisible by 11 can be reformulated as 3x + 7y = 0 (mod 11). Solve this equation for x by multiplying it with the multiplicative inverse of 3 (mod 11) which is 4 to obtain 12x + 28y = 0 (mod 11) or x = -28y = 5y (mod 11). Plug this into the given choices to see: (a) 4*5y + 6y = 26y = 4y (mod 11) -> not necessarily 0 (b) 6y + 5 (mod 11) -> not necessarily 0 (c) 45y + 4y = 49y = 5y (mod 11) -> not necessarily 0 (d) 20y - 9y = 11y = 0 (mod 11) -> this is it (e) 5y + y - 1 = 6y - 1 (mod 11) -> not necessarily 0 Peter === Subject: Re: algebra (?) number theory > I am preparing for GRE math, and I dont understand how this problem is > solved (the book im preparing with does not have solutions). > Let x and y be positive integers such that 3x + 7y is divisible by 11. Which > of the following must also be divisible by 11? > (a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 > The answer in the book is (d), but dunno how they got it? > (I was looking in my number theory book and read about the linear > Diophantine equation, but im not sure if I should apply it in here!?) Maybe this is too brute force. But if you multiply the original equation by 5 you don't change the divisibility by 11. So 0 = 3x + 7y = 15x + 35y mod 11 which is just 4x - 9y mod 11. The rationale for multiplying by 5? I just looked for the smallest positive number such that 3k = 4 (11) because of (a). Everything then fell into place for (d). If it hadn't worked, I would have done the same for (b): the smallest k such that 3k = 1 (11). This would have lead to x + 6y When I used to worry about these kind of things, this was my standard don't want to think today approach. Rick === Subject: Re: algebra (?) number theory > I am preparing for GRE math, and I dont understand how this problem is > solved (the book im preparing with does not have solutions). > Let x and y be positive integers such that 3x + 7y is divisible by 11. > Which of the following must also be divisible by 11? > (a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 > The answer in the book is (d), but dunno how they got it? > (I was looking in my number theory book and read about the linear > Diophantine equation, but im not sure if I should apply it in here!?) (b) and (e) can be instantly eliminated (x=y=11 fails). (3x+7y)+(8x+12y)=11x+19y is not necessarily 0 modulo 11. (a) fails (3x+7y)+(9x+4y)=12x+11y is not necessarily 0 modulo 11. (c) fails. That leaves (d) by elimination! OK, if not pressed for time, note that (3x+7y)+(8x-18y)=11x-11y is identically 0 modulo 11. (d) passes. Just using the fact that if each expression is zero modulo 11, then so is any sum or difference; and, doing very quick by inspection sums (didn't even need to try differences) such that the coefficients of one or the other term was a multiple of 11. Is the approach sound? Lynn Killingbeck === Subject: Re: algebra (?) number theory >> I am preparing for GRE math, and I dont understand how this problem is >> solved (the book im preparing with does not have solutions). >> Let x and y be positive integers such that 3x + 7y is divisible by 11. >> Which of the following must also be divisible by 11? >> (a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 >> The answer in the book is (d), but dunno how they got it? >> (I was looking in my number theory book and read about the linear >> Diophantine equation, but im not sure if I should apply it in here!?) >(b) and (e) can be instantly eliminated (x=y=11 fails). >(3x+7y)+(8x+12y)=11x+19y is not necessarily 0 modulo 11. (a) fails >(3x+7y)+(9x+4y)=12x+11y is not necessarily 0 modulo 11. (c) fails. >That leaves (d) by elimination! OK, if not pressed for time, note that >(3x+7y)+(8x-18y)=11x-11y is identically 0 modulo 11. (d) passes. >Just using the fact that if each expression is zero modulo 11, then so >is any sum or difference; and, doing very quick by inspection sums >(didn't even need to try differences) such that the coefficients of one >or the other term was a multiple of 11. Is the approach sound? >Lynn Killingbeck Ok, here's another method, pairing the first part of your agument with another idea ... Since 3x+7y is a multiple 11 of whenever x and y are congruent to 0 mod 11 (for example x=11, y=11), the non-homogeneous candidates are eliminated. Now, since Z_11 is a field, we can use linear algebra. Simply consider the system of 2 equations in 2 unknowns, where the first equation is 3x +7y = 0 and the second is one of the homogeneous candidates. Then take the determinant of the 2 by 2 coefficient matrix mod 11. So for example, for (d), we have the system 3x + 7y = 0 4x - 9y = 0 The 2 by 2 matrix of coefficients is: 3 7 4 -9 with determinant 0 (mod 11) Hence 4x - 9y is linearly dependent on 3x + 7y over the field Z_11. Thus, 3x + 7y = 0 implies 4x - 9y = 0, so already this tells you that if there's only one answer, it has to be (d). To eliminate the others, use the trick noted by MuTsun Tsai -- namely, that 3x + 7y = 0 mod 11 has the obvious integer solution x=7, y=-3. In Z_11, -3 = 8, hence x=7, y=8 must also be a solution. In particular, x=7, y=8 is a nonzero solution in Z_11, and also a positive integer solution to the congruence 3x + 7y = 0 mod 11. So consider the candidate (a) ... The coefficient matrix is 3 7 4 -9 which has determinant = 1, hence, the system of equations 3x + 7y = 0 4x - 9y = 0 has the unique solution x=0, y=0 in Z_11. So the solution (7,8) which solves the first equation can't also be a solution of the the second. Thus, candidate (a) is eliminated. The same reasoning eliminates (c), which then completes the analysis. The Chinese Remainder Theorem is possibly hiding in this argument somewhere, recast in linear algebraic form. quasi === Subject: Re: algebra (?) number theory > I am preparing for GRE math, and I dont understand how this problem is > solved (the book im preparing with does not have solutions). > > Let x and y be positive integers such that 3x + 7y is divisible by 11. > Which of the following must also be divisible by 11? > (a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 > > The answer in the book is (d), but dunno how they got it? > (I was looking in my number theory book and read about the linear > Diophantine equation, but im not sure if I should apply it in here!?) > > > >>(b) and (e) can be instantly eliminated (x=y=11 fails). >>(3x+7y)+(8x+12y)=11x+19y is not necessarily 0 modulo 11. (a) fails >>(3x+7y)+(9x+4y)=12x+11y is not necessarily 0 modulo 11. (c) fails. >>That leaves (d) by elimination! OK, if not pressed for time, note that >>(3x+7y)+(8x-18y)=11x-11y is identically 0 modulo 11. (d) passes. >>Just using the fact that if each expression is zero modulo 11, then so >>is any sum or difference; and, doing very quick by inspection sums >>(didn't even need to try differences) such that the coefficients of one >>or the other term was a multiple of 11. Is the approach sound? >>Lynn Killingbeck >Ok, here's another method, pairing the first part of your agument with >another idea ... >Since 3x+7y is a multiple 11 of whenever x and y are congruent to 0 >mod 11 (for example x=11, y=11), the non-homogeneous candidates are >eliminated. >Now, since Z_11 is a field, we can use linear algebra. Simply consider >the system of 2 equations in 2 unknowns, where the first equation is >3x +7y = 0 and the second is one of the homogeneous candidates. >Then take the determinant of the 2 by 2 coefficient matrix mod 11. >So for example, for (d), >we have the system >3x + 7y = 0 >4x - 9y = 0 >The 2 by 2 matrix of coefficients is: > 3 7 > 4 -9 >with determinant 0 (mod 11) >Hence 4x - 9y is linearly dependent on 3x + 7y over the field Z_11. >Thus, 3x + 7y = 0 implies 4x - 9y = 0, so already this tells you that >if there's only one answer, it has to be (d). >To eliminate the others, use the trick noted by MuTsun Tsai -- namely, >that 3x + 7y = 0 mod 11 has the obvious integer solution x=7, y=-3. In >Z_11, -3 = 8, hence x=7, y=8 must also be a solution. In particular, >x=7, y=8 is a nonzero solution in Z_11, and also a positive integer >solution to the congruence 3x + 7y = 0 mod 11. >So consider the candidate (a) ... >The coefficient matrix is > 3 7 > 4 -9 The above matrix should have been written as: 3 7 4 6 >which has determinant = 1, hence, the system of equations >3x + 7y = 0 >4x - 9y = 0 The above system should have been written as: 3x + 7y = 0 4x + 6y = 0 >has the unique solution x=0, y=0 in Z_11. >So the solution (7,8) which solves the first equation can't also be a >solution of the the second. Thus, candidate (a) is eliminated. >The same reasoning eliminates (c), which then completes the analysis. >The Chinese Remainder Theorem is possibly hiding in this argument >somewhere, recast in linear algebraic form. >quasi Aside from those 2 corrections, the argument should be ok now. quasi === Subject: Re: algebra (?) number theory >> I am preparing for GRE math, and I dont understand how this problem is >> solved (the book im preparing with does not have solutions). >> Let x and y be positive integers such that 3x + 7y is divisible by 11. >> Which of the following must also be divisible by 11? >> (a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 >> The answer in the book is (d), but dunno how they got it? >> (I was looking in my number theory book and read about the linear >> Diophantine equation, but im not sure if I should apply it in here!?) >(b) and (e) can be instantly eliminated (x=y=11 fails). >(3x+7y)+(8x+12y)=11x+19y is not necessarily 0 modulo 11. (a) fails >(3x+7y)+(9x+4y)=12x+11y is not necessarily 0 modulo 11. (c) fails. >That leaves (d) by elimination! OK, if not pressed for time, note that >(3x+7y)+(8x-18y)=11x-11y is identically 0 modulo 11. (d) passes. >Just using the fact that if each expression is zero modulo 11, then so >is any sum or difference; and, doing very quick by inspection sums >(didn't even need to try differences) such that the coefficients of one >or the other term was a multiple of 11. Is the approach sound? Very sound, and simple too. But if there are no easily visible eliminations by inspection, then you would need to bring some kind of reduction method into play as well. quasi === Subject: Re: algebra (?) number theory > Let x and y be positive integers such that 3x + 7y is divisible by 11. > Which of the following must also be divisible by 11? > (a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 > The answer in the book is (d), but dunno how they got it? > (I was looking in my number theory book and read about the linear > Diophantine equation, but im not sure if I should apply it in here!?) Ok, now let me give a full theoretic explanation for this problem In general we may ask the following question : Let x and y be integer such that ax+by==0 (mod m). Find all linear equations of x and y such that it must be congruent to zero under the assumption. By CRT ( Chinese Remainder Theorem ), we know that we may only consider the case where m is a prime number, say p. Now, the linear congruence equation ax+by==0 (mod p) has the general solution (x, -b^(-1)ax) in Z/pZ. In the case we've encounter above, the solution is (x, -2x), since 7^(-1)==8 and 8*3==2 (mod 11). So our problem is to find all A, B and C such that Ax-2Bx+C==0 (mod 11) for any x. Obviously C must be congruent to zero, otherwise take x==0 and you'll have a contradiction. So the equation becomes x(A-2B)==0 (mod 11). Take any invertible x, and you'll have A-2B==0 (mod 11). Obviously this is also sufficient, so our conclusion is : Such equations are of the form Ax+By+C, where A-2B==0 (mod 11) and C==0 (mod 11). Only (d) fits the condition. === Subject: Re: algebra (?) number theory On Sat, 19 Nov 2005 20:39:32 -0500, Someonekicked >I am preparing for GRE math, and I dont understand how this problem is >solved (the book im preparing with does not have solutions). >Let x and y be positive integers such that 3x + 7y is divisible by 11. Which >of the following must also be divisible by 11? >(a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 >The answer in the book is (d), but dunno how they got it? >(I was looking in my number theory book and read about the linear >Diophantine equation, but im not sure if I should apply it in here!?) It's a trick question. The key word in the problem is the word positive. The problem can be done in your head -- no sophisticated math required. quasi === Subject: Re: algebra (?) number theory > The key word in the problem is the word positive. Why? The statement holds for negative integers x and y as well, right? === Subject: Re: algebra (?) number theory >On Sat, 19 Nov 2005 20:39:32 -0500, Someonekicked >>I am preparing for GRE math, and I dont understand how this problem is >>solved (the book im preparing with does not have solutions). >>Let x and y be positive integers such that 3x + 7y is divisible by 11. Which >>of the following must also be divisible by 11? >>(a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 >>The answer in the book is (d), but dunno how they got it? >>(I was looking in my number theory book and read about the linear >>Diophantine equation, but im not sure if I should apply it in here!?) >It's a trick question. >The key word in the problem is the word positive. >The problem can be done in your head -- no sophisticated math >required. >quasi Try 11. Fails, right? Try 22. Works with x=3, y=1. Substituting these must eliminate all but one answer, assuming the problem has exactly one correct answer. Ok, but is there a more analytical way? Sure -- here's a simple, unsophisticated approach, just a kind of playing around with the problem usng congruences. Rewrite the problem this way: 3x + 7y = 11z This could be viewed as a congruence in any of 3 ways: 3x + 7y = 0 mod 11 11z - 3y = 0 mod 7 11z - 7y = 0 mod 3 If you can satisfy any of these congruences with integer values, you at least get integer solutions to the original equation. So stay with integers for now -- we'll worry about positivity later. The last one is attractive since the modulus is small, and the other coefficients easily reduce mod 3. In other words, mod 3, you can replace 11 by 2, or even better by -1, and you can replace 7 by 1. Thus, you get the simple congruence -z - y = 0 mod 3, or equivalently, z + y = 0 mod 3. So any integers z,y satisfying this congruence will yield integer solutions, but since you need positive integer solutions, you also need to insure that x is positive, so you want 11z - 7y > 0. Equivalently, z > (7/11)*y. Thus choose y=1, z=2. Solving for x gives x=3. Then proceed as before, substituting these values into the multiple choice alternatives. quasi === Subject: Re: algebra (?) number theory I don't get it. What am I missing here? The question puts no conditions on x and y other than they be positive integers, so if x = 11, and y = 11, then answers a, c, and d are all divisible by 11, though d is negative; that objection can be removed by choosing x = 44, right? I contend that, given the phrasing of the question, my answer is correct; in fact, it is better than all the other replies because I got more corrct answers than anyone else 8-) Show me where I am wrong, indulging an amateur.. cfe === Subject: Re: algebra (?) number theory >I don't get it. What am I missing here? The question puts no >conditions on x and y other than they be positive integers, so if x = >11, and y = 11, then answers a, c, and d are all divisible by 11, >though d is negative; that objection can be removed by choosing x = 44, >right? I contend that, given the phrasing of the question, my answer >is correct; in fact, it is better than all the other replies because I >got more corrct answers than anyone else 8-) Show me where I am wrong, >indulging an amateur.. >cfe The one word in the problem which you are not taking into account is the word must. Try reading the problem again with special emphasis on that word. quasi === Subject: Re: algebra (?) number theory I stand corrected!!! Hmmm, how does one grovel while standing? Connie The Groveler Eaton === Subject: Re: algebra (?) number theory >>On Sat, 19 Nov 2005 20:39:32 -0500, Someonekicked >I am preparing for GRE math, and I dont understand how this problem is >solved (the book im preparing with does not have solutions). >Let x and y be positive integers such that 3x + 7y is divisible by 11. Which >of the following must also be divisible by 11? >(a) 4x + 6y (b) x + y + 5 (c) 9x + 4y (d) 4x - 9y (e) x + y - 1 >The answer in the book is (d), but dunno how they got it? >(I was looking in my number theory book and read about the linear >Diophantine equation, but im not sure if I should apply it in here!?) As MuTsun Tsai's reply makes clear, positivity can't really be relevant, and hence, a natural way to deal with this problem is to use the Chinese Remainder Theorem to characterize all integer solutions. But my previous, more primitive attack can still be salvaged, with corrections, as follows ... Rewrite the problem this way: 3x + 7y = 11z This could be viewed as a congruence in any of 3 ways: 3x + 7y = 0 mod 11 11z - 3x = 0 mod 7 11z - 7y = 0 mod 3 Any complete solution to any these congruences will yield all integer solutions to the original equation. The last one is attractive since the modulus is small, and the other coefficients easily reduce mod 3. In other words, mod 3, you can replace 11 by 2, or even better by -1, and you can replace 7 by 1. Thus, you get the simple congruence -z - y = 0 mod 3, or equivalently, z + y = 0 mod 3. This is equivalent to z = 3t - y, where t is an arbitrary integer. Substituting z = 3t - y for z in the original equation 3x + 7y = 11z, and then solving for x yields x = 11t - 6y So the general integer solution to 3x + 7y = 11z is x = 11t - 6y, y = y, z = 3t-y Then, substituting x = 11t - 6y into the multiple choices gives (a) 4x + 6y simplifies to -18y + 44t (b) x + y + 5 simplifies to -5y + 11t + 5 (c) 9x + 4y simplifies to -50y + 99t (d) 4x - 9y simplifies to -33y + 44t (e) x+y-1 simplifies to -5y+11t Viewed mod 11, the t terms all drop out, but then it's clear, since y is arbitrary, that only (d) is always a multiple of 11. So my prior argument is now repaired, but still, I readily admit that the Chinese Remainder Theorem is the power tool here, and the right one. By comparison, what I did was just fooling around. quasi === Subject: Re: algebra (?) number theory > Try 11. Fails, right? > Try 22. Works with x=3, y=1. I think you mean x=5, y=1. > Substituting these must eliminate all but one answer, assuming the > problem has exactly one correct answer. Not necessarily of course. Some choice might be correct by coincidence. For example, if you take x=5 and y=1, then (b) is also correct. You'll probably have to try many solutions to have only one left. Well actually since the adjective positive is not important, you can try an even trivial solution, which is x=7 and y=-3. This well indeed eliminate all but one choice. === Subject: Re: algebra (?) number theory On Sun, 20 Nov 2005 10:54:24 +0800, MuTsun Tsai >> Try 11. Fails, right? >> Try 22. Works with x=3, y=1. >I think you mean x=5, y=1. >> Substituting these must eliminate all but one answer, assuming the >> problem has exactly one correct answer. >Not necessarily of course. Some choice might be correct by coincidence. For >example, if you take x=5 and y=1, then (b) is also correct. You'll probably >have to try many solutions to have only one left. >Well actually since the adjective positive is not important, you can try >an even trivial solution, which is x=7 and y=-3. This well indeed eliminate >all but one choice. Yes, you're right, positivity is more of a diversion. It's natural to try to characterize the integer solutions first, by viewing the requirements in terms of congruences. I looked at the problem too quickly and thought it was solved instantly by inspection, seeing that 22 works with x=5, y=1 (by the In fact, I never even tested the alternatives -- I just assumed they would be eliminated. Had I actually tested the alternatives, I would have realized that that I had made a conceptual error, since two of the choices would have survived my test. Also, as you point out, if all I wanted was integer solutions, x=7 and y=-3 are instant solutions which I overlooked. For that matter, x=-7 and y=3 are also instant. I jumped too quickly on this problem. quasi === Subject: Re: More of an Algorithems question 4ax.com: > >> > > >On 15 Nov 2005 14:55:51 -0800, Filter > >> >> >>How can I proove that no Algorithm can compress every file of >>length 10^6? > >However there is an algorithm that will losslessly compress every >file of length 10^6 except one. > > >> I don't believe that. > >> The pigeonhole principle applies easily to invalidate such a >> claim ... > >> Let n=10^6. > >> There are 2^n possible files with n bits, but only 2^(n-1) files >> with n-1 bits. Hence, at most half the files can be compressed. > >There are 2^n-1 files with fewer than n bits. Hence, the claim >stands. > >> Not by my count. > >> For example, 16 possible files with 4 bits but only 8 possible >> files with 3 bits, no? > >Are there any other files with fewer than 4 bits that you have >overlooked in your count? >> >>Ok, I see my confusion. >> >>I failed to take into account files with less than (n-1) bits. >> >> >>quasi > > Ok, but then along the lines of what I was thinking, the following > claim is valid: > > In any lossless compression scheme, less than half of the files with n > or fewer bits can be compressed. > > quasi >>Isn't less than half even stronger as none? A 1-bit file cannot be >>compressed, since the only smaller file (0 bits) is needed to >>represent itself. Then a 2-bit file cannot be compressed, since all of >>the 0- and 1-bit files are needed to represent themselves. Induction. > Yes, I saw that the two 1-bit files were a problem as far compression, > but I was assuming that a compression scheme is allowed to compress > some files while potentially expanding others. Thus, a compression > scheme could map some files of <=n bits to more than n bits, and, as > you point out, something like that is necessary. If anything gets > smaller, something has to get bigger -- I guess it's a kind of > conservation law for bits. > quasi Can one conserve the total size, over all files? For example, suppose we have all 5-bit and smaller files, and some 5-bit file compresses to a 2-bit file. The original 2-bit file has to become larger - and there is a natural empty slot where the uncompressed 5-bit file came from. The net effect is that the 2-bit and the 5-bit file are simply interchanged, with one compressing and the other growing, while the total size over all files remains constant. If the above holds, and if there is a theorem that any random permutation of the integers 1,2,3,...,N can be re-ordered (or, even, placed into any other order) by pair-wise interchanges, then the result should be a theorem that the total size can be retained (or, of course made bigger). The truth of the pair-wise swap idea seems obvious: works trivially for N=2; and for N>2 do a swap that moves any number into its final position, reducing the remaining numbers to the same problem with only (N-1) to re-order. Lynn Killingbeck === Subject: Re: More of an Algorithems question >4ax.com: >> > >> >> >>On 15 Nov 2005 14:55:51 -0800, Filter >> > > >How can I proove that no Algorithm can compress every file of >length 10^6? >> >>However there is an algorithm that will losslessly compress >every >>file of length 10^6 except one. >> >> > I don't believe that. >> > The pigeonhole principle applies easily to invalidate such a > claim ... >> > Let n=10^6. >> > There are 2^n possible files with n bits, but only 2^(n-1) files > with n-1 bits. Hence, at most half the files can be compressed. >> >>There are 2^n-1 files with fewer than n bits. Hence, the claim >>stands. >> > Not by my count. >> > For example, 16 possible files with 4 bits but only 8 possible > files with 3 bits, no? >> >>Are there any other files with fewer than 4 bits that you have >>overlooked in your count? > >Ok, I see my confusion. > >I failed to take into account files with less than (n-1) bits. > > >quasi >> >> Ok, but then along the lines of what I was thinking, the following >> claim is valid: >> >> In any lossless compression scheme, less than half of the files with >> or fewer bits can be compressed. >> >> quasi >Isn't less than half even stronger as none? A 1-bit file cannot be >compressed, since the only smaller file (0 bits) is needed to >represent itself. Then a 2-bit file cannot be compressed, since all of >the 0- and 1-bit files are needed to represent themselves. Induction. >> Yes, I saw that the two 1-bit files were a problem as far compression, >> but I was assuming that a compression scheme is allowed to compress >> some files while potentially expanding others. Thus, a compression >> scheme could map some files of <=n bits to more than n bits, and, as >> you point out, something like that is necessary. If anything gets >> smaller, something has to get bigger -- I guess it's a kind of >> conservation law for bits. >> quasi >Can one conserve the total size, over all files? For example, suppose we >have all 5-bit and smaller files, and some 5-bit file compresses to a >2-bit file. The original 2-bit file has to become larger - and there is a >natural empty slot where the uncompressed 5-bit file came from. The net >effect is that the 2-bit and the 5-bit file are simply interchanged, with >one compressing and the other growing, while the total size over all >files remains constant. >If the above holds, and if there is a theorem that any random >permutation of the integers 1,2,3,...,N can be re-ordered (or, even, >placed into any other order) by pair-wise interchanges, then the >result should be a theorem that the total size can be retained (or, >of course made bigger). The truth of the pair-wise swap idea seems >obvious: works trivially for N=2; and for N>2 do a swap that moves >any number into its final position, reducing the remaining numbers >to the same problem with only (N-1) to re-order. >Lynn Killingbeck Yes, as you suggest, any permutation on the set of n-or-less bit files can be used to define a compression scheme. Note that if such a compression scheme is used on the full set of n-or-less bit files, less than half of the files will get smaller. Of course, as I pointed out earlier in this thread, the pigeonhole principle implies that in any compression scheme, less than half of the full set of n-or-less bit files will get smaller. Also, for the permutation scheme, at least one set must also end up with exactly the same number of bits, so might as well let the permutation fix that one. quasi === Subject: prove about ceil function I have some question. But I can't find any solution. Is there anyone who can help me? Prove that if n is odd, then [n^2/4] = (n^2+3)/4 === Subject: Re: prove about ceil function >I have some question. But I can't find any solution. Is there anyone >who can help me? >Prove that if n is odd, then [n^2/4] = (n^2+3)/4 Hint: If n is odd, what can you say about the remainder when n^2 is divided by 4? quasi === Subject: Re: prove about ceil function >I have some question. But I can't find any solution. Is there anyone > who can help me? > Prove that if n is odd, then [n^2/4] = (n^2+3)/4 Write n=2k+1, then n^2/4=(4k^2+4k+1)/4. The term 4k^2+4k is divisible by 4 but the constant 1 does not. So to get the ceiling, you have to add 3 to get a multiple of 4. Hence [n^2/4] = (n^2+3)/4. === Subject: Boolean function problem. It seems easy, but difficult to me. Let p be a Boolean variable. How many different Boolean functions of p are there? How many different Boolean functions of two Boolean variables are there? I'm confusing. Is there anyone who can help me? === Subject: Re: Boolean function problem. It seems easy, but difficult to me. >Let p be a Boolean variable. How many different Boolean functions of p >are there? How many different Boolean functions of two Boolean >variables are there? 1 input variable 2^1 possible inputs 2^(2^1) possible mappings from the set of inputs to the set of outputs {0,1} Now try 2 input variables. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Boolean function problem. It seems easy, but difficult to me. >Let p be a Boolean variable. How many different Boolean functions of p >are there? How many different Boolean functions of two Boolean >variables are there? >I'm confusing. Is there anyone who can help me? First, do you understand the possible values for a boolean variable? What are they? Next. a boolean function of one variable means what? What are the possible inputs? What are the possible outputs? Start with that. When you solve that, you'll see that the 2 variable case is not conceptually different. It's still a question of possible inputs and possible outputs. quasi === Subject: How many bijections are there from A to A? If a set A has n elements, how many bijections are there from A to A? === Subject: Re: How many bijections are there from A to A? > If a set A has n elements, how many bijections are there from A to A? It is easy to prove that if n is countably infinite, the number of permutations is equal to 2^n, which is equal to c (the number of real numbers). Every permutation is a sequence with each natural number appearing once, for example ... 5, 1, 3, 17 .... Where 1 -> 5, 2->1, 3->3, 4->17 etc. There is an injection from this into all sequences comprising natural numbers, where each natural number can appear more than once, for example: 2, 1, 2, 1, 1, 4 ... (Because all sequences without repeats appear in the list of all sequences that allow repeats.) But these sequences with repeats can be put into a one to one correspondence with Real numbers through the use of continued fractions (http://en.wikipedia.org/wiki/Continued_fraction); the example I gave is the start of the continued fraction representation of e. Accordingly, there is an injection from permutations over N to R. The injection from R to permutations of N is as follows. Consider an arbitrary R as a binary number, say 0.110011100... Map this onto a sequence where the mth number is m iff the mth digit is a 1, and otherwise it is the lowest unused number that isn't needed for an m=1. For example, 0.110011100 -> 1,2,x,x,5,6,x,x -> 1,2,8,9,5,6,7,3,4 Thus there is an injection from R to permutations. As there is an injection in both directions, it must be a bijection. This works for countably infinite, but I had to use arithmetic to prove it (continued fractions) so it won't work for larger sets. I don't know of a purely set theoretic argument that can be used to explicitly form a bijection for larger sets. === Subject: Re: How many bijections are there from A to A? >If a set A has n elements, how many bijections are there from A to A? But if you didn't see the answer right away, you should have at least tried it for particular values of n. Try actually counting bijections for n=1, n=2, n=3. Then, see if you can explain the count by organizing the list of bijections in a natural way. If the explanation works in general, you're done, but if there's any doubt, try it for n=4, and verify. You don't want to have to verify it for n=5 (you'll see why), so if you reach that stage, it becomes critical to try to think through the logic so as to get it to work for all n. quasi === Subject: Re: How many bijections are there from A to A? > If a set A has n elements, how many bijections are there from A to A? n! === Subject: Re: How many bijections are there from A to A? > If a set A has n elements, how many bijections are there from A to A? > n! What if A is infinite? 2^|A| ? === Subject: Re: How many bijections are there from A to A? On Sat, 19 Nov 2005 23:10:49 -0800, William Elliot >> If a set A has n elements, how many bijections are there from A to A? >> n! >What if A is infinite? 2^|A| ? It's |A| factorial, what else would you expect? (just kidding) But I'm pretty sure it's worse than 2^|A|. Even for finite sets, if n>3, 2^n is too small, and the disparity gets worse as n-->infinity so it's unlikely that it would suddenly catch up at the infinite level. I think for infinite sets, the answer is |A|^|A|, the same cardinality as the set of all functions from A to A. quasi === Subject: Re: How many bijections are there from A to A? |On Sat, 19 Nov 2005 23:10:49 -0800, William Elliot | |>> If a set A has n elements, how many bijections are there from A to A? |>> n! |>What if A is infinite? 2^|A| ? | |It's |A| factorial, what else would you expect? | |(just kidding) | |But I'm pretty sure it's worse than 2^|A|. | |Even for finite sets, if n>3, 2^n is too small, and the disparity gets |worse as n-->infinity so it's unlikely that it would suddenly catch up |at the infinite level. This is a very misleading intuition. |I think for infinite sets, the answer is |A|^|A|, the same cardinality |as the set of all functions from A to A. True, but |A|^|A|=2^|A| for infinite sets A. We can show that 2^|A|<=|Aut(A)|<=|A|^|A|<=2^|A|, so they are actually equal. (The axiom of choice is needed for much of this.) For the first step, start by considering an injection of 2^A into Aut(Ax{0,1}). For each subset of A, consider the bijection of Ax{0,1} that swaps (a,0) and (a,1) for a in the subset, but otherwise leaves everything fixed. Using the axiom of choice, we can prove that Ax{0,1} is in one- to-one correspondence with A (i.e. 2|A|=|A|) so that |Aut(Ax{0,1})|=|Aut(A)|. (We can for instance well-order A, and rearrange it so that it has no last element. If we pair off the even and odd elements, the pairs are in one-to-one correspondence with the original elements of A.) The second step is easy, since each bijection of A is a function from A to A, and the functions from A to A have cardinality |A|^|A|. For the third step, note first that the functions from A to A are a subset of the binary relations on A, i.e. the subsets of AxA. The cardinality of the set of relations on A is therefore 2^(|A|^2). Using the axiom of choice again, we can prove that for infinite A, |A|^2=|A| and hence 2^(|A|^2)=2^|A|. I'm afraid I've forgotten how one proves that |A|^2=|A| for infinite A. I don't remember it as having been a hard proof, once you know consequences of the axiom of choice like the well-ordering theorem. Keith Ramsay === Subject: Re: How many bijections are there from A to A? >|On Sat, 19 Nov 2005 23:10:49 -0800, William Elliot >|>> >|>> If a set A has n elements, how many bijections are there from A >to A? >|>> >|>> n! >|>> >|>What if A is infinite? 2^|A| ? >|It's |A| factorial, what else would you expect? >|(just kidding) >|But I'm pretty sure it's worse than 2^|A|. >|Even for finite sets, if n>3, 2^n is too small, and the disparity gets >|worse as n-->infinity so it's unlikely that it would suddenly catch up >|at the infinite level. >This is a very misleading intuition. >|I think for infinite sets, the answer is |A|^|A|, the same cardinality >|as the set of all functions from A to A. >True, but |A|^|A|=2^|A| for infinite sets A. >We can show that 2^|A|<=|Aut(A)|<=|A|^|A|<=2^|A|, >so they are actually equal. (The axiom of choice is >needed for much of this.) Ok, so without the axiom of choice, what is the provable relationship? It seems that the inequality, 2^|A|<=|Aut(A)|<=|A|^|A| should still be true, even without the axiom of choice. Presumably, AC is needed to get 2^|A|=|A|^|A|, thus trapping |Aut(A)| between them. Are the above claims true? If so, what happens without AC? Does |Aut(A)| round up to A|^|A|, round down to 2^|A|, or can we have the strict double inequality 2^|A|<|Aut(A)|<|A|^|A|? quasi === Subject: Re: How many bijections are there from A to A? <8ci0o1lj35n9sa5qodmajonaevp24jalc5@4ax.com> >We can show that 2^|A|<=|Aut(A)|<=|A|^|A|<=2^|A|, > 2^|A|<=|Aut(A)|<=|A|^|A| should still be true, even without the axiom > double inequality 2^|A|<|Aut(A)|<|A|^|A|? Nocomment,toomucheyestrain,ohsigh,forenjoyablereading. === Subject: Re: How many bijections are there from A to A? >On Sat, 19 Nov 2005 23:10:49 -0800, William Elliot > If a set A has n elements, how many bijections are there from A to A? > n! >>What if A is infinite? 2^|A| ? >It's |A| factorial, what else would you expect? >(just kidding) >But I'm pretty sure it's worse than 2^|A|. >Even for finite sets, if n>3, 2^n is too small, and the disparity gets >worse as n-->infinity so it's unlikely that it would suddenly catch up >at the infinite level. >I think for infinite sets, the answer is |A|^|A|, the same cardinality >as the set of all functions from A to A. >quasi On the other hand, these infinite sets are very devious. I have a sneaky feeling that for infinite sets, |A|^|A| is actually the same as 2^|A|, in which case the cardinality of the set of bijections would, in fact, be 2^|A|. quasi === Subject: Re: How many bijections are there from A to A? days. My association with the Department is that of an alumnus. >>But I'm pretty sure it's worse than 2^|A|. >>Even for finite sets, if n>3, 2^n is too small, and the disparity gets >>worse as n-->infinity so it's unlikely that it would suddenly catch up >>at the infinite level. >>I think for infinite sets, the answer is |A|^|A|, the same cardinality >>as the set of all functions from A to A. >On the other hand, these infinite sets are very devious. I have a >sneaky feeling that for infinite sets, |A|^|A| is actually the same as >2^|A|, in which case the cardinality of the set of bijections would, >in fact, be 2^|A|. Indeed. For any ordinals alpha, beta, if alpha<= beta, then 2^{ aleph_{beta}} = (aleph_{alpha})^{aleph_{beta}}. In particular, if A is any infinite set, and k is any cardinal with 2<=k<=|A|, then |A|^|A| = k^|A|. -- 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: How many bijections are there from A to A? >> >> If a set A has n elements, how many bijections are there from A to A? >> n! >> >What if A is infinite? 2^|A| ? > But I'm pretty sure it's worse than 2^|A|. > Even for finite sets, if n>3, 2^n is too small, and the disparity gets > worse as n-->infinity so it's unlikely that it would suddenly catch up > at the infinite level. 2^1, 2^2, 2^3, ... 2^n, ... 2^Aleph_0, oh ouch what a huge jump. Better use ordinals, lim(n->oo) 2^n = 2^omega_0 is countable ordinal. Oh, and don't forget Aleph_0 = omega_0. ;-) > I think for infinite sets, the answer is |A|^|A|, the same cardinality > as the set of all functions from A to A. |A|^|A| = 2^|A| for infinite A. Thus number of bijections is between A and 2^|A|. How many sequences of integers without repetitions are there? Divide them odd and even and let Cj = { 2j, 2j + 1 } and cj = a choice from Cj. Thus one sees that there are at least 2^|N| sequences without repetitions. Thus the number of bijections N -> N is c. To generalize, well order A and divide them odd and even, whoops ... make the problem intricate, don't assume GCH. === Subject: Re: How many bijections are there from A to A? > |A|^|A| = 2^|A| for infinite A. > Thus number of bijections is between A and 2^|A|. The number of bijections between A and 2^|A| is always zero. === Subject: Re: How many bijections are there from A to A? , > |A|^|A| = 2^|A| for infinite A. > Thus number of bijections is between A and 2^|A|. > The number of bijections between A and 2^|A| is always zero. :-o Smileys aren't required when one is joking? :-) --Ron Bruck === Subject: Re: How many bijections are there from A to A? > The number ... is always zero. Wow! ;-) === Subject: Re: How many bijections are there from A to A? On Sat, 19 Nov 2005 21:12:10 -0700, Virgil >> If a set A has n elements, how many bijections are there from A to A? (but not an emphatic answer of n) === Subject: Re: How many bijections are there from A to A? > On Sat, 19 Nov 2005 21:12:10 -0700, Virgil >> If a set A has n elements, how many bijections are there from A to A? >n! > (but not an emphatic answer of n) Right! (This is an emphatic anwser!) === Subject: PSL(2,7) Hi How do I figure out what are the Sylow 2 subgroups of PSL(2,7)? They must be of order 8, and there are either 3, 7 or 21 of them. But how do I describe them? === Subject: Re: PSL(2,7) >How do I figure out what are the Sylow 2 subgroups of >PSL(2,7)? They must be of order 8, and there are either 3, 7 or 21 of them. But how do I describe them? It is hard to help with problems like this without knowing about how much you know already about groups like SL(2,q). I would prefer to work in SL(2,7). Since PSL(2,7) = SL(2,7)/< -I >, where <-I > is the central subgroup of order 2, the Sylow 2-subgroups of PSL(2,7) will just be the quotients of those of SL(2,7) by < -I >. In general, you can define elements of order q^2 - 1 in GL(2,q) which represent multiplication in GF(q^2) by a primitive element w of GF(q^2). The determinant of t is w^(q+1), and so the (q-1)-st power x of t has order q+1 and lies in G := SL(2,q). Since acts irreducibly, by Schur's Lemma, its centralizer in GL(2,q) is isomorphic to the multiplicative group of GF(q^2), so the centralizer of in GL(2,q) is , and C_G() = . The eigenvalues of in GF(q^2) have the form lambda and lambda^-1, and they can only be fixed or interchanged by an element of N_G(), so |N_G(): | <= 2. In fact, we have equality. In the case of SL(2,7), we can take, for example x = [ 0 1 ] and y := [ 1 5 ] normalizes but does not centralize , so [ 6 3 ] [ 1 6 ] S := < x, y > has order 16 and is a Sylow 2-subgroup of G. < x > is the unique cyclic subgroup of index 2 in S (which is a generalized quaternion group), so < x > is characteristic in S, and hence N_G(S) = N_G(), and we have 21 Sylow 2-subgroups. I am not really sure how best to describe them all other than as the conjugates in G of S. Hope this helps a little! Derek Holt. === Subject: Re: PSL(2,7) On 19-Nov-2005, royy23 <27014782.1132465337532.JavaMail.jakarta@nitrogen.mathforum.org>: > Hi > How do I figure out what are the Sylow 2 subgroups of > PSL(2,7)? They must be of order 8, and there are either 3, 7 or 21 of > them. But how do I describe them? Can't be 3 of them, otherwise PSL(2,7) would be a subgroup of S_3. I suspect people aren't rushing to answer because: (a) Over the years his problem's been covered here several times before--try Googling for it in sci.math; (b) Depending on how much, or rather how little group theory you know, it's rather tedious. An outline of one approach using only fairly elementary group theory: What's the structure of a normalizer of a Sylow 7-subgroup? What does that tell you about the (WDTTYAt) number of Sylow 3-subgroups? WDTTYAt structure of a normalizer of a Sylow 3-subgroup? WDTTYAt structure of any subgroups of order 24? WDTTYAt number of Sylow 2-subgroups? WDTTYAt size(s) of the intersections of the Sylow 2-subgroups? What's the maximum size of a subgroup of PSL(2,7)? WDTTYAt structure of a Sylow 2-subgroup? -- Jim Heckman === I am stuck with the transformation of the following discrete time equations into continuous time: x(t) = f(k(t-1)) - c(t) and k(t) = (1-d)*k(t-1) + (1-[(x(t)-x(t-1))/x(t-1)]^2))x(t) You can take c(t) as an exogenous function. Ali === I am stuck with the transformation of the following discrete time equations into continuous time: x(t) = f(k(t-1)) - c(t) and k(t) = (1-d)*k(t-1) + (1-[(x(t)-x(t-1))/x(t-1)]^2))x(t) You can take c(t) as an exogenous function. Ali === I am stuck with the transformation of the following discrete time equations into continuous time: x(t) = f(k(t-1)) - c(t) and k(t) = (1-d)*k(t-1) + (1-[(x(t)-x(t-1))/x(t-1)]^2))x(t) You can take c(t) as an exogenous function. Ali === Subject: trig differentiation Hi all. Say we have the relationship... tan(theta) = (b/a)tan(phi).....a,b are constants How would we differentiate to get deltatheta ? TIA Mick. === Subject: Re: trig differentiation > Say we have the relationship... > tan(theta) = (b/a)tan(phi).....a,b are constants > How would we differentiate to get deltatheta ? Huh? d/dphi tan theta = (1 + tan^2 theta) dtheta/dphi d/dphi (b/a)tan phi = (b/a)(1 + tan^2 phi) dtheta/dphi = (b/a)(1 + tan^2 phi)/(1 + tan^2 theta) -- To Google and MathForum users: Reply only if adequate context is included _within_ the reply. Use the quote feature. Otherwise, if you insist upon using quick reply, don't reply, as all the contexts are removed from view, the flow of thought messed up and chaos reigns. === Subject: Re: trig differentiation >> Say we have the relationship... >>tan(theta) = (b/a)tan(phi).....a,b are constants >>How would we differentiate to get deltatheta ? > Huh? > d/dphi tan theta = (1 + tan^2 theta) dtheta/dphi > d/dphi (b/a)tan phi = (b/a)(1 + tan^2 phi) > dtheta/dphi = (b/a)(1 + tan^2 phi)/(1 + tan^2 theta) > -- > To Google and MathForum users: > Reply only if adequate context is included _within_ the reply. > Use the quote feature. Otherwise, if you insist upon using > quick reply, don't reply, as all the contexts are removed > from view, the flow of thought messed up and chaos reigns. You just did his homework for him. === Subject: Re: trig differentiation >> Say we have the relationship... >>tan(theta) = (b/a)tan(phi).....a,b are constants >>How would we differentiate to get deltatheta ? > Huh? > d/dphi tan theta = (1 + tan^2 theta) dtheta/dphi > d/dphi (b/a)tan phi = (b/a)(1 + tan^2 phi) > dtheta/dphi = (b/a)(1 + tan^2 phi)/(1 + tan^2 theta) > You just did his homework for him. Oh well, my revenge will come when I miss the exam. === Subject: Re: trig differentiation > Hi all. > Say we have the relationship... > tan(theta) = (b/a)tan(phi).....a,b are constants > How would we differentiate to get deltatheta ? > TIA Mick. The derivative of tan(y) with respect to x, d(tan(y))/dx, = (sec(y))^2.dy/dx Using this formula, you can differentiate both sides of your equation wrt theta. On the left side (where y = theta) you'll have a factor d(theta)/d(theta), which is 1. On the right side you'll have a factor d(phi)/d(theta). You can rearrange and express trig functions of theta in terms of tan(theta), then eliminate theta (by substituting your initial equation), or alternatively you can eliminate phi. You'll arrive at an expression for d(phi)/d(theta) as a function of theta or of phi, whichever you prefer. This you can rearrange to give delta(theta) in terms of delta(phi). === Subject: What is this called? Suppoes we have an equation like y''y'-3y'''=0 (except a lot harder) where the sum of the number of the derivative is the same all across. What is that called? Is there a general method to solve such equations? === Subject: Re: What is this called? >Suppoes we have an equation like y''y'-3y'''=0 (except a lot harder) >where the sum of the number of the derivative is the same all across. >What is that called? I doubt that it has a special name. >Is there a general method to solve such equations? I doubt it. Nonlinear equations are hard. ************************ David C. Ullrich === Subject: Re: What is this called? >>Suppoes we have an equation like y''y'-3y'''=0 (except a lot harder) >>where the sum of the number of the derivative is the same all across. >>What is that called? >I doubt that it has a special name. >>Is there a general method to solve such equations? >I doubt it. Nonlinear equations are hard. For example, I doubt that there's a known method that will solve y'' y' + y''' + x y (y')^3 = 0 (where x is the independent variable) in closed form, or even reduce it to a second-order equation. Maple can't do it, and finds no symmetries. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: What is this called? What about solving it in terms of an infinite sum? Like a_0+a_1x+a_2x^2+...+a_nx^n+... ? === Subject: Re: What is this called? >What about solving it in terms of an infinite sum? Like >a_0+a_1x+a_2x^2+...+a_nx^n+... ? Yes, of course series solutions are possible, as with any ODE. But there's no closed-form expression for the coefficients. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: What is this called? > Suppoes we have an equation like y''y'-3y'''=0 (except a lot harder) > where the sum of the number of the derivative is the same all across. > What is that called? Is there a general method to solve such equations? Is homogenuous the answer you're looking for? There surely are good methods to solve such equations... look up any text on ODE please. === Subject: Re: What is this called? <43802e04$0$24482$892e7fe2@authen.yellow.readfreenews.net> I don't think it's homogenous. A homogenous equation would be something like y-2y'+y'', because there are the same number of ys, differentiated or not, in each term. === Subject: Re: What is this called? >I don't think it's homogenous. A homogenous equation would be something > like y-2y'+y'', because there are the same number of ys, differentiated > or not, in each term. Hum. I think you're probably right. Then I don't know what will others call such equations. === Subject: Re: What is this called? > Suppoes we have an equation like y''y'-3y'''=0 (except a lot harder) > where the sum of the number of the derivative is the same all across. > What is that called? Is there a general method to solve such equations? let y' = z and your equation simplifies to z' z -3 z'' = 0, which does not have that same property, so unless the equation contains an undifferentiated variable, it may often be reduced to an equation not of that same type. === Subject: Line formula Hello all, I have a line (L1) of slope y-intercept form : y=m*x+b (where m is finite) and a known point P1(x1,y1) outside L1. Can anyone please tell me what is the line formula (in slope y-intercept form) of the perpedicular line to L1 passing through P1? === Subject: Re: Line formula > Hello all, > I have a line (L1) of slope y-intercept form : y=m*x+b (where m is finite) > and a known point P1(x1,y1) outside L1. Can anyone please tell me what is > the line formula (in slope y-intercept form) of the perpedicular line to > L1 > passing through P1? === Subject: Re: Line formula For m != 0, we have y = -1/m * x + c. === Subject: Re: Line formula > I have a line (L1) of slope y-intercept form : y=m*x+b > (where m is finite) and a known point P1(x1,y1) outside > L1. Can anyone please tell me what is the line formula > (in slope y-intercept form) of the perpedicular line to > L1 passing through P1? Hint: If L1 has the form y = mx + b and L2 is perpendicular to L1, then L2 has a slope of -1 / m. Kyle === Subject: Re: Line formula On Sun, 20 Nov 2005 10:27:28 +0200, Peter S. >Hello all, >I have a line (L1) of slope y-intercept form : y=m*x+b (where m is finite) >and a known point P1(x1,y1) outside L1. Can anyone please tell me what is >the line formula (in slope y-intercept form) of the perpedicular line to L1 >passing through P1? You should be able to work this out. Do you know the relationship between the slopes of 2 perpendicular lines? Once you know the slope, can you find the equation, given that you know a point on the line? Once you have the equation, to express it in slope-intercept form, just solve for y. But before you tackle this problem, you should try some examples where m,b,x1,y1 are all known numbers as opposed to unknowns. Any precalculus book (or even earlier levels of algebra) will almost certainly have problems of this type, including worked examples. quasi === Subject: Re: Line formula Okay, I misread that. Okay, so if we have point x1, y1, then we get (y-y1)=-1/m * (x-x1). === Subject: Can anybody tell me which are the most current hottest research topics in statistics and probability theory? Hi all, I am trying to decide which research direction shall I take... Please point out to me which are the most current research topics in statistics and probability theory? Let's discuss about them: I am mostly interested in the most current/up-to-date/hottest research topics in the following areas: 1. Data Mining 2. Machine Learning 3. Theoratical and Computational Statistics 4. Math Finance I want to do some theoratical research such that later I can become a professor and I love doing academic research. === Subject: Re: Can anybody tell me which are the most current hottest research topics in statistics and probability theory? Please have a look at Experimental Datasets from Chemical Thermodynamics http://evgenii.rudnyi.ru/doc/misc/ExperimentalDatasets.html In my view, there are many interesting unsolved problems here. Best wishes, Evgenii Rudnyi === Subject: Re: Can anybody tell me which are the most current hottest research topics in statistics and probability theory? >I am trying to decide which research direction shall I take... Please >point out to me which are the most current research topics in >statistics and probability theory? Let's discuss about them: >I am mostly interested in the most current/up-to-date/hottest research >topics in the following areas: >1. Data Mining >2. Machine Learning >3. Theoratical and Computational Statistics >4. Math Finance >I want to do some theoratical research such that later I can become a >professor and I love doing academic research. If you love doing academic research, do it! What does it matter what the hottest research topics are? What matters is what topics interest you and what topics you can contribute to. dave === Subject: Re: Can anybody tell me which are the most current hottest research topics in statistics and probability theory? Theoratically you are right! It does not matter... Realisitically it does matter... I have to find up-to-date areas so I can find job to make a living at least; I have to find frontiers so I can quickly catch up... All of the above topics interest me.. I just want to find their latest development so I can catch up... as I am a new comer now... === Subject: Re: Can anybody tell me which are the most current hottest research topics in statistics and probability theory? > Theoratically you are right! It does not matter... > Realisitically it does matter... I have to find up-to-date areas so I > can find job to make a living at least; I have to find frontiers so I > can quickly catch up... > All of the above topics interest me.. I just want to find their latest > development so I can catch up... as I am a new comer now... Your best approach would be to look at recent issues of some of the major statistical journals. You can probably find contents-lists online to get an idea of topics but you may well need to look at actual paper copies to get a real idea of what is being discussed. See http://www.stata.com/links/journals4.html for links to statistical journals. If you are at a higher ed. place, try either the main library or maths or statistics department libraries for journals themselves, of see if a friend subscribes. Also, consider joining a statistical society for access to their journals and meetings: this is usually better/cheaper than subscribing to a journal without joining. Try to go to statistical seminars and meetings if there are only local to you. And consider attending a major statistical conference if you have funding for this. David Jones === Subject: Re: Can anybody tell me which are the most current hottest research topics in statistics and probability theory? <43819ebe$1@news.nwl.ac.uk> Hi David, Do you have the impact ratings for the journals? I guess I should start with the most influential journals in the fields... Could anybody give some pointers about the most influential journals in the following fields? Data mining; AI/Machine Learning; Theoreatical & Computation Statistics; Game Theory; Probability Theory; Optimization === Subject: Re: Can anybody tell me which are the most current hottest research topics in statistics and probability theory? > Hi all, > I am trying to decide which research direction shall I take... Please > point out to me which are the most current research topics in > statistics and probability theory? Let's discuss about them: > I am mostly interested in the most current/up-to-date/hottest research > topics in the following areas: > 1. Data Mining > 2. Machine Learning > 3. Theoratical and Computational Statistics > 4. Math Finance > I want to do some theoratical research such that later I can become a > professor and I love doing academic research. Dunno of it's up-to-date or a hot research area, but this paper looks like it might interest someone in probability and statistics combined with information theory: http://eeweb.poly.edu/~onur/publish/contoured.pdf Information Theoretic Inequalities for Contoured Probability Distributions === Subject: Abstract questions I'm currently working on some homework for my abstract algebra class but I can't figure some things out that I was hoping someone could help me with: 1. H=<5>, K=<7>. How would I prove that Z=HK, where Z is the set of all integers. 2. H is a normal subgroup of G and a is in G. |aH|=3 in group G/H (G is a factor group of H) and |H|=10. What are the possibilities of the order of a? 3. Determine the order of (Z*Z)/<(2,2)> (* means the direct product)...is it cyclic? === Subject: Re: Abstract questions > I'm currently working on some homework for my abstract > algebra class but I can't figure some things out that > I was hoping someone could help me with: Abstract algebra is, well, rather abstract, and I see you are doing a lot of quotient groups material, which I say from experience is the worst of it. If you can understand what quotient groups are about your life will be much easier. > 1. H=<5>, K=<7>. How would I prove that Z=HK, where Z > is the set of all integers. This might be confusing if you read it literally, since it seems to be using multiplicative notation when the standard binary operation for the group of integers is addition. Here's a hint: Z is cyclic, so if you can get its generator in HK (which is H + K in additive notation) then you are done. > 2. H is a normal subgroup of G and a is in G. |aH|=3 > in group G/H (G is a factor group of H) and |H|=10. > What are the possibilities of the order of a? The notation is unfortunate here too, since |aH| could mean either the size of the coset aH or the order of the group element aH in G/H. get around it since these notations are both standard). Here's a hint: if (aH)^3 = H (in G/H), then a^3 is in H (in G). Now forget about G and remember what you know about the orders of subgroups of finite groups. > 3. Determine the order of (Z*Z)/<(2,2)> (* means the direct product)..is it cyclic? I would write direct product as x if I were you; Z*Z means free product (which you may not have heard of), which is completely different. One way to at least start analyzing the structure of a quotient group is to analyze the order of each element. So, pick some element g = (a,b) in Z x Z, and ask yourself: what's the smallest exponent of its image in (Z x Z)/<(2,2)> which is the identity there? -- Ryan Reich ryan.reich@gmail.com === Subject: Re: Abstract questions > Abstract algebra is, well, rather abstract, and I see you are doing a > lot of quotient groups material, which I say from experience is the > worst of it. If you can understand what quotient groups are about your > life will be much easier. yeah, I'm taking algebra I, we use Fraleigh as a textbook. After we went through ch 14-15 (factor groups), roughly half of the class has disappeared. Then I looked at the Artin's Algebra, and one of his recommendations to instructor is to de-emphasize the factor groups (or something like that). Go figure. === Subject: Re: Abstract questions > 1. H=<5>, K=<7>. How would I prove that Z=HK, > where Z > is the set of all integers. > This might be confusing if you read it literally, > since it seems to be > using multiplicative notation when the standard > binary operation for > the group of integers is addition. Here's a hint: Z > is cyclic, so if > you can get its generator in HK (which is H + K in > additive notation) > then you are done. Using the idea from another post earlier, I got the 5n+7m=1 when n is 10 and m is -7. How do I use this to get the generator? > 2. H is a normal subgroup of G and a is in G. > |aH|=3 > in group G/H (G is a factor group of H) and |H|=10. > What are the possibilities of the order of a? > The notation is unfortunate here too, since |aH| > could mean either the > size of the coset aH or the order of the group > element aH in G/H. > way, there's no way to > get around it since these notations are both > standard). Here's a hint: > if (aH)^3 = H (in G/H), then a^3 is in H (in G). Now > forget about G > and remember what you know about the orders of > subgroups of finite > groups. Going through this question, by Lagrange's Theorem, the order of a can be either 1,2,5, or 10. I see that a^3 is in G but how is it in H. Even knowing that a^3 is in H, what should I do now to narrow it further down? === Subject: Re: Abstract questions days. My association with the Department is that of an alumnus. >> 1. H=<5>, K=<7>. How would I prove that Z=HK, >> where Z >> is the set of all integers. >> This might be confusing if you read it literally, >> since it seems to be >> using multiplicative notation when the standard >> binary operation for >> the group of integers is addition. Here's a hint: Z >> is cyclic, so if >> you can get its generator in HK (which is H + K in >> additive notation) >> then you are done. >Using the idea from another post earlier, I got the 5n+7m=1 when n is >10 and m is -7. How do I use this to get the generator? To prove that Z = HK, you must show that every element of Z can be written as an element of H plus an element of K. Since H= <5>, H contains every multiple of 5; likewise, K contains every multiple of 7. Since you have shown that (10)5 + (-7)7 = 1, that means that you can write 1 as a multiple of 5 (i.e., an element of H), plust a multiple of 7 (that is, an element of K). Now that you have 1, can you show that EVERY integer can be written as an element of H (i.e., a multiple of 5) plus an element of K (a multiple of 7)? >> 2. H is a normal subgroup of G and a is in G. >> |aH|=3 >> in group G/H (G is a factor group of H) and |H|=10. >> What are the possibilities of the order of a? >> The notation is unfortunate here too, since |aH| >> could mean either the >> size of the coset aH or the order of the group >> element aH in G/H. >> way, there's no way to >> get around it since these notations are both >> standard). Here's a hint: >> if (aH)^3 = H (in G/H), then a^3 is in H (in G). Now >> forget about G >> and remember what you know about the orders of >> subgroups of finite >> groups. >Going through this question, by Lagrange's Theorem, >the order of a can be either 1,2,5, or 10. I see that a^3 is in G > but how is it in H. The operation in G/H is the following: the result of multiplying the element xH by yH is (xH)(yH) = (xy)H. And, by definition of congruent modulo H, we know that xH = wH if and only if w^{-1}x is in H. Now, the identity element of G/H is the element eH = H. So, waht does it mean for (aH) to be of order 3? It means that (aH)^3 = H in G/H. Which means that a^3H = eH in G/H. Which means that e^{-1}a^3 is an element of H. Which says that a^3 is an element of H. > Even knowing that a^3 is in H, what should I > do now to narrow it further down? You know that EVERY element of H is of order 1, 2, 5, or 10. Since a^3 is in H, a^3 is of order 1, 2, 5, or 10. What does each of them tell you about the order of ->a<-? -- 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: Abstract questions Firgured out the first one >> 2. H is a normal subgroup of G and a is in G. >> |aH|=3 >> in group G/H (G is a factor group of H) and > |H|=10. >> What are the possibilities of the order of a? >> >> The notation is unfortunate here too, since |aH| >> could mean either the >> size of the coset aH or the order of the group >> element aH in G/H. >> way, there's no way to >> get around it since these notations are both >> standard). Here's a hint: >> if (aH)^3 = H (in G/H), then a^3 is in H (in G). > Now >> forget about G >> and remember what you know about the orders of >> subgroups of finite >> groups. >> >Going through this question, by Lagrange's Theorem, >the order of a can be either 1,2,5, or 10. I see > that a^3 is in G > but how is it in H. > The operation in G/H is the following: the result of > multiplying the > element xH by yH is (xH)(yH) = (xy)H. > And, by definition of congruent modulo H, we know > that xH = wH if > and only if w^{-1}x is in H. > Now, the identity element of G/H is the element eH = > H. So, waht does > it mean for (aH) to be of order 3? It means that > (aH)^3 = H in > G/H. Which means that a^3H = eH in G/H. Which means > that e^{-1}a^3 is > an element of H. Which says that a^3 is an element of > H. > Even knowing that a^3 is in H, what should I > do now to narrow it further down? > You know that EVERY element of H is of order 1, 2, 5, > or 10. Since a^3 > is in H, a^3 is of order 1, 2, 5, or 10. What does > each of them tell > you about the order of ->a<-? For this one, if order of a^3 were: 1 - order of a could only be 1 2- order of a could be 1 or 2 5- order of a could be 1 or 5 10- order of a could be 1, 2, 5, 10 (I got these results by multiplying a^3 by the corresponding order and setting that equal to a.) Does this mean that the order could be 1, 2, 5 or 10 === Subject: Re: Abstract questions days. My association with the Department is that of an alumnus. >Firgured out the first one Or so you hope... >> You know that EVERY element of H is of order 1, 2, 5, >> or 10. Since a^3 >> is in H, a^3 is of order 1, 2, 5, or 10. What does >> each of them tell >> you about the order of ->a<-? >For this one, if order of a^3 were: >1 - order of a could only be 1 No. If the order of a^3 is 1, then the order of a could be any divisor of 3. >2- order of a could be 1 or 2 No. If the order of a^3 is 2, that means that a^3 is not equal to e (since a^3 is not of order 1), but (a^3)^2 = a^6 is equal to e. If a were of order 1, then a = e, so a^3 = e, so a^3 would be of order 1; but it is not: it is of order 2. So the order must be a divisor of 6, it cannot be 1, and it cannot be 3 (since a^3 is not equal to e). Can it be 2? Can it be 6? >5- order of a could be 1 or 5 No. It cannot be 1. You know that a^{15} = (a^3)^5 = e, and that a^3 is NOT equal to e. The order of a must be a divisor of 15, and it is not 3; and it cannot be 1. What can it be? >10- order of a could be 1, 2, 5, 10 No. >(I got these results by multiplying a^3 by the >corresponding order and setting that equal to a.) WHY would you set it equal to a? The order of an element is NOT the smallest positive integer n such that a^n = a. It is the smallest positive integer n such that a^n = e (the identity). >Does this mean that the order could be 1, 2, 5 or 10 No, because you messed up the definition of order and did not solve the problem correctly anyway. -- 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: Abstract questions > 1. H=<5>, K=<7>. How would I prove that Z=HK, where Z > is the set of all integers. Hint: 4*5 - 3*7 = 1 === Subject: Re: Abstract questions > Hint: 4*5 - 3*7 = 1 Sure about that? Jose Carlos Santos === Subject: Re: Abstract questions <3ubhpqF10mg7rU1@individual.net> > Hint: 4*5 - 3*7 = 1 > Sure about that? Ah er... consider it a trick hint. ;-) === Subject: Re: Abstract questions On Sun, 20 Nov 2005 16:15:57 -0800, William Elliot >> Hint: 4*5 - 3*7 = 1 >> Sure about that? >Ah er... consider it a trick hint. ;-) That's been known to happen to those who spend too much time analyzing Boolean rings -- they lose the ability to distinguish between 1 and -1. === Subject: Re: Abstract questions <3ubhpqF10mg7rU1@individual.net> <7p42o1pul9b7elc71bg9fd1ua9e197vgp9@4ax.com> >> >> Hint: 4*5 - 3*7 = 1 >> Sure about that? >> >Ah er... consider it a trick hint. ;-) > That's been known to happen to those who spend too much time analyzing > Boolean rings -- they lose the ability to distinguish between 1 and > -1. LOL. Forwarded Boolean counter example to professor for review and haven't got even a Boo from him. I'll have to remind him after ample wait about the problem. What's happening in academia now? Like when's the next time the pace of teaching and exams will slow down to less than breath taking? === Subject: Re: Abstract questions On Sun, 20 Nov 2005 17:37:01 -0800, William Elliot >>> >>> Hint: 4*5 - 3*7 = 1 >>> Sure about that? >>> >>Ah er... consider it a trick hint. ;-) >> That's been known to happen to those who spend too much time analyzing >> Boolean rings -- they lose the ability to distinguish between 1 and >> -1. >LOL. Forwarded Boolean counter example to professor for review and >haven't got even a Boo from him. I'll have to remind him after >ample wait about the problem. What's happening in academia now? >Like when's the next time the pace of teaching and exams will slow >down to less than breath taking? By January, the pace should slow down. quasi === Subject: Re: Abstract questions > I'm currently working on some homework for my abstract > algebra class but I can't figure some things out that why don't you just drop out? === Subject: Re: L^1 question >> Im doing some work on the various spaces (normed/banach/hilbert/inner >> product space) >> and am having some difficulty proving >> the set of all cts functions on [a.b] is dense in L^1([a,b]) >> The book im using thinks its quite easy,...but im a little lost! > Do you know the definition of L^1([a,b])? > Given an element of L^1([a,b]), can you find a sequence > of continuous functions which converges to that element? yes i know the definition on L^1. but with this hint i still cannot see where i am going === Subject: Re: L^1 question >> Im doing some work on the various spaces (normed/banach/hilbert/inner >> product space) >> and am having some difficulty proving >> the set of all cts functions on [a.b] is dense in L^1([a,b]) >> The book im using thinks its quite easy,...but im a little lost! > Do you know the definition of L^1([a,b])? > Given an element of L^1([a,b]), can you find a sequence > of continuous functions which converges to that element? > yes i know the definition on L^1. but with this hint i still cannot see > where i am going What's the name of your book? I'd like to avoid it when I teach measure theory/functional analysis again. There are many ways to do this, but none can be reasonably LTR. (Although they can be left to an exercise.) Someone has already suggested Lusin's theorem. I'll suggest three others: (a) [Essentially, this is a reproof of Lusin] Show the simple functions, the linear span of the chi_A, where A is measurable with finite measure, are dense. (That's trivial--it's more or less the DEFINITION of integral.) Now use the fact that A can be approximated from inside by a closed set F and from the outside by an open set G so that the measure of GF is as small as we please. Thus chi_A is well-approximable in L^1 by chi_F. But chi_F is well-approximable in L^1 by some continuous function f. (You can even take f to be piecewise linear, with chi_F <= f <= chi_G.) Now, instead of approximating a general L^1 function f by a linear combination of chi_A, use the same linear combination of the continuous functions which approximate chi_A. (b) Since you're in [a,b] you can use the fact that the characteristc functions of intervals [c,d] span a dense subset of L^1. This requires a little work (it's not perfectly trivial). Then use the fact that chi_[c,d] is approximable by a piecewise linear function, e.g. ___________ / / __________/ ____________ ^ ^ ^ ^ | | | | c-e c d d+e where e is small. (c) Use the convolution f*phi_e(x) = int_-infty^infty f(x-t)phi_e(t) dt for an approximate identity phi_e. This can even be taken to be C^infty. I guess first you have to set f = 0 outside [a,b]. --Ron Bruck === Subject: searching for phd position Hi everybody, I am now searching for a Professor to supervise my PHD research work, and I would like to introduce myself briefly here. I am a postgraduate student in China, majoring in functional analysis. My research interest during my postgraduate study is mainly on the generalized orthogonalities in normed linear spaces, such as Birkhoff orthogonality, isosceles orthogonality, and some related problems. I have now one paper published in Chinese and another to appear in the journal of mathematical analysis and applications(as the second author). Contact me if you would like to accept me as a phd student or you have some information related to it please. === Subject: Re: searching for phd position > Hi everybody, I am now searching for a Professor to supervise my PHD > research work, and I would like to introduce myself briefly here. > I am a postgraduate student in China, majoring in functional analysis. > My research interest during my postgraduate study is mainly on the > generalized orthogonalities in normed linear spaces, such as Birkhoff > orthogonality, isosceles orthogonality, and some related problems. I > have now one paper published in Chinese and another to appear in the > journal of mathematical analysis and applications(as the second > author). Contact me if you would like to accept me as a phd student or > you have some information related to it please. Is this a joke? === Subject: lim inf and lim sup this is quite an obvious query, i can see it is true, but i cannot seem to prove it with ease lim sup a_n > or = lim inf a_n n n === Subject: Re: lim inf and lim sup > this is quite an obvious query, i can see it is true, but i cannot > seem to prove it with ease > lim sup a_n > or = lim inf a_n > n n Straight from the definition, by contradiction. Suppose lim inf> lim sup for some sequence. Then there is an N s.t. sup a_nN). This is impossible, since sup a_n > a_{N+1} > inf a_n. Peter === Subject: Re: lim inf and lim sup > this is quite an obvious query, i can see it is true, but i cannot seem to > prove it with ease > lim sup a_n > or = lim inf a_n > n n Problem is, there's more than one definition of limsup and liminf. The proofs will vary, depending on which definition you use. The USUAL definition, and the one for which limsup and liminf are named, is this: put M_n = sup {a_i : i >= n} m_n = inf {a_i : i >= n}. Then it is clear that m_n <= M_n (true for the inf and sup of ANY set), and it is also clear that m_n is increasing (= non-decreasing) and M_n is decreasing (= non-increasing). Thus {M_n} and {m_n} converge in the extended real numbers (the limits may be infinite). The limsup a_n is defined to be the limit of M_n and the liminf is the limit of m_n, and it is trivial that m_n <= limsup_i a_i <= M_n thus liminf a_i <= limsup_i a_i. It might be useful to your understanding to first prove the lemmas: if A is an extended real number and a_n <= A for all n, then limsup a_n <= A; while if A <= a_n for all n, then A <= liminf a_n. A secondary definition relies on the set S of all subsequential limits of {a_n} (in the extended reals). Then limsup a_n is defined to be the largest element of S [which is closed, hence has a max] while liminf is the smallest. This is usually proved as a theorem from the first definition [and it's very useful, since this is the way we usually COMPUTE the limsup and liminf of a sequence], but I have seen it used liminf <= limsup, since the smallest of anything is <= the biggest (as long as the anything is nonempty :-) --Ron Bruck === Subject: Re: lim inf and lim sup > this is quite an obvious query, i can see it is true, but i cannot seem to > prove it with ease > lim sup a_n > or = lim inf a_n > n n Would you mind telling what's the definition of lim sup and of lim inf that you're using? The answer depends upon that. Jose Carlos Santos === Subject: Re: lim inf and lim sup > this is quite an obvious query, i can see it is true, but i > cannot seem to prove it with ease > lim sup a_n > or = lim inf a_n > n n I went around in circles a couple times myself. Assuming lim sup a_n < lim inf a_n (*) and showing a contradiction seems to works pretty easily. Jim Burns === Subject: Re: lim inf and lim sup > this is quite an obvious query, i can see it is true, but i > cannot seem to prove it with ease > lim sup a_n > or = lim inf a_n > n n > I went around in circles a couple times myself. > Assuming > lim sup a_n < lim inf a_n (*) > and showing a contradiction seems to works pretty easily. You can do it directly, you just have to avoid one particular catch. We have by definition limsup_n a_n = inf_N S_N, where S_N = sup { a_n | n >= N } liminf_n a_n = sup_N I_N, where I_N = inf { a_n | n >= N } Pick any N. Then by definition I_N <= a_n for all n >= N; then for any M > N surely this holds for n >= M as well, so I_N <= S_M for all such M, and hence I_N <= inf_{M > N} S_M. Now observe that if P < N and M > N, { a_n | n >= P } contains { a_n | n >= M } and hence has at least as high a supremum; i.e. S_P >= S_M. Thus limsup a_n = inf_{M > N} S_M, for any N (i.e. the first N terms do not add anything to the eventual supremum). Thus we have shown I_N <= limsup a_n. Then surely liminf a_n = sup_N I_N <= limsup a_n by definition of supremum. The catch was that it is very tempting to say that clearly I_N <= S_N, so taking limits, QED. But of course you are not taking limits, but suprema and infima, and on top of that the indices on each side of the inequality are varying independently, so that this argument completely dodges the point of the question. -- Ryan Reich ryan.reich@gmail.com === Subject: Re: lim inf and lim sup On 20 Nov 2005 06:47:45 -0800, Ryan Reich >> >> this is quite an obvious query, i can see it is true, but i >> cannot seem to prove it with ease >> >> lim sup a_n > or = lim inf a_n >> n n >> I went around in circles a couple times myself. >> Assuming >> lim sup a_n < lim inf a_n (*) >> and showing a contradiction seems to works pretty easily. >You can do it directly, you just have to avoid one particular catch. >We have by definition >limsup_n a_n = inf_N S_N, where S_N = sup { a_n | n >= N } >liminf_n a_n = sup_N I_N, where I_N = inf { a_n | n >= N } >Pick any N. Then by definition I_N <= a_n for all n >= N; then for any >M > N surely this holds for n >= M as well, so I_N <= S_M for all such >M, and hence >I_N <= inf_{M > N} S_M. >Now observe that if P < N and M > N, { a_n | n >= P } contains { a_n | >n >= M } and hence has at least as high a supremum; i.e. S_P >= S_M. >Thus >limsup a_n = inf_{M > N} S_M, for any N >(i.e. the first N terms do not add anything to the eventual >supremum). Thus we have shown >I_N <= limsup a_n. >Then surely >liminf a_n = sup_N I_N <= limsup a_n >by definition of supremum. The catch was that it is very tempting to >say that clearly I_N <= S_N, so taking limits, QED. But of course you >are not taking limits, but suprema and infima, and on top of that the >indices on each side of the inequality are varying independently, so >that this argument completely dodges the point of the question. There is no catch - the proof is entirely trivial. In fact it's clear that I_N <= S_N, right? It's also clear that the sequence S(N) is monotone, which implies that lim sup a_n = lim_N S_N (actually lim_N S_N is the usual definition, don't know where you got yours.) And lim inf a_n = lim_N I_N, so there you are. ************************ David C. Ullrich === Subject: Re: lim inf and lim sup <4hd1o19sqsmojirkiu9pijnunhtlouulvt@4ax.com> > On 20 Nov 2005 06:47:45 -0800, Ryan Reich >You can do it directly, you just have to avoid one particular catch. >We have by definition >limsup_n a_n = inf_N S_N, where S_N = sup { a_n | n >= N } >liminf_n a_n = sup_N I_N, where I_N = inf { a_n | n >= N } >Pick any N. Then by definition I_N <= a_n for all n >= N; then for any >M > N surely this holds for n >= M as well, so I_N <= S_M for all such >M, and hence >I_N <= inf_{M > N} S_M. >Now observe that if P < N and M > N, { a_n | n >= P } contains { a_n | >n >= M } and hence has at least as high a supremum; i.e. S_P >= S_M. >Thus >limsup a_n = inf_{M > N} S_M, for any N >(i.e. the first N terms do not add anything to the eventual >supremum). Thus we have shown >I_N <= limsup a_n. >Then surely >liminf a_n = sup_N I_N <= limsup a_n >by definition of supremum. The catch was that it is very tempting to >say that clearly I_N <= S_N, so taking limits, QED. But of course you >are not taking limits, but suprema and infima, and on top of that the >indices on each side of the inequality are varying independently, so >that this argument completely dodges the point of the question. > There is no catch - the proof is entirely trivial. > In fact it's clear that I_N <= S_N, right? It's also > clear that the sequence S(N) is monotone, which implies > that lim sup a_n = lim_N S_N (actually lim_N S_N is > the usual definition, don't know where you got yours.) > And lim inf a_n = lim_N I_N, so there you are. I agree the proof is easy, but in fact yours is the same as mine. We both started essentially in the same place, though I gave a slightly broader statement. We then both invoked monotonicity and derived an equivalent expression for the limsup or liminf, and from there the answer was trivial. All I did differently was use a slightly different definition and explain the work in greater detail, with more whitespace. And the catch is still a catch if you are using my definition, unless you invoke monotonicity again. That is, clearly, the key point here. Here's an alternate proof that tries to proceed along the lines of the catch: since I_N <= S_N for all N, and since the I_N are increasing and the S_N decreasing, in fact the set { I_N } lies entirely below the set { S_N }. Then clearly the sup of the former is below the inf of the top of that, the first sentence essentially establishes what I established in the first part of my original proof. I guess my point is that there is only one proof of this statement, though you can say it slightly differently in many ways. The definition I gave is the definition I learned (I think it's in Spivak's Calculus, though that's not where I learned it. I suspect it's due to Darboux, since in defining the integral by Darboux sums one uses supremum and infimum extensively). It's equivalent to yours since, as we both noted, S_N is a decreasing sequence and I_N an increasing sequence, so the limits are the infimum and supremum, respectively. My definition seems preferable (this is of course subjective) since one of the benefits of using limsup and liminf is that they always exist (at least, if you count infinities as reasonable values) and, if they are equal, then the limit itself exists. Defining limsup in terms of limit has the appearance of circularity from this perspective (actually, this is a rather attractive way to define the limit in the first place, as it completely elides the epsilons). In fact, in some ways it is actually circular, in that one proves that a monotone (say, increasing) sequence has a limit specifically by exhibiting the supremum of that sequence as the limit. -- Ryan Reich ryan.reich@gmail.com === Subject: New combinatorial coder: tighter & faster than Arithmetic coding There is a new algorithm for entropy coding, constrained coding and fast combinatorial enumeration described in the arXiv preprint cs.IT/0511057 (also submitted to DCC 2006): http://arxiv.org/abs/cs.IT/0511057 It is coding based on enumerative combinatorics (walks on Pascal triangle, counting of integer lattice paths). The new algorithm significantly improves on the previously known method enumerative coding (which is the regular un/ranking procedure in combinatorics), increasing its speed and reducing memory size by factor O(n), where n is number of input symbols). It is the algorithm Rissanen was looking for when he found a partial solution, arithmetic coding (in 1976). The new algorithm, even the exploratory unoptimized prototype runs much faster (see the table below, copied from the of paper, with speedup factors over 200) and always compresses better than the best of arithmetic coders (such as Moffat et al 1998). Besides its usefulness in entropy coding (for all types of file, image,... compressors where Huffman, arithmetic or similar coding is used in the final phase) and for encoding of complex objects (e.g. permutations, trees, graphs, self-delimiting sequences,.. etc) the new algorithm extends significantly the domain of enumerative combinatorics accessible to computer explorations. Here is the abstract: --------------------------------------------------------- Title: Quantized Indexing: Beyond Arithmetic Coding Author: Ratko V. Tomic Quantized Indexing is a fast and space-efficient form of enumerative (combinatorial) coding, the strongest among asymptotically optimal universal entropy coding algorithms. The present advance in enumerative coding is similar to that made by arithmetic coding with respect to its unlimited precision predecessor, Elias coding. The arithmetic precision, execution time, table sizes and coding delay are all reduced by a factor O(n) at a redundancy below log(e)/2^(g-1) bits/symbol (for n input symbols and g-bit QI precision). Due to its tighter enumeration, QI output redundancy is below that of arithmetic coding (which can be derived as a lower accuracy approximation of QI). The relative compression gain vanishes in large n and in high entropy limits and increases for shorter outputs and for less predictable data. QI is significantly faster than the fastest arithmetic coders, from factor 6 in high entropy limit to over 100 in low entropy limit (`typically' 10-20 times faster). These speedups are result of using only 3 adds, 1 shift and 2 array lookups (all in 32 bit precision) per less probable symbol and no coding operations for the most probable symbol. Further, the exact enumeration algorithm is sharpened and its lattice walks formulation is generalized. A new numeric type with a broader applicability, sliding window integer, is introduced. --------------------------------------------------------- Below are few test results of QI vs. AC (AC: Moffat et al. 1998, all buffers prealloced, no slow streams; only inner coding loops timed via sub-microsecond timers) for input sizes N (K=1024 bits) and for given number of 1's randomly distributed (except for the input Vary which represents a non-stationary source). Both coders tested were binary order 0 coders (i.e. they consider input data as a sequence of uncorrelated bits). Each result is an average obtained from 500 random inputs. Column Speed: coding times ratio TimeAC/TimeQI (e.g. QI is up to 247.5 times faster than AC, cf. top right) Column N: output size percent using (A/Q-1)*100 (e.g. AC output is up to 110% larger than QI's, cf. bottom left) Input array Vary = array of int32 {...,-2,-1,0,+1,+2,...}. --------------------------------------------------------------- #1's N:4K Speed N:8K Speed N:32K Speed N:128K Speed --------------------------------------------------------------- 8 6.846 68.3 6.421 112.8 5.447 199.6 5.966 247.5x 16 4.175 59.7 3.830 78.5 3.389 138.1 3.730 168.0 32 2.297 49.7 2.090 58.9 2.096 95.9 2.220 117.2 N/64 1.370 40.3 0.606 41.0 0.186 41.7 0.073 42.5 N/32 0.897 30.8 0.343 33.7 0.123 34.2 0.049 34.5 N/16 0.505 21.8 0.197 25.3 0.084 24.6 0.040 24.8 N/8 0.359 14.4 0.155 16.7 0.069 16.8 0.045 16.8 N/4 0.288 9.2 0.138 10.8 0.083 10.6 0.068 10.5 N/2 0.509 6.6 0.445 6.6 0.367 6.4 0.332 6.4 Vary 110.899 21.9 96.736 19.6 71.308 16.5 52.580 14.1 --------------------------------------------------------------- === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > http://arxiv.org/abs/cs.IT/0511057 I looked quickly through the paper, and here's my impression of what it's about. The idea of enumerative coding is that you divide the set of possible unencoded messages into disjoint subsets with the property that all of the messages in a particular subset are equally likely to occur (according to some model). Then the coded message is a code identifying the subset which contains the message, followed by a code identifying an element of that subset. For example, suppose that your messages are the bit strings [01]*, and you divide them into subsets according to how many zeroes and ones they contain: {}, {0}, {1}, {00}, {01,10}, {11}, {000}, {001,010,100}, ... If our message has K ones and L zeroes, we can encode the index of the appropriate subset by encoding K and L at a cost of log K + log L bits. Encoding the index of the appropriate element of that subset takes log C(K,K+L) = log ((K+L)! / (K! L!)) bits. Compare this to arithmetic coding with a static order-0 model. Here we first encode the length of the message (K+L) and p(one) = K/(K+L), which again takes log K + log L bits. Then for each one bit in the message we encode log ((K+L)/K) bits, and for each zero bit we encode log ((K+L)/L) bits, for a total of K log (K+L)/K + L log (K+L)/L = log ((K+L)? / (K? L?)) bits, where I write N? to mean N^N. Arithmetic coding is asymptotically as efficient as enumerative coding, since O(N?) = O(N!), but enumerative coding will always use fewer bits, since N! < N? for interesting values of N. I think that this paper describes an efficient algorithm for enumerative coding for the order-0 model I used above, but generalized to arbitrarily many symbols. If so, it won't replace coders based on sophisticated models with arithmetic-coding back ends, but it may find a niche in applications that use static order-0 models, such as video compression. -- Ben === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > If so, it won't replace coders based on sophisticated models > with arithmetic-coding back ends,... The AC modeling paradigm ('probability of next single symbol' modeling interface bottleneck) was created following path of 'virtue out of necessity'. The necessity being the particular entanglement of the inner most coding steps with the model parameters p(a). The EC/QI can code the same way, too, except they can also code a much better way (Kolmogorov modeling paradigm). There is a bit on the modeling on the page 9 in the paper and a much longer discussion in the tech report cited: TR05-0625 Quantized indexing: Background information http://www.1stworks.com/ref/TR/tr05-0625a.pdf As you may have noted in the row Vary of the test, which is a non-probabilistic sequence for the order-0 coders, one can easily provide inputs which will surprise catastrophically the arithmetic 0-order coder, but you can't provide inputs on which AC works well and EC fails as dramatically (as does AC on the data Vary). The same phenomenon exists for the higher order coders, as soon as the input falkls outside of the predictable at a given order. The adaptive AC fails badly, while EC merely comes out slighly sub-optimally (since the effect of a suprise is always much greater on the predictor scheme than on the 'descriptor' scheme; with EC only the frequency table encoding may turn out suboptimal, while the bulk of output, the index is still optimal). The essential difference between the adaptive AC and EC/QI is that AC subjects the modeling engine to an aritificial constraint (the memoryless coder allowed to have only a single symbol at a time, a limitation largely not relevant in practical coding since Morse telegraph days). The EC can do that, too, but it normally doesn't. Trying to predict furture, as AC modeling paradigm insists, is inherently harder than describing the present or past. Even the practical forms of AC modeling recognize the problem and allow a backdoor cheating against the ideal pure prediction (via the fragile and ad hoc escape mechanisms, where the coder is allowed to look ahead and clue the decoder as to what is coming). As another aspect of the distinction, note that the EC mdeling engine normally needs to know only whether, say messages X and Y have the same probabilities (whatever their values might be), while the AC modeling engine needs to know what the actual values of the probabilities are, which is a much harder task. The upshot is that the native EC/QI modeling (the Kolmogorov/MDL scheme) is much simpler and more robust than the predictive AC paradigm (EC/QI can reuse the existent AC modeling engine, though, e.g. by splitting the output into different probability classes, cf p. 9, [N7]). === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding >The AC modeling paradigm ('probability of next single symbol' modeling >interface bottleneck) was created following path of 'virtue out of >necessity'. [etc, etc, etc] Look, I understand all of this. I understand the weaknesses of PPM and of arithmetic coding, and I understand that an enumerative coder can do better in principle. None of that matters. What matters is this: what is your contribution? Can you, in practice, do a better job than state-of-the-art compressors based on PPM+ari? Going on about theoretical benefits is a waste of time. Theoretically, the best way to write a chess-playing program is to use minimax on the complete game tree. Actual chess-playing algorithms are gross hacks which attempt, in a highly inelegant way, to approximate that ideal. Pointing out that that's true accomplishes nothing. Everyone versed in the art knows it already. >As you may have noted in the row Vary of the test, which is a >non-probabilistic sequence for the order-0 coders, one can easily >provide inputs which will surprise catastrophically the arithmetic >0-order coder, but you can't provide inputs on which AC works well and >EC fails as dramatically (as does AC on the data Vary). Two questions. First, do you understand the difference between adaptive order-0 and static order-0 models? Second, how would the QI and AC in your tests perform if the input consisted of N/2 zeroes followed by N/2 ones? -- Ben === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding The most relevant papers for this discussion are available online at: Quantized Indexing References http://www.1stworks.com/ref/tr/tr05-0625r.htm Specifically of interest here are the papers: 5. J. Rissanen Generalised Kraft inequality and arithmetic coding IBM J. Res. Dev. 20, 198-203, 19 76 27. J. Rissanen Arithmetic codings as number representations Acta Polyt. Scand., Math. & Comp. Sc. Vol. 31 pp 44-51, 1979 http://www.1stworks.com/ref/ariNumRepr.pdf 28. L.D. Davisson Universal noiseless coding IEEE Trans. Inform. Theory IT-19 (6), 783-795, 1973 http://cg.ensmp.fr/%7Evert/proj/bibli/local/Davisson1973Universal.pdf 29. J. Rissanen, G.G. Langdon Universal Modeling and Coding IEEE Trans. Inform. Theory IT-19 (1), 12-23, 1981 http://cg.ensmp.fr/%7Evert/proj/bibli/local/Rissanen1981Universal.pdf 34. J.G. Cleary, I.H. Witten A Comparison of Enumerative and Adaptive Codes, IEEE Trans. Inform. Theory IT-30 (2), 306-315, 1984 http://www.1stworks.com/ref/Cleary84Enum.pdf === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > Can you, in practice, do a better job than state-of-the-art compressors > based on PPM+ari? QI is a very new algorithm (and a first truly practical form of genuine EC). A better BWT variant, without mangling of its contexts (as the MTF phase presently does), is certainly high on the list of things to apply QI to. It is still only one among many interesting possibilities opened by the existence of practical EC. The whole field of practical EC modeling is presently in a very primitive form and most people have many misconceptions what it is (or generally on modeling, identifying it entirely with the AC modeling paradigm). Hopefully a publication of a clean, well performing reference implementation of QI (on which I am working, and which will perform even better than the tests results of the crude research prototype shown in the preprint) in the forthcoming months, should bring more people to try out new ways of modeling and answer those questions. > First, do you understand the difference between adaptive order-0 and > static order-0 models? Second, how would the QI and AC in > your tests perform if the input consisted of N/2 zeroes followed by N/2 > ones? The sequence Vary was a sequence similar to that, since the negative half of that sequence is mostly 1's and the positive part mostly 0's. The rest of the answer obviously depends on how the QI segments the sequence, what are the rules of the game. If you insist on requiring that QI/EC models ...000111... sequence in a single block of size N, then obviously it will produce output which is N + log*(N) + 1/2 log(N) bits (since it needs to encode the length of a sequence in log*(N) = log(log(...N) bits and the count of 1's in log(N) bits, while the index is C(N,N/2) which is Catalan number with approx N - 1/2 log(N) bits). Of course, if you wish to compare apples to apples, then if the predictive AC coder is allowed some adaptation, one would need to allow QI to also have some kind of corresponding capabilities (i.e. not to function as a static order-0 coder), such as allowing the encoder to select the input segmentation, in which case it would produce output smaller than adaptive AC (as array Vary illustrates). Comparing static vs adaptive order-0 coders is comparing apples and oranges. There is no instrinsic algorithmic constraint for EC/QI to code static order-0 way (i.e. a la Lynch-Davisson 1966 EC coder). If you take AC and EC/QI and make both code static order-0 way with the same segmentation, the EC/QI will give you slighly smaller output and code much faster. Then if you make AC adaptive, with adaptation rate faster than N/2, you need to decide what is the corresponding EC/QI mode of coding? Whatever it is, it certainly cannot be the same scheme that was just tested as a comparison of the static order-0 coders, which is what you seem to be hinting it ought to be. The most natural EC/QI counterpart to AC adapting faster than N/2 is to allow EC/QI coder the segmentation choice to at least 2 parts, in which case it will come out on top again (as the array Vary showed; even though the QI coder tested was somewhat more primitive than that: it had simply used fixed 1Kbit blocks). In any case, this particular tangent is more about verbal pigeonholing, rather than about the substance of the difference. The basic difference between AC and QI coding algorithms is that AC enumerates messages via cummulative probabilities and EC/QI via exact ranking of combinations of symbol sequences. The latter way is more accurate and it is much faster (since much of the coder work is with the universal sequence properties, thus it is precomputed in the quantized binomial tables, outside of the coding loop). The modeling scheme is an entirely unrelated issue. You can select a modeling scheme where both coders servicing that modeler perform equally poorly (e.g. by forcing QI to do lattice jumps a la AC, in which case it will have to work exactly as hard per symbol as AC; with adaptive output splitting QI will compress well as adaptive AC, but code much faster). Generally, though, given the same modeling scheme, QI will always perform at least as well as AC, in speed and in accuracy (but mostly better, especially in speed). Anything else is comparing apples and oranges (note that the test in the row Vary is of that apple-vs-orange kind, except that QI was the one slightly handicapped since it wasn't really as adaptive as the EC counterpart of adaptive AC ought to be i.e. free select at least one input partition for AC allowed to adapt faster than N/2; QI/EC produces 0 length index for sequence of all 1's or all 0's). === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > First, do you understand the difference between adaptive order-0 and > static order-0 models? Second, how would the QI and AC in > your tests perform if the input consisted of N/2 zeroes followed by N/2 > ones? > The sequence Vary was a sequence similar to that, since the negative > half of that sequence is mostly 1's and the positive part mostly 0's. What do you mean by negative half and positive part? Applying the concept of sign to sequences of bits is, erm, unconventional. Phil -- If a religion is defined to be a system of ideas that contains unprovable statements, then Godel taught us that mathematics is not only a religion, it is the only religion that can prove itself to be one. -- John Barrow === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding >> First, do you understand the difference between adaptive order-0 >> and static order-0 models? Second, how would the QI and AC in >> your tests perform if the input consisted of N/2 zeroes followed by >> N/2 ones? >> The sequence Vary was a sequence similar to that, since the >> negative half of that sequence is mostly 1's and the positive part >> mostly 0's. > What do you mean by negative half and positive part? > Applying the concept of sign to sequences of bits is, erm, > unconventional. > Phil I guess that means if your given say a file with say one thousand bytes of zero followed by say one thousand bytes of all ones 0xFF he has no idea how his compressor or some arbitrary arithmetic would compress it. It really should not be that hard to test. But maybe he can't test it with his method yet. David A. Scott -- My Crypto code http://bijective.dogma.net/crypto/scott19u.zip http://www.jim.com/jamesd/Kong/scott19u.zip old version My Compression code http://bijective.dogma.net/ **TO EMAIL ME drop the roman five ** Disclaimer:I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged. As a famous person once said any cryptograhic system is only as strong as its weakest link === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding <87wtj2ncpl.fsf@megaspaz.fatphil.org> >> The sequence Vary was a sequence similar to that, since the negative >> half of that sequence is mostly 1's and the positive part mostly 0's. > What do you mean by negative half and positive part? > Applying the concept of sign to sequences of bits is, erm, unconventional. The negative 32 bit integer, say -1 has 32 1's. The idea was to give both coders (which were order 0 binary coders, thus they view data as sequence of uncorrelated bits) some simple sequence which has regularity invisible to order-0 (but visible to some higher order coder). It was meant to illustrate the robustness of the universal way of coding of the EC/QI coder and the fragility of the predictive/adaptive way of the AC coder. The origin of the greater robustness in the descriptive way of modeling is in the fact that effect of the surprise affects chiefly the count component of the output, which for the 1K input block is on average a 10 bit number (e.g. specifying # of 1's in the block). The rest is a pure combinatorial index, giving the shortest possible decription, absent any further knowledge by the coder, on how that number of bits is placed in that block. With the adaptive AC, each coding step and the entire output is affected by the incorrectly conjectured probabilities. Thus it produces as much as twice the output size of the QI. Obviously, the AC can be made to code the 'universal'/descriptive way, in which case the error would be much smaller (similar to the rest of tables, for the matching density of 1's and input size i.e. few percent extra output), while the speed ratio would still remain dramatically in QI favor. The example was meant to help demythologize a bit the 'adaptive/predictive coding mystique' that lots of people are apparently mesmerized by (as the previous discussion in this thread illustrates). The 'predictive coding', at any modeling order, is a handicap (a historical fluke of bending the entire modeling engine to the peculiarities of the particular interpretation of the AC computational algorithm, the perceived need to have probability of next symbol to perform the encoding), not a strength as often touted, and a 'descriptive'/universal type algorithm at the same order of modeling will always come out ahead. When AC is made to code the universal/descriptive way, the probabilities used for the computation are simply interpreted as normalized plain counts (instead of the presumed infinite n limits of such ratios, the 'probabilities, which is the adaptive AC interpretation of the AC algorithm and which contains a much bigger implicit assumption about the infinite rest of the sequence, that in the case of the array Vary takes it to a catastrophically wrong track). But in this universal mode, AC uses the plain normalized counts in the exactly same way as the counts of EC/QI, slightly less accurately (by 1/2 log(n) excess bits in the AC output, for the sequence of length n) and encoded much more slowly. The Rissanen's 1976 coder (which was the first finite precision arithmetic coder) was coding this way. See the reference: 5. J. Rissanen Generalised Kraft inequality and arithmetic coding IBM J. Res. Dev. 20, 198-203, 19 76 where the quantities Phi (in eq. 5) are merely the approximate binomial coefficients of regular EC (see footnote on page 2 in the preprint and ref [23] for details). Although Rissanen suggests in that paper a table based faster coding alternative (which allows skipping of the most frequent symbol as in QI), for the AC one needs different tables of O(n^2) size for each source probability, while QI needs only a single table of O(n^2), the quantized binomials C(n,k), which is valid for all quasi-stationary sources and all probabilities. That's why the AC algorithm was a doubly partial solution of the precision problem of the exact enumeration -- it not only produced an approximate (and a larger) enumerative index, but it also left unsolved the O(n^3) table size problem of the exact enumeration (which it then worked around by trading the table size for quite a few extra CPU cycles, as illustrated in the tests chart). The QI solves both problems - it takes the factor O(n) out of the exact EC computation speed, as did AC, and it takes the factor O(n) out of the table sizes, which AC missed (thus QI doesn't need to trade off the coding speed to reduce the table sizes as AC does; although QI can do that kind of tradeoffs too, cf. page 9, [N2] in the preprint, but from a much better starting point than AC). Rissanen was pleasantly surprised to see that great difficulty (see the preprint pages 1-2, for citations) finaly solved optimally and commented on the solution Very clever! === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding ) The negative 32 bit integer, say -1 has 32 1's. The idea was to give ) both coders (which were order 0 binary coders, thus they view data as ) sequence of uncorrelated bits) some simple sequence which has ) regularity invisible to order-0 (but visible to some higher order ) coder). It was meant to illustrate the robustness of the universal way ) of coding of the EC/QI coder and the fragility of the ) predictive/adaptive way of the AC coder. If an adaptive order-0 AC coder performs badly on a sequence that has A good adaptive model views symbols that are nearer with more weight. It's no wonder that badly written code is outperformed. SaSW, Willem -- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding <87wtj2ncpl.fsf@megaspaz.fatphil.org> > If an adaptive order-0 AC coder performs badly on a sequence > model. > A good adaptive model views symbols that are nearer with more weight. Making AC adapt faster to more variable inputs, costs in extra redundancy for the stationary inputs (which don't need a waste in code space equivalent to an explicitly reserved escape code space used to change the statistics on the fly). You can't have it both ways, optimal for the stationary or slow varying source yet adapting fast to the dynamic source, while coding within the adaptive AC scheme. The faster AC adaptation implies fewer significant bits in the probabilities p(a). You will have roughly 1/2 log(n) bits in p(a) if you count exactly, with full weight, the n past bits, i.e. p(a) has only half the precision bits of the precision of n. Any forgetting of the older stuff (be it via sharp cutoff or some decaying law) simply shortens the effective n, thus the precision of p(a). The fast changing AC thus comes down to the similar approximation that quasi-arithmetic coders do when they approximate p(a) with numbers which have only few bits set, so they can scale the AC interval quicker via few shifts and adds instead of multiplies. Here it would make AC more robust to the sudden drastic changes of statistics, but at the cost of extra 5-10 percentage points compression loss for the stationary inputs. Thus, this is a matter of a tradeoff you need to select. In the extreme preference for the robustness to the input dynamics, you can easily have AC code the 'universal' / descriptive way instead of the usual adaptive / predictive way, i.e. make it code exactly as QI does, which is extremely robust to statistical variability and which with QI is optimal, while the AC coding this way will still produces an excess of O(log(n)) output bits (where n is the input size or indexing block size; the regular adaptive AC has this funadmental excess vs QI as well, cf. preprint footnote on page 2). The O(log(n)) bits AC output excess becomes pretty significant compared to the output size if the source entropy is low (see top rows in the earlier chart) or if the n is small (which may be required for very dynamic sources, in order to segment the input into quasi-static sections), as illustrated by the leftmost column in the test chart Moffat et al 1998, which is the AC we used, explain their particular tradeoff in their paper (hyperlinked earlier). You can tell them what you think of their order 0 binary coder if you think you have a case. I don't think you do. The problem is inherent in the adaptive / predictive coding scheme -- they could choose to look Ok in the last row of the chart, while looking much worse in the top rows (when just few bits are set), or they could choose what they did, look bad in the last row and look Ok in the top rows. On the other hand, there is no such an input where they would look Ok, while QI would look bad (not just at order-0, but at any order source, provided both coders code in the same order model, obviously). It is a trivial observation (but which you nevertheless keep bringing up in different wordings), that you can make the two coders use inadequate order model for some source and have both look very bad on such source, compared to some other coder using a higher order model. As long as you compare apples to apples and oranges to organges, thus let them use the same order model (adequate or inadequate to the order of the source), the adaptive AC will always do worse than QI, and very much so for the very unpredicatble sources. That is the fundamental weakness or fragility of any predictive scheme when faced with unpredictable (for the model order that coder has) inputs. The universal / descriptive schemes don't have such fatal fragility, which was the whole point of the Davisson's minimax universal codes (constructed using enumerative coding) -- they may code slightly suboptimally, when given the inadequate order model for a given source, but their worst case is never farther than about O(log(n)/n) bits per symbol from the output of the best specialized coder which was tuned specifically to the particular source. (And when they do know or model the proper source parameters, the EC/QI code optimally.) The point of the array Vary in the test was to illustrate how an adaptive AC, when tuned to perform nearly optimally on the stationary inputs, thus to appear fairly competitive with QI on such inputs (with no more than about 6-7% worse output in the rest of the chart), is very fragile and it breaks down catastrophically if the statistics suddenly changes. The array Vary has only one drastic change of statistics, in the very middle of the array when -1 goes to 0, hence the negative effects on AC decrease as the array size increases. Had the statistics changed again in the longer arrays such as in 128k input, the AC performance would have been reset down, to the worse performance figures from the shorter arrays. === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding )> If so, it won't replace coders based on sophisticated models )> with arithmetic-coding back ends,... ) ) The AC modeling paradigm ('probability of next single symbol' modeling ) interface bottleneck) was created following path of 'virtue out of ) necessity'. The necessity being the particular entanglement of the ) ... ) ) As you may have noted in the row Vary of the test, which is a ) non-probabilistic sequence for the order-0 coders, one can easily ) provide inputs which will surprise catastrophically the arithmetic ) 0-order coder, You don't seem to grasp the existence of higher-order models. Would something like PPM be easy to implement using this QI coder ? SaSW, Willem -- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding >>) As you may have noted in the row Vary of the test, which is a >> non-probabilistic sequence for the order-0 coders, one can easily >> provide inputs which will surprise catastrophically the arithmetic >> 0-order coder, > You don't seem to grasp the existence of higher-order models. That was a test of the same order 0 coders. At any order, you can suprise the predictive scheme catastrophically (unless it cheats via some ad hoc esc mechanism, allowing it to use temporarily universal way of coding), while you can't do so to a descriptive/universal scheme. > Would something like PPM be easy to implement using this QI coder ? There are couple ways, at least, to do that. QI can code (via lattice jumps) to the 'probability of the next single symbol' exactly as AC, but that costs it much of speed advantage (since mantissa scaling used by the jumps, requires multiplications & divisions). But it can also code from the AC modeling engine instructions much faster by splitting the outputs into the probability classes (which needs also an error estimates on the probabilities given to the coder; the scheme grows the classes in the manner of Context Tree Weighing modeling scheme). OTOH, it could do much better, with any given a priori knowledge about the input that is fed to PPM, by coding the native EC way, which is by splitting the input into proper enumerative classes. Such segmentation can be done based on the last s symbols (e.g. for order-s static Markov source; a method similar to Context Tree Weighing to select the output stream), or dynamically via BWT-like segmentation which doesn't assume a priory idealized finite order Markov source (you can check the discussion of this in the Ref [23] cited earlier). Even with the kludgy ad hoc entropy coding schemes (such as MTF variants which crudely mangle the enumerative classes present in the final BWT matrix), BWT is already surprisingly competitive with PPM in compression quality, while executing much quicker. With the proper handling of these presently mangled BWT enumerative classes, which QI is naturally suited to do quickly and optimally, BWT should consistently outperform PPM not just in speed but in compression quality. In short, AC is a particular historical approximation of QI, hence anything AC can do, QI can do but generally much faster (or at least as fast) and generally more accurately (or at least as accurately). Similarly, the PPM or generally the predictive way of modeling via the 'probability of the next single symbol' is another historical accident (tailored to a particular hardwired coding algorithm of AC), by no means the only (and certainly not the best) way to model the properties of finite sequences of symbols. AC algorithm happens to require that all properties must be translated into exactly such form (as 'the probabilities of the single next symbol') so it can encode it. That is more of a handicap than a strength for the modeling engine to bend and distort everything else around the AC computational peculiarities, to have to translate and funnel all the conceivable properties of symbol sequences available to modeling engine through that particular coder interface bottleneck. So, instead of questing anyones grasp of 'higher order models' you might wish to reaxamine first, how do some manage to convince themselves to perceive, with all their sincerity, a plain, gross handicap as a great strength. === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > That was a test of the same order 0 coders. At any order, you can > suprise the predictive scheme catastrophically (unless it cheats via > some ad hoc esc mechanism, allowing it to use temporarily universal way > of coding), while you can't do so to a descriptive/universal scheme. Look is your so called coded able to actaully compress some files. Do you have test files to test the order 0 codings. So one can campare it to real world order 0 encoders. With out real examples one can test I don't belive you. David A. Scott -- My Crypto code http://bijective.dogma.net/crypto/scott19u.zip http://www.jim.com/jamesd/Kong/scott19u.zip old version My Compression code http://bijective.dogma.net/ **TO EMAIL ME drop the roman five ** Disclaimer:I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged. As a famous person once said any cryptograhic system is only as strong as its weakest link === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > How about you stating what you feel a typical binary zero order > based arthmetic coder would code one thousand bytes of all zero > followed by one thousand bytes of all ones. That question (besides being trivial) was already discussed in this thread... see the post: === Subject: [very OT] Re: New combinatorial coder: tighter & faster than Arithmetic coding I'm confused - what on earth is the point of such a header in news-posting context? And why is a cantabrigian prioritising en-us above en?!?!? Phil -- If a religion is defined to be a system of ideas that contains unprovable statements, then Godel taught us that mathematics is not only a religion, it is the only religion that can prove itself to be one. -- John Barrow === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > The new algorithm, even the exploratory unoptimized > prototype runs much faster (see the table below, copied > from the of paper, with speedup factors over 200) and > always compresses better than the best of arithmetic > coders (such as Moffat et al 1998). Besides its usefulness > in entropy coding (for all types of file, image,... > compressors where Huffman, arithmetic or similar coding > is used in the final phase) and for encoding of complex objects > (e.g. permutations, trees, graphs, self-delimiting sequences,.. > etc) the new algorithm extends significantly the domain of > enumerative combinatorics accessible to computer explorations. Sounds like hype. Especially the statement always compresses better than the best of arithmetic coders This is obviouly a false statement. It is impossible for any compressor to always compress better than every other compressor. First of all the best compressors would have to be bijective and then the best they can do is change the mappings from one set to another. Take any compressor that could make one stream or file smaller. Then the identity compressor could beat it on files that the previous compressor lengthened since its a loose loose situation if you make something else smaller or better compressed then it has to make something else longer. Second if you actaully took the time to test the online versions of code that use Moffat style code you would soon relize it not the best. David A. Scott -- My Crypto code http://bijective.dogma.net/crypto/scott19u.zip http://www.jim.com/jamesd/Kong/scott19u.zip old version My Compression code http://bijective.dogma.net/ **TO EMAIL ME drop the roman five ** Disclaimer:I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged. As a famous person once said any cryptograhic system is only as strong as its weakest link === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > Second if you actaully took the time to test the online versions of code that > use Moffat style code you would soon relize it not the best. One can take Moffat et al code (version 3, 2000) and select some other 'accuracy vs speed' tradeoffs. If you know of a version of that coder which is faster and produces smaller output (the figures I listed refer to binary 0-order coders test), you are welcome present the comparison figures between the two and I will gladly check it out. Keep in mind, the speed factors in the paper show QI is 200+ times faster than AC on some inputs, and typically 10-20 times faster (this was a generic QI coder, i.e. the same loop and the same steps coding the dense and the sparse arrays, and not something tuned to run fast for sparse arrays, such as run length coder, which otherwise blows on size and speed for denser inputs). There are dime a dozen quasi-arithmetic coders, which sacrifice some compression quality (say another 5-10% larger output) to gain some speed (a factor 2-5). The new algorithm does much less work (and a simpler work: just an add of the table addend, a machine word) per coding step and always does fewer coding steps, thus it has a fundamental reason for a much better performance than AC. === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > the speed factors in the paper show QI is 200+ times faster than AC on The question this raises for me is why AC implementations are 200x wouldn't surprise me), then there's no way on earth that QI could be 200 times faster in _any_ circumstances. Phil -- If a religion is defined to be a system of ideas that contains unprovable statements, then Godel taught us that mathematics is not only a religion, it is the only religion that can prove itself to be one. -- John Barrow === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding <871x1bnczp.fsf@megaspaz.fatphil.org> > The question this raises for me is why AC implementations are 200x The AC has to process and encode each bit, 32 per machine word, for which QI needs only to establish if there is any of the less frequent symbols (e.g. 1). Even though that AC implementation uses large loop unrolling and other speed optimizations, it only needs 8 times more work per one bit than QI needs to fetch the dword and answer is it only the more frequent symbols in this word (e.g. w==0 or w==(-1)) to get the speed factor of 256. If you look their code (binary coder, no contexts): http://www.cs.mu.oz.au/%7Ealistair/arith_coder/ you will see that they certainly do that much more work per encoding of each bit, in fact even more than the ratios show, if you count the numbers of instructions. But, as you note, the fast processors with lots of cache mask out some of the speed difference, penalizing more the new memory reads, which both coders have to do. On an older 400Mhz Celeron laptop the speed ratios are 1.5-2 times larger in favor of QI than on the recent VAIO lapop where those figures in the preprint came from. === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding How about comparing QI with a faster and more efficient binary arithmetic coder such as the fpaq one (see http://www.cs.fit.edu/~mmahoney/compression/ for more info). Malcolm >>The question this raises for me is why AC implementations are 200x > The AC has to process and encode each bit, 32 per machine word, for > which QI needs only to establish if there is any of the less frequent > symbols (e.g. 1). Even though that AC implementation uses large loop > unrolling and other speed optimizations, it only needs 8 times more > work per one bit than QI needs to fetch the dword and answer is it > only the more frequent symbols in this word (e.g. w==0 or w==(-1)) to > get the speed factor of 256. If you look their code (binary coder, no > contexts): > http://www.cs.mu.oz.au/%7Ealistair/arith_coder/ > you will see that they certainly do that much more work per encoding of > each bit, in fact even more than the ratios show, if you count the > numbers of instructions. But, as you note, the fast processors with > lots of cache mask out some of the speed difference, penalizing more > the new memory reads, which both coders have to do. On an older 400Mhz > Celeron laptop the speed ratios are 1.5-2 times larger in favor of QI > than on the recent VAIO lapop where those figures in the preprint came > from. === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding <871x1bnczp.fsf@megaspaz.fatphil.org> <1132541691.187691@drone2-svc-skyt.qsi.net.nz> > How about comparing QI with a faster and more efficient binary > arithmetic coder such as the fpaq one (see > http://www.cs.fit.edu/~mmahoney/compression/ for more info). QI is a fundamental algoritm not a program (although several research variants of the algoritham are implemented). If your algorithm needs to do more work than about 3 adds, 1 shift and two 1-dim array reads, per less probably symbol, and any operation at all (other than traversing the array to examine what the symbols are) per more probable symbol, then it is inherently slower algorithm than QI. Any arithmetic coder falls in this category, since a general purpose AC needs to encode 0's and 1's explicitly. Check the simplicity of the generic EC/QI coding loop (from p. 8 in [23]) 23. TR05-0625 Quantized indexing: Background information http://www.1stworks.com/ref/TR/tr05-0625a.pdf I=0; // init cumulative path index for(k=n=0; n QI is a fundamental algoritm not a program (although several research > variants of the algoritham are implemented). If your algorithm needs to > do more work than about 3 adds, 1 shift and two 1-dim array reads, per > less probably symbol, and any operation at all (other than traversing > the array to examine what the symbols are) per more probable symbol, > then it is inherently slower algorithm than QI. Any arithmetic coder > falls in this category, since a general purpose AC needs to encode 0's > and 1's explicitly. Check the simplicity of the generic EC/QI coding > loop (from p. 8 in [23]) This is incorrect. Check the MQ and QM coder family. Typical implementations have about the same complexity and are fine implementations of arithmetic coding. So long, Thomas === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding <871x1bnczp.fsf@megaspaz.fatphil.org> <1132541691.187691@drone2-svc-skyt.qsi.net.nz> >> This is incorrect. Check the MQ and QM coder family. Typical >> implementations have about the same complexity and are fine >> implementations of arithmetic coding. It is correect if you restrict the statement to full precision arithmetic coders, as I did, (thus the Moffat et al 1998 coder is the best of full AC implementations). The quasi-arithmetic coders which sacrifice 5-10% of compression for speed are dime a dozen. Moffat98 paper shows some benchmarks against the best Q coder implementation they found and it runs only 2-3 times faster than their binary coder. That is still up to 100 times slower than QI, at the great compression cost in the low entropy range of the inputs (it would do Ok only on medium to high entropy inputs, just a point or two below the full AC). But, even the quasi-AC still needs to either code each symbol, 0 and 1, or they need an O(n^2) size table for each different probability (they may just use a few of such tables, with very coarse quantization of probabilities) -- you can also check Rissanen's discussion of these two tradeoffs he already outlined in his 1976 AC paper. An AC coder may also use a multiplication table of O(n^2) size, thus do no multiplications, but it still needs to code each symbol 0 and 1. To avoid coding each symbol, like QI, there is no way for AC around having a different O(n^2) table for each different probability (however many quantized p() levels the pick). If you think someone can do it, code via AC only the less frequent symbol and just skip over the most frequent symbol (without renormalization) and still use the single O(n^2) table for all probabilities, post a link to a paper showing that. Basically, the AC enumerative addends which avoid the coding on the more probable symbol are of the form A(n,k) = p^k q^(n-k) (where p=p(1) and q=1-p = p(0) >= p(1), k=count of 1's, n=number of coded symbols so far; you get these 1-step addends by expanding the regular AC scaling, cf. [23] pp. 19-22). They have the probability p hardwired into the addend. You can have a table of A(n,k) for AC, just as you have the quantized binomial table C(n,k) for QI, and then each coder will do exactly the same amount of work per coding step, but the entries in the AC table work only for the given p & q used to compute that table, while the entries in the single QI table work for all p & q. The AC is simply an approximation (of QI), which in the approximation process (cf. preprint footnote on p.2 ) has mangled the delicate combinatorics of the binomial coefficients, entangling them with the p & q, and has lost the addend universality. When I post the reference implementation of QI on the coder page (in the next 2-3 months), anyone is welcome to test it against different kinds of quasi-arithmetic coders. I would be curious to see the results. [this may a duplicate, since I already posted this reply, but it didn't show up somehow.] === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding <871x1bnczp.fsf@megaspaz.fatphil.org> <1132541691.187691@drone2-svc-skyt.qsi.net.nz> > This is incorrect. Check the MQ and QM coder family. Typical > implementations have about the same complexity and are fine > implementations of arithmetic coding. It is correect if you restrict the statement to full precision arithmetic coders, as I did, (thus the Moffat et al 1998 coder is the best of full AC implementations). The quasi-arithmetic coders which sacrifice compression for speed are dime a dozen. Moffat98 paper shows some benchmarks against the best Q coder implementation they found and it runs only 2-3 times faster than their binary coder. That is still up to 100 times slower than QI, at the great compression cost in the low entropy range of the inputs (it would do Ok only on medium to high entropy inputs, just a point or two below the full AC). But, even the quasi-AC still needs to either code each symbol, 0 and 1, or they need an O(n^2) size table for each different probability (they may just use a few of such tables, with very coarse quantization of probabilities) -- you can also check Rissanen's discussion of these two tradeoffs he already outlined in his 1976 AC paper. An AC coder may also use a multiplication table of O(n^2) size, thus do no multiplications, but it still needs to code each symbol 0 and 1. To avoid coding each symbol, like QI, there is no way for AC around having a different O(n^2) table for each different probability (however many quantized p() levels the pick). If you think someone can do it, code via AC only the less frequent symbol and just skip over the most frequent symbol and still use the single O(n^2) table for all probabilities, post a link to a paper showing that and I will show you where the error in their proof was. Basically, the AC enumerative addends which avoid the coding on the more probable symbol are of the form A(n,k) = p^k q^(n-k) (where p=p(1) and q=1-p = p(0) >= p(1), k=count of 1's, n=number of coded symbols so far; you get these 1-step addends by expanding the regular AC scaling, cf. [23] pp. 19-22). They have the probability p hardwired into the addend. You can have a table of A(n,k) for AC, just as you have the quantized binomial table C(n,k) for QI, and then each coder will do exactly the same amount of work per coding step, but the entries in the AC table work only for the given p & q used to compute that table, while the entries in the single QI table work for all p & q. The AC is simply an approximation (of QI), which in the approximation process (cf. preprint footnote on p.2 ) has mangled the delicate combinatorics of the binomial coefficients, entangling them with the p & q, and has lost the addend universality. When I post the reference implementation of QI on the coder page (in the next 2-3 months), anyone is welcome to test it against different kinds of quasi-arithmetic coders. I would be curious to see the results. === Subject: Re: New combinatorial coder: WHERE'S THE BEEF > If you think someone can do it, code via AC only the less frequent > symbol and just skip over the most frequent symbol and still use the > single O(n^2) table for all probabilities, post a link to a paper > showing that and I will show you where the error in their proof was. How about you stating what you feel a typical binary zero order based arthmetic coder would code one thousand bytes of all zero followed by one thousand bytes of all ones. Acatually a good zero order binary bit compressor would as a limit compress any permutation of the above bit string to the same length file give or take one byte. How good can your coder do. Or is it that you can't do something this simple? David A. Scott -- My Crypto code http://bijective.dogma.net/crypto/scott19u.zip http://www.jim.com/jamesd/Kong/scott19u.zip old version My Compression code http://bijective.dogma.net/ **TO EMAIL ME drop the roman five ** Disclaimer:I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged. As a famous person once said any cryptograhic system is only as strong as its weakest link === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding ) ... ) tradeoffs he already outlined in his 1976 AC paper. An AC coder may ) also use a multiplication table of O(n^2) size, thus do no ) multiplications, ... Are you aware of the fact that a multiplication table is a very stupid idea on any recent CPU ? SaSW, Willem -- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding <871x1bnczp.fsf@megaspaz.fatphil.org> <1132541691.187691@drone2-svc-skyt.qsi.net.nz> > Are you aware of the fact that a multiplication table is > a very stupid idea on any recent CPU ? Depends what you call any recent CPU. For battery powered, embedded processors (and there are more of those than deskotops; and they're as recent), the absence of multiplications and divisions in a coder is quite an important element for stretching the battery life. What may be a stupid idea for a desktop, may be a smart idea for a cell phone or a camera. I don't wish to appear condenscending or anything, but scientific papers are written on this very topic, you know, and there are people who specialize in how to transform the existent algorithms so they consume less power. Multiplications & divisions are often the key targets for elimination in these transformations. Since your conversational manners in this thread haven't been exactly exemplary, I will skip the relevant links and let you enlighten yourself on that subject on your own. === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding ) ) ... Since your ) conversational manners in this thread haven't been exactly exemplary, I ) will skip the relevant links and let you enlighten yourself on that ) subject on your own. I assume that answering simple questions with several paragraphs of in-depth analysis of the QI coder, most of which is irrelevant to said question, is exemplary conversational manner ? I do admit that politicians do this all the time, but I personally would not take example from them. SaSW, Willem -- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding )> How about comparing QI with a faster and more efficient binary )> arithmetic coder such as the fpaq one (see )> http://www.cs.fit.edu/~mmahoney/compression/ for more info). ) ) QI is a fundamental algoritm not a program (although several research ) variants of the algoritham are implemented). If your algorithm needs to ) do more work than about 3 adds, 1 shift and two 1-dim array reads, per How large are these arrays ? SaSW, Willem -- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding <871x1bnczp.fsf@megaspaz.fatphil.org> <1132541691.187691@drone2-svc-skyt.qsi.net.nz> >> two 1-dim array reads, per > How large are these arrays ? In the code used for the test results in the preprint, the two arrays were n^2/4 of 32 bit entries (just mantissa was stored; exponent was computed cf. report page 9, [N2]) in the main quantized binomial table and n pointers in the row pointer array (pointing into the rows of binomial table). The n used was 1024, hence the big table was 256K entries i.e. 1 Mbyte, which was about 0.05% of the memory (2G) in the machine it was tested on. There would have been no visible difference in the compression ratios for the the range of the tests, had only 16 bit mantissas been used by QI, thus the table could have been 1/2 Mbyte. But the research prototype code wasn't really optimized, it simply kept table space for the maximum mantissa it may ever use and there was no need to waste time coding separate 16-bit mantissa routines to save a bit of table memory. If one were aiming to merely match the AC on compression (and not to be uniformly ahead), a much smaller table could have been used such as n=256, which needs for the big table only 16K entries of 8 bits each to produce an output excess of less than 1 bit per 256 bit input block (the QI worst case redundancy is below log(e)/2^(g-1) bits/symbol for precision g). One can also trade off the table size for some execution time (cf. preprint, p. 9, [N2]), by skipping the rows of binomials, e.g. by skipping every 2nd row in Pascal triangle, you need an extra add to compute a missing binomial, hence on average it has 1/2 add per step more work, but the big table size drops in half, e.g. to 8k for the n=256 fixed block coder. Similarly, the same tables of log(x!) used for the exponent computations (see p.9 [N2]), are accurate enough to give you all but the lowest 10-11 bits of the 32 bit mantissa for the quantized binomials, although, again, this kind of saving (to cut down the big table to 1/3rd) wasn't a great priority. For an embedded production coder, yes, one would certainly take advantage of such savings. For a desktop coder, e.g. in a video codec, it probably wouldn't matter much. === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding <871x1bnczp.fsf@megaspaz.fatphil.org> <1132541691.187691@drone2-svc-skyt.qsi.net.nz> >> two 1-dim array reads, per > How large are these arrays ? In the code used for the test results in the preprint, the two arrays were n^2/4 of 32 bit entries (just mantissa was stored; exponent was computed cf. report page 9, [N2]) in the main quantized binomial table and n pointers in the row pointer array (pointing into the rows of binomial table). The n used was 1024, hence the big table was 256K entries i.e. 1 Mbyte, which was about 0.05% of the memory (2G) in the machine it was tested on. There would have been no visible difference in the compression ratios for the the range of the tests, had only 16 bit mantissas been used by QI, thus the table could have been 1/2 Mbyte. But the research prototype code wasn't really optimized, it simply kept table space for the maximum mantissa it may ever use and there was no need to waste time coding separate 16-bit mantissa routines to save a bit of table memory. If one were aiming to merely match the AC (and not to be uniformly ahead) on compression, a much smaller table could have been used, e.g. n=256 needs for the big table only 16K entries of 8 bits each to produce an output excess below 1 bit per 256 bit input block (the QI worst case redundancy is below log(e)/2^(g-1) bits/symbol for precision g). One can also trade off the table size for some execution time (cf. preprint, p. 9, [N2]), by skipping the rows of binomials, e.g. by skipping evey 2nd row in Pascal triangle, you need an extra add to compute a missing binomial, hence on average it has on average 1/2 add per step more work, but the big table size drops in half, e.g. to 8k for the n=256 fixed block coder. Similarly, the same tables of log(x!) used for the exponent computations (see p.9 [N2]), are accurate enough to give you all but the lowest 10-11 bits of the 32 bit mantissa for the quantized binomials, although, again, this kind of saving (to cut down the big table to 1/3rd) wasn't a great priority. For an embedded production coder, yes, one would certainly take advantage of such savings. For a desktop coder, e.g. in a video codec, it probably wouldn't matter much. === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding )>> two 1-dim array reads, per )> How large are these arrays ? ) ) In the code used for the test results in the preprint, the two arrays ) were n^2/4 of 32 bit entries (just mantissa was stored; exponent was ) computed cf. report page 9, [N2]) in the main quantized binomial table ) and n pointers in the row pointer array (pointing into the rows of ) binomial table). The n used was 1024, hence the big table was 256K ) entries i.e. 1 Mbyte, which was about 0.05% of the memory (2G) in the ) machine it was tested on. ... Are you familiar with the concepts of CPU caches ? A cache miss on most CPUs is easily as costly as a multiply. On modern CPUs it may get near the cost of a division. SaSW, Willem -- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding <871x1bnczp.fsf@megaspaz.fatphil.org> <1132541691.187691@drone2-svc-skyt.qsi.net.nz> > Are you familiar with the concepts of CPU caches ? > A cache miss on most CPUs is easily as costly as a multiply. > On modern CPUs it may get near the cost of a division. Of course, that's why n=1K was used for the tests. Using larger n made the active parts of the table (which are roughly the sqrt(n) wide band around the straight line k = p*n for given probability of 1, p) fall out of the cache on some machines, reducing the speed ratios against AC. The 1k seems to be just about the right size for most of todays PC's laptops & desktops. Now, with the older CPUs, e.g. 300-400Mhz old Celerons from 6-7 years ago, the cache is small, but the ratio of the in the chart (QI vs AC) look even better on such machines by about factor 1.5-2, even with larger n (e.g. 2 or 4k). With larger n, the compression is usually better for very low entropy, and it also runs faster since the inter-block housekeeping may get costly if the blocks are too small (due to the mixed radix coding used to eliminate bit fraction losses at the block boundaries; in practice one may ignore this tiny compression improvement, but it does matter for accuracy benchmarks). The production version of QI targeted to some CPU range, could easily optimize the tradeoff by using the right kind of row skipping & mantissa interpolation, to reduce the table size if the cache gain offsets the extra calculation for the interpolations. Other simple improvement (which wasn't used in the tests, bit which does work well in another prototype coder) is to use variable n for different sections of the table. Normally, for fixed block coding, the table is triangular (Pascal triangle). But one can get much more bang per K of tables, by using V shaped table, which has much longer n range near the lattice axes (addends fo sparse densities) and shorter n near the center of symmetry of Pascal triangle (the high entropy limit). In the simple fixed block coding used in the test, the accuracy was a large overkill in the high entropy regions, which could have easily (with more complicated code) been cut in 1/2 or 1/3rd without affecting the compression ratios. You can check Lawrence & Tjalkens et al papers on this kind of enumerative coding: 12. J. C. Lawrence A new universal coding scheme for the binary memoryless source IEEE Trans. Inform. Theory IT-23 (4), 466-472, 1977 http://citeseer.ist.psu.edu/lawrence77new.html 13. T.J. Tjalkens, F.M.J. Willems A universal variable-to-fixed length source code based on Lawrence's algorithm IEEE Trans. Inform. Theory IT-38 (2), 247-253, 1992 http://citeseer.ist.psu.edu/585782.html === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > Sounds like hype. Especially the statement always compresses better > than the best of arithmetic coders This is obviouly a false statement. > It is impossible for any compressor to always compress better than > every other compressor. The two are not arbitrary unrelated algorithms -- arithmetic coding (AC) and QI are both approximations of the exact enumerative coding (the exact enumeration of messages). The difference is that AC first approximates the enumeration problem itself (via stirling approximation and ignoring some factors, cf. the paper p.2), then it applies additional precision reduction (truncation to r-bit arithmetic). The QI doesn't do the 1st AC approximation at all, i.e. it works from exact enumeration and it performs only the second part of the AC approximation and even here it selects precision truncation which has 4 times smaller redundancy than that of AC's truncation. So, the AC is a lower accuracy approximation of QI, hence the AC always produces larger or at best equal output as QI. What you're referring to is trivial observation that for any given algorithm A, one can create another algorithm B which will compress a given input X better than A. It is trivial since you can select a message X for which A produces an output of more than one bit, then you simply create an algorithm B which outputs 0 for message X and outputs 1 followed by otherwise. The relation between AC and QI is not of this kind, though. The AC message index is the same number that QI produces, but multiplied with a number which is strictly greater than 1 (cf. footnote on p.2 on double index majorization by AC). This may or may not (depending on how fractions round to next whole bit) result in a greater number of whole bits output by AC, but mathematically it cannot ever result in a smaller number of whole bits in the AC output. When Rissanen was looking to solve the precision problem of the exact enumeration (in 1976), he ran into problem which seemed intractable (due to a missing piece of formalism in the conventional EC, which the paper above constructs as a natural setting for QI). Hence he approximated the enumeration first (by majorizing the index), which allowed him to construct a decodable index -- this was the arithmetic coding. Rissanen (who is the original creator of arithmetic coding and of MDL modeling principle) recognized instantly the solution in this paper, his old nemesis conquered at last, commenting: I understand what you have done. Very clever! === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding )> Sounds like hype. Especially the statement always compresses better )> than the best of arithmetic coders This is obviouly a false statement. )> It is impossible for any compressor to always compress better than )> every other compressor. ) ) The two are not arbitrary unrelated algorithms -- arithmetic coding ) (AC) and QI are both approximations of the exact enumerative coding ) (the exact enumeration of messages). AC is not only enumerative coding. It is also used with adaptive models and higher-order statistics. SaSW, Willem -- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > AC is not only enumerative coding. It is also used with adaptive models > and higher-order statistics. The coding aspects in both cases are message enumerations -- with EC in the exact lists of messages (performed using the virtual dictionary of combinations), with AC in the cummulative probablity space (which only asymptotically for large n approaches the exact EC coding, while being uniformly suboptimal for any finite n). It is a historic fluke that EC was presented in conjunction with the 'descriptive'/universal way of modeling (MDL/Kolmogorov), while AC with the predictive way (such as PPM). Both coding methods can work with modeling engines either way, though. The coding algorithm of AC does not separate cleanly the model parameters from the coder arithmetic (by virtue of hardwiring 'probabilities of next single symbol' into its coding addends). Thus, its native mode is the 'predictive' way, as a virtue out of necessity. EC has a much cleaner separation between the modeling parameters and the coding computations -- the modeling engine of EC defines the enumerative classes (the subsets of equiprobable messages), while coder computations deal only with the universal combinatorial properties of sequences (belonging to a given class). The result is a much faster operation, since the 'universal properties of sequences' are independent of the particular source parameters, thus they can be precomputed and optimized away, outside of the coding loop, as it were. If AC were to code via precomputed values, it would need a separate table for each set of source probabilities, which is impractical even for a static binary source. === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding )> AC is not only enumerative coding. It is also used with adaptive models )> and higher-order statistics. ) ) The coding aspects in both cases are message enumerations -- with EC in ) the exact lists of messages (performed using the virtual dictionary of ) combinations), with AC in the cummulative probablity space (which only ) asymptotically for large n approaches the exact EC coding, while being ) uniformly suboptimal for any finite n). Not true. Adaptive coding is trivially capable of outperforming a model that uses only the cumulative probability, in cases where there are local variations in the probability distribution. Also, adaptive coding works very well with higher-order models, where it is intractable to transmit the model because of its exponential size. SaSW, Willem -- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT === Subject: Re: New combinatorial coder: tighter & faster than Arithmetic coding > Adaptive coding is trivially capable of outperforming a > model that uses only the cumulative probability, in cases > where there are local variations in the probability distribution. You are comparing different modeling schemes (adaptive vs static), while I was comparing the bare coding algorithms (both can be attached to the same model). At any given point AC gives a position (an interval) on the cummulative probability line for the sequence seen so far, while EC/QI gives the exact ordinal of the same sequence in the same list of sequences (the list is what model produces, we assume the same model and look at the coder difference). The only coder level difference is that the cummulative probabilty index is an approximation (a uniform majorization via the Stirling approx. of binomials and dropping of the two factors < 1, cf p. 2 and ref [23] for more details) of the exact ordinal index and that the QI algorithm can compute it generally much faster than AC algorithm can (and always at least as fast). The reason for the speed difference is discussed in ref [23] -- it is basically that the QI algorithm has much cleaner separation between the coding computations and the specific instance properties of the source, than the AC alogotithm, hence QI can shift much of the coding computation outside of the instance specific coding loop. > Also, adaptive coding works very well with higher-order models, where > it is intractable to transmit the model because of its exponential size. This is an interesting point, but outside of the comparison of QI and AC coding algorithms. Taken on its own, though, your statement is a bit self-contradictory, since the model information obviously can be transmitted in sub-exponential size (e.g. via Adaptive AC scheme). What you really mean is that the first simple-minded encoding of the model parameters that you can came up with, off the top of your head, is exponential. Well, ok, so what? For example, the BWT, which is a bit less obvious way to extract and transmit a higher order models, is not exponential in size (and it is also not an adaptive PPM scheme). In fact, even when coupled with the crude MTF obliteration of the context boundaries, it performs nearly as well as PPM in compression quality (while running much faster). === Subject: How the heck do PAR files work so effectively? I recently downloaded a rar file in about ten parts, each 15 Mbytes, and the first part was missing. I also downloaded the associated par files, whose total size amounted to about 10 MBytes, and running the Techsono par repair utility, available at http://www.techsono.com/news/?source=newztoolz managed to recreate the entire first part. My question is, how the hell (in general terms) can a file be recvered from a set of par files whose size is _less_ than even the size of that file let alone the whole set of files? It seems like an extreme case of fitting a quart into a pint pot! === Subject: Re: How the heck do PAR files work so effectively? There's been a lot of speculation on this thread about how the PAR format works. I happen to know how it works, having written a paper on the subject (not yet published), so I'll describe it briefly. The idea is very simple, much simpler than Reed-Solomon coding (as seen on CD). It's based on the following familiar theorem: If x_1, ..., x_n are distinct elements of some algebraic field F, and y_1, ..., y_n are elements of the same field (not necessarily distinct), then there is exactly one polynomial P of degree less than n such that P(x_i) = y_i for all i in {1,...,n}. In other words, two points determine a line, three points determine a parabola, and so on. If you have a multivolume archive that looks like this: infringement.part001.rar infringement.part002.rar ... infringement.partNNN.rar you can think of each volume as an (x,y) pair, with x being an arbitrary index that's different for each file (e.g. x=### for part###.rar) and y being the actual contents of the file, considered as a single huge number. Now by the theorem above, you can find a polynomial P of degree less than N which yields the contents of part###.rar when evaluated at x=###. The recovery data in the PAR files consists of that polynomial evaluated at other values of x (distinct from the values assigned to the original files). Now suppose that you have some combination of original files and PAR files. If the total number of files that you have is at least N, then you can pick any N of those files and find a polynomial which generates those files when evaluated at appropriate values of x. This will be the same polynomial that was used to generate the PAR files in the first place (this follows from the interpolation theorem). Therefore, by evaluating it at x=1,...,x=N, you can recover all of the original data. The only problem with what I described above is that the coefficients of the interpolation polynomial are huge! Worse, the interpolated values used as PAR blocks won't necessarily be the image of any file under whatever mapping you used to represent the original files as numbers. The solution is to divide the original files into their constituent bytes and treat each byte as an element of the finite field GF(256). A side effect of this is that you can only have 256 different blocks (files) in total, because there are only 256 distinct x coordinates in GF(256). PAR2 is the same as PAR, except that it uses GF(65536) instead of GF(256) (so the block limit is much higher), and it decouples the block size from the file size. One caveat: all of the above is how it's *supposed* to work. Actually the PAR and PAR2 specifications are broken, because they're based on a broken paper by Plank. This manifests itself as very slow operation and occasional recovery failures. -- Ben === Subject: Re: How the heck do PAR files work so effectively? > I recently downloaded a rar file in about ten parts, each 15 Mbytes, > and the first part was missing. I also downloaded the associated > par files, whose total size amounted to about 10 MBytes, and > running the Techsono par repair utility, available at > http://www.techsono.com/news/?source=newztoolz > managed to recreate the entire first part. > My question is, how the hell (in general terms) can a file be > recvered from a set of par files whose size is _less_ than > even the size of that file let alone the whole set of files? > It seems like an extreme case of fitting a quart into a > pint pot! Can PAQ* compression beat RAR's compression of the original - 1 chunk + PAR size ? If so, then RAR is a highly redundant format, and you've simply been fortunate that it has some error-correcting ability. Which means that everyone who doesn't lose chunks is paying unnecessarily. There are possibly better compressors than PAQ*, but they are certainly more esoteric; it can be found at main author Matt Mahoney's site: http://www.cs.fit.edu/~mmahoney/ Check out 'BARF' while you're there... :-) Phil -- If a religion is defined to be a system of ideas that contains unprovable statements, then Godel taught us that mathematics is not only a religion, it is the only religion that can prove itself to be one. -- John Barrow === Subject: Re: How the heck do PAR files work so effectively? > I recently downloaded a rar file in about ten parts, each 15 Mbytes, > and the first part was missing. I also downloaded the associated > par files, whose total size amounted to about 10 MBytes, and > running the Techsono par repair utility, available at > http://www.techsono.com/news/?source=newztoolz > managed to recreate the entire first part. > My question is, how the hell (in general terms) can a file be > recvered from a set of par files whose size is _less_ than > even the size of that file let alone the whole set of files? > It seems like an extreme case of fitting a quart into a > pint pot! Way back when the mass memory for the Space Shuttle was a magnetic tape, and an inspection of the carefully stored spares (temperature controlled, humidity controlled, hermetically sealed, etc. expensive ways) showed that entire sections of the magnetic coating were flaking off (on the order of 1/4 inch, IIRC) that would give massive block errors, a very clever mathematician at the mass memory manufacturer came up with the idea to use the parity bits for each 16-bit word in the 512-bit blocks as a very effective error detection and correction code that would recover the missing data. The key provision: The errors were all together, not randomly scattered through the block. [Is this called the Hamming distance?] It was never implemented, though, since another custom tape manufacturer was contracted to make new tapes. I won't even pretend to begin to start to understand the math involved. But, I suspect the same idea is involved, for reconstructing isolated blocks of missing data by use of a super-parity EDC algorithm. By the way, I believe the extension *.PAR is because these are PARity files. For a much simpler example, suppose you have an 8-bit byte for which you want to detect any 1-bit error. Then, the extension to a 9th bit (the parity bit) is sufficient to detect any 1-bit error. If you want also want to correct the error, more parity-type bits are needed. Going from memory, it takes 3 such bits (or is it 5?) to recover - including an error in the parity bits themselves. It is -not- necessary to have entire copies of the byte (e.g., the very old TMR (triple modular redundancy) circuitry used in some mission-critical computers). So, with far less than entire copies of the files, and with the presumption that any errors are in blocks (missing sub-files, or missing parts of sub-files, for example), recovery is possible. A real mathematician will have to point you to any literature. For myself, this falls under Heinlein's(?) comment: Any sufficiently advanced technology is indistinguishable from magic. This is magic. Lynn Killingbeck === Subject: Re: How the heck do PAR files work so effectively? > Way back when the mass memory for the Space Shuttle was a magnetic tape, > and an inspection of the carefully stored spares (temperature controlled, > humidity controlled, hermetically sealed, etc. expensive ways) showed > that entire sections of the magnetic coating were flaking off (on the > order of 1/4 inch, IIRC) that would give massive block errors, a very > clever mathematician at the mass memory manufacturer came up with the > idea to use the parity bits for each 16-bit word in the 512-bit blocks as > a very effective error detection and correction code that would > recover the missing data. The key provision: The errors were all > together, not randomly scattered through the block. [Is this called > the Hamming distance?] It was never implemented, though, since another > custom tape manufacturer was contracted to make new tapes. > I won't even pretend to begin to start to understand the math involved. > But, I suspect the same idea is involved, for reconstructing isolated > blocks of missing data by use of a super-parity EDC algorithm. By the > way, I believe the extension *.PAR is because these are PARity files. [...] I think that PAR files are generated using the Reed-Solomon algorithm. === Subject: Re: How the heck do PAR files work so effectively? > Way back when the mass memory for the Space Shuttle was a magnetic tape, > and an inspection of the carefully stored spares (temperature controlled, > humidity controlled, hermetically sealed, etc. expensive ways) showed > that entire sections of the magnetic coating were flaking off (on the > order of 1/4 inch, IIRC) that would give massive block errors, a very > clever mathematician at the mass memory manufacturer came up with the > idea to use the parity bits for each 16-bit word in the 512-bit blocks as > a very effective error detection and correction code that would > recover the missing data. The parity bits for 16-bit words in 512-bit blocks will not give the information you wish. For instance, when all bits in a 16-bit word are lost, there is *no* way to recover them if you did only use the parity bits in your error correction. > I think that PAR files are generated using the Reed-Solomon algorithm. That is possible. I *do* know that Reed-Solomon error correcting codes are used on CD's. On data CD's even two levels. But what this means is that for *each* bit of data, three bits of information are written on the CD (at least that was it the last time I looked at it). The second level is only used for data CD's to get still more reliability. That one means that a data block of the same size containse 2300+ bytes of audio but only 2048 bytes of data. The first level is more extensive and consists of shuffling bits around blocks and adding Reed-Solomon to it. (The shuffling around blocks means that many continuous errors will show up as single errors after shuffling.) And finally sequences of bits are encoded so that no long consecutive 1 or 0 bits are written (this can give problems with synchronisation). If I remember right, it is a 8/13 encoding. Also, if I remember right (it is long ago), CDC Cyber computers did not use parity on their memory, but used a system that was able to detect three errors and was able to correct two errors. In memory a 60 bit word was actually 73 bits. But if there are more than three errors, all odds are off. than a single XOR of the bits of the files at the first level. Hence we have only to consider how many additional bits are needed to correct an error in n bits (from a total of m bits). To correct a single error (one missing file) we only need parity, so one additional file is all we need. It gets a bit more complicated if we want to account for multiple errors (missing files), moreover apparently there is the requirement that the first level is a plain parity, which makes it not straightforward Reed-Solomon. But as Lynn Killingbeck noted, Hamming distance has all to do with it. Now I have a problem. Given an arbitrary number of k bits. How many bits should be tagged on to correct n (n <= k) errors, with the additional requirement that the first bit is able to correct a single error, and the number of errors you are able to correct increases when you add bits. ? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: How the heck do PAR files work so effectively? > Way back when the mass memory for the Space Shuttle was a magnetic > tape, and an inspection of the carefully stored spares > (temperature controlled, humidity controlled, hermetically sealed, > etc. expensive ways) showed that entire sections of the magnetic > coating were flaking off (on the order of 1/4 inch, IIRC) that > would give massive block errors, a very clever mathematician at > the mass memory manufacturer came up with the idea to use the > parity bits for each 16-bit word in the 512-bit blocks as a very > effective error detection and correction code that would recover > the missing data. > The parity bits for 16-bit words in 512-bit blocks will not give the > information you wish. For instance, when all bits in a 16-bit word > are lost, there is *no* way to recover them if you did only use the > parity bits in your error correction. > I think that PAR files are generated using the Reed-Solomon > algorithm. > That is possible. I *do* know that Reed-Solomon error correcting > codes are used on CD's. On data CD's even two levels. But what this > means is that for *each* bit of data, three bits of information are > written on the CD (at least that was it the last time I looked at it). > The second level is only used for data CD's to get still more > reliability. That one means that a data block of the same size > containse 2300+ bytes of audio but only 2048 bytes of data. The first > level is more extensive and consists of shuffling bits around blocks > and adding Reed-Solomon to it. (The shuffling around blocks means > that many continuous errors will show up as single errors after > shuffling.) And finally sequences of bits are encoded so that no long > consecutive 1 or 0 bits are written (this can give problems with > synchronisation). If I remember right, it is a 8/13 encoding. > Also, if I remember right (it is long ago), CDC Cyber computers did > not use parity on their memory, but used a system that was able to > detect three errors and was able to correct two errors. In memory a > 60 bit word was actually 73 bits. But if there are more than three > errors, all odds are off. > more than a single XOR of the bits of the files at the first level. > Hence we have only to consider how many additional bits are needed to > correct an error in n bits (from a total of m bits). To correct a > single error (one missing file) we only need parity, so one additional > file is all we need. It gets a bit more complicated if we want to > account for multiple errors (missing files), moreover apparently there > is the requirement that the first level is a plain parity, which makes > it not straightforward Reed-Solomon. But as Lynn Killingbeck noted, > Hamming distance has all to do with it. > Now I have a problem. Given an arbitrary number of k bits. How many > bits should be tagged on to correct n (n <= k) errors, with the > additional requirement that the first bit is able to correct a single > error, and the number of errors you are able to correct increases when > you add bits. I'm having a problem with using the right words, most likely. Let me try to start over, and try to use the name 'parity bit' better. Imagine a tape containing just one track, along which the bits are written sequentially. [This is not a traditional 7-track or 9-track magnetic tape drive, which has 6- or 8-bit bytes crosswise of the tape, with a parity bit per byte.] The original organization has 16 data bits [call it a half-word] followed by a parity bit for those 16 bits. With a fixed block size of 512 such half-words, each block has 512*16=8192 data bits and 512 parity bits, for a total of 8704 bits per block. When read, there is a sequential stream of bits, with each 16 data bits followed by its parity bit. Later, it turns out the tape is subject to mechanical problems (the magnetic coating flaking off), which will show up as a sequencial group of bits being erroneous. They are not missing, but they are unreliable. (There is also a parallel timing track to provide tape position information - and if that timing track also fails, we are really SOL for that block. Critical data such as the 'get me home!' software is recorded multiple places on multiple tape drives.) Now, instead of using the 512 'extra' non-data bits as individual parity bits per half-word, is there something better that can be done? Given that the erroneous data is in a single block (measured from the first bad bit to the last bad bit, which might be what is called the Hamming distance), how long a sequence of bad bits can be corrected? The total block size remains at 8704 bits. It will be a new design, so the 8192 data bits and 512 EDC bits can be interleaved in whatever way is most advantageous. I don't know if this description is any better, or changes the comment about being impossible to recover a missing 16-bit half-word. I am somewhat at a loss for the proper terminology. Hopefully equivalent: Suppose there is a single 8192 bit data word, and 512 bits available for EDC purposes. How long a sequence of bits can be corrected? Lynn Killingbeck === Subject: Re: How the heck do PAR files work so effectively? >> I recently downloaded a rar file in about ten parts, each 15 Mbytes, >> and the first part was missing. I also downloaded the associated >> par files, whose total size amounted to about 10 MBytes, and >> running the Techsono par repair utility, available at >> http://www.techsono.com/news/?source=newztoolz >> managed to recreate the entire first part. >> My question is, how the hell (in general terms) can a file be >> recvered from a set of par files whose size is _less_ than >> even the size of that file let alone the whole set of files? >> It seems like an extreme case of fitting a quart into a >> pint pot! >Way back when the mass memory for the Space Shuttle was a magnetic tape, >and an inspection of the carefully stored spares (temperature controlled, >humidity controlled, hermetically sealed, etc. expensive ways) showed >that entire sections of the magnetic coating were flaking off (on the >order of 1/4 inch, IIRC) that would give massive block errors, a very >clever mathematician at the mass memory manufacturer came up with the >idea to use the parity bits for each 16-bit word in the 512-bit blocks as >a very effective error detection and correction code that would >recover the missing data. The key provision: The errors were all >together, not randomly scattered through the block. [Is this called >the Hamming distance?] It was never implemented, though, since another >custom tape manufacturer was contracted to make new tapes. >I won't even pretend to begin to start to understand the math involved. >But, I suspect the same idea is involved, for reconstructing isolated >blocks of missing data by use of a super-parity EDC algorithm. By the >way, I believe the extension *.PAR is because these are PARity files. >For a much simpler example, suppose you have an 8-bit byte for which you >want to detect any 1-bit error. Then, the extension to a 9th bit (the >parity bit) is sufficient to detect any 1-bit error. If you want also >want to correct the error, more parity-type bits are needed. Going from >memory, it takes 3 such bits (or is it 5?) to recover - including an >error in the parity bits themselves. It is -not- necessary to have >entire copies of the byte (e.g., the very old TMR (triple modular >redundancy) circuitry used in some mission-critical computers). >So, with far less than entire copies of the files, and with the >presumption that any errors are in blocks (missing sub-files, or >missing parts of sub-files, for example), recovery is possible. >A real mathematician will have to point you to any literature. For >myself, this falls under Heinlein's(?) comment: Any sufficiently >advanced technology is indistinguishable from magic. This is magic. >Lynn Killingbeck But that's a nice explanation of the essence of the science behind the magic. === Subject: Re: How the heck do PAR files work so effectively? > I recently downloaded a rar file in about ten parts, each 15 Mbytes, > and the first part was missing. I also downloaded the associated > par files, whose total size amounted to about 10 MBytes, and > running the Techsono par repair utility, available at > http://www.techsono.com/news/?source=newztoolz > managed to recreate the entire first part. > My question is, how the hell (in general terms) can a file be > recvered from a set of par files whose size is _less_ than > even the size of that file let alone the whole set of files? > It seems like an extreme case of fitting a quart into a > pint pot! See http://www.slyck.com/ng.php?page=6 Dirk Vdm === Subject: Re: How the heck do PAR files work so effectively? >I recently downloaded a rar file in about ten parts, each 15 Mbytes, >and the first part was missing. I also downloaded the associated >par files, whose total size amounted to about 10 MBytes, and >running the Techsono par repair utility, available at >http://www.techsono.com/news/?source=newztoolz >managed to recreate the entire first part. >My question is, how the hell (in general terms) can a file be >recvered from a set of par files whose size is _less_ than >even the size of that file let alone the whole set of files? >It seems like an extreme case of fitting a quart into a >pint pot! Seems like one of two possibilities: (i) the par decoder contains the contents of every par file in the universe, hard-wired. (Is it more than a few terabytes in size?) (ii) some sort of compression. ************************ David C. Ullrich === Subject: Re: How the heck do PAR files work so effectively? >I recently downloaded a rar file in about ten parts, each 15 Mbytes, >and the first part was missing. I also downloaded the associated >par files, whose total size amounted to about 10 MBytes, and >running the Techsono par repair utility, available at >http://www.techsono.com/news/?source=newztoolz >managed to recreate the entire first part. >My question is, how the hell (in general terms) can a file be >recvered from a set of par files whose size is _less_ than >even the size of that file let alone the whole set of files? >It seems like an extreme case of fitting a quart into a >pint pot! > Seems like one of two possibilities: > (i) the par decoder contains the contents of > every par file in the universe, hard-wired. > (Is it more than a few terabytes in size?) Funny you should say that - Downloading it I noticed the loop animation showed a globe arching across to my folder instead of a file. It seemed to take forever, but I got impatient and hit cancel after a couple of years. Seriously, the explanation at the link Dirk quoted makes it plausible how par2 files work so well given their sizes relative to the original files. But it's still seems amazing how much they can recover. > (ii) some sort of compression. > ************************ > David C. Ullrich === Subject: Re: How the heck do PAR files work so effectively? > Funny you should say that - Downloading it I noticed the > loop animation showed a globe arching across to my > folder instead of a file. It seemed to take forever, but > I got impatient and hit cancel after a couple of years. Since the repair program that you're using (which I'd never heard of) seems to be bundled with a premium news server, my guess is that it's simply downloading as much extra PAR data as it needs, over the net, from the news server. That would explain why you could recover 15MB of missing data with only 10MB of predownloaded PAR files. -- Ben === Subject: Re: How the heck do PAR files work so effectively? > I recently downloaded a rar file in about ten parts, each 15 Mbytes, > and the first part was missing. I also downloaded the associated > par files, whose total size amounted to about 10 MBytes, and [...] > managed to recreate the entire first part. > My question is, how the hell (in general terms) can a file be > recvered from a set of par files whose size is _less_ than > even the size of that file let alone the whole set of files? What you describe is impossible in general, since you could use it to compress every file (just generate 10MB of par files, then throw away the 15MB original). It's also impossible even in specific cases with PAR and PAR2, since these formats don't use compression. So some part of your story must be wrong. :-) -- Ben === Subject: Digital Geometry Workshop The Digital Geometry Workshop (DGW 2005) is planned to be the first of a series of international meetings which will take place periodically. DGW are initiated and co-organized by the mathematics departments of the University of Messina, Messina, Sicily, and SUNY Buffalo State College, Buffalo, NY, USA. The main objective is to provide graduate and senior undergraduate students with the most essential background for their research in certain modern areas of contemporary applied mathematics and theoretical computer science. Although the essential part of the audience is expected to consist mainly of students, participants from the industry or professional researchers are very welcome. Thus DGW will appear as short-term, intensive schools through which the participants will obtain useful knowledge in certain research areas, learn about challenging open problems, and establish useful contacts for future collaboration. The first edition of the workshop that will take place in Messina (Italy) from 21st to 24th November. The lectures will be held at the Aula Majorana of the Physic Department at Science Faculty in Papardo (Messina). Further information on www.digitalgeometryworkshop.it === Subject: Re: Convergence of a function sequence in L^infty(R) On Fri, 18 Nov 2005 16:05:21 +0000, Jos.8e Carlos Santos >First of all, I want to say that I thought about all this and that I now >agree with you about this: if the notation is not ambiguous within the >context where it appeared, it is silly to say that, in a more general >context, it is ambiguous. Note, however, that I *do* think that there >was already an ambiguity at the original post. >> You misread the question yet again. That sum does _not_ converge >> in L^infinity. >Can't see why. First of all, I remind you that your original question >was: > What's an example of a sequence of functions f_n (n in Z) which have > pairwise disjoint supports, and two notions of sum_{-infty}^infty > such that sum f_n converges in L^infinity with one notion but not > with the other? >So, here's my example: f_n (from R into R) is the characteristic >function of [n,n + 1[. I hope that you agree that: >1) f_n belongs to L^oo(R); >2) the f_n's have pairwise disjoint supports. >Now, consider these two notions of sum_{-oo}^{+oo} in L^oo. >1) sum_{-oo}^{+oo}f_n = f iff f is in L^oo and, for each real _x_, >f(x) = lim_n sum_{k = -n}^n f_k(x). That's not convergence _in L^infinity_. At least it's not an example of what I meant by the phrase. We were talking about ambiguities arising from reordering the terms, or rather talking about different sorts of partial sums. (2) below is certainly an example of such a thing, and considering convergence _in L^infinity_ is such a notion, but this is not. >Then, whith this notion, >sum_{-oo}^{+oo}f_n(x) = identity function. >2) sum_{-oo}^{+oo}f_n = f iff f is in L^oo and, for each r > 0 there's >some finite set F of integers such that, whenever F' is a finite set of >integers containg F, you have ||f - sum_{n in F'} f_n||_{oo} < r. Then, >with this notion, sum_{-oo}^{+oo}f_n does not converge. >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: Convergence of a function sequence in L^infty(R) >>1) sum_{-oo}^{+oo}f_n = f iff f is in L^oo and, for each real _x_, >>f(x) = lim_n sum_{k = -n}^n f_k(x). > That's not convergence _in L^infinity_. > At least it's not an example of what I meant by the > phrase. We were talking about ambiguities arising > from reordering the terms, or rather talking > about different sorts of partial sums. (2) below > is certainly an example of such a thing, and > considering convergence _in L^infinity_ is such > a notion, but this is not. All right! Now I see what you meant by convergence _in L^infinity_. But I hope that you agree that the expression is ambiguous (no joke intended! :-) ). Jose Carlos Santos === Subject: Galois Extensions Is it true that A/B and B/C are Galois extensions implies A/C is Galois? My guess is a no, does anyone have a proof/counterexample? === Subject: Re: Galois Extensions days. My association with the Department is that of an alumnus. >Is it true that A/B and B/C are Galois extensions implies A/C is >Galois? >My guess is a no, does anyone have a proof/counterexample? Since any extension of degree 2 over Q is Galois, try letting A/C be a non-Galois extension of degree 4. -- 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: Galois Extensions > Is it true that A/B and B/C are Galois extensions > implies A/C is Galois? > My guess is a no, does anyone have a proof/counterexample? I believe you can get an example with C = the rational numbers, B = the field generated the roots of a cubic polynomial all three of whose roots are real and fall into the casus irreducibilis case (e.g. x^3 - 6x + 2), and A = the reals. (This specific example comes from the 1998 Isaacs/Moulton paper in my post below.) The following may also help: sci.math post by Dave L. Renfro (Sept. 26, 2005) Dave L. Renfro === Subject: Re: Galois Extensions >> Is it true that A/B and B/C are Galois extensions >> implies A/C is Galois? >> My guess is a no, does anyone have a proof/counterexample? > I believe you can get an example with C = the rational numbers, > B = the field generated the roots of a cubic polynomial all > three of whose roots are real and fall into the casus > irreducibilis case (e.g. x^3 - 6x + 2), and A = the reals. > (This specific example came from the 1998 Isaacs/Moulton > paper in my post below.) > The following may also help: > sci.math post by Dave L. Renfro (Sept. 26, 2005) On second thought, I think I might be working with the converse, and maybe not even the converse of what you're asking! I'll bow out and let someone who knows what they're talking about (regarding Galois theory) step in. But if it helps in your search, Galois extensions are sometimes called normal extensions. Dave L. Renfro === Subject: Re: Coprime Row/Column Sum Grid-Puzzle > (Here is a puzzle that is similar to other puzzles I have posted in the > past. > Maybe I have actually posted this very same puzzle in the past. > If so, I apologize.) > Place 1 through n^2 (in order) in an n-by-n grid, one integer per > square, > so that: > *Each integer (after 1) is placed in an empty square immediately > adjacent (in the direction of either up, down, left, or right) > to a square already filled with a (lower) integer. > *When m is placed in the grid, the SUM of the lower integers already > in m's ROW is coprime to m. > *When m is placed in the grid, the SUM of the lower integers already > in m's COLUMN is coprime to m. > So for example, if we have the unfinished grid: > 1 3 > 2 # > 4 5 * > 6 7 > 8 cannot be put at # because 2 isn't coprime to 8. > (Although 3+5+7 is coprime to 8.) > 8 can be put at *, however, because 4+5 is coprime to 8. > I have found by-hand a solution for the 4-by-4 and 5-by-5 grids. > Can you find solutions for bigger grids? > Leroy Quet I know it has been a while since I originally posted this puzzle, but I might as well post my 4-by-4 and 5-by-5 solutions. Spoiler: | | V | | V | | V | | V | | V | | V 7 12 15 13 6 5 9 11 14 4 3 8 16 10 2 1 9 8 20 22 25 10 7 6 18 19 11 13 5 4 14 17 15 12 3 2 24 23 21 16 1 Leroy Quet === Subject: Re: Prove f(x) = f(1/x) (complicated) You guys make me feel really really stupid - I think I'll go to rec.4dummies.com instead >I will post here 2 related math problems, the first easier than the > last. > In both problems zeta(n,x) = sum{k>=0} 1/(k+x)^n, > the Hurwitz zeta function. > Also, H(x) = Euler's constant + d(ln(Gamma(x+1)))/dx > = sum{k=1 to x} 1/k (if x is a positive integer). > Problem 1) > Let > f(x) = > n-1 > 1 --- H(kx) --- --- zeta(j+1,xk+1) > ---- ------- - --------------- > n/2 / n+1 / / n+1-j n/2 -j > x --- k --- --- k x > k>=1 j=1 k>=1 > (View with fixed-width font, of course.) > In linear-mode: > f(x) = > (1/x^(n/2)) sum{k>=1} H(kx)/k^(n+1) > - sum{j=1 to n-1} sum{k>=1} zeta(j+1,xk+1)/(k^(n+1-j) x^(n/2 -j)) > Prove, for x > 0, for all n = integer >= 1, > f(x) = f(1/x). > Problem 2) > Let > f(x) = > ---- / > n/2 H(j) * ( n zeta(n+1,jx) + 2*H(jx -1) zeta(n,jx) > x > / --- > / > ---- - / zeta(k,jx) zeta(n+1-k,jx) ) > j>=1 --- / > 2<=k<=n-1 > In linear-mode: > f(x) = > x^(n/2) Sum{j>=1} H(j) * ( > n zeta(n+1,jx) + 2*H(jx -1) zeta(n,jx) > - sum{2<=k<=n-1} zeta(k,jx) zeta(n+1-k,jx) ) > Prove, for x > 0, for all n = integer >= 3, > f(x) = f(1/x). > Leroy Quet === Subject: Re: Prove f(x) = f(1/x) (complicated) > You guys make me feel really really stupid - I think I'll go to > rec.4dummies.com instead Yes but you are a girl so that's ok. McC === Subject: Prove f(x) = f(1/x) (complicated) I will post here 2 related math problems, the first easier than the last. In both problems zeta(n,x) = sum{k>=0} 1/(k+x)^n, the Hurwitz zeta function. Also, H(x) = Euler's constant + d(ln(Gamma(x+1)))/dx = sum{k=1 to x} 1/k (if x is a positive integer). Problem 1) Let f(x) = n-1 1 --- H(kx) --- --- zeta(j+1,xk+1) ---- ------- - --------------- n/2 / n+1 / / n+1-j n/2 -j x --- k --- --- k x k>=1 j=1 k>=1 (View with fixed-width font, of course.) In linear-mode: f(x) = (1/x^(n/2)) sum{k>=1} H(kx)/k^(n+1) - sum{j=1 to n-1} sum{k>=1} zeta(j+1,xk+1)/(k^(n+1-j) x^(n/2 -j)) Prove, for x > 0, for all n = integer >= 1, f(x) = f(1/x). Problem 2) Let f(x) = ---- / n/2 H(j) * ( n zeta(n+1,jx) + 2*H(jx -1) zeta(n,jx) x / --- / ---- - / zeta(k,jx) zeta(n+1-k,jx) ) j>=1 --- / 2<=k<=n-1 In linear-mode: f(x) = x^(n/2) Sum{j>=1} H(j) * ( n zeta(n+1,jx) + 2*H(jx -1) zeta(n,jx) - sum{2<=k<=n-1} zeta(k,jx) zeta(n+1-k,jx) ) Prove, for x > 0, for all n = integer >= 3, f(x) = f(1/x). Leroy Quet === Subject: weird equivalent hello all, I have a problem with an equivalent. I'm struggling with it for days but I still don't have the answer. How can one compute an equivalent of the function (well defined by Abel criterion for example) f(x)=sum [ Log(n)*sin(n*x) ] / n , n >=1, when x tends towards 0+. I know how to find an equivalent of series like g(x)=sum [ sin(n*x) ] / n , n >=1 This is the imaginary part of g(x)=sum z^n / n , n >=1 when z is on the unit circle.And one can verify that: sum z^n / n = sum int( [z*exp(-t)]^n,t=0...infty) ] = int( (zexp(-t))/(1-(zexp(-t)),t=0...infty) which is more easy. But in this case I've no idea ?? anyone has already dealed with such equivalent ? thanx in advance, Fedor === Subject: Re: weird equivalent no one ? === Subject: Re: weird equivalent On 21 Nov 2005 03:16:50 -0800, Fedor >no one ? One problem is that it's not at all clear what you're asking for. What is an equivalent of a sum? (You want a closed-form expression for the sum? The limit of the sum as x -> 0? A closed-form expression that's asymptotic to the sum as x -> 0?) ************************ David C. Ullrich === Subject: Re: weird equivalent ho sorry for my poor english... In fact I am looking for (and i know it exists since this is an exercice) a function h(x) such that LIM ( f(x)/h(x), x->0+)=1 Is it clear now ? Thanx :-) === Subject: Re: weird equivalent I thnik one can prove that: Lim ( x*(Sum [ Log(nx)*sin(n*x) ] / nx , x=1..infinity) = int( ln(t)*sin(t)/t, y=0 ..infinity) If it is right, the problem is greatly simplified ! === Subject: Another Sum = to pi^2/6, for all x (I posted the first {and simpler} such sum equal to pi^2/6 for all x to sci.math a while ago. See result 2 at 3 simple infinite sums with harmonic numbers at ) Is this true? For all x where the double sum converges, pi^2/6 = sum{j>=1} (x^j/(j!j)) ( - H(j)B(j) + sum{k>=1} k^(j-1)/e^(xk)), where H(j) = sum{k=1 to j} 1/k, and B(j) is the jth Bernoulli number (sum{j>=0} B(j) x^j/j! = x/(e^x -1)). Leroy Quet === Subject: Equivalence of two compactness definitions for R^n For example, let compatct subset to be a ball in R^n or a closed interval on R. How does one show the equivalence of 'closed and bounded' to 'every open cover has finite subcover' ? I think that major stumbling block for me is understanding the concept of an open cover, i.e. how can you come up with an open cover on [0,1] that includes points 0 and 1? === Subject: Re: Equivalence of two compactness definitions for R^n > For example, let compatct subset to be a ball in R^n or a closed > interval on R. > How does one show the equivalence of 'closed and bounded' to 'every > open cover has finite subcover' ? > I think that major stumbling block for me is understanding the concept > of an open cover, i.e. how can you come up with an open cover on [0,1] > that includes points 0 and 1? Example of an open cover: {[0,0.6),(0.4,1]}. Why is this an open cover of [0,1]? Because the open sets on [0,1] are the intersections of open sets in R with [0,1]. If you prefer - the complement of an open set should be closed. The complement of [0,0.6) in [0,1] is [0.6,1] which is certainly closed. Anyway. Suppose you have a set X, for which every open cover has a finite subcover. Certainly the sets X intersect B(0,r) (the open ball radius r centre 0) form an open cover of X, hence they have a finite subcover, hence there is an R s.t. X is contained in B(0,R), i.e. X is bounded. Now suppose you have a point x in R^n - X. Then the sets X-B_c (x, 1/n), where B_c(x,1/n) is the closed ball radius 1/n about x, are certainly open, and certainly cover X. Hence they have a finite subcover, so there is an N such that no point of X is within distance 1/N of x. So any point not in X is not a limit point of X, i.e. X is closed. Note this was basically the same argument repeated twice; if you'd worked in the one-point compactification of R^n and X a topologically compact set not including infinity then you wouldn't have needed even to repeat the argument. Now suppose you have a c+b set X, and an open cover. Suppose there is no finite subcover, then take a minimal (in the sense of no set can be removed and keep it a cover) open cover {C_1,..}. Let S_1=X-C_1, S_n=S_{n-1}-C_n for n>1. Then S_n is always a closed set with at least one point. So choose s_n in S_n, for each n. Observe that certainly s_n is in s_N for all n>N. Suppose the set of s_n is finite: then there is some x which is s_n for infinitely many n, hence x is in S_n for infinitely many n, so x is not in C_n for any n and the C_n do not form a cover. Then the set of s_n is infinite, but then by Bolzano-Weierstrass they have a limit point s, and s is in S_n for all n. But then s is not in any C_n and the C_n do not form a cover. This is a contradiction, so there must be a finite subcover. (note: I seem to be assuming that the minimal cover is countable, but rewrite with an uncountable index set and everything works fine; this simply avoids notation which doesn't carry well in plain text). HTH Peter === Subject: Re: Equivalence of two compactness definitions for R^n I am (obviously) having conceptual troubles with the topology and open sets. My first reaction was to respond - so, taken 'in itself', any set is an open set and, in a way, even a point in R can be an open set, and it all seemed quite confusing and counterinituitive. Also, I had a hard time understanding the concept of an open set in spaces that are not metric - without distance concept all my intuition and visualization goes through the window. Then I read few mathematica pages and realized that, given the broadest definition of topology via subsets that contain intersections and unions, the above can be true (for ex. a point can be an open set, but not in the topology of R) and there is no contradiction. Thus, when talking about open (or closed) sets one has to specify the topology w.r.t. which some set is open or closed.. >.... > HTH === Subject: Re: Equivalence of two compactness definitions for R^n > I am (obviously) having conceptual troubles with the topology and open > sets. > My first reaction was to respond - so, taken 'in itself', any set is > an open set and, in a way, even a point in R can be an open set, and > it all seemed quite confusing and counterinituitive. Except the point 'in R' is only an open set when you're considering the topology on a subset X of R in which the point is isolated. So by defnition you can find an epsilon small enough that the ball radius epsilon about your point is just the point. This is generally true - if you have a metric space, and you take a subset with the subspace topology, then a point of the subset is only an open set in the subspace topology if it's isolated in the metric. > Also, I had a hard time understanding the concept of an open set in > spaces that are not metric - without distance concept all my intuition > and visualization goes through the window. This is true for most people when you're in a space which really isn't a metric space (i.e. the topology cannot come from a metric). You get used to it. > Then I read few mathematica pages and realized that, given the > broadest definition of topology via subsets that contain > intersections and unions, the above can be true (for ex. a point can > be an open set, but not in the topology of R) and there is no > contradiction. > Thus, when talking about open (or closed) sets one has to specify the > topology w.r.t. which some set is open or closed.. Yes. For example, the point 0 is open in the reals - if I use the discrete topology. The ball {x: |x|<1} is not open - if I use the indiscrete topology. They are respectively closed and open, if I use the usual topology. Peter === Subject: Re: Equivalence of two compactness definitions for R^n > For example, let compatct subset to be a ball in R^n or a closed > interval on R. > How does one show the equivalence of 'closed and bounded' to 'every > open cover has finite subcover' ? > I think that major stumbling block for me is understanding the concept > of an open cover, i.e. how can you come up with an open cover on [0,1] > that includes points 0 and 1? {[0,1]} is an open cover of [0,1] in the relative topology. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Equivalence of two compactness definitions for R^n |For example, let compatct subset to be a ball in R^n or a closed |interval on R. | |How does one show the equivalence of 'closed and bounded' to 'every |open cover has finite subcover' ? | |I think that major stumbling block for me is understanding the |concept of an open cover, i.e. how can you come up with an open cover |on [0,1] that includes points 0 and 1? by remembering that the open sets only have to be relatively open. -- [e-mail address jdolan@math.ucr.edu] === Subject: derivative of arctan(tan(x)) Hello All If y = arctan(tan(x)) what would be dy/dx and how do we arrive at it? TIA Mick. === Subject: Re: derivative of arctan(tan(x)) > If y = arctan(tan(x)) what would be dy/dx and how > do we arrive at it? When -pi/2 < x < pi/2, y = x and y' = 1. We also can use the Chain Rule, which works for all x for which the function is defined. y' = (1/(1+(tan(x))^2))(d/dx(tan(x))) = (1/(sec(x))^2)(sec(x))^2 = 1. The function is periodic. On -pi/2 < x < pi/2, we get y = x. On other intervals, we get the same graph translated left or right by integer multiples of pi. === Subject: Re: derivative of arctan(tan(x)) > If y = arctan(tan(x)) what would be dy/dx and how > do we arrive at it? > When -pi/2 < x < pi/2, y = x and y' = 1. > We also can use the Chain Rule, > which works for all x for which the function > is defined. > y' = (1/(1+(tan(x))^2))(d/dx(tan(x))) > = (1/(sec(x))^2)(sec(x))^2 = 1. > The function is periodic. On -pi/2 < x < pi/2, > we get y = x. On other intervals, we get the > same graph translated left or right by integer > multiples of pi. what about complex x ? === Subject: Re: derivative of arctan(tan(x))- Think of complex-differentable functions and their inverses. Johan E. Mebius >If y = arctan(tan(x)) what would be dy/dx and how >do we arrive at it? >>When -pi/2 < x < pi/2, y = x and y' = 1. >>We also can use the Chain Rule, >>which works for all x for which the function >>is defined. >>y' = (1/(1+(tan(x))^2))(d/dx(tan(x))) >> = (1/(sec(x))^2)(sec(x))^2 = 1. >>The function is periodic. On -pi/2 < x < pi/2, >>we get y = x. On other intervals, we get the >>same graph translated left or right by integer >>multiples of pi. >what about complex x ? === Subject: Re: derivative of arctan(tan(x)) On Mon, 21 Nov 2005 06:53:38 +1100, Michael Harrington >Hello All >If y = arctan(tan(x)) what would be dy/dx and how >do we arrive at it? >TIA Mick. tan(y) = tan(x) y = x dy/dx = 1 === Subject: Re: derivative of arctan(tan(x)) On Mon, 21 Nov 2005 06:53:38 +1100, Michael Harrington >Hello All >If y = arctan(tan(x)) what would be dy/dx and how >do we arrive at it? >TIA Mick. tan(y) = tan(x) y = x dy/dx = 1 Would that simply be a/b ? TIA Mick. === Subject: Re: derivative of arctan(tan(x)) On Mon, 21 Nov 2005 07:43:01 +1100, Michael Harrington >On Mon, 21 Nov 2005 06:53:38 +1100, Michael Harrington >>Hello All >>If y = arctan(tan(x)) what would be dy/dx and how >>do we arrive at it? >>TIA Mick. > tan(y) = tan(x) > y = x > dy/dx = 1 >Would that simply be a/b ? >TIA Mick. No. For this new problem you could proceed as shown in the post by N. Silver, using the chain rule. If we let c = a/b, the derivative is [ c sec(x)^2 ] / [ c^2 tan(x)^2 + 1 ] Look at Silver's post. It the matter is still unclear, post again here. === Subject: Help with integration I've got a problem on a rather easy integration. I use Mathematica notation for the equations. let I = Integrate[Sqrt[1-x^2],x] to solve this I make the substitution: u = Arcsin[x] => x = Sin[u] => x' = (Sin[u])' => 1=Cos[u] * u' => dx = Cos[u] du So, I = Integrate[ Sqrt[1-(Sin[u])^2] * Cos[u] ,u] = Integrate[ ( Cos[u] )^2 ,u] But the first results to (1/2)*( x * Sqrt[1 - x^2] + ArcSin[x] ) in Mathematica and the second results to (1/2)*(Sin[x]*Cos[x] + x) These functions are different, one can see it immediately by ploting them. What am I doing wrong??? === Subject: Re: Help with integration > I've got a problem on a rather easy integration. I use Mathematica > notation for the equations. > let I = Integrate[Sqrt[1-x^2],x] > to solve this I make the substitution: u = Arcsin[x] => x = Sin[u] => > x' = (Sin[u])' => 1=Cos[u] * u' => dx = Cos[u] du > So, I = Integrate[ Sqrt[1-(Sin[u])^2] * Cos[u] ,u] = Integrate[ ( > Cos[u] )^2 ,u] > But the first results to (1/2)*( x * Sqrt[1 - x^2] + ArcSin[x] ) in > Mathematica and the second results to (1/2)*(Sin[x]*Cos[x] + x) > These functions are different, one can see it immediately by ploting > them. > What am I doing wrong??? Probably what you did wrong, was using Mathematica as a replacement for mathematics :-) I don't have this Mathematica, but for the second integral it should produce 1/2 ( sin(u) cos(u) + u ) and when you insert your u = arcsin(x) into that, you get 1/2 ( sin(arcsin(x)) cos(arcsin(x)) + arcsin(x) ) = 1/2 ( x sqrt(1-x^2) + arcsin(x) ) Dirk Vdm === Subject: Re: Help with integration anyway === Subject: Bezier curve - sinus function I have a question. I know that it is not possible to draw exactly sinus or cosinus functions using bezier curves. But here is my question: why ? :) I guess it is the same problem for circle. Bartek === Subject: Re: Bezier curve - sinus function On Sun, 20 Nov 2005 21:24:51 +0100, Feniks >I have a question. I know that it is not possible to draw exactly sinus or >cosinus functions using bezier curves. But here is my question: why ? :) I >guess it is the same problem for circle. >Bartek Because each section of a Bezier curve is a cubic polynomial. --Lynn === Subject: Re: Bezier curve - sinus function > I have a question. I know that it is not possible to draw exactly sinus or > cosinus functions using bezier curves. But here is my question: why ? :) I Becouse Bezier curve is a parametric rational or polynomial (so special kind of rational) curve. The sinus is transcendental. > guess it is the same problem for circle. No, it's not. The circle cannot be drawn with *polynomial* Bezier curve but it can be pretty well drawn with a *rational* Bezier curve. Anyway, you may also wish to ask the question at comp.graphics.algorithms where you may find even more info on Bezier curves. Przemek -- Beauty is the first test: there is no permanent place in the world for ugly mathematics. G.H. Hardy === Subject: Invertible elements of a coordinate ring of a real curve Welcome, Take a real closed field K and a real irreducible affine curve C over K Csubset K^2. Let the curve C be associated with a polynomial fin K[X,Y]. Let further R= K[C]= K[x,y]= K[X,Y]/(f) be the ring of polynomial functions on C. What are the invertible elements of R? I believe this question must have been solved decades ago. Yet I cannot find anything useful on neither MR nor Zbl. This surely means I'm using wrong criteria/keyword in my queries. So can anyone point me toward the right papers/books? Any useful keywords to query the databases? Przemek P.S.1 The ``easy'' answer: U(R)= K is also *wrong* (consider e.g. a hyperbola). P.S.2 For some special types of C (e.g. C of Abhyankar-Moh type, or C semi-algebraically compact), the answer is truelly easy. So probably, spending enough time on that I should be able to get the answer myself. Yet, as said before, I'm sure it has been bitten to death long time ago. -- Beauty is the first test: there is no permanent place in the world for ugly mathematics. G.H. Hardy === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime {11,41,89,101,111,167,179,181} It took less than two minutes with this C++ program: #include #include bool p[100000000]; int x[8]; bool check_all_fives(int i){ for(int a=3;a=0;--k) std::cout << x[k] << ' '; std::cout << std::endl; std::exit(0); } for(x[i]=x[i-1]-2;x[i]>=0;x[i]-=2){ if(i>=4 && !check_all_fives(i)) continue; check_sets_of_five(i+1); } } int main(){ for(int i=0;i<100000000;++i) p[i]=true; p[0]=p[1]=false; for(int k=2;k<10000;++k){ if(!p[k]) continue; for(int j=k*2;j<100000000;j+=k) p[j]=false; } for(x[0]=1;x[0]<20000000;x[0]+=2){ check_sets_of_five(1); std::cout << x[0] << std::endl; } } === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime > {11,41,89,101,111,167,179,181} See description of the bug below. Ignoring the 181 (and if there is no other bug), you found a set of 7 positive integers, the sum of any 5 of which is prime. > It took less than two minutes with this C++ program: > #include > #include > bool p[100000000]; > int x[8]; > bool check_all_fives(int i){ > for(int a=3;a int s2=x[i]+x[a]; > for(int b=2;b int s3=s2+x[b]; > for(int c=1;c int s4=s3+x[c]; > for(int d=1;d int s5=s4+x[d]; > if(!p[s5]) > return false; > } > } > } > } > return true; > void check_sets_of_five(int i){ > if(i==8){ > for(int k=7;k>=0;--k) > std::cout << x[k] << ' '; > std::cout << std::endl; > std::exit(0); > } > for(x[i]=x[i-1]-2;x[i]>=0;x[i]-=2){ > if(i>=4 && !check_all_fives(i)) > continue; > check_sets_of_five(i+1); > } > int main(){ > for(int i=0;i<100000000;++i) > p[i]=true; > p[0]=p[1]=false; > for(int k=2;k<10000;++k){ > if(!p[k]) > continue; > for(int j=k*2;j<100000000;j+=k) > p[j]=false; > } > for(x[0]=1;x[0]<20000000;x[0]+=2){ > check_sets_of_five(1); > std::cout << x[0] << std::endl; > } === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime >{11,41,89,101,111,167,179,181} This has five elements congruent to 1 mod 5, and cannot be a solution. 11 + 41 + 101 + 111 + 181 = 445 is not prime. Another poster suspected the mod 5 pattern must be 0, 0, 0, 0, c, c, c, c where c <> 0. Also possible is c, c, c, c, d, d, d, d where c <> d. -- The 2nd John Adams president lost to the winner of the Battle of New Orleans. The 2nd George Bush president lost the Battle of New Orleans. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime > {11,41,89,101,111,167,179,181} [snip program] This isn't right. 11 + 41 + 101 + 111 + 181 is divisible by 5. === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime >Does there exist a set of 8 positive integers, the sum of any 5 of >which is prime? As has been noted, there is the simple answer with all x_i = 1. Well, what if all eight of the x_i are supposed to be distinct? That will give as many as (8 choose 5) = 56 numbers which must be prime. That seems hard to arrange (and gets statistically harder as the numbers grow larger). However, I believe that a solution exists and I suspect that finding one is just within the limits of what one can accomplish with a brute-force search, using the particular arrangements I will now outline. (I did not stumble across an example after a brief search.) The trick is to reduce the number 56. For example, if the x_i form an arithmetic progression, then there will be only 16 sums-of-5 --- and they will also form an arithmetic progression. As it turns out, there do exist sets of 16 (or more!) primes in arithmetic progression, e.g. the numbers 1275290173428391 + 30030 i, for i = 0, 1, ..., 15. We can turn that example around to get a set of x_i, but unfortunately this requires that the eight x_i be non-integral --- the x_i would be 1275290173428391/5 + 30030 i, for i = -2, -1, ..., 5 . It is impossible to avoid the fractions here since there can be no A.P. of primes starting with 5 and having more than five terms. Still, maybe this gives a framework in which we can find a set of eight distinct x_i : look for numbers of the form x_i = a + b n_i where the n_i are chosen from among small integers so that there are relatively few distinct five-fold sums s ; then we need only have all the numbers 5a + s b be prime. Well, they can't all be prime if any of them is even. This is ensured iff all the x_i are odd, and in my setting that just means choosing an odd a and an even b. To avoid having a multiple of 3 among the sums-of-5, the residues x_i mod 3 must be one of {1,1,1,1,1,1,1,1}, {2,2,2,2,2,2,2,2}, {1,1,1,1,1,1,1,0}, or {2,2,2,2,2,2,2,0}. I can force one of these arrangments by taking a to be congruent to 1 mod 3 and b to be a multiple of 3. For p >= 7, there are many sets of 8 residues mod p for which none of the 56 possible sums of 5 is a multiple of p, but as for p=2 and p=3 it seems simpler for some of these primes (p = 7,13,17) just to choose b (and not a) to be a multiple of p; then none of the five-fold sums 5a + s b will be a multiple of p. (For other small p we will have a small set of residue classes for (a,b) mod p which will avoid multiples of p.) The prime 5 gives a more complicated problem. Having excluded the simplest answer (all x_i = 1), we can never get the sum of five x_i to EQUAL 5, so if we want primes, we have to make sure we never get a MULTIPLE of 5 either, so if as above we write the typical five-fold sum as 5a + s b, then we need to choose the n_i so that none of the 5-fold sums s is a multiple of 5. (Of course we must also avoid having b be a multiple of 5.) If you try all possible 8-tuples of elements in GF(5), you will find that there is usually at least one subset of 5 which sums to 0. I haven't catalogued all the possible ways to avoid the existence of a subset of cardinality 5 which sums to 0, but they do exist, e.g. if n_i = 1 mod 5 for i=1,2,3,4 and n_i = 2 mod 5 for i = 5,6,7,8 We want the eight x_i to be distinct so I might use, say, { n_i } = { 1, 2, 6, 7, 11, 12, 16, 17 }. Then the 5-fold sums are the numbers 5a + s b where s ranges over the 24 values 27, 28, 32, 33, 36, 37, 38, 39, 41, 42, 43, 44, 46, 47, 48, 49, 51, 52, 53, 54, 57, 58, 62, 63 none of which is a multiple of 5. Thus as long as b is not a multiple of 5, none of our sums-of-five-x_i will be a multiple of 5. (After a pretty extensive brute-force search I conjecture that a set of eight residues n_i which produces fewer than 25 sums-of-five, none of which is a multiple of 5, must be of the form { n1 + 5 m k | k = 0, 1, 2, 3 } union { n2 + 5 m k | k = 0, 1, 2, 3 } for some n1, n2, m. That is: I've already made this as simple as possible!) To summarize, as long as b is divisible by 2, 3, 7, 13, and 17 but not 5, then there is no prime p for which this set of 24 numbers 5a + s b __must_ contain a multiple of p. I believe the terminology here is that this is a potential prime constellation, and the conjecture is that there exist values of a and b for which all 24 values are prime. Indeed, there are, conjecturally, infinitely many values of a for which all 24 are prime even for the specific value b = 2*3*7*13*17 = 9282. (Think of that as a jazzed-up version of the Twin Prime Conjecture.) If you would like to do a search note that a cannot be a multiple of 2,3,7,13,or 17; and it must be congruent to 7 mod 11, and to 21 mod 23, i.e. a = 205 + 2*11*23*c for some c. However, I note that what we are hoping for is something comparable to finding an arithmetic progression of 24 primes, and that has never yet been done, although there is no reason to think that none exists (the longest known prime AP at this writing is of length 23). So I would expect it would take quite a bit of computational power to find values of a and b which give us a set of 24 primes. For another variation, one might insist that not only the eight x_i but also the 56 five-fold sums all be distinct. I suppose it follows from the same conjectures I have cited already that there will be such things, but finding them may be well beyond the capacities of simple exhaustive computer searches. dave === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime > Here's an open question from the BENT of Tau Beta Pi: > Does there exist a set of 8 positive integers, the sum of any 5 of > which is > prime? > Ray Steiner Apparently you guys know your stuff, since if we pick any set of 9 integers, then there is a subset of 5 integers whose sum is divisible by 5. So your task is impossible for a set of 9 positive integers. Now let's go on to your question. (1) Clearly all integers in your set are odd. That cuts your search space by 1/256. (2) It is easy to show: all integers are congruent mod 3. That cuts your search space by 2/6561. (3) I believe (but can't be sure) that modulo 5, we must have 0,0,0,0,c,c,c,c (where c=1,2,3,4). That cuts it by 280/5^8 = 56/5^7. By considering various cases of c, a Pentium can probably find a solution overnight. My apologies if I've contributed nothing to your problem. === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime >> Here's an open question from the BENT of Tau Beta Pi: >> Does there exist a set of 8 positive integers, the sum of any 5 of >> which is >> prime? >> Ray Steiner >Apparently you guys know your stuff, since if we pick any set of 9 >integers, then there is a subset of 5 integers whose sum is divisible >by 5. So your task is impossible for a set of 9 positive integers. >Now let's go on to your question. >(1) Clearly all integers in your set are odd. That cuts your search >space by 1/256. >(2) It is easy to show: all integers are congruent mod 3. That cuts >your search space by 2/6561. >(3) I believe (but can't be sure) that modulo 5, we must have >0,0,0,0,c,c,c,c (where c=1,2,3,4). That cuts it by 280/5^8 = 56/5^7. >By considering various cases of c, a Pentium can probably find a >solution overnight. >My apologies if I've contributed nothing to your problem. I'm currently searching through all the sets (a, a+c, a+2c, a+3c, b, b+c, b+2c, b+3c) with a>b and whrer all the candidate primes are <10^8 There are only 4610 different combinations of: (a mod 30030, b mod 30030, c mod 30030) that are possible. These numbers have 24 different sums: 4a+b+kc with 6<=k<=9 3a+2b+kc with 4<=k<=11 2a+3b+kc with 4<=k<=11 a+4b+kc with 6<=k<9 After about 20 minutes I've found 1 combination with 22 primes and 5 with 21 primes. I hope to get a solution in a few days. -- Wim Benthem === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime >> Here's an open question from the BENT of Tau Beta Pi: >> >> Does there exist a set of 8 positive integers, the sum of any 5 of >> which is >> prime? >> >> Ray Steiner >Apparently you guys know your stuff, since if we pick any set of 9 >integers, then there is a subset of 5 integers whose sum is divisible >by 5. So your task is impossible for a set of 9 positive integers. >Now let's go on to your question. >(1) Clearly all integers in your set are odd. That cuts your search >space by 1/256. >(2) It is easy to show: all integers are congruent mod 3. That cuts >your search space by 2/6561. >(3) I believe (but can't be sure) that modulo 5, we must have >0,0,0,0,c,c,c,c (where c=1,2,3,4). That cuts it by 280/5^8 = 56/5^7. >By considering various cases of c, a Pentium can probably find a >solution overnight. >My apologies if I've contributed nothing to your problem. > I'm currently searching through all the sets > (a, a+c, a+2c, a+3c, b, b+c, b+2c, b+3c) with a>b and whrer all the > candidate primes are <10^8 > There are only 4610 different combinations of: > (a mod 30030, b mod 30030, c mod 30030) that are possible. I'm not sure that I follow this I get that mod 2, (a,b,c) == (1,1,0), mod 3, (a,b,c) in {(1,1,0),(2,2,0)} mod 5, c ==0 mod 5 and a!=b mod 5 mod 7: c==0 and either precisely one of a,b ==0 or a==+-b That is 960 combinations of (a mod 210,b mod 210,c mod 210) > These numbers have 24 different sums: > 4a+b+kc with 6<=k<=9 > 3a+2b+kc with 4<=k<=11 > 2a+3b+kc with 4<=k<=11 > a+4b+kc with 6<=k<9 > After about 20 minutes I've found 1 combination with 22 primes > and 5 with 21 primes. I hope to get a solution in a few days. > -- > Wim Benthem === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime The problem should have said 8 DISTINCT positive integers. My humble apologies! Ray Steiner === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime >> For the 8/5 question, I would start by making 4 of the numbers >> EQUAL. >> This might make a brute-force search tractable. >> Actually, one could consider (1,1,1,1,1, x, y, z). >> The problem did not say distinct >Bingo! >(1,1,1,1,1, 7, 9, 19) Why bother? (1,1,1,1,1,1,1,1) satisfies that any 5 add up to a prime. -- 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: set of 8 positive integers, the sum of any 5 of which is prime >> For the 8/5 question, I would start by making 4 of the numbers >> EQUAL. >> This might make a brute-force search tractable. >> >> Actually, one could consider (1,1,1,1,1, x, y, z). >> >> The problem did not say distinct >Bingo! >(1,1,1,1,1, 7, 9, 19) > Why bother? (1,1,1,1,1,1,1,1) satisfies that any 5 add up to a prime. But can it be done with 8 primes, where 1 does not count as a prime? === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime >(1,1,1,1,1, 7, 9, 19) > Why bother? (1,1,1,1,1,1,1,1) satisfies that any 5 add up to a prime. Yep!!! A more interesting problem is to find 8 distinct numbers with the required property. Also, a more general question: If we require adding a subset of k numbers chosen from a set of size n, n>k, Is there a maximum value of n as a function of k? Does every value of k have a number n that works? Clearly k=2 is impossible. In fact, any even k is impossible. === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime <3u6lg3Fvr83rU1@individual.net> > Actually, one could consider (1,1,1,1,1, x, y, z). > The problem did not say distinct > But the set {1,1,1,1,1,x,y,z} is the same set as > {1,x,y,z}. It has at most 4 elements (and maybe fewer, > depending on x, y and z). Nowhere is it written that the elements of a set must be distinct. The question say 8 numbers, not 8 distinct numbers === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime >Actually, one could consider (1,1,1,1,1, x, y, z). >The problem did not say distinct >>But the set {1,1,1,1,1,x,y,z} is the same set as >>{1,x,y,z}. It has at most 4 elements (and maybe fewer, >>depending on x, y and z). > Nowhere is it written that the elements of a set must be distinct. Eh? You think that {1} is a different set from {1,1}? As far as I can see, every element of {1} is an element of {1,1} and vice versa. So they're the same set. You want to claim that {1,1} has cardinality 2? > The question say 8 numbers, not 8 distinct numbers It says a set of eight numbers. 1,1,1,1,1,1,1,1 are eight numbers that have the property that that sum of any five of them is a prime, but {1,1,1,1,1,1,1,1} is a set of one number. === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime <3u6lg3Fvr83rU1@individual.net> >> Actually, one could consider (1,1,1,1,1, x, y, z). >> The problem did not say distinct >> But the set {1,1,1,1,1,x,y,z} is the same set as >> {1,x,y,z}. It has at most 4 elements (and maybe fewer, >> depending on x, y and z). > Nowhere is it written that the elements of a set must be distinct. > The question say 8 numbers, not 8 distinct numbers Nowhere is it written, you say. Actually, it's written in Wikipedia and in Mathworld that a set consists of distinct objects. This is also implied by the axiom in chapter 1 of Halmos' Naive Set Theory. Indeed, the word set in mathematics usually means a collection of distinct things (in contrast to multiset or n-tuple, for example). -- J. H. Palmieri Associate Professor of Mathematics University of Washington Box 354350, Seattle, WA 98195-4350 palmieri@math.washington.edu http://www.math.washington.edu/~palmieri/ === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime <3u6lg3Fvr83rU1@individual.net> We had this very debate in Math 55 many years ago. It was a somewhat long and heated debate..... Andrew Gleason was leading the class. Perhaps you would like to debate this with him? I will use the same example he gave: You have a set of 500 pieces of paper. You would like to argue that the set has size = 1 because the pieces are indistinguishable? Ridiculous... Or take another set of pieces of paper; this time with (say) 300 sheets. Call them sets A and B. They are both finite. I take a piece from set A and a piece from set B. I put them together and then place them aside. I continue until set A is exhausted. Look Ma! There are still sheets left in set B. We can't put them into 1-1 correspondence! They must be different sizes! Yet you would argue that they are both the same size (1). I will take Andrew Gleason's teaching over what is written in Wikipedia. Here's another example: Consider the set of Fibonacci numbers. F = (F_0, F_1, F_2, F_3,.....) We have F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, etc. You want to argue that the set consists of F = (1,2,3,5,8,....)? Where is F_0? It is in the set you say? It equals 1 you say? OK, then where is F_1? It too is in the set you say?? I don't see it! F_0 and F_1 are different elements, even though they have the same VALUE. === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime <3u6lg3Fvr83rU1@individual.net> > We had this very debate in Math 55 many years ago. > It was a somewhat long and heated debate..... > Andrew Gleason was leading the class. Perhaps you would like to > debate this with him? I will use the same example he gave: > You have a set of 500 pieces of paper. You would like to argue > that the set has size = 1 because the pieces are indistinguishable? > Ridiculous... > Or take another set of pieces of paper; this time with (say) 300 > sheets. > Call them sets A and B. They are both finite. I take a piece from > set A > and a piece from set B. I put them together and then place them aside. > I continue until set A is exhausted. Look Ma! There are still sheets > left in set B. We can't put them into 1-1 correspondence! They must > be different sizes! Yet you would argue that they are both the same > size (1). > I will take Andrew Gleason's teaching over what is written in > Wikipedia. Wow, amazing. You're rejecting not just what's said in wikipedia, but everywhere, in every book on elementary set theory (or even first-order logic) I've read about. I've heard of people rejecting the Continuum Hypothesis or even the Axiom of Choice, but this is the first time I've heard of someone rejecting the Axiom of Extensionality. I give up. It is beyond me. === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime >> We had this very debate in Math 55 many years ago. >> It was a somewhat long and heated debate..... >> Andrew Gleason was leading the class. Perhaps you would like to >> debate this with him? I will use the same example he gave: >> You have a set of 500 pieces of paper. You would like to argue >> that the set has size = 1 because the pieces are indistinguishable? >> Ridiculous... >> Or take another set of pieces of paper; this time with (say) 300 >> sheets. >> Call them sets A and B. They are both finite. I take a piece from >> set A >> and a piece from set B. I put them together and then place them aside. >> I continue until set A is exhausted. Look Ma! There are still sheets >> left in set B. We can't put them into 1-1 correspondence! They must >> be different sizes! Yet you would argue that they are both the same >> size (1). Indistinguishable does not mean identical. If we have 8 indistinguishable balls in an urn, we can draw them out one at a time until the urn is exhausted, and place them in 8 distinguishable positions. At that time, we can refer to any specific one by its position. >> I will take Andrew Gleason's teaching over what is written in >> Wikipedia. >Wow, amazing. You're rejecting not just what's said in wikipedia, but >everywhere, in every book on elementary set theory (or even first-order >logic) I've read about. I've heard of people rejecting the Continuum >Hypothesis or even the Axiom of Choice, but this is the first time I've >heard of someone rejecting the Axiom of Extensionality. >I give up. It is beyond me. -- 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: set of 8 positive integers, the sum of any 5 of which is prime > We had this very debate in Math 55 many years ago. > It was a somewhat long and heated debate..... > Andrew Gleason was leading the class. Perhaps you would like to > debate this with him? I will use the same example he gave: > You have a set of 500 pieces of paper. You would like to argue > that the set has size = 1 because the pieces are indistinguishable? What does that have to do with claiming that {1,1,1,1,1,1,1,1} has eight elements? I prefer my sets to have properties described by the axioms of ZFC, given the option. > Here's another example: Consider the set of Fibonacci numbers. F = > (F_0, F_1, F_2, F_3,.....) > We have F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, etc. You want to > argue > that the set consists of F = (1,2,3,5,8,....)? Assuming that it's a set, and not a sequence or a multiset, I have little option. You might want to consider the advantages of not abusing terminology, and only using the word 'set' when you mean 'set'. For cases where you want to have elements with multiplicity greater than 1, the usual name is 'multiset'. But feel free to give your alternative axioms and insist that everybody in the world except you is out of step. === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Actually, one could consider (1,1,1,1,1, x, y, z). >> The problem did not say distinct >> But the set {1,1,1,1,1,x,y,z} is the same set as >> {1,x,y,z}. It has at most 4 elements (and maybe fewer, >> depending on x, y and z). > Nowhere is it written that the elements of a set must be distinct. Wrong. You are confusing sets and multisets. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: set of 8 positive integers, the sum of any 5 of which is prime >> Actually, one could consider (1,1,1,1,1, x, y, z). >> The problem did not say distinct >> But the set {1,1,1,1,1,x,y,z} is the same set as >> {1,x,y,z}. It has at most 4 elements (and maybe fewer, >> depending on x, y and z). > Nowhere is it written that the elements of a set must be distinct. > The question say 8 numbers, not 8 distinct numbers Have you heard of the axiom of extensionality? Given A = {1} and B = {1,1,1,1}, can you name an element of B that is not an element of A (or conversely)? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: triangle inequality equvalent to set convexity Hi everybody I have a problem with one theory. I have to prove that triangle inequality in norm definition ||x+y|| <= ||x|| + ||y|| is equivalent to convexity of the set ||a||<=1. It should probably looks like ||x*lambda + y(1-lambda) || but what can be done with that ? I checked it for norm maximum and it`s true but don`t know how to generalize it :( Jesper === Subject: Re: triangle inequality equvalent to set convexity > I have a problem with one theory. I have to prove that triangle inequality > in norm definition ||x+y|| <= ||x|| + ||y|| is equivalent to convexity of > the set ||a||<=1. How to prove it ? It should probably looks like ||x*lambda > + y(1-lambda) || but what can be done with that ? > Bartek Damn, I answer this same question at sci.math. Learn about cross posting and use it!!! === Subject: Re: triangle inequality equvalent to set convexity > I have a problem with one theory. I have to prove that triangle inequality > in norm definition ||x+y|| <= ||x|| + ||y|| is equivalent to convexity of > the set ||a||<=1. Sloopy thinking gives naught but sloopy results. ||a|| <= 1 isn't a set, it's an inequality. A = { a : ||a|| <= 1 } is a set. If a,b in A, then for all t in [0,1], ||a + t(b - a)|| = || (1 - t)a + tb || <= (1 - t)||a|| + t||b|| <= 1 - t + t = 1 If A convex, x /= 0 /= y, then let r = ||x|| + ||y|| x1 = x/||x||, y1 = y/||y|| in A (1 - ||y||/r) x1 + (||y||/r) y1 = x/r + y/r in A etc... === Subject: Re: triangle inequality equvalent to set convexity > Hi everybody > I have a problem with one theory. I have to prove that triangle inequality > in norm definition ||x+y|| <= ||x|| + ||y|| is equivalent to convexity of > the set ||a||<=1. It should probably looks like ||x*lambda > + y(1-lambda) || but what can be done with that ? I checked it for norm > maximum and it`s true but don`t know how to generalize it :( I don't understand this last sentence. If you have ANY norm (satisfying the triangle inequallity and the condition ||c x|| = c ||x|| for any positive real c), then you are done. Given ||x||, ||y|| <= 1, we have (for real 0 < r < 1) that ||rx + (1-r)y|| <= ||r x|| + ||(1-r) y|| = r ||x|| + (1-r)||y||, etc. R.G. Vickson > Jesper === Subject: RV and CLT I am considering a problem about central limit theorm. Given a sequence of random variables {X_n} with E(X_n)=1 and Var(X_n) tends to infinity as n grows. Is it possible to find some constants a_n and b_n such that (X_n - a_b)/b_n converges to a standard normal distribution? I think not. Could someone have a suggestion? I have no idea to prove or disprove === Subject: Re: RV and CLT >I am considering a problem about central limit theorm. >Given a sequence of random variables {X_n} with E(X_n)=1 and Var(X_n) >tends to infinity as n grows. Is it possible to find some constants a_n >and b_n such that (X_n - a_b)/b_n converges to a standard normal >distribution? >I think not. >Could someone have a suggestion? I have no idea to prove or disprove If b_n goes to infinity so that b_(n+1)/b_n -> 1, a necessary and sufficient condition is the Lindeberg-Feller condition that for every e > 0, there exist c_n such that sum^nint_(|x -c_k| <= eb_n) (x -c_k)^2 dF_k(x) /b_n -> 1. This is needed no matter what, unless some of the X's are close to normal. Variances going to 0 are as bad a problem. -- 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: RV and CLT >I am considering a problem about central limit theorm. >Given a sequence of random variables {X_n} with E(X_n)=1 and Var(X_n) >tends to infinity as n grows. Is it possible to find some constants a_n >and b_n such that (X_n - a_b)/b_n converges to a standard normal >distribution? >I think not. >Could someone have a suggestion? I have no idea to prove or disprove By the way, I don't see any tie to the CLT since you are just taking limits -- nothing is being averaged. quasi === Subject: Re: RV and CLT >I am considering a problem about central limit theorm. >Given a sequence of random variables {X_n} with E(X_n)=1 and Var(X_n) >tends to infinity as n grows. Is it possible to find some constants a_n >and b_n such that (X_n - a_b)/b_n converges to a standard normal >distribution? >I think not. >Could someone have a suggestion? I have no idea to prove or disprove Your intution is correct -- it can't always be done. Hint: Let X_n be uniformly distributed on the interval (-n+1,n+1). quasi === Subject: Re: RV and CLT >>I am considering a problem about central limit theorm. >>Given a sequence of random variables {X_n} with E(X_n)=1 and Var(X_n) >>tends to infinity as n grows. Is it possible to find some constants a_n >>and b_n such that (X_n - a_b)/b_n converges to a standard normal >>distribution? >>I think not. >>Could someone have a suggestion? I have no idea to prove or disprove >Your intution is correct -- it can't always be done. >Hint: >Let X_n be uniformly distributed on the interval (-n+1,n+1). I think what was meant was that the sum of the first n X's has this property. It does in your example. >quasi -- 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: RV and CLT On 20 Nov 2005 21:01:02 -0500, hrubin@odds.stat.purdue.edu (Herman >I am considering a problem about central limit theorm. >Given a sequence of random variables {X_n} with E(X_n)=1 and Var(X_n) >tends to infinity as n grows. Is it possible to find some constants a_n >and b_n such that (X_n - a_b)/b_n converges to a standard normal >distribution? >I think not. >Could someone have a suggestion? I have no idea to prove or disprove >>Your intution is correct -- it can't always be done. >>Hint: >>Let X_n be uniformly distributed on the interval (-n+1,n+1). >I think what was meant was that the sum of the first n X's >has this property. It does in your example. Could someone then restate the problem correctly. You can't use the same symbol X_n for both the original sequence of random variables and, at the same time, for their sum or average. quasi === Subject: Abstract A couple more hw problems I was hoping people could give me hints on. 1. Order of G is pq where p and g are primes (not nexcessarily distinct). Prove that the order of the center of G is either 1 or pq. 2. If H is a normal subgroup of G and order of H is 2, prove H is contained in the center of G. === Subject: Re: Abstract > A couple more hw problems I was hoping people could > give me hints on. > 1. Order of G is pq where p and g are primes (not nexcessarily distinct). > Prove that the order of the center of G is either 1 or pq. In addition to what others have said, there is a lovely quick solution if you know what the class equation is. > 2. If H is a normal subgroup of G and order of H is 2, > prove H is contained in the center of G. Hint: if g is an element of G, then g is in the center of G if and only if for all x in G, x g x^{-1} = g. -- Ryan Reich ryan.reich@gmail.com === Subject: Re: Abstract days. My association with the Department is that of an alumnus. >A couple more hw problems I was hoping people could >give me hints on. >1. Order of G is pq where p and g are primes (not nexcessarily >distinct). Prove that the order of the >center of G is either 1 or pq. Prove that if N is a subgroup of Z(G), and G/N is central, then G is abelian (and therefore Z(G)=G). The center of G must have order dividing pq. It is either 1, p, q, or pq. If it were equal to p, what would be the order of G/Z(G)? And what kind of group would it have to be? Reach a contradiction. A similar argument will do if Z(G) is of order q. >2. If H is a normal subgroup of G and order of H is 2, >prove H is contained in the center of G. If H = {1,a}, and g in G, you know that gag^{-1} must be in H. What can it be equal to? -- 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: Abstract > A couple more hw problems I was hoping people could > give me hints on. > 1. Order of G is pq where p and g are primes (not nexcessarily distinct). Prove that the order of the > center of G is either 1 or pq. The order of Z(G) can be 1, p, q, or pq. If it is p, G/Z(G) is a group of order q, thus cyclic. It is calssical to prove that if G/Z(G) is cyclic then G is abelian. This is a contradiction with Z(G) = G The same argument shows that Card(Z(G)) = q. === Subject: Re: Abstract How do you conclude that G is not abelian? === Subject: Re: Abstract days. My association with the Department is that of an alumnus. >How do you conclude that G is not abelian? Don't use the Reply button at the bottom of the posts. Click on More options at the top, then use the reply button there. That will allow you to quote the messages you are replying to in order to provide context. If G is a group of order pq, and the center is NOT trivial, then the conclusion is in fact that G ->IS<- abelian, via the lemma I mentioned (if G/N is cyclic, and N is contained in the center, then G is abelian). -- 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: question about non-uniform sampling? <0-WdnSIUeY_HOuHeRVn-ow@rcn.net> > Ummm... If the second half isn't sampled, how is it fed to the filter? The actual sampler has to go after the low pass filter unless the data is already bandlimited. Any samples before the filter aren't bandlimited samples, just intermediate voltages or data in the pre-sampler low pass filter. > I note that you said samples from the first half, not during the > first half. They're different due to the low pass filter delay. > The output of the filter may be delayed, but there's no delay at the > input. Sampled _from_ the first half are put in _during_ the first half > with at most trivial delay. But for band-limited samples, you have to sample after the filter at the filter output, and after waiting for the filter lag if you want synchronized time-stamping. -- rhn === Subject: Re: question about non-uniform sampling? >>Ummm... If the second half isn't sampled, how is it fed to the filter? > The actual sampler has to go after the low pass filter unless > the data is already bandlimited. Any samples before the filter > aren't bandlimited samples, just intermediate voltages or data > in the pre-sampler low pass filter. Naturally. I don't see the connection except, of course, that you don't need to low pass an interval that won't be sampled. >I note that you said samples from the first half, not during the >first half. They're different due to the low pass filter delay. >>The output of the filter may be delayed, but there's no delay at the >>input. Sampled _from_ the first half are put in _during_ the first half >>with at most trivial delay. > But for band-limited samples, you have to sample after the filter at > the filter output, and after waiting for the filter lag if you want > synchronized time-stamping. Sure, but once you get to the part that you don't intend to sample, however long that takes, you can stop filtering. Keep fixed on the structure of the counterexample I proposed: one oversamples the first N minutes of a 2N-minute signal by a factor of more than two, then reconstructs all N minutes worth from those samples. Reconstruction can begin as soon as the requisite number of samples have been collected. There are still knowledgeable people here who think I am recalcitrant for claiming that this won't work. I can only conclude that I haven't adequately communicated the proposed conditions. Shall I try again? Jerry -- Engineering is the art of making what you want from things you can get. øøøøøøøø[OS lash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøø === Subject: Re: question about non-uniform sampling? <0-WdnSIUeY_HOuHeRVn-ow@rcn.net> >>Ummm... If the second half isn't sampled, how is it fed to the filter? > The actual sampler has to go after the low pass filter unless > the data is already bandlimited. Any samples before the filter > aren't bandlimited samples, just intermediate voltages or data > in the pre-sampler low pass filter. > Naturally. I don't see the connection except, of course, that you don't > need to low pass an interval that won't be sampled. >I note that you said samples from the first half, not during the >first half. They're different due to the low pass filter delay. >> >>The output of the filter may be delayed, but there's no delay at the >>input. Sampled from the first half are put in during the first half >>with at most trivial delay. > But for band-limited samples, you have to sample after the filter at > the filter output, and after waiting for the filter lag if you want > synchronized time-stamping. > Sure, but once you get to the part that you don't intend to sample, > however long that takes, you can stop filtering. Keep fixed on the > structure of the counterexample I proposed: one oversamples the first N > minutes of a 2N-minute signal by a factor of more than two, then > reconstructs all N minutes worth from those samples. Reconstruction can > begin as soon as the requisite number of samples have been collected. > There are still knowledgeable people here who think I am recalcitrant > for claiming that this won't work. I can only conclude that I haven't > adequately communicated the proposed conditions. Shall I try again? You don't need to, since what you're doing isn't signal samplimg. it's infornmation sampling. And you conveniently picked the onle application that is does work well, with telecommunications equipment. i.e. you're essentially not sampling anything physical, but rather than Goedel's Theroem. And what you're doing is usually called reverese-engineering, rather than reconstruction. > Jerry > -- > Engineering is the art of making what you want from things you can get. > øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøø === Subject: Re: question about non-uniform sampling? <0-WdnSIUeY_HOuHeRVn-ow@rcn.net> >>Ummm... If the second half isn't sampled, how is it fed to the filter? > The actual sampler has to go after the low pass filter unless > the data is already bandlimited. Any samples before the filter > aren't bandlimited samples, just intermediate voltages or data > in the pre-sampler low pass filter. > Naturally. I don't see the connection except, of course, that you don't > need to low pass an interval that won't be sampled. >I note that you said samples from the first half, not during the >first half. They're different due to the low pass filter delay. >> >>The output of the filter may be delayed, but there's no delay at the >>input. Sampled _from_ the first half are put in _during_ the first half >>with at most trivial delay. > But for band-limited samples, you have to sample after the filter at > the filter output, and after waiting for the filter lag if you want > synchronized time-stamping. > Sure, but once you get to the part that you don't intend to sample, > however long that takes, you can stop filtering. And since the filter lag is the length of the entire signal you cannot stop filtering until after the entire signal has been fed into the filter. >Keep fixed on the > structure of the counterexample I proposed: one oversamples the first N > minutes of a 2N-minute signal by a factor of more than two, then > reconstructs all N minutes worth from those samples. Reconstruction can > begin as soon as the requisite number of samples have been collected. But the vaiues of the requisite number of samples cannot be determined until after the entire signal has been processed. So reconstruction cannot begin until after the entire signal. There is no causality paradox. > There are still knowledgeable people here who think I am recalcitrant > for claiming that this won't work. I can only conclude that I haven't > adequately communicated the proposed conditions. Shall I try again? Perhaps you should to figure out what the conditions that others are talking about are (the fact that you cannot reconstruct a real world signal with samples taken in the first half is accepted by everyone). The simple point is that you can only predict the entire signal from samples taken in the first half if this signal is band limited. And you cannot find a band limited signal that adequately approximates your real world signal without processing the entire signal. So there is no causality paradox. -William Hughes === Subject: Re: question about non-uniform sampling? ... > But the vaiues of the requisite number of samples cannot be determined > until after the entire signal has been processed. So reconstruction > cannot begin until after the entire signal. There is no causality > paradox. I can't imagine how causality entered into this discussion. As I understood the original assertion about nonuniform sampling, reconstruction is possible so long as the average sampling interval (or was it rate?) meets the Nyquist criterion. For uniform sampling of a signal of bandwidth B Hz and duration D seconds, at least 2BD samples are needed. The assertion stated that 2BD samples also suffices for nonuniform sampling, and that the location of the samples within the signal is irrelevant. >>There are still knowledgeable people here who think I am recalcitrant >>for claiming that this won't work. I can only conclude that I haven't >>adequately communicated the proposed conditions. Shall I try again? Consider a two-second telephone monolog. B = 3.5 kHz; D = 2. By sampling at 8 KHz, there is some processing margin. With uniform sampling, there will be 16,000 samples, and according to the assertion, 16,000 samples taken anywhere in the monolog suffice to characterize it. Before sampling, the signal is passed through a very sharp lowpass filter that cuts off at 3.5 KHz. The cutoff of the analog filter is so sharp and deep -- its impulse response is so long -- that the delay through it is several days, but we're patient. At the appropriate time, sampling begins; not at 8 KHz, but 16. After one second of sampling, all of the necessary 16,000 samples have been recorded. Although the filter continues to produce output, the sampler can stop. I claim that the 16,000 samples so obtained meet the conditions of the assertion but do not support the reconstruction of the entire monolog. Do think my claim is false? Jerry -- Engineering is the art of making what you want from things you can get. øøøøøøøø[OS lash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøø === Subject: Re: question about non-uniform sampling? <0-WdnSIUeY_HOuHeRVn-ow@rcn.net> > ... > But the vaiues of the requisite number of samples cannot be determined > until after the entire signal has been processed. So reconstruction > cannot begin until after the entire signal. There is no causality > paradox. > I can't imagine how causality entered into this discussion. Your counterexamples have been reductio ad absurdum based on causality (you cannot reconstruct this because you don't know what it is yet). >As I > understood the original assertion about nonuniform sampling, > reconstruction is possible so long as the average sampling interval (or > was it rate?) meets the Nyquist criterion. For uniform sampling of a > signal of bandwidth B Hz and duration D seconds, at least 2BD samples > are needed. The assertion stated that 2BD samples also suffices for > nonuniform sampling, and that the location of the samples within the > signal is irrelevant. Yes, this was the original claim. (Note that we are assuming that all samples are at exact times and of infinite precision). >>There are still knowledgeable people here who think I am recalcitrant >>for claiming that this won't work. I can only conclude that I haven't >>adequately communicated the proposed conditions. Shall I try again? > Consider a two-second telephone monolog. B = 3.5 kHz; D = 2. By sampling > at 8 KHz, there is some processing margin. With uniform sampling, there > will be 16,000 samples, and according to the assertion, 16,000 samples > taken anywhere in the monolog suffice to characterize it. Not quite, the signal must be band limited. The assertion applies only after the filtering (but see below). > Before sampling, the signal is passed through a very sharp lowpass > filter that cuts off at 3.5 KHz. The cutoff of the analog filter is so > sharp and deep -- its impulse response is so long -- that the delay > through it is several days, but we're patient So we start sampling long after the monologue is finished. > At the appropriate time, > sampling begins; not at 8 KHz, but 16. After one second of sampling, all > of the necessary 16,000 samples have been recorded. Although the filter > continues to produce output, the sampler can stop. Correct. (Note that the values of the samples we obtain depend on the whole signal). > I claim that the 16,000 samples so obtained meet the conditions of the > assertion but do not support the reconstruction of the entire monolog. > Do think my claim is false? Yes. All your arguments so far have been of the form we can't reconstruct the signal yet, we don't know what it is going to be. However, in the above case we do know what the signal is going to be and that information was used in creating the samples. Do you have another argument? [assume the first second was silence. Will the samples obtained be 0. No Are they too small for practical work. Yes.] - William Hughes === Subject: Re: question about non-uniform sampling? <0-WdnSIUeY_HOuHeRVn-ow@rcn.net> > The simple point is that you can only predict the entire > signal from samples taken in the first half if this > signal is band limited. And you cannot find a band > limited signal that adequately approximates your real > world signal without processing the entire signal. So > there is no causality paradox. Yes, but there is one thing that seems to be generally ignored. Not only does the signal have to be bandlimited - it has to be bandlimited rather precisely and you have to possess knowledge of exactly how it is bandlimited. Take the concert example: if you have a set of uniform samples of an entire concert you then have a FFT of the entire concert as well as the capability to produce a continuous precisely bandlimited function of the concert. You can now re-sample that continuous function with a non-uniform scheme where you end up with a new set of samples of the same number as before clustered in the first half of the concert. Can you now reconstruct the entire concert? No, you can't unless you know the exact time interval of the whole concert. It's not enough to simply say the samples represent a bandlimited signal, you need to know what the actual frequency limits are. -jim http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- === Subject: Re: question about non-uniform sampling? <0-WdnSIUeY_HOuHeRVn-ow@rcn.net> <1132581472_17779@spool6-east.superfeed.net> > The simple point is that you can only predict the entire > signal from samples taken in the first half if this > signal is band limited. And you cannot find a band > limited signal that adequately approximates your real > world signal without processing the entire signal. So > there is no causality paradox. > Yes, but there is one thing that seems to be generally ignored. Not only > does the signal have to be bandlimited - it has to be bandlimited rather > precisely and you have to possess knowledge of exactly how it is > bandlimited. > Take the concert example: if you have a set of uniform samples of an > entire concert you then have a FFT of the entire concert as well as the > capability to produce a continuous precisely bandlimited function of the > concert. Nit. You can only produce a bandlimited approximation (unless the concert was bandlimited (unlikely)) > You can now re-sample that continuous function with a > non-uniform scheme where you end up with a new set of samples of the > same number as before clustered in the first half of the concert. Can > you now reconstruct the entire concert? No, you can't unless you know > the exact time interval of the whole concert. It's not enough to simply > say the samples represent a bandlimited signal, you need to know what > the actual frequency limits are. Indeed. (There is of course a similar problem in the uniform case, but it is not as extreme as the fact that you know the samples are uniform gives you a lot of information, so it is possible to calculate a frequency limit below which there is no ambiguity. In the non uniform case we cannot do this without knowing the length of the signal to be reconstructed.) -William Hughes === Subject: Re: question about non-uniform sampling? <0-WdnSIUeY_HOuHeRVn-ow@rcn.net> > Nit. You can only produce a bandlimited approximation (unless > the concert was bandlimited (unlikely)) Whether you think it resembles the concert or not is not relevant. The important thing is we would have a bandlimited function which we can hypothetically sample for the purpose of analyzing the process of non-uniform sampling. > You can now re-sample that continuous function with a > non-uniform scheme where you end up with a new set of samples of the > same number as before clustered in the first half of the concert. Can > you now reconstruct the entire concert? No, you can't unless you know > the exact time interval of the whole concert. It's not enough to simply > say the samples represent a bandlimited signal, you need to know what > the actual frequency limits are. > Indeed. (There is of course a similar problem in the uniform case, > but it is not as extreme as the fact that you know the samples > are uniform gives you a lot of information, so it is possible > to calculate a frequency limit below which there is no ambiguity. > In the non uniform case we cannot do this without knowing > the length of the signal to be reconstructed.) Yes, but that's far from a trivial problem. In the uniform case you may not know what the original sample rate was but its easy to imagine that the concert samples could be analyzed and enough clues found to reconstruct and play back where the listener would not even notice. With the non-uniform case if you don't pretty much know the frequency content of the signal before hand, its game over. In the example I gave above of non-uniform sampling, you couldn't even do a good job of reconstructing the half of the concert that the samples were taken from. Knowing that the samples came from a bandlimited signal is pretty much useless info. -jim http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- === Subject: Re: question about non-uniform sampling? <0-WdnSIUeY_HOuHeRVn-ow@rcn.net> <1132581472_17779@spool6-east.superfeed.net> <1132591624_17985@spool6-east.superfeed.net> > Nit. You can only produce a bandlimited approximation (unless > the concert was bandlimited (unlikely)) > Whether you think it resembles the concert or not is not relevant. > The important thing is we would have a bandlimited function which we can > hypothetically sample for the purpose of analyzing the process of > non-uniform sampling. > You can now re-sample that continuous function with a > non-uniform scheme where you end up with a new set of samples of the > same number as before clustered in the first half of the concert. Can > you now reconstruct the entire concert? No, you can't unless you know > the exact time interval of the whole concert. It's not enough to simply > say the samples represent a bandlimited signal, you need to know what > the actual frequency limits are. > > Indeed. (There is of course a similar problem in the uniform case, > but it is not as extreme as the fact that you know the samples > are uniform gives you a lot of information, so it is possible > to calculate a frequency limit below which there is no ambiguity. > In the non uniform case we cannot do this without knowing > the length of the signal to be reconstructed.) > Yes, but that's far from a trivial problem Dividing the length of the signal by the number of samples is non-trivial? -William Hughes === Subject: Re: question about non-uniform sampling? ... > Indeed. (There is of course a similar problem in the uniform case, > but it is not as extreme as the fact that you know the samples > are uniform gives you a lot of information, so it is possible > to calculate a frequency limit below which there is no ambiguity. > In the non uniform case we cannot do this without knowing > the length of the signal to be reconstructed.) reduces the information they can yield. I've been saying that if the nonuniformity is great enough, they become essentially useless. We're on the same track. No problem becomes no way. Jerry -- Engineering is the art of making what you want from things you can get. .88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N 210N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N21 0N.88N.88N.88N.88N === Subject: Re: question about non-uniform sampling? <0-WdnSIUeY_HOuHeRVn-ow@rcn.net> <1132581472_17779@spool6-east.superfeed.net> > ... > Indeed. (There is of course a similar problem in the uniform case, > but it is not as extreme as the fact that you know the samples > are uniform gives you a lot of information, so it is possible > to calculate a frequency limit below which there is no ambiguity. > In the non uniform case we cannot do this without knowing > the length of the signal to be reconstructed.) > reduces the information they can yield. Only to the extent that given samples taken at an aribtrary time you cannot determine the length of the sampled signal from the number of samples. Given uniform samples and the sampling interval you can. The difference in information does not depend on the extent of the nonuniformity. I've been saying that if the > nonuniformity is great enough, they become essentially useless. We're on > the same track. No problem becomes no way. No. Making the samples less uniform does not reduce the amount of information they hold. - William Hughes === Subject: Re: question about non-uniform sampling? ... > No. Making the samples less uniform does not reduce the > amount of information they hold. It certainly reduces the information you can extract with practical means. Careful preparation of the signal before it's sampled is really a red herring. The usual case is something like a log of recorded measurements, each with time of measurement noted. Suppose them to be the level of a river or the speed of a ship. On most days, there are two measurements, some have three or more, and for a few days there are none. Given these data, How well can the continuous function be deduced? ... Jerry -- Engineering is the art of making what you want from things you can get. .88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N 210N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N21 0N.88N.88N.88N.88N === Subject: Re: question about non-uniform sampling? (snip) >> Given all the samples of his work and the first three as input, one >> might compute in some weeks the most probable fourth movement. > Including the choral parts, which were unprecedented for a Beethoven > symphony? I was thinking about the people who analyze written work to identify the author. I believe such analysis is also done on musical compositions. It seems, then, that the actual fourth movement is not the most probable one. -- glen === Subject: Re: question about non-uniform sampling? <1Tdef.5604$wc.714955@wagner.videotron.net> <_MadnV2tPbflv-feRVn-vg@rcn.net> > (snip) >> There is nothing wrong with any >> extreme of non-uniformity in a purely mathematical sense. That is in a >> world with infinite sampling precision and no noise due to the >> converter itself. > It's also a world where signals exist for all time. I doesn't matter how > precisely one can sample and how often, nothing can be known about a > speech yet to be given, even if the mathematics of nonuniform and highly > clumped sampling shows that it can. > I am not so sure what quantum mechanics says about this. The only thing quantum mechanics says about anything that is not quantum mechanics is the scattering strength of even use QM are mathematicians, since both Giga's and eV's are stastical propertites, not physical properties. Since there is no conversion from an eV to any other observalble macroscopic property that isn't first filtered through General Relavity, Protons, and The Eiffel Tower, rather than sampling theory. Much can even said about speeches yet t be given, since the only people who even listen to them are the speaker's speahwriter rather than the speaker. Which is why blow-up barbie dolls still confuse Behaviourists, more than they do Barbie. === Subject: Re: question about non-uniform sampling? <4375FA3F.921CBB9A@acm.org> > 11/12/2005 09:32: >> Can non-uniform sampled signal be used to perfectly reconstruct the >> original continuous time signal? > Yes, but it isn't easy. >> What is the Nyquist sampling rate in the non-uniform case? > Believe it or not, it's the same as the uniform case ... the number > of samples over the time interval must exceed twice the bandwidth > of the signal. That sounds like a sensible conclusion but how do we define bandwidth? For instance, if you have one peak in the frequency domain, you can measure the bandwidth as the width of the peak. But what if you have more then one peek? If they were the same amplitude I would assume you could add the peaks up. If they were we different amplitudes I would think you would want to weight them somehow. === Subject: Re: question about non-uniform sampling? <4375FA3F.921CBB9A@acm.org> > What is the Nyquist sampling rate in the non-uniform case? > Believe it or not, it's the same as the uniform case ... the number > of samples over the time interval must exceed twice the bandwidth > of the signal. > That sounds like a sensible conclusion but how do we define bandwidth? > For instance, if you have one peak in the frequency domain, you can > measure the bandwidth as the width of the peak. But what if you have > more then one peek? If they were the same amplitude I would assume you > could add the peaks up. If they were we different amplitudes I would > think you would want to weight them somehow. The bandwidth is defined as the bandwidth of the perfect brick-wall low-pass or band-pass antialias filter you ran the signal through *prior* to sampling it. The impulse response of this filter is the interpolation function used in the system of equations you have to solve as shown in my writeup. -- Dave Tweed === Subject: Re: question about non-uniform sampling? <4375FA3F.921CBB9A@acm.org> <437FA36A.9F16DF19@enpoint.com> > The bandwidth is defined as the bandwidth of the perfect brick-wall > low-pass or band-pass antialias filter you ran the signal through > *prior* to sampling it. The impulse response of this filter is the > interpolation function used in the system of equations you have to > solve as shown in my writeup. > -- Dave Tweed The bandwidth of the signal could be less then output of the filter could be less then the bandwidth of the filter and I think such problems are also of interest. Do we lose generality by dealing with the case they are the same? Anyway the link you gave wasn't much of a proof. It was more like two sample problems with perhaps some had waving in the solutions. I'm trying to recall why we interpolate with sinc functions. I believe it is a consequence of sign x of x compensation which effectively reduces the distortion caused by a sample an hold output. Also why are all the sinc functions the same width. Shouldn't the width be dependent upon the time interval the sample is held for? You mentioned something about the error in the matrix inversion but I think only vaguely in the since that small elements in the matrix lead to a large inverse, which cause a large error. Perhaps, with the use of singular value decomposition you could develop a better expression for the error in the inverse. === Subject: Re: question about non-uniform sampling? <4375FA3F.921CBB9A@acm.org> <437FA36A.9F16DF19@enpoint.com> > The bandwidth is defined as the bandwidth of the perfect brick-wall > low-pass or band-pass antialias filter you ran the signal through > *prior* to sampling it. The impulse response of this filter is the > interpolation function used in the system of equations you have to > solve as shown in my writeup. > -- Dave Tweed > The bandwidth of the signal could be less then output of the filter > could be less then the bandwidth of the filter and I think such > problems are also of interest. Do we lose generality by dealing with > the case they are the same? > Anyway the link you gave wasn't much of a proof. It was more like two > sample problems with perhaps some had waving in the solutions. I'm > trying to recall why we interpolate with sinc functions. I believe it > is a consequence of sign x of x compensation which effectively reduces > the distortion caused by a sample an hold output. Well, it has more to do with that the definite integral of sinc has a value on every interval. The compensation isn't really a compensation of anything. It's better known as Gibb's phenomenon in physics. Siince the sample and hold isn't actually a distortion. Musicians call it a distortion, basically since the only thing any of them knew about sampling or physics is the directions to the nearest mall. > Also why are all the sinc functions the same width. Shouldn't the > width be dependent upon the time interval the sample is held for? You > mentioned something about the error in the matrix inversion but I think > only vaguely in the since that small elements in the matrix lead to a > large inverse, which cause a large error. Perhaps, with the use of > singular value decomposition you could develop a better expression for > the error in the inverse. === Subject: Re: question about non-uniform sampling? <4375FA3F.921CBB9A@acm.org> <437FA36A.9F16DF19@enpoint.com> > What is the Nyquist sampling rate in the non-uniform case? > > Believe it or not, it's the same as the uniform case ... the number > of samples over the time interval must exceed twice the bandwidth > of the signal. > That sounds like a sensible conclusion but how do we define bandwidth? > For instance, if you have one peak in the frequency domain, you can > measure the bandwidth as the width of the peak. But what if you have > more then one peek? If they were the same amplitude I would assume you > could add the peaks up. If they were we different amplitudes I would > think you would want to weight them somehow. > The bandwidth is defined as the bandwidth of the perfect brick-wall > low-pass or band-pass antialias filter you ran the signal through > *prior* to sampling it. The impulse response of this filter is the > interpolation function used in the system of equations you have to > solve as shown in my writeup. But, what mathematicians don't seem to understand about sampling theory is that the perfect brickwall low-pass filter isn't made of bricks. Its a forest made of trees, leaves, and lakes, rather than bricks and mrrors. Since the rocks in the forest need no interpolation, the lake surface needs no idiiot QM chemists and thermostats, and they automatically generate generalized sampling theory, both non-uniform sampling theory, uniform sampling, and non-Gaussian interpolation functions. > -- Dave Tweed === Subject: Re: question about non-uniform sampling? >>11/12/2005 09:32: >>Can non-uniform sampled signal be used to perfectly reconstruct the >>original continuous time signal? >Yes, but it isn't easy. >>What is the Nyquist sampling rate in the non-uniform case? >Believe it or not, it's the same as the uniform case ... the number >of samples over the time interval must exceed twice the bandwidth >of the signal. > That sounds like a sensible conclusion but how do we define bandwidth? > For instance, if you have one peak in the frequency domain, you can > measure the bandwidth as the width of the peak. But what if you have > more then one peek? If they were the same amplitude I would assume you > could add the peaks up. If they were we different amplitudes I would > think you would want to weight them somehow. The bandwidth if a signal is simply the frequency range needed to encompass all of the energy in it. In practice, when we say a signal has a certain bandwidth, we mean that the energy outside that range is small enough not to bollux up what the signal is intended to accomplish. Jerry -- Engineering is the art of making what you want from things you can get. øøøøøøøø[OS lash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøø === Subject: Re: question about non-uniform sampling? 14:21: >> 11/12/2005 09:32: >> Can non-uniform sampled signal be used to perfectly reconstruct the >> original continuous time signal? > > Yes, but it isn't easy. > >> What is the Nyquist sampling rate in the non-uniform case? > > Believe it or not, it's the same as the uniform case ... the number > of samples over the time interval must exceed twice the bandwidth > of the signal. quote ostensibly attributed to me. > That sounds like a sensible conclusion but how do we define bandwidth? > For instance, if you have one peak in the frequency domain, you can > measure the bandwidth as the width of the peak. But what if you have > more then one peek? If they were the same amplitude I would assume you > could add the peaks up. If they were we different amplitudes I would > think you would want to weight them somehow. -- r b-j rbj@audioimagination.com Imagination is more important than knowledge. === Subject: Re: question about non-uniform sampling? <4375FA3F.921CBB9A@acm.org> > 14:21: >> 11/12/2005 09:32: >> >> Can non-uniform sampled signal be used to perfectly reconstruct the >> original continuous time signal? > > Yes, but it isn't easy. > >> What is the Nyquist sampling rate in the non-uniform case? > > Believe it or not, it's the same as the uniform case ... the number > of samples over the time interval must exceed twice the bandwidth > of the signal. > quote ostensibly attributed to me. To me all it looks like is you quoted someone else. And I believe that is all I attributed to you. So I would say it is correct but perhaps unnecessary. === Subject: Re: question about non-uniform sampling? > Hi all, > Can non-uniform sampled signal be used to perfectly reconstruct the > original continuous time signal? > What is the Nyquist sampling rate in the non-uniform case? > -L I read about half the posts here so I hope I am not missing something that is already said. The original post here seems to be about perfectly reconstructing the signal, which would imply perfect sampling. Other posts deal with how practical it is to reconstruct the original signal in terms of error. The problem described by these later posts would be dependent on the type of signal. For instance someone gave the example of trying to sample music every half hour. Another easier problem which was not addressed but similar is only trying to estimate the power spectrum and not the entire signal. Posters have pointed out that the ability to cheat Nyquest depends on how stationary the signal is. A stationary signal is a signal where the Fourier transform is described by a sum of impulse functions (AKA delta functions). As the signal becomes less stationary the bandwidth of these spikes will widen. To the extreme of white noise where the power spectrum is a constant. Computationally, if the signal spectrum is fairly board, I would suggest the singular value based pseudo inverse. If there is a lot of noise in the system then this won't give good result. To see how good a fit is there is something called the Akaike information. It is an information quantity that helps evaluate the tradeoff between the order of fit and the goodness of fit. If there are only a few frequency a better approach would be to use an order recursive method of fitting the signal. With an order recursive approach you can obtain an order 1 fit all the way up to an order N fit with the same amount of computation it takes to do an order N fit. You can use singular value decomposition to see how statically significant each parameter you add to your fit is and you can use the method of leaps and bounds to reduce the amount of fits you need to try. There is a technique of model identification that uses all these techniques called CVA (canonical vector analysis) which I believe was created by Wallace E. Larimore. He crated software the runs in MATLAB that uses these techniques which is called ADAPTx. Sine the software is typically applied to control systems the models identified are typically dynamic models. I assume he published some papers about this technique but has not yet published a book. === Subject: Re: question about non-uniform sampling? > Can non-uniform sampled signal be used to perfectly reconstruct the > original continuous time signal? > Yes, but it isn't easy. > What is the Nyquist sampling rate in the non-uniform case? > Believe it or not, it's the same as the uniform case ... the number > of samples over the time interval must exceed twice the bandwidth > of the signal. > See here (questions 2 and 3) for a little more detail: > http://www.circuitcellar.com/library/eq/136/index.asp > hi Dave, > could you take a look at the paper that Bob Adams did in 1992 that i > reference here: > i haven't cracked your brief analysis, but does that accomplish what > i was hoping would be shown that if your average sample rate is more > than twice the bandwidth, then random sampling will also be sufficient > for reconstruction? Sorry for not replying sooner -- things got kind of busy around here this past week. I can't easily get at the paper itself, but your description in that message leads me to believe that it does. Obviously, the samples have to be unique. I just browsed down through the rest of the thread -- it's amazing causality and how it relates to strict band-limiting. -- Dave Tweed === Subject: Re: question about non-uniform sampling? 11/18/2005 23:41: >> Can non-uniform sampled signal be used to perfectly reconstruct the >> original continuous time signal? > > Yes, but it isn't easy. > >> What is the Nyquist sampling rate in the non-uniform case? > > Believe it or not, it's the same as the uniform case ... the number > of samples over the time interval must exceed twice the bandwidth > of the signal. > > See here (questions 2 and 3) for a little more detail: > http://www.circuitcellar.com/library/eq/136/index.asp >> hi Dave, >> could you take a look at the paper that Bob Adams did in 1992 that i >> reference here: >> i haven't cracked your brief analysis, but does that accomplish what >> i was hoping would be shown that if your average sample rate is more >> than twice the bandwidth, then random sampling will also be sufficient >> for reconstruction? > Sorry for not replying sooner -- things got kind of busy around here > this past week. > I can't easily get at the paper itself, but your description in that > message leads me to believe that it does. Obviously, the samples have > to be unique. > I just browsed down through the rest of the thread -- it's amazing > causality and how it relates to strict band-limiting. well, i generally identify with Jerry both technically and in terms of attitude. i took a brief look at your circuitcellar.com answer and i think i understand it. it seems to me that you need to solve a system of an infinite number of equations that have an infinite number of unknowns, unless you consider a windowed sinc(), and then the theory is not perfect. i think what you did is a generalization of what Bob Adams did (his paper involved the periodic removal of a sample and reconstructing) and, if i can find the paper, i'll post a synopsis of it here. -- r b-j rbj@audioimagination.com Imagination is more important than knowledge. === Subject: Re: question about non-uniform sampling? > it seems to me that you need to solve a system of an infinite > number of equations that have an infinite number of unknowns, No, you only have one equation for each nonuniform sample you took. The solution is the corresponding set (same number) of uniform samples. You have to evaluate the sinc function N^2 times to get the coefficients for the equations. -- Dave Tweed === Subject: Re: question about non-uniform sampling? >> it seems to me that you need to solve a system of an infinite >> number of equations that have an infinite number of unknowns, >No, you only have one equation for each nonuniform sample you took. >The solution is the corresponding set (same number) of uniform samples. >You have to evaluate the sinc function N^2 times to get the coefficients >for the equations. Are you saying that you can exctly reproduce any band-limited signal with finitely many samples? That is not only amazing, but also to tell if you are trying to exactly reproduce the original band you can plug them into the reconstruction equation given previously to reconstruct the original signal. Since you don't say approximately or exactly, it is hard to tell the intent. If you are approximating the original signal, it would be nice to state that and what the error is. Certainly it is possible to construct a band-limited function that agrees with the sampled function at finitely many sampled points. There are infinitely many functions with the same band limit that agree with the sampled function at the same sample points. However, this is different than reconstructing the sampled signal. Rob Johnson take out the trash before replying === Subject: Re: question about non-uniform sampling? ... > I just browsed down through the rest of the thread -- it's amazing > causality and how it relates to strict band-limiting. I didn't inject causality into this thread. Causality has nothing to do with the point I tried to make. I'll try again. A telephone conversation is sampled 8000 times a second, or 480,000 times a minute. A 4-minute monolog uses 1,920,000 samples. If it is required only that this rate be met as an average, how may the samples be distributed in time? Will 1,920,000 samples obtained at any time during the monolog suffice? Suppose the sampler runs for one minute, collecting all 1,920,000 samples in that time; can the entire monolog be reconstructed? Suppose the sampler runs at 16,000 per second in each even second and not at all in the odd ones? How non-uniform can the sampling be and still support reconstruction? Jerry -- Engineering is the art of making what you want from things you can get. .88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N 210N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N21 0N.88N.88N.88N.88N === Subject: Re: question about non-uniform sampling? > How non-uniform can the sampling be and still support reconstruction? It doesn't matter. I won't get drawn into this debate; you've already conceded the point. I just don't understand why you continue to pound away at it. If you're making a different point, you're not making yourself clear enough to the rest of us here. > I'm talking about precisely what you are. A signal that is truly > bandlimited isn't time limited, and vice versa. In the real world, > signals with finite duration can bandlimited well enough so that > we can deal with them. But when one becomes pedantic about what is > theoretically possible, on must be likewise aware of what is > theoretically impossible. But in all the subsequent posts, you never explained exactly what it is you feel is theoretically impossible. You keep talking in allegories about symphonies and forests. Do you understand the dual of what you said above? It goes like this: Signals with finite bandwidth can be time-limited well enough so that we can deal with them. Think about it. -- Dave Tweed === Subject: Re: question about non-uniform sampling? >>How non-uniform can the sampling be and still support reconstruction? > It doesn't matter. > I won't get drawn into this debate; you've already conceded the point. > I just don't understand why you continue to pound away at it. If you're > making a different point, you're not making yourself clear enough to > the rest of us here. >>I'm talking about precisely what you are. A signal that is truly >>bandlimited isn't time limited, and vice versa. In the real world, >>signals with finite duration can bandlimited well enough so that >>we can deal with them. But when one becomes pedantic about what is >>theoretically possible, on must be likewise aware of what is >>theoretically impossible. > But in all the subsequent posts, you never explained exactly what it > is you feel is theoretically impossible. You keep talking in allegories > about symphonies and forests. > Do you understand the dual of what you said above? It goes like this: > Signals with finite bandwidth can be time-limited well enough so that > we can deal with them. Think about it. I understand, Dave. I believe I made that point early on. (Read the approximations are all we can do, and that they are adequate. It is impossible to reconstruct a signal from a set of samples that have large gaps in time, even if the total number of samples is larger than the number of uniformly spaced samples that are enough. Although mathematics tells us that it is possible if the signal is perfectly bandlimited, mathematics also tells us that such perfectly bandlimited signals don't exist. The larger the gaps in the sampling sequence, the worse the reconstruction will be. Where the samples are matters. Jerry -- Engineering is the art of making what you want from things you can get. øøøøøøøø[OS lash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøø === Subject: Re: question about non-uniform sampling? <4375FA3F.921CBB9A@acm.org> <437EAA05.352D9355@acm.org> <437F34E6.61E16836@acm.org> >>How non-uniform can the sampling be and still support reconstruction? ... > It is impossible to reconstruct a signal from a set of samples that have > large gaps in time, even if the total number of samples is larger than > the number of uniformly spaced samples that are enough. Although > mathematics tells us that it is possible if the signal is perfectly > bandlimited, mathematics also tells us that such perfectly bandlimited > signals don't exist. The larger the gaps in the sampling sequence, the > worse the reconstruction will be. Where the samples are matters. We don't get mathematically perfectly bandlimited signals in the real world. However, engineers succeed quite well at resampling signals using imperfectly bandlimited signals. How well one can resample depends on how close the real world bandlimiting gets to the ideal. The better the pre-sampling bandlimit filter, the better one can resample that almost bandlimited signal, including some ability to deal with wider gaps in the uniformity of the sampling. Higher quality bandlimiting filters which allow gap replacement also have a longer delay (before sampling) so that causality is not violated. It's not a binary decision. Your thought experiment fails because you failed to hypothesize a good enough pre-sampling filter. IMHO. YMMV. -- rhn A.T nicholson d.O.t C-o-M === Subject: Re: question about non-uniform sampling? JERRY> It is impossible to reconstruct a signal from a set of samples that have > large gaps in time, even if the total number of samples is larger than > the number of uniformly spaced samples that are enough. Although > mathematics tells us that it is possible if the signal is perfectly > bandlimited, mathematics also tells us that such perfectly bandlimited > signals don't exist. The larger the gaps in the sampling sequence, the > worse the reconstruction will be. Where the samples are matters. RON> We don't get mathematically perfectly bandlimited signals in the > real world. However, engineers succeed quite well at resampling > signals using imperfectly bandlimited signals. The trouble is that as Jerry's sample gap increases in size, the reconstruction of the signal in that gap becomes much more sensitive to sampling filter leakage and other out-of-band noise. The growth in sensitivity is exponential, so improving the sampling filter and quantizer resolution by some given number of dB will only increase your usable gap size by some fixed time. Jerry's examples of impossible stuff with large gaps really are impossible. To get feel for how quickly these things become impossible, try this thought experiment: Imagine a signal, filtered and sampled non-uniformly with a big gap. Now, take the impulse response of your sampling filter, representing a spike in the original singal, and add it to the filtered signal, centered in the gap. How much does that addition change the sample values, relative to its amplitude? How big do you think you can make the gap before those perturbations are so small that you can't reconstruct the difference in the signal? -- Matt === Subject: Re: question about non-uniform sampling? <4375FA3F.921CBB9A@acm.org> <437EAA05.352D9355@acm.org> <437F34E6.61E16836@acm.org> > JERRY> It is impossible to reconstruct a signal from a set of samples that > have > large gaps in time, even if the total number of samples is larger than > the number of uniformly spaced samples that are enough. Although > mathematics tells us that it is possible if the signal is perfectly > bandlimited, mathematics also tells us that such perfectly bandlimited > signals don't exist. The larger the gaps in the sampling sequence, the > worse the reconstruction will be. Where the samples are matters. > RON> We don't get mathematically perfectly bandlimited signals in the > real world. However, engineers succeed quite well at resampling > signals using imperfectly bandlimited signals. > The trouble is that as Jerry's sample gap increases in size, the > reconstruction of the signal in that gap becomes much more sensitive to > sampling filter leakage and other out-of-band noise. The growth in > sensitivity is exponential, so improving the sampling filter and quantizer > resolution by some given number of dB will only increase your usable gap > size by some fixed time. > Jerry's examples of impossible stuff with large gaps really are > impossible. To get feel for how quickly these things become impossible, try > this thought experiment: > Imagine a signal, filtered and sampled non-uniformly with a big gap. > Now, take the impulse response of your sampling filter, representing a spike > in the original singal, and add it to the filtered signal, centered in the > gap. > How much does that addition change the sample values, relative to its > amplitude? How big do you think you can make the gap before those > perturbations are so small that you can't reconstruct the difference in the > signal? I think the gap can be made quite wide before the arithmetic and in the universe, or the filter width even comes close to the length of time since the big-bang. Of course, I am speaking theoretically. I don't actually have a PC with that much memory. But I assume Jerry's example was just a thought experiment. IMHO. YMMV. -- rhn A.T nicholson d.O.t C-o-M === Subject: Re: question about non-uniform sampling? ... > I think the gap can be made quite wide before the arithmetic and > in the universe, or the filter width even comes close to the length > of time since the big-bang. > Of course, I am speaking theoretically. I don't actually have a PC > with that much memory. But I assume Jerry's example was just a > thought experiment. Not exactly. I was trying to show that from a set of real-world non-uniform samples in which the nonuniformity (i.e., the ratio of the longest interval to the shortest) exceeds a fairly small number, you can't satisfactorily deduce the underlying continuous function. Jerry -- Engineering is the art of making what you want from things you can get. .88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N 210N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N.88N21 0N.88N.88N.88N.88N === Subject: Re: question about non-uniform sampling? >>How non-uniform can the sampling be and still support reconstruction? > ... >>It is impossible to reconstruct a signal from a set of samples that have >>large gaps in time, even if the total number of samples is larger than >>the number of uniformly spaced samples that are enough. Although >>mathematics tells us that it is possible if the signal is perfectly >>bandlimited, mathematics also tells us that such perfectly bandlimited >>signals don't exist. The larger the gaps in the sampling sequence, the >>worse the reconstruction will be. Where the samples are matters. > We don't get mathematically perfectly bandlimited signals in the > real world. However, engineers succeed quite well at resampling > signals using imperfectly bandlimited signals. How well one can > resample depends on how close the real world bandlimiting gets to > the ideal. The better the pre-sampling bandlimit filter, the better > one can resample that almost bandlimited signal, including some > ability to deal with wider gaps in the uniformity of the sampling. > Higher quality bandlimiting filters which allow gap replacement > also have a longer delay (before sampling) so that causality is > not violated. > It's not a binary decision. > Your thought experiment fails because you failed to hypothesize > a good enough pre-sampling filter. OK, do it then. Sample a signal uniformly at four times the prescribed Nyquist rate, then discard a contiguous group consisting of half of them. There remains twice the minimum number needed for reconstruction. Reconstruct. Display the result and the original. When I see (or hear) it, I'll believe that the locations of the samples in time can be arbitrary. Jerry -- Engineering is the art of making what you want from things you can get. øøøøøøøø[OS lash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøøøøøøø[OSl ash]øøøøøøøø øøø === Subject: Re: question about non-uniform sampling? <4375FA3F.921CBB9A@acm.org> <437EAA05.352D9355@acm.org> <437F34E6.61E16836@acm.org> >> > > >>How non-uniform can the sampling be and still support reconstruction? > ... >>It is impossible to reconstruct a signal from a set of samples that have >>large gaps in time, even if the total number of samples is larger than >>the number of uniformly spaced samples that are enough. Although >>mathematics tells us that it is possible if the signal is perfectly >>bandlimited, mathematics also tells us that such perfectly bandlimited >>signals don't exist. The larger the gaps in the sampling sequence, the >>worse the reconstruction will be. Where the samples are matters. > We don't get mathematically perfectly bandlimited signals in the > real world. However, engineers succeed quite well at resampling > signals using imperfectly bandlimited signals. How well one can > resample depends on how close the real world bandlimiting gets to > the ideal. The better the pre-sampling bandlimit filter, the better > one can resample that almost bandlimited signal, including some > ability to deal with wider gaps in the uniformity of the sampling. > Higher quality bandlimiting filters which allow gap replacement > also have a longer delay (before sampling) so that causality is > not violated. > It's not a binary decision. > Your thought experiment fails because you failed to hypothesize > a good enough pre-sampling filter. > OK, do it then. Sample a signal uniformly at four times the prescribed > Nyquist rate, then discard a contiguous group consisting of half of > them. There remains twice the minimum number needed for reconstruction. > Reconstruct. Display the result and the original. When I see (or hear) > it, I'll believe that the locations of the samples in time can be arbitrary. Take 8 samples of 1 second of a 1 Hz sine signal plus DC. Throw away the last 4 samples. Find any reconstruction using just those 4 samples which has absolutly no frequency content above 1 Hz when repeating those 4 samples every 1 second over an infinite time extent. Use closet fit if needed to overcome quantization error. Take any 8 contiguous samples for 1 second at an 8 Hz sampling rate of any signal. Perfectly filter those 8 samples so that they contain no content above 1 Hz (using either an FFT/IFFT or a symmetric FIR of width much greater than 16 samples). Call that signal X. Continue as above to reconstruct X from any 4 samples, including only the first 4 samples Repeat for any N*8 samples plus perfect prefiltering to below N Hz before discarding N*4 samples. etc. Use an arbitrary precision math package so that the tails of a FIR prefilter or FFT are never discarded in quantization noise before reconstruction of the 0 to N Hz signal. As N doubles, increase the precision of the of the coefficients and filtered samples as required. Increase N to 4 * 44100 and time from 1 second to the length of a symphony. Filter. Throw away the last half of the samples after perfect filtering. Reconstruct. The number of bits of precision required is left as an exercise for the student. IMHO. YMMV. -- rhn A.T nicholson d.O.t C-o-M === Subject: math help The million dollar getaway Main bank was robbed this morning. Videotape from the security camera shows a lone robber carrying the loot out of the building in a big leather bag. The teller agrees that only one robber was involved. The bank manager says the robber got away with approximately $1 million dollars in small bills. Most of the bills were ones, fives, and tens. The editor for the local news is wondering if this is possible. 1. How difficult would it be for a single robber to carry so much money in small bills? 2. Analyze this situation to determine what might be possible. 3. Present your results and your conclusions in a report that will help the editor decide what to say about the robbery for tonight's broadcast. Explain what math concepts you learned in this problem? 4. Describe in detail the solution path you took to solve this problem. 5. Give examples of ways this problem could be extended? I was given this sheet to help me figure it out in Excel with formulas: Please use this chart below with the formulas place in it to solve this math question. Make at least three different charts of to This activity explores whether one million dollars will fit into a standard suitcase. If so, how large would the suitcase need to be? How heavy would it be? Involve students in formulating and exploring questions to investigate the truth of this claim. For example: á Can $1,000,000 in one-dollar bills fit in a standard-sized suitcase? If not, what is the smallest denomination of bills you could use to fit the money in a suitcase? á Could you lift the suitcase if it contained $1,000,000 in one-dollar bills? Estimate its weight. Note: The dimensions of a one-dollar bill are approximately 6 inches by 2.5 inches. Twenty one-dollar bills weigh approximately 0.7 ounces. You may wish for students to locate these facts about dollar bills on their own, using internet or other appropriate resources. The students will also need to determine the dimensions of a standard suitcase. === Subject: Re: math help > The million dollar getaway > Main bank was robbed this morning. Videotape from the security camera > shows a lone robber carrying the loot out of the building in a big > leather bag. The teller agrees that only one robber was involved. The > bank manager says the robber got away with approximately $1 million > dollars in small bills. Most of the bills were ones, fives, and tens. > The editor for the local news is wondering if this is possible. > 1. How difficult would it be for a single robber to carry so much money > in small bills? > 2. Analyze this situation to determine what might be possible. One of the one-dollar bills has George Washington slightly off center. It's an extremely rare bill; easily regarded by collectors as being worth one million dollars. That's the only bill the robber took. She didn't need a big leather bag, she just put it in her little purse and walked out. === Subject: Re: math help On Sun, 20 Nov 2005 16:35:36 -0800, fishfry >> The million dollar getaway >> Main bank was robbed this morning. Videotape from the security camera >> shows a lone robber carrying the loot out of the building in a big >> leather bag. The teller agrees that only one robber was involved. The >> bank manager says the robber got away with approximately $1 million >> dollars in small bills. Most of the bills were ones, fives, and tens. >> The editor for the local news is wondering if this is possible. >> 1. How difficult would it be for a single robber to carry so much money >> in small bills? >> 2. Analyze this situation to determine what might be possible. >One of the one-dollar bills has George Washington slightly off center. >It's an extremely rare bill; easily regarded by collectors as being >worth one million dollars. >That's the only bill the robber took. She didn't need a big leather bag, >she just put it in her little purse and walked out. haha === Subject: Re: math help On 20 Nov 2005 16:05:35 -0800, ashmangat@yahoo.com >The million dollar getaway >Main bank was robbed this morning. Videotape from the security camera >shows a lone robber carrying the loot out of the building in a big >leather bag. The teller agrees that only one robber was involved. The >bank manager says the robber got away with approximately $1 million >dollars in small bills. Most of the bills were ones, fives, and tens. >The editor for the local news is wondering if this is possible. >1. How difficult would it be for a single robber to carry so much money >in small bills? >2. Analyze this situation to determine what might be possible. >3. Present your results and your conclusions in a report that will help >the editor decide what to say about the robbery for tonight's >broadcast. Explain what math concepts you learned in this problem? This teaches you -- if you going to rob a bank, either demand larger denominations or bring a bigger suitcase. === Subject: Good book (on/offline) on WAVE equation with VARIABLE coefficients. I understand standard PDEs with constant coefficients pretty well. I now want to better understand them with VARIABLE coefficients. Is there a good book? For example, I vaguely remember that if the propagation speed varies as 1/x (or similar), then there is no reflection. What is the solution? Also, are there Green's-function-like solutions that deal with reflection when the speed v varies? === Subject: optimal policy There are two binary {0, 1} stochastic processes X1(nT) and X2(nT) (n=0, 1, 2...) where value changes only at the edge of each time slot. Moreover, assuming each stochastic process can be modeled as a 2-state Markov Chain problem with parameters (pi1, pi2) (cross-over transition probabilities and i = 1, 2 for each Markov Chain). My purpose is to design a probabilistic policy to choose either X1 or X2 (at the boundary of the time slot) such that the probability I got the value of 1 is maximized. My question is that what kind of statistic information should I know to design such a policy? Is there any known problem close to it? or Is there any theory just for such kind of problem? Any comment is welcome! I really appreciate! === Subject: Re: optimal policy > There are two binary {0, 1} stochastic processes X1(nT) and X2(nT) > (n=0, 1, 2...) where value changes only at the edge of each time slot. > Moreover, assuming each stochastic process can be modeled as a 2-state > Markov Chain problem with parameters (pi1, pi2) (cross-over transition > probabilities and i = 1, 2 for each Markov Chain). > My purpose is to design a probabilistic policy to choose either X1 or > X2 (at the boundary of the time slot) such that the probability I got > the value of 1 is maximized. > My question is that what kind of statistic information should I know to > design such a policy? Is there any known problem close to it? or Is > there any theory just for such kind of problem? > Any comment is welcome! I really appreciate! I think you need a better definiton of the problem. First, some questions: (1) Are the processes completely observable; that is, if the system is in state 1, can you know this for sure; and if you are using process 1, do you know it? (2) Do you know the initial state at n=0? Assuming the answers to question (1) are both YES, you still need to better define the problem. Assume the answer to question (2) is also YES. So, are you choosing, at time n=0, a policy that maximizes the probability the system will be in state 1 at time n=1? Why can't a policy that maximizes the state-1 prob. at n=1 be a poor one for achieving state-1 at time n=2? Don't you somehow need to consider something like the time-average probability of being in state-1, or its discounted value, or something like that? I think that with the appropreate re-definition of the problem, it becomes a standard Markov Decision problem. R.G. Vickson Adjunct Professor, University of Waterloo === Subject: Volume without volume 'Hooft-Susskind' It's well known that for constant magnitude of ODLRO VEV in V = G(disordered)/H(ordered) coset space: Zero homotopy group non-trivial has 2D domain wall defects surrounding space of defect is S0 i.e. points -1, +1. First homotopy group non-trivial has 1D string (vortices) with S1 loop surrounding space and G/H = S1 also. Second homotopy group non-trivial has 0D point defect monopole (not necessarily magnetic, the geometrodynamic neutral monopole gives S/k = AREA/4Lp^2 + World Hologram + maybe NASA Pioneer Anomaly as dark energy hedgehog with wrapping number -1 defect in center of Sun?) with surrounding space = S2 and also G/H = S2. Third homotopy group non-trivial has texture defect with G/H = S3. Internal symmetries have no non-trivial tetrad valued in the Lie algebra of the locally gauged symmetry group. Space-time symmetries DO and that is the DIFFERENCE from the EQUIVALENCE PRINCIPLE AKA EEP. 2 independent ideas here 1. Local gauging 2. Spontaneous vacuum symmetry breaking. In my theory of emergent gravity i.e. my version of Andrei Sakharov's metric elasticity I. I SPONTANEOUS BREAK INTERNAL SYMMETRY at Planck scale -> GUT vacuum phase transition II. I locally gauge at least 10-parameter Poincare group - ultimately 16 parameter GL(4,R). Note HOW TO GET GENNADY SHIPOV's 10 MANIFOLD (same as string theory) is to put the entire Lie algebra of the Poincare group into the TETRAD FIELD directly. This makes it a Kaluza-Klein theory. However, I DO NOT DO THIS - but I can later. I keep the tetrad field 4D from T4 and that's why I need the indirect quantities such as B^I a 1-index 1-form curved tetrad from locally gauging T4 e^I = 1^I + B^I W^IJ a 2-index 1-form CURVATURE spin-CONNECTION in D = d + W^IJ/ And also S^I a 1-index 1-form in the 4D tetrad from locally gauging O(1,3) e'^I = e^I + S^I Z^IJ a 2-index 1-form TORSION CONNECTION in D' = D + Z^IJ/ Where we have the 2-forms T^I = De = 0 R^IJ = DR^IJ T'I = D'e' R'IJ = D'(W^IJ + Z^IJ) With Bianchi identities (analog of dF = 0 of Maxwell internal U(1) EM field in flat space-time without gravity & torsion) DR^IJ = 0 D'R'^IJ = 0 D'T'^I = 0 Source equations D'*W^IK = *J'(T4)^IK D'*Z^IK = *J(O(1,3)^IK D'*T^I = *j(O(1,3)^I Local covariant current density conservation D'*J'(T4)^IK = 0 D'*J(O(1,3))^IK = 0 D'*j(O(1,3)^I = 0 OK, vortices have degenerate ODLRO macro-quantum manifold V = G/H of local order parameters with topology of the circle. That is we hold the Higgs amplitude fixed and we only have ONE independent Goldstone phase factor e^iTheta where the order parameter has 2 real scalar components. Non-integrable Goldstone phase factors e^iQ^a(Theta)a, for Lie algebra [Q^a,Q^b] = f^a^bcQ^c give both the global and local properties of Yang-Mills internal gauge theories as well as gravity and torsion & non-metricity, dilaton gauge theories both classical and quantum via Feynman path integrals. Point defects has V = S2 with 2 independent Goldstone phase factors e^iTheta and e^iPhi, i.e. polar and azimuthal angles in V = G/H space where the order parameter has 3 real scalar components. The geometrodynamical fields. I will only do T4 for simplicity. CASE 1 V = G/H = S1 with ONE GOLDSTONE PHASE Theta that sweeps out a unit circle. B^I(S1) = (LpP^I(Theta)/ih Tetrad e^I = 1^I + B^I Einstein invariant is ds^2 = B^I(Minkowski)IJB^J QED! Here we expect STRING DEFECTS! In simpler completely invariant notation B(S1) = Lpd(Theta) Case 2 V = G/H = S2 with TWO INDEPENDENT GOLDSTONE PHASES Theta & Phi that sweep out a unit 2D spherical surface that is G/H topology. So what do we do? Simple! I was a DUMMY not to see this weeks ago! But I woke up this morning hundreds of miles north of SF on the beach with the idea crystal clear! Define the 0-form (Theta)(Phi) these are local functions of Einstein's local coincidences P of course. Our new geometrodynamic 1-form is obviously via product rule of Newton's fluxions, i.e. ghosts of departed quantities (Bishop Berkeley) B^I(S2) = (LpP^I(ThetaPhi)/ih i.e. 2 -terms B^I(S2) = (Lp(P^I(Theta)/ih)Phi + Theta(LpP^I(Phi)/ih = LENGTH OPERATOR i.e. in completely invariant notation B(S2) = Lp{d(Theta)Phi + Thetad(Phi)} using d^2 = 0 We have the 2-form AREA OPERATOR (forget Loop Quantum Gravity & Spin Foams!) Area = Lp^2dTheta/dPhi We also have Closed surface integral of Area operator = Volume integral of d(Area) = Wrapping Number when the closed surface surrounds the point defect since second homotopy group for S2 is Z This is obviously essentially both Hawking-Bekenstein BITS and 't Hooft-Susskind hologram idea that is simply VOLUME WITHOUT VOLUME as an example of Bohm-Aharonov's Flux without flux in which the gauge potentials have direct non-classical physical effects on phenomena. We can continue this to case G/H = S3 (textures) where now we have 3 independent Goldstone phases Theta, Phi & Chi with a VOLUME OPERATOR. If we continue into higher dimensional hyperspace, I suppose V = G/H are what? Calabi-Yau spaces? I will check. Here we need to extend the EEP tetrad idea missing in conventional internal symmetry Yang-Mills gauge force theories. Basically we have S1xS1x... for possible V = G/H broken vacuum symmetries. === Subject: Re: Volume without volume 'Hooft-Susskind' > It's well known that for constant magnitude of ODLRO VEV in > V = G(disordered)/H(ordered) coset space: > Zero homotopy group non-trivial has 2D domain wall defects > surrounding space of defect is S0 i.e. points -1, +1. > First homotopy group non-trivial has 1D string (vortices) > with S1 loop surrounding space and G/H = S1 also. > [...] > VOLUME WITHOUT VOLUME as an example of Bohm-Aharonov's Flux without > flux in which the gauge potentials have direct non-classical physical > effects on phenomena. > We can continue this to case > G/H = S3 (textures) where now we have 3 independent Goldstone phases > Theta, Phi & Chi with a VOLUME OPERATOR. > If we continue into higher dimensional hyperspace, I suppose V = G/H are > what? Calabi-Yau spaces? I will check. > Here we need to extend the EEP tetrad idea missing in conventional > internal symmetry Yang-Mills gauge force theories. > Basically we have S1xS1x... for possible V = G/H broken vacuum symmetries. I'd love to know if any of this makes sense. It sounds vaguely like my idea of localized broken symmetries of a sea of torus branes in phase space, or some kind of abstract space, whereby they polarize (line up parallel), elongate, and wrap themselves round a torus and join to form a larger-scale torus by foliation. Their dynamic evolutions, which are to maximize the mutual parallelism of branes, are constrained only by the requirement of an everywhere constant moduli field. In particular, as they approach parallelism there is some kind of flow (parallelism evolution of neighboring smaller-scale branes - this is a recursive model) which increases in intensity until blocked by conflicting flows nearby. Spacetime is an emergent property of the relative orientations of branes of the predominant brane size at any given scale. Possibly this brane orientation aspect can be formalized by symplectic geometry, and spacetime somehow corresponds to an associated cotangent bundle. I'm still looking into this (and learning a lot of differential geometry, even if the whole idea is a load of tosh ;-P) It makes clear, to me at any rate, why Einstein's later attempts at a Unified Field theory foundered, as did his ideas on a concept called teleparallelism: Considering smooth manifolds gets one only so far. Instead, my branes have an infinitely complex banded structure because each sub-brane forming a foliation, being itself a stretched and reconnected (at the poles) torus is also a foliation of sub-sub-branes, and is thus banded by the latter. So, assuming the sub-brane foliation makes at least one twist round the major axis of its torus, this requires (for local parallelism at the sub-brane level within the brane foliation) sub-brane strings to be arranged with sub-brane bands alternating. In turn this is neatly provided by the fact that a stretched and reconnected torus becomes two tori with opposite bandings. In 3D these are nested, and cannot cross without causing havoc with sub-parallelism; but presumably in higher dimensions they can float apart without crossing. Three aligned strings each with one of two possible mirror-image bandings cannot be joined in a solid 3D bundle, any more than three cog wheels can move compatibliy in mutual triangular contact. So branes should always be 2D structures even in higher dimensions. This perhaps accounts for the recently expressed idea that our Universe is fundamentally 2D and only masquerades as 3D (the hologram principle). All sub-brane bands in any given foliation must obviously have the same winding number. But assuming all this goes on in 4D or some higher dimension, a limiting parallel structure can comprise a whole sheaf (using that word in a generic sense) of nested toroidal branes, which can form without sub-strings having to cross, and each with a different foliation winding number and presumably (to maximize overall parallelism between layers) each rotating relative to the others like turbine blades and with high-energy tangential moduli flows sandwiched between. So these beasties have a rich topological and dynamic structure. In the past, when I read about theories of branes in higher dimensions, in particular the idea that two 5D branes could hover close to each other and cause a Big Bang by colliding, it seemed ridiculously contrived - Why would they bother to hover so close and yet rarely touch, and what are the chances of them staying so parallel? But with a parallelism principle it makes perfect sense. (I don't buy the 5D colliding brane idea though - It seems far more likely to me that the Big Bang was the result of a mass-inflated toroidal black hole singularity disrupted assymetrically by evaporation of its containing black hole due to accelerated expansion of the universe containing it. This accounts in general terms for the early inflation epoch, (which has now been established moreorless beyond all doubt), because in my model moduli flows are at their most intense when branes are _almost_ parallel.) The brane parallelism model also indicates that the expansion of our (visible and causally connected) Universe will continue and accelerate, as sub-branes, corresponding to the spacetime vacuum, slowly but surely increase their overall relative parallelness. The limiting state, one assumes, would be an infinitely large hypersphere in which branes are elongated and aligned radially. But the good news is that there are plenty of universes in waiting inside black holes, ready to burst forth in new Big Bangs (as explained above) when the expansion rate of our universe increases sufficiently to make its causal horizon shrink down to black hole size. John R Ramsden (jhnrmsdn@yahoo.com.uk) * Remove m from com to reply === Subject: Re: Volume without volume 'Hooft-Susskind' welcome to the kill file Living in a Fantasy world will kill you - HeavenGate * PLONK * === Subject: notation For a hw question for abtract algebra, something is written as <(2,2)>. Does this notation mean all even numbers in the direct product? === Subject: Re: notation >For a hw question for abtract algebra, something is written as <(2,2)>. Does this notation mean all even numbers in the direct product? That notation could mean lots of things. Don't you think you should give us the benefit of seeing the complete problem? Ok, so let's see the actual problem, correctly stated. quasi === Subject: Re: notation >>For a hw question for abtract algebra, something is written as <(2,2)>. Does this notation mean all even numbers in the direct product? >That notation could mean lots of things. >Don't you think you should give us the benefit of seeing the complete >problem? >Ok, so let's see the actual problem, correctly stated. >quasi Direct product of what? Are we talking direct product of groups, or direct product of rings? If then, after you answer that, what are the factors? If I had to guess, I might assume Z x Z but why should I have to guess? Besides, I still wouldn't know if you're talking groups or rings. Also, your phrase even numbers in the direct product is probably meaningless. Does the direct product you are analyzing even have numbers as elements? quasi === Subject: Re: notation The (ZxZ) denotes the external direct product where Z is the set of all integers. === Subject: Re: notation Accidentaly clicked post before writing product of a group. The (ZxZ) denotes the external direct product of a group where Z is the set of all integers. === Subject: Re: notation The question was to determine the order of (ZxZ)/<(2,2>). x denotes the direct product. And then it asked to determine if it was cyclic. === Subject: Re: notation >The question was to determine the order of (ZxZ)/<(2,2>). >x denotes the direct product. And then it asked to determine if it was cyclic. Ok, so that's makes it clearer as to what is meant. Presumably we are viewing Z as an abelian group under addition. ZxZ denotes the direct product (or direct sum) of 2 copies of Z. (2,2) is an ordered pair, and element of ZxZ. <(2,2)> is the subgroup of ZxZ generated by the element (2,2). The HW problem asks about the quotient group ZxZ/<(2,2)>. Ok, so now that it's clear what you're talking about, let's consider your original question. Your original question is essentially asking what are the elements in the subgroup <(2,2)>. Ok, try generating some elements of <(2,2)>. After that, see if you can specify all of them in some general way. Characterizing all of them is not hard, but it would help to see a few first. For any group G, the group is just the cyclic subgroup generated by g. Thus, = {g^n | n in Z}. For an additive group, you could write it as = {ng | n in Z}. As far as the HW problem, try to see what the cosets are, how many, and how do they multiply. Alternatively, see which elements are equivalent mod <(2,2)>, and then choose distinct representatives to be regarded as representative elements in the quotient group. Then see how those representatives multiply. quasi === Subject: Re: notation >>The question was to determine the order of (ZxZ)/<(2,2>). >>x denotes the direct product. And then it asked to determine if it was cyclic. >Ok, so that's makes it clearer as to what is meant. >Presumably we are viewing Z as an abelian group under addition. >ZxZ denotes the direct product (or direct sum) of 2 copies of Z. >(2,2) is an ordered pair, and element of ZxZ. ><(2,2)> is the subgroup of ZxZ generated by the element (2,2). >The HW problem asks about the quotient group ZxZ/<(2,2)>. >Ok, so now that it's clear what you're talking about, let's consider >your original question. >Your original question is essentially asking what are the elements in >the subgroup <(2,2)>. >Ok, try generating some elements of <(2,2)>. After that, see if you >can specify all of them in some general way. Characterizing all of >them is not hard, but it would help to see a few first. >For any group G, the group is just the cyclic subgroup generated >by g. >Thus, = {g^n | n in Z}. >For an additive group, you could write it as = {ng | n in Z}. >As far as the HW problem, try to see what the cosets are, how many, >and how do they multiply. Alternatively, see which elements are >equivalent mod <(2,2)>, and then choose distinct representatives to be >regarded as representative elements in the quotient group. Then see >how those representatives multiply. >quasi Of course, after you get a sense of what the quotient group is like, you can solve the HW problem in a slightly more sophisticated way by defining a homorphism from ZxZ to some other group G, then showing that the homomorphism is surjective and that the kernel is <(2,2)>. That would prove that ZxZ/<(2,2)> is isomorphic to G. The whole idea is that the group G would be a group for which you can immediately answer questions like what is the order of G, or is G cyclic. Having shown that the quotient group is isomorphic to G, any questions about the quotient group can be answered by asking them instead about G. quasi === Subject: Re: notation > On Sun, 20 Nov 2005 23:51:57 -0500, quasi >On Sun, 20 Nov 2005 23:18:44 EST, Eric >>The question was to determine the order of > (ZxZ)/<(2,2>). >>x denotes the direct product. And then it asked to > determine if it was cyclic. >Ok, so that's makes it clearer as to what is meant. >Presumably we are viewing Z as an abelian group > under addition. >ZxZ denotes the direct product (or direct sum) of 2 > copies of Z. >(2,2) is an ordered pair, and element of ZxZ. ><(2,2)> is the subgroup of ZxZ generated by the > element (2,2). >The HW problem asks about the quotient group > ZxZ/<(2,2)>. >Ok, so now that it's clear what you're talking > about, let's consider >your original question. >Your original question is essentially asking what > are the elements in >the subgroup <(2,2)>. >Ok, try generating some elements of <(2,2)>. After > that, see if you >can specify all of them in some general way. > Characterizing all of >them is not hard, but it would help to see a few > first. >For any group G, the group is just the cyclic > subgroup generated >by g. >Thus, = {g^n | n in Z}. >For an additive group, you could write it as = > {ng | n in Z}. >As far as the HW problem, try to see what the cosets > are, how many, >and how do they multiply. Alternatively, see which > elements are >equivalent mod <(2,2)>, and then choose distinct > representatives to be >regarded as representative elements in the quotient > group. Then see >how those representatives multiply. In this question, is <(2,2)> just {(2,2),(4,4),...(2n,2n)} for all n in Z. Is the order 2 then since everything is either (1,1) or (2,2) mod (2,2)? === Subject: Re: notation [ Question is about (Z x Z)/<(2,2)> ] : In this question, is <(2,2)> just {(2,2),(4,4),...(2n,2n)} for all n in Z. Yes. : Is the order 2 then since everything is either (1,1) or (2,2) mod (2,2)? No, everything is not (1,1) or (2,2). For example: (1,3) is not equivalent to (1,1) mod <(2,2)>, since (1,3)-(1,1) = (0,2) which is not in <(2,2)>. Ted === Subject: Send unlimited postcard for free while travel Send unlimited postcard for free while travel www.myepostcard.com Traditional postcard, it's possible in electronic === Subject: Re: Send unlimited postcard for free while travel > Send unlimited postcard for free while travel [snipped] Write your address as the return address, don't put a stamp on it. Or did I miss something? Nicholas Sherlock === Subject: Re: Operations Research - shadow price & marginalvalues reduced-size problem without spending a bit of time writing it out. I've sort of figure out what I was asking initially and I'll read some text on basic OR to get a better handle of this topic. === Subject: when does induction fail? well-ordering property? Hi all, My understanding of induction proof goes as follows: 1. First prove P(1) true; 2. Prove P(n) implies P(n+1) for any n in set of Natural numbers ... I vaguely heard that it works for finite number of cases, and works for countably infinitely many of cases; and it does not work for uncoutably infinitely many of cases, because the n is in the Natural numbers set... Is my understanding right? -------------------- How does it related to the well-ordering property? What is well-ordering property exactly? To me, it sounds like just another version of a set has a infnimum inside this set is called to have the well-ordering property... or a set has a minimum is defined to have the well-ordering property... am I right? The well-ordering thoerem says that any set can be made to possess the well-ordering property... is that true? === Subject: Re: when does induction fail? well-ordering property? > Hi all, > My understanding of induction proof goes as follows: > 1. First prove P(1) true; > 2. Prove P(n) implies P(n+1) for any n in set of Natural numbers ... This is induction as applied to the naturals in their usual ordering (which is a well-ordering). You can use a slightly better form: (2) Prove P(k) for all k I vaguely heard that it works for finite number of cases, and works > for countably infinitely many of cases; and it does not work for > uncoutably infinitely many of cases, because the n is in the Natural > numbers set... > Is my understanding right? Sort of... see below. > -------------------- > How does it related to the well-ordering property? > What is well-ordering property exactly? To me, it sounds like just > another version of a set has a infnimum inside this set is called to > have the well-ordering property... or a set has a minimum is defined > to have the well-ordering property... am I right? > The well-ordering thoerem says that any set can be made to possess the > well-ordering property... is that true? If you assume the axiom of choice, yes. It's one of the odd results that comes out of that assumption, but since it seems to be necessary to make a bunch of things work in general, most people do accept it. Logicians argue about it, everyone else quietly uses the axiom of choice as necessary and hopes the logicians will make the whole issue go away sometime, or at least not cause it to blow up in our faces. I'm safe, I don't need it for what I do anyway. So the way it relates is, if you have a set and a well-ordering on that set, then you can reformulate the principle of induction as: 1) Prove P(a) where a is the first element in the well-ordering. 2) Prove that P(t) for all tSo the way it relates is, if you have a set and a well-ordering on that set, >then you can reformulate the principle of induction as: >1) Prove P(a) where a is the first element in the well-ordering. >2) Prove that P(t) for all t P(t))) --> P(s) has to be done in cases, depending on whether the set {t : t My understanding of induction proof goes as follows: > 1. First prove P(1) true; > 2. Prove P(n) implies P(n+1) for any n in set of Natural numbers ... > I vaguely heard that it works for finite number of cases, and works for > countably infinitely many of cases; and it does not work for uncountably > infinitely many of cases, because the n is in the Natural numbers > set... > Is my understanding right? Understanding of what? What works? Be clear and precise or nothing mathematical will be understood. Induction Induction is (1) and (2) implies for all n in N, P(n) where N is the natural numbers or positive integers. > How does it related to the well-ordering property? > What is well-ordering property exactly? To me, it sounds like just > another version of a set has a infnimum inside this set is called to > have the well-ordering property... or a set has a minimum is defined > to have the well-ordering property... am I right? A set is well ordered when every nonnul subset has a least element. Traditionally, well ordered also included totally ordered. If N is well ordered, then Induction holds. Visa versa, if Induction holds, then N is well ordered. > The well-ordering theorem says that any set can be made to possess the > well-ordering property... is that true? No, it's an axiom of much dispute. The Axiom of Choice, AxC or AC has even become a paramount issue in politics. Should the Axiom of Choice be taught in schools? Yea say prochoice mathematicians. Nay, say the constructionist, viewed upon some as obstructionists, intelligent choice should be taught. Others think intelligent choice should be taught in sex education. Perhaps they're right, have you ever been bother or haunted by an unintelligent choice? However the axiom of choice isn't just about you or me making a choice or two, it's about us simultaneously making infinitely many choices. This bothers some, as for example, with it you can cut a sphere into a few pieces and rearrange them into another sphere with a different surface area. === Subject: Re: when does induction fail? well-ordering property? > Hi all, > My understanding of induction proof goes as follows: > 1. First prove P(1) true; > 2. Prove P(n) implies P(n+1) for any n in set of Natural numbers ... The inductin allows one to conclude that P(n) is true for all naturals. > I vaguely heard that it works for finite number of cases, and works for > countably infinitely many of cases; and it does not work for uncoutably > infinitely many of cases, because the n is in the Natural numbers > set... > Is my understanding right? I think so. The principle, or axiom, of induction states that when your (1) and (2) above are true then P(n) is true for all naturals, n. > -------------------- > How does it related to the well-ordering property? > What is well-ordering property exactly? A set is well ordered when every non-empty subset has a first element relative to that ordering. Every finite ordered set is well ordered by any ordering. The set of naturals are well ordered by the usual ordering. There are other well-orderings of countable sets. For example, if one splits the naturals into evens and odds, keeping the normal ordering within each but reordering to put all odds before all evens, the result is still well ordered but the first even has no immediate predecessor in the new ordering. > To me, it sounds like just > another version of a set has a infnimum inside this set is called to > have the well-ordering property... or a set has a minimum is defined > to have the well-ordering property... am I right? Not quite. if every non-empty subset of an ordered set then it is well ordered. > The well-ordering thoerem says that any set can be made to possess the > well-ordering property... is that true? That every set can be well ordered is not a theorem in most set theories it is not a theorem, but is an axiom in some of them. One can even build set theories in which it is false. === Subject: Re: when does induction fail? well-ordering property? Well-ordering leads to problems, for example in well-ordering the reals. A well-order is a binary relation, that is in computer talk a function with two arguments, that returns less than, or greater than or equal. It might instead admit greater than, or less than or equal. Then, also there exists an element for the set that is less than, for any other element of the set. The symbol for that ordering is similar to the < angle bracket, but the sides of the two legs of the triangle are curved in, concave, say <{ in ASCII, it's a different symbol to describe less than in terms besides the normal ordering, of for example numbers. http://en.wikipedia.org/wiki/ASCII In ZFC, the Zermelo Fraenkel, for Ernst Zermelo and A.A. Fraenkel, set theory with the axiom of Choice or AC, ZF+C, or ZFC, a theorem of that axiom set is that any set is well-orderable, that there exists a choice function for a set, and for any subset of the set. The choice function basically returns the least element in a well-ordering for the set. Now, imagine the open interval of the reals (0,1). The endpoints would be zero and one but that's an open interval so they're not elements of the set of each real number r for 0 < r < 1. While that is so, the interval has two endpoints. How can that endpoint be addressed? If there's a well-ordering of the reals, then for some subset of R(0,1), the choice functions returns an endpoint. If it doesn't, then it's not a well-ordering of the reals, because for no subset would it be the least element. The element exists, because it's a complete ordered field. To thus address it demands another way of looking at the reals, which I consider to be similar to Leibniz's infinitesimals, integral iota-multiples, as well being the complete ordered field, Tim's pseudo-reals, R-bar-umlaut, dually complete ordered field and partially ordered ring, in the real numbers. I can tell you, that's _not_ a standard viewpoint, and many do not accept it. Care, I do not. Ross === Subject: Re: when does induction fail? well-ordering property? > Well-ordering leads to problems, for example in well-ordering the > reals. > A well-order is a binary relation, that is in computer talk a function > with two arguments, that returns less than, or greater than or equal. > It might instead admit greater than, or less than or equal. Then, also > there exists an element for the set that is less than, for any other > element of the set. You clearly do not know what a well-ordering is, that is the definition Tony Orlow uses, who also does not know it. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: when does induction fail? well-ordering property? > > Well-ordering leads to problems, for example in well-ordering the > > reals. > > > > A well-order is a binary relation, that is in computer talk a function > > with two arguments, that returns less than, or greater than or equal. > > It might instead admit greater than, or less than or equal. Then, also > > there exists an element for the set that is less than, for any other > > element of the set. > You clearly do not know what a well-ordering is, that is the definition > Tony Orlow uses, who also does not know it. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ I'll dispute that. Not that I don't mind Tony, I find Tony refreshing, but that's a property of a well-ordering, as applied a well-ordered set, it has a least element. That applies to non-empty sets, which was an omission. The existence of a well-ordering on a set implies the existence of a choice function that given the set as an argument returns the unique least element of the set by the well-ordering. Why do you not agree with that? I have not been following Tony's definition of what a well-ordering is. Maybe you are thinking of a total ordering. - Tapio Hurme The existence of a well-ordering is convenient for quantifying over the elements of a set, for example for purposes of induction. Ross === Subject: Need help 1. What is the sum of: 1/3 +(1/3)2+(1/3)3+(1/3)4+.... +(1/3) n +....= Please explain clearly how you got the answer step by step what === Subject: Need help 1. What is the sum of: 1/3 +(1/3)2+(1/3)3+(1/3)4+.... +(1/3) n +....= Please explain clearly how you got the answer step by step what === Subject: Re: Need help > 1. What is the sum of: > 1/3 +(1/3)2+(1/3)3+(1/3)4+.... +(1/3) n +....= > Please explain clearly how you got the answer step by step what If your (1/3)2 means 1/3 squared, the usual notation for exponents uses the caret symbol, ^, so that x squared is written x^2, rather than x2. On that assumption, the answer to your question is based on geometric series, for which there are lots of sources on the internet. A Google search came up with over 200,000 hits. === Subject: Re: Need help >> 1. What is the sum of: >> 1/3 +(1/3)2+(1/3)3+(1/3)4+.... +(1/3) n +....= >> Please explain clearly how you got the answer step by step what > If your (1/3)2 means 1/3 squared, the usual notation for exponents uses > the caret symbol, ^, so that x squared is written x^2, rather than x2. > On that assumption, the answer to your question is based on geometric > series, for which there are lots of sources on the internet. > A Google search came up with over 200,000 hits. 0.333333... 0.111111....... 0.037037.... 0.012345....... 0.0041152......... 0.0013717.... 0.0004572.... 0.0001524.... --------------- 0.499(9)..... I would guess 1/2 or 0.50....., but check using the geometric series recursavily, a(0) = 1/3, a(n+1) = a(n)/3 === Subject: Q: Interpolating Wavelet Transform for Plotting? I am new to wavelet transform. I could use the subroutines of Numerical Recipes in FORTRAN on pages 587-588 to transform a TIME SERIES of data. With the results, I would like to produce a contour plot, or a colored field of scatters, in which the horizontal axis is time, and the vertical axis is either scale, or period. And the contours, or colors would indicate the amplitudes of the wavelets. In order to generate a decent plot, I need an evenly grided field of data, or so it seems to me, but the wavelet transform generated by the subroutines increases the SCALE exponentially (a factor of 2, to be exact), and the number of amplitudes at each scale decreases by a factor of 2. Consequently, I got a very uneven field of data, which is NOT ideal for plotting. Now the question: Am I supposed to INTERPOLATE the uneven field to an even field before plotting the results? If so, is linear interpolation good enough? If not, what is supposed to be the right way to plot the data? Or should I forget the subroutines in Numerical Recipes and write my own code in which I make the scale increases LINEARLY instead of EXPONENTIALLY? --Roland === Subject: Birthday problem for such non-uniform birthday probabilities Consider the birthday problem: There are n randomly chosen person in a room. What is the proabability that there exist k persons who have birthday on the same day. I know how to compute the probability assuming a uniform birthday distribution. It is reasonable that this probability increases for a non-uniform birthday distribution. Where can I find a proof for this result? Helmut === Subject: Re: Birthday problem for such non-uniform birthday probabilities > Consider the birthday problem: > There are n randomly chosen person in a room. What is the proabability > that there exist k persons who have birthday on the same day. > I know how to compute the probability assuming a uniform birthday > distribution. > It is reasonable that this probability increases for a non-uniform > birthday distribution. > Where can I find a proof for this result? > Helmut You won't, as it is also reasonable that this probability decreases for a non-uniform distribution. You need to specify the non-uniform distributions to determine if it increases/decreases. === Subject: Re: Birthday problem for such non-uniform birthday probabilities <4381e0cc$0$84151$892e7fe2@authen.yellow.readfreenews.net> > Consider the birthday problem: > There are n randomly chosen person in a room. What is the proabability > that there exist k persons who have birthday on the same day. > I know how to compute the probability assuming a uniform birthday > distribution. > It is reasonable that this probability increases for a non-uniform > birthday distribution. > Where can I find a proof for this result? Blom, D. (1973), A birthday problem, American Mathematical Monthly, vol. 80, pp. 1141-1142 > Helmut > You won't, as it is also reasonable that this probability decreases for a > non-uniform > distribution. > You need to specify the non-uniform distributions to determine if it > increases/decreases. I would be very interested in seeing a distribution that decreases the probability of a match. === Subject: Re: Birthday problem for such non-uniform birthday probabilities >> Consider the birthday problem: >> >> There are n randomly chosen person in a room. What is the proabability >> that there exist k persons who have birthday on the same day. >> >> I know how to compute the probability assuming a uniform birthday >> distribution. >> >> It is reasonable that this probability increases for a non-uniform >> birthday distribution. >> >> Where can I find a proof for this result? > Blom, D. (1973), A birthday problem, American Mathematical Monthly, > vol. 80, pp. 1141-1142 >> >> Helmut >> You won't, as it is also reasonable that this probability decreases for a >> non-uniform >> distribution. >> You need to specify the non-uniform distributions to determine if it >> increases/decreases. > I would be very interested in seeing a distribution that decreases the > probability of > a match. simple- define distribution as sequential days beginning on Jan 1 === Subject: Re: Birthday problem for such non-uniform birthday probabilities > Consider the birthday problem: > > There are n randomly chosen person in a room. What is the proabability > that there exist k persons who have birthday on the same day. > > I know how to compute the probability assuming a uniform birthday > distribution. > > It is reasonable that this probability increases for a non-uniform > birthday distribution. > > Where can I find a proof for this result? >> Blom, D. (1973), A birthday problem, American Mathematical Monthly, >> vol. 80, pp. 1141-1142 > > Helmut > You won't, as it is also reasonable that this probability decreases for a > non-uniform > distribution. > You need to specify the non-uniform distributions to determine if it > increases/decreases. >> I would be very interested in seeing a distribution that decreases the >> probability of >> a match. >simple- define distribution as sequential days beginning on Jan 1 Wrong. Remember, no matter what distribution you specify, we are selecting randomly from it, so if k>1, repetitions are always possible. If the distribution is not uniform, the probability of a match in k sample elements, where n>=k>1, is always greater than it would be if the distribution was uniform, It's not hard to prove this in general, but as an example, assume a 2 day year and let n>=k=2. Number the days of the year 1,2. (Note that for this type of year, every day is either New Year's Day or New Year's Eve) Let p1, p2 be the probabilities of having a birthday on days 1,2 respectively. Since n>=k=2, the probability of a match can be calculated exactly as p1^2+p2^2. Since p1, p2 are nonnegative, p1^2+p2^2 >= 2*p1*p2 with equality iff p1=p2. Then, adding p1^2+p2^2 to both sides and factoring, we get 2(p1^2+p2^2) >= (p1+p2)^2 Since p1+p2=1, we get p1^2+p2^2 >= 1/2 with equality iff p1=p2. Hence if the distribution is uniform, the probability of a match is exactly 1/2, otherwise it's strictly greater than 1/2. quasi === Subject: Re: Birthday problem for such non-uniform birthday probabilities >> Consider the birthday problem: >> >> There are n randomly chosen person in a room. What is the >> proabability >> that there exist k persons who have birthday on the same day. >> >> I know how to compute the probability assuming a uniform birthday >> distribution. >> >> It is reasonable that this probability increases for a non-uniform >> birthday distribution. >> >> Where can I find a proof for this result? > Blom, D. (1973), A birthday problem, American Mathematical Monthly, > vol. 80, pp. 1141-1142 >> >> Helmut >> >> You won't, as it is also reasonable that this probability decreases for >> a >> non-uniform >> distribution. >> >> You need to specify the non-uniform distributions to determine if it >> increases/decreases. > I would be very interested in seeing a distribution that decreases the > probability of > a match. >>simple- define distribution as sequential days beginning on Jan 1 > Wrong. > Remember, no matter what distribution you specify, we are selecting > randomly from it, so if k>1, repetitions are always possible. > If the distribution is not uniform, the probability of a match in k > sample elements, where n>=k>1, is always greater than it would be if > the distribution was uniform, > It's not hard to prove this in general, but as an example, assume a 2 > day year and let n>=k=2. > Number the days of the year 1,2. > (Note that for this type of year, every day is either New Year's Day > or New Year's Eve) > Let p1, p2 be the probabilities of having a birthday on days 1,2 > respectively. > Since n>=k=2, the probability of a match can be calculated exactly as > p1^2+p2^2. > Since p1, p2 are nonnegative, > p1^2+p2^2 >= 2*p1*p2 with equality iff p1=p2. > Then, adding p1^2+p2^2 to both sides and factoring, we get > 2(p1^2+p2^2) >= (p1+p2)^2 > Since p1+p2=1, we get > p1^2+p2^2 >= 1/2 with equality iff p1=p2. > Hence if the distribution is uniform, the probability of a match is > exactly 1/2, otherwise it's strictly greater than 1/2. > quasi nice show, but still wrong. You rely on a continuous non-uniform distribution, not a non-uniform discrete distribution, or a distribution that is dependent upon previous selections, like removing a ball at a time from a container. Leap year is less. Assume a 1000 day year. Also far less than original problem. a good learning lesson for you. === Subject: Re: Birthday problem for such non-uniform birthday probabilities >> > Consider the birthday problem: > > There are n randomly chosen person in a room. What is the > proabability > that there exist k persons who have birthday on the same day. > > I know how to compute the probability assuming a uniform birthday > distribution. > > It is reasonable that this probability increases for a non-uniform > birthday distribution. > > Where can I find a proof for this result? >> >> Blom, D. (1973), A birthday problem, American Mathematical Monthly, >> vol. 80, pp. 1141-1142 >> > > Helmut > > You won't, as it is also reasonable that this probability decreases for > a > non-uniform > distribution. > > You need to specify the non-uniform distributions to determine if it > increases/decreases. >> >> I would be very interested in seeing a distribution that decreases the >> probability of >> a match. >> >simple- define distribution as sequential days beginning on Jan 1 >> Wrong. >> Remember, no matter what distribution you specify, we are selecting >> randomly from it, so if k>1, repetitions are always possible. >> If the distribution is not uniform, the probability of a match in k >> sample elements, where n>=k>1, is always greater than it would be if >> the distribution was uniform, >> It's not hard to prove this in general, but as an example, assume a 2 >> day year and let n>=k=2. >> Number the days of the year 1,2. >> (Note that for this type of year, every day is either New Year's Day >> or New Year's Eve) >> Let p1, p2 be the probabilities of having a birthday on days 1,2 >> respectively. >> Since n>=k=2, the probability of a match can be calculated exactly as >> p1^2+p2^2. >> Since p1, p2 are nonnegative, >> p1^2+p2^2 >= 2*p1*p2 with equality iff p1=p2. >> Then, adding p1^2+p2^2 to both sides and factoring, we get >> 2(p1^2+p2^2) >= (p1+p2)^2 >> Since p1+p2=1, we get >> p1^2+p2^2 >= 1/2 with equality iff p1=p2. >> Hence if the distribution is uniform, the probability of a match is >> exactly 1/2, otherwise it's strictly greater than 1/2. >> quasi >nice show, but still wrong. >You rely on a continuous non-uniform distribution, not a non-uniform >discrete distribution, or a distribution that is dependent upon previous >selections, like removing a ball at a time from a container. >Leap year is less. >Assume a 1000 day year. Also far less than original problem. >a good learning lesson for you. Unfortunately, I have to run, so I can't reply in full until later. But no, I'm not assuming a continuous distribution. It's a discrete distribution with probabilities p_1,p_2,...,p_m where m is the number of days in the year. For a uniform distribution, all probabilities are = 1/m. If non-uniform, then the probabilities add to 1, but are not all equal. The parameters of the problem are: m = the number of days in the year p_1, p_2, ..., p_m = the probability that a randomly chosen element has birthday 1,2,...,m respectively (in other words -- a discrete distribution). n = size of the population k = size of the sample I just proved a special case, namely, that if m=2 n>=2 k=2 then the probability of a match is 1/2 iff p1=p2 and strictly greater otherwise. However it can be proved in general that as long as m>1 and n>=k>1, the probability of a match is strictly least when distribution is uniform. If you don't believe the math, try a simulation. I have to go -- I'm late -- back later. quasi === Subject: On k kth powers equal to a kth power Hello all, Two questions: 1. Is it true that given a primitive integral solution to k kth powers equal to a kth power, then all the addends except one must be divisible by k+1, IF k+1 is prime? (Of course this implies k is an even exponent. I know the statement is true for k = 2 and 6, probably also for k = 4, though I don't know if this holds in general when k+1 is prime. With k = 8, a solution is known but here k+1 is not prime. And indeed only one term is divisible by 9.) 2. Given a primitive solution to a^4+b^4+c^4+d^4=e^4. If (1) is true then we really have 5^4(a^4+b^4+c^4)+d^4=e^4. See the known solutions in Mathworld. Are the ff true?: a) If d+e or -d+e is divisible by 5, then in fact it is divisible by 5^4. Also, d^2-e^2 is divisible by 5^4. b) If d+e or -d+e is not divisible by 5, then d^2+e^2 is divisible by 5^4. True or not? --Titus === Subject: How can I simplify Cos(Arcsin(x)) ??? Can somebody explain to me, how can I simplify Cos(Arcsin(x)) ??? === Subject: Re: How can I simplify Cos(Arcsin(x)) ??? days. My association with the Department is that of an alumnus. >Can somebody explain to me, how can I simplify Cos(Arcsin(x)) ??? You were given one answer. Here is another that can be used in similar manner for expressions using other trig and inverse trig functions. Let theta = arcsin(x); theta is between -pi/2 and pi/2, so the cosine will be nonnegative. And we want to figure out cos(theta). If we can find any right triangle that has an angle theta, then calculating the cosine of theta in that triangle will give you the value. How can we find a triangle that will have the angle theta? Let's assume first that 0 < = theta < = pi/2 (so 0<= x <= 1). We want a right triangle that will look like (use monospace font to see the following): /| / | / | / | / | / | / | / | / theta | /---------| Now, the sine is the length of the opposite side divided by the length of the hypothenuse. We know that sin(theta)=x. So the easiest triangle that has this is one where the opposide side is length x and the hypothenuse is length 1. If this is the case, then by the Pythagorean theorem, the base of the triangle has to have length sqrt(1^2 - x^2) = sqrt(1-x^2): /| / | / | 1 / | / | / | x / | / | / theta | /---------| sqrt(1-x^2) In which case, the cosine of theta is sqrt(1-x^2)/1 = sqrt(1-x^2). If -pi/2 <= theta < 0, the triangle goes down, but you get the same answer because the cosine is positive in any case. This has the advantage that we can figure out other trig. functions of arcsin(x) relatively easy: for example, tan(arcsin(x)) = x/sqrt(1-x^2) cot(arcsin(x)) = sqrt(1-x^2)/x etc. -- 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: How can I simplify Cos(Arcsin(x)) ??? > Can somebody explain to me, how can I simplify Cos(Arcsin(x)) ??? Use cos x = sqrt(1-sin^2 x) Helmut === Subject: Re: How can I simplify Cos(Arcsin(x)) ??? : Use cos x = sqrt(1-sin^2 x) cos(x) = +/- sqrt(1-sin^2(x)) Justin === Subject: Re: How can I simplify Cos(Arcsin(x)) ??? > : Use cos x = sqrt(1-sin^2 x) > cos(x) = +/- sqrt(1-sin^2(x)) In the context of the considered question? Helmut === Subject: Re: How can I simplify Cos(Arcsin(x)) ??? >> : Use cos x = sqrt(1-sin^2 x) >> cos(x) = +/- sqrt(1-sin^2(x)) >In the context of the considered question? In this example, arcsin(x) is in the interval [-pi/2,pi/2] (assuming he's working over the reals) so the - case doesn't arise. But it's a bad idea to forget the +/- because there are plenty of examples where it is needed. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: How can I simplify Cos(Arcsin(x)) ??? Cool, thank you very much === Subject: Direct Product, Free Product, Direct Sum, Categorical Product, Categorical Coproduct Could someone kindly provide some examples to distinguish them clearly? I find the definitions really abstract, and it is really hard to distinguish them without seeing some examples / counter-examples. For example, why is Direct Sum different from Direct Product in non-abelian groups? why is Direct Sum different from Direct Product when we have infinite factors of abelian groups? Why do we use Free Product instead of Direct Sum for the coproduct of non-abelian groups? I thought this should be in a FAQ, but to my surprise I could not find it. === Subject: Re: Direct Product, Free Product, Direct Sum, Categorical Product, Categorical Coproduct days. My association with the Department is that of an alumnus. >Could someone kindly provide some examples to distinguish them clearly? >I find the definitions really abstract, and it is really hard to >distinguish them without seeing some examples / counter-examples. >For example, why is Direct Sum different from Direct Product in >non-abelian groups? why is Direct Sum different from Direct Product >when we have infinite factors of abelian groups? Why do we use Free >Product instead of Direct Sum for the coproduct of non-abelian groups? Okay, let me be a bit more expansive on this reply... What we want from a product construction is the following: Given objects {A_i}, an object P, called the product of the A_i, and maps p_i:P --> A_i, called the projections such that If G is any object, and we have maps f_i:G-->A_i for each i, then there exists a unique map f:G --> P such that p_i o f = f_i for each i. This is in analogy to the way the cartesian product works for sets: if you have a family {A_i} of sets, then the cartesian product of the A_i has projections onto each A_i, (project onto the i-th coordinate) and given any family of maps f_i: S-->A_i, we map s in S to the elements (f_i(s)) of the cartesian product. E.g., if the sets are A_1 ={ 1, 2} A_2 = {a,b,c} A_3 = {5,9} and S = {r,s} and we have maps f_1:S->A_1 given by r|->1, s|->2 f_2:S->A_2 given by r|->a, s|->a f_3:S->A_3 given by r|->9, s|->5 then we have a unique map from S to A_1 x A_2 x A_3, namely r|-> (1, a, 9) s|-> (2, a, 5) such that if we first map to the product and then down to A_i, we get the same thing as the original map. For each class of algebraic objects, one has to figure out what object, if any, has these properties. In *most* cases (but not in all) the object in question is just the usual direct product, with the operations defined componentwise. In that respect, for most classes of algebraic objects, the direct product is the 'obvious' thing. For an example where it does not, consider the category of all fields; this category does NOT have products at all: in general, not even for two fields can you find a field P which will have the desired maps and properties. There are other classes where this works as well: for example, in topological spaces, the direct product with the product topology has the desired properties, so it is the product of the category. The idea of the coproduct is to reverse the conditions. Given a family of objects {A_i}, a coproduct is an object C, together with maps e_i: A_i --> C, such that For every object G, if we have a family of maps g_i:A_i --> G for each i, then there exists a unique map g: C-->G such that for all i, e_i o g = g_i. If we work with sets, then the coproduct is the disjoint union. Taking the disjoint union of the A_i and defining the map G to be the 'union' of the maps g_i (the maps e_i are just the natural embeddings of each set into the disjoint union). If we use A_1, A_2, A_3 from above, the coproduct is C = {1, 2, a, b, c, 5, 9} with the obvious embeddings. Given maps from A_1, A_2, and A_3 into a set G, we can map C into G by following the same rules as the ones for A_1, A_2, A_3. Now, going to other categories, we need to look for an object that has the right conditions for this to work. In topological spaces, for example, we can take the disjoint union of the spaces with the obvious topology (open in each disjoint part); there are categorical reasons why in topology we can take the disjoint union (they have to do with the fact that the underlying set functor has adjoints on both sides, but ignore that for the moment). Alas, this does not hold for most categories of algebraic objects. Let's take abelian groups. The direct sum has all the necessary properties: if you have the direct sum of the A_i, and you have maps g_i:A_i --> G, you can map an element of the direct sum, which is of the form (a_i) i in I, all but finitely many of the a_i equal to 0 to the element sum f_i(a_i) because the sum is well defined (since the sum is commutative, and all but finitely many of the summands are equal to 0). When the number of A_i is finite, this is the same as the direct product, but when the number of A_i is infinite, it is a smaller object. Note that in general you may not have maps FROM the direct product to G that agree with the g_i, since there may be no way to interpret an infinite sum inside of G: groups only have finite sums in the absence of a lot more structure. The direct sum, however, does not work for arbitrary groups, not even for finitely many objects. For example, consider the case when we have two grups, C_2 and C_3 (the cyclic groups of orders 2 and 3), and we map each of them into S_3, the symmetric group on three letters. We map C_2 into S_3 by mapping the generator to the permutation (1|->2,2|->1) (I use this notation because in the next paragraph I'm going to talk about elements of the direct sum of C_2 and C_3, and it might be confusing then); we map C_3 into S_3 by mapping the generator to the permutation (1|->2,2|->3,3|->1). If we try to define a map from the direct sum/direct product (only two factors, so they are the same) of C_2 and C_3 into S_3 that agrees with this, we run into a problem. Let C_2 = {1, x}, and C_3 = {1,y,y^2}. Let C = C_2 x C_3. We know that if we want the map from C to S_3 to agree with the maps from C_2 and from C_3 to S_3, we need to send (x,1) to the permutation (1|->2,2|->1); and we need to map (1,y) to the permutation (1|->2, 2|->3, 3|->1). But in C_2 x C_3, we have that (x,1)(1,y) = (x,y) = (1,y)(x,1), and in S_3 we do not have that the two permutations commute. So there is no group morphism from C_2 x C_3 to S_3 that will send (x,1) to (1|->2, 2|->1) and will send (1,y) to (1|->2, 2|->3, 3|->1). So the direct sum does NOT have the properties of the coproduct for arbitrary groups. It turns out that the group that DOES have the properties is the FREE PRODUCT (which had already been defined and studied before the development of the notion of category, categorical product, and categorical coproduct). So the free product is the coproduct for the category of all groups. If you consider smaller classes of groups, for example, all metabelian groups, then the coproduct will turn out to be something which is not the usual free product, but somewhere between the free product and the direct sum (we can think of the direct sum as a quotient of the free product, and the coproduct there will be the free product modulo a smaller normal subgroup). In the category of commutative rings with 1, the direct product works as a product, but the direct sum does not work as a coproduct (for arbitrarily many rings, the direct sum is not even a ring with 1); there are other constructions that give you a coproduct, though, namely the tensor product. In the category of fields, there is no coproduct. Does that help somewhat? -- 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: Direct Product, Free Product, Direct Sum, Categorical Product, Categorical Coproduct days. My association with the Department is that of an alumnus. >Could someone kindly provide some examples to distinguish them clearly? >I find the definitions really abstract, and it is really hard to >distinguish them without seeing some examples / counter-examples. >For example, why is Direct Sum different from Direct Product in >non-abelian groups? Because they are defined differently? They are isomorphic when the number of factors is finite. Why would we define them differently? Because for an infinite number of factors, the direct sum has a left universal property (given any family of maps fi: A_i --> B, there is a unique map from the direct sum of the A_i to B which commutes with the canonical embeddings of the A_i into the direct sum) but not a right universal property (in general, given a family of maps gi:B-->A_i, there is no unique map from B to the direct product of the A_i which will commute with the givne map and the canonical projections); and the product has the right universal property but not the left one. >why is Direct Sum different from Direct Product >when we have infinite factors of abelian groups? Because we want the sum to have the universal property of the coproduct, and we want the direct product to have the universal property of the product. > Why do we use Free >Product instead of Direct Sum for the coproduct of non-abelian groups? Because the direct sum does not have the universal property in the category of all groups; you need the free product, which is the one that has the universal property. -- 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: New mathematics/physical sciences positions at http://jobs.phds.org, November 21, 2005 New job listings at http://jobs.phds.org - Jobs for PhDs List your job at no cost! http://jobs.phds.org/jobs/post * Senior C++ Quantitative Research Developer, Hedge Fund: NJF Search International, New York. Premier Hedge Fund is seeking a talented individual to join their growing team in the capacity of Quantitative Research Developer. They desire a candidate with strong quantitative abilities combined with the willingness to develop systems. 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Define K=Fp(t) and L=K(a) where a satisfies a^p=t. Prove that any inseparable extension M of degree p of K is isomorphic to L. 3. Let p(t) be a polynomial of degree n>=3 over the rationals Q. Let a1,a2,...an indicate its roots at the complex field C. also , the following conditions hold: * a1+a2 = a3. * there isn't any other relation of that kind for other roots. meaning that if a_i + a_ j= a_k then k=3 and i,j = 1,2 that way or another. prove that a3 belong to the rationals field Q, and that there exist a quadraic extension L that contains a1 and a2. Ori. === Subject: Re: Three Questions Relating Algebraic Structures > Hello All, > A few days ago I took the final test of a course called algebraic > structures 2. There were 3 questions there i didn't, and still don't > know how to answer. Though I've gotten already the results, i'm very > interested in knowing how to solve them. any help will be appriciated. > 1. Prove that the polynomial x^64+x^16+x^4+x+1 has a root in F_256. Presumably, the polynomial is over F_2. Then rewrite the equation as: x^64 = x^16 + x^4 + x + 1, and so taking the 4-th power gives x^256 = (x^16 + x^4 + x + 1)^4 = x^64 + x^16 + x^4 + 1 = .... = x. So all roots of your polynomial are roots of x^256 - x. > 2. Define K=Fp(t) and L=K(a) where a satisfies a^p=t. Prove that any > inseparable extension M of degree p of K is isomorphic to L. Let a be in M-K, and q(x) be its minimal polynomial. Since M = K(a), a must be inseparable, and so q(x) is inseparable. What must q(x) look like? > 3. Let p(t) be a polynomial of degree n>=3 over the rationals Q. Let > a1,a2,...an indicate its roots at the complex field C. also , the > following conditions hold: > * a1+a2 = a3. > * there isn't any other relation of that kind for other roots. > meaning that if a_i + a_ j= a_k > then k=3 and i,j = 1,2 that way or another. > prove that a3 belong to the rationals field Q, and that there exist a > quadraic extension L that contains a1 and a2. Let K = Q(a1, a2, ...) which is a Galois extension of Q. Then each s in Gal(K/Q) takes a_1, a_2, a_3 to some a_i, a_j, a_k. Then a_i + a_j = a_k, and so by the conditions in the problem, k=3, (i,j) = (1,2) or (2,1). Thus s(a_3) = a_3 for each s. This means a_3 is in Q. Likewise, s(a_1) = a_1 or a_2, so the subgroup of Gal(K/Q) fixing Q(a_1) is a subgroup of index 1 or 2. So Q(a_1) is Q or quadratic. > Ori. === Subject: =?gb2312?B?0ru2qNPQxLPQqb3hwtvU2tXi0KnHsMzhz8LO0sPHzt63qNakw/fL/NX9yLfT67fxoa M=?=