A147 I have the following question: suppose a is primitive in GF(4), > then how to find a^4+a^2, why it is equ to 1 ? > a^4+a^3, why it is equ to a^2 ? Anybody can tell me what is the trick? so, in GF(4), or any GF(q), is l the -a equ to +a, that's to say, there is no minus sign GF(q), and once we see - in computation, we just change it to +? am I right? ==== > > I have the following question: suppose a is primitive in GF(4), > then how to find > a^4+a^2, why it is equ to 1 ? > a^4+a^3, why it is equ to a^2 ? GF(4) has 4 elements. One of them is 0, another is 1. Let a be a third. The fourth has to be a + 1, but it so has to be a^2. So a^2 = a + 1. Then a^3 = a(a + 1) = a^2 + a = (a + 1) + a = 1. Etc. > so, in GF(4), or any GF(q), is l the -a equ to +a, that's to say, there is no minus sign in > GF(q), and once we see - in computation, we just change it to +? am I right? No. Only if q is even. By the way, it's spelled Gois. -- ==== > > I have the following question: suppose a is primitive in GF(4), > then how to find > a^4+a^2, why it is equ to 1 ? > a^4+a^3, why it is equ to a^2 ? GF(4) has 4 elements. One of them is 0, another is 1. Let a be a third. > The fourth has to be a + 1, but it so has to be a^2. ------------>> why The fourth has to be a + 1, but it so has to be a^2? > So a^2 = a + 1. > Then a^3 = a(a + 1) = a^2 + a = (a + 1) + a = 1. > Etc. > so, in GF(4), or any GF(q), > is l the -a equ to +a, that's to say, there is no minus sign in > GF(q), and once we see - in computation, we just change it to +? > am I right? No. Only if q is even. when q is even, say 10, then in GF(10), -a^4 == a^4, right? when q is odd, say 225, why in GF(225), -a^4 =/= a^4? thank you ve much! ==== >when q is even, say 10, then in GF(10), -a^4 == a^4, right? >when q is odd, say 225, why in GF(225), -a^4 =/= a^4? What GF(10) and GF(225)? A finite field's cardinity has to be a power of a prime. Department of Mathematics http://www.math.ubc.ca/~israel ==== > GF(4) has 4 elements. > One of them is 0, another is 1. Let a be a third. > The fourth has to be a + 1, but it so has to be a^2. ------------> why The fourth has to be a + 1, but it so has to be a^2? GF(4) has 4 elements. Three have been accounted for: 0, 1, a. Note that they are distinct since a does not equ 0 or 1 etc. Consider a + 1. It is certainly in GF(4). Could it be equ to one of the accounted three elements so far? No, for a + 1 = 0 says a = 1; a + 1 = 1 says a = 0; and a + 1 = a says 1 = 0. Thus, a + 1 must be the fourth unaccounted element. As an aside, you can approach your origin question as follows. GF(4) has 4 elements and the element a is primitive. Since the non-zero elements of GF(4) form a cyclic group of order 3, the distinct elements of GF(4) are: 0, a, a^2, a^3 = 1. This is so the way that a primitive element is used for other GF(q), q not 4. You want to find out what a^4 + a^2 is. First note that a^4 = a^3 * a = 1 * a = a. Thus, a^4 + a^2 = a + a^2. You have four choices for a + a^2: a + a^2 = 0 a + a^2 = a a + a^2 = a^2 a + a^2 = 1 The second and third are easy to rule out since they imply the fse statements a^2 = 0 and a = 0. The first is so easy to rule out since you get 1 + a = 0 by dividing both sides by a. But, 1 + a = 0 implies the fse statement a = 1. Thus, you are left with just the last statement, which is what you want. They may be the trick you mentioned: just set the expression you want to evuate to the four possible choices and eliminate the three that lead to fse statements by simple gebraic manipulations. ==== one can make a dystinction beeen a) omniprescience, b) broad predictions, c) vuable forecasts, d) promoting **** that you're in on. which do you claim? >thank you Jedi Knight Noostradangit; > [blah, blah,blah] 8 instances of references from the same source on record in USENET > made to things that hadn't happened yet counter the assertion > that the future can't be known in advance. QED. Whining or ridiculing about it is irrelevant. The origin > assertion (by Popper) is proven wrong. Period. --les Dicks d'Enron! ==== > I've cled the factoring of polynomi into non-polynomi factors > advanced for a reason.... > To move beyond factoring polynomis into polynomi factoyou need > to have a *thorough* understanding of coprimeness, ring operations, > and logic argument. LOL. www.crank.net/harris.html ==== Does anyone know how to convert .JMP data to comma delimited format (or better yet, MINITAB format)? I rely, rely need to convert this data because I 't own JMP--only MINITAB.. any points to conversion utilities (or if someone would be so kind as to convert it), I would be ve grateful. Data (2 files): http://www-stat.wharton.upenn.edu/~lzhao/stat431/DataFiles/Cl%20Center%20A rrivs.JMP http://www-stat.wharton.upenn.edu/~lzhao/stat431/DataFiles/Philadelphia%20(S ==== I found a program that does conversions.. thanks. ex > Does anyone know how to convert .JMP data to comma delimited format (or > better yet, MINITAB format)? I rely, rely need to convert this data > because I 't own JMP--only MINITAB.. any points to conversion utilities > (or if someone would be so kind as to convert it), I would be ve grateful. ex Chavez > ex@divide0.net Data (2 files): http://www-stat.wharton.upenn.edu/~lzhao/stat431/DataFiles/Cl%20Center%20A rrivs.JMP http://www-stat.wharton.upenn.edu/~lzhao/stat431/DataFiles/Philadelphia%20(S ==== what a beautiful evocation of rationity, that old report! a lot more than I can say for Jack's hearsay about Roswell, which had a couple of interesting associations around WW2, which Art Bell (e.g.) never mentions (though he must know about one of them). what Art's promoted is the thesis of the late Lt.Col. Corso, that this little pile of **** was parcelled-out to the big companies, and we reverse-engineered fiber-optics, the transistor, RADAR ... you name it; I forgot. (fortunately, Corso's co-author exposed him, though I 't know if it's in the book; he did in the promotion tour !-) > The Relius Experience of Philip K. Dick by R. Crumb > http://www.philipkdick.com/weirdo.htm > Yes! There is no relin, no Way of God, no Way of Divine > Reization, no Way of Enlightenment, and no Way of > .... Therefore, Reity (Itself) Is Truth, > and Reity (Itself) Is the Only Truth. > -- Adi Da Samraj, a.k.a. Bubba Free John, > a.k.a. Da Free John, a.k.a. Da Kki, > a.k.a. the Ruchira Avatar, Adi Da Love-Ananda Samraj, > a.k.a. lin Jones > http://adidam.org/ > http://www.daplastique.com/home.html > [1968] SCIENTIFIC STUDY OF UNIDENTIFIED FLYING OBJECTS Conducted by the University of Colorado > Under contract No. 44620-67-C-0035 With > the United States Air Force Dr. Edward U. Con, Scientific Director [1968] > C O N T E N T S > http://ncas.sawco.com/con/text/contents.htm Section I Conclusions and Recommendations Edward U. Con > http://ncas.sawco.com/con/text/sec-i.htm > This formulation carries with it the corolla that we do not > think that at this time the feder government ought to set up > a major new agency, as some have suggested, for the scientific > study of UFOs. This conclusion may not be true for l time. > If, by the progress of research based on new ideas in this field, > it then appears worthwhile to create such an agency, the decision > to do so may be taken at that time. We find that there are important areas of atmosph optics, > including radio wave propagation, and of atmosph electricity > in which present knowledge is quite incomplete. These topics came > to our attention in connection with the interpretation of some > UFO reports, but they are so of fundament scientific interest, > and they are relevant to practic problems related to the > improvement of safety of milita and civilian flying. > As the reader of this report will readily judge, we have focussed > attention most entirely on the physic sciences. This was in part > a matter of determining priorities and in part because we found > rather less than some persons may have expected in the way of > psychiatric problems related to belief in the reity of UFOs as > craft from remote gactic or intergactic civilizations. We believe > that the rigorous study of the beliefs--unsupported by vid > evidence--held by individus and even by some groups might prove of > scientific vue to the soci and behavior sciences. There is no > implication here that individu or group psychopathology is a > princip area of study. Reports of UFOs offer interesting chges > to the student of cognitive processes as they are affected by > individu and soci variables. By this connection, we conclude that > a content-anysis of press and television coverage of UFO reports > might yield data of vue both to the soci scientist and the > communications speciist. The lack of such a study in the present > report is due to a judgment on our part that other areas of > investigation were of much higher priority. We do not suggest, however, > that the UFO phenomenon is, by its nature, more amenable to study in > these disciplines than in the physic sciences. On the contra, we > conclude that the same specificity in proposed research in these areas > is as desirable as it is in the physic sciences. > It has been contended that the subject has been shrouded in > offici secrecy. We conclude otherwise. We have no evidence > of secrecy concerning UFO reports. What has been miscled > secrecy has been no more than an intelligent policy of delay in > releasing data so that the public does not become confused by > premature publication of incomplete studies of reports. The subject of UFOs has been widely misrepresented to the public > by a sml number of individus who have given sensationized > presentations in writings and public lectures. So far as we can > judge, not many people have been misled by such irresponsible > behavior, but whatever effect there has been has been bad. > Therefore we strongly recommend that teachers refrain from > giving students credit for school work based on their reading > Teachers who find their students strongly motivated in this > direction should attempt to channel their interests in the > direction of serious study of astronomy and meteorology, and > in the direction of critic anysis of arguments for fantastic > propositions that are being supported by appes to flacious > reasoning or fse data. We hope that the results of our study will prove useful to > scientists and those responsible for the formation of public > policy generly in deing with this problem which has now > been with us for 21 years. The Con Report - 1968 CE > http://ncas.sawco.com/con/index.html > http://www.csicop.org/klassfiles/posner_klass.html > Parlel Universes http://www.sciam.com/issue.cfm > Not just a staple of science fiction, other universes > are a direct implication of cosmologic observations ==== >By the way, what Le Roux cls an application is usuly cled a >> >> >Oops, sor. That's because there is a difference beeen fonction and >application in French, and I thought there was the same difference in >English. >> >That's interesting. What's the difference? I have not seen >>application used this way in English, but I am not a profession >>mathematician. >> >Could it be the same distinction as beeen function and map? Marc translate to the English map. However, I ways thought that map and function were synonymous. -- ==== >>Oops, sor. That's because there is a difference beeen fonction and >>application in French, and I thought there was the same difference in >>English. >> >That's interesting. What's the difference? I have not seen >application used this way in English, but I am not a profession >mathematician. > >Could it be the same distinction as beeen function and map? >>Marc > translate to the English map. However, I ways thought that map > and function were synonymous. It could be just a hard dying habit (cf. Funktion and Abbildung), from Anysis, where maps with the res as codomain are more equ than the rest Marc ==== > Of interest perhaps are these formulas: > When U subset A subspace S > cl_A (U) = A / cl_S (U) > int_A (U) = A / int_S (SA / U) Prove the first and for the second use. int_A(U) = A - cl_A (A-U) As adapted from the theorem S - cl U = int (S-U) ==== Of interest perhaps are these formulas: >>When U subset A subspace S >>cl_A (U) = A / cl_S (U) >>int_A (U) = A / int_S (SA / U) >> And thank *you* for using my preferred plur of formula! ==== Many of you may be confused by the use by of an overly complicated cubic polynomi and unneeded variables. Here is a somewhat simpler version. Lemma 1: Let: P(m) be a polynomi with coefficients in A, the gebraic integers. Let each coefficient be divisible (in A) by the gebraic integer f. Let P(m) = g_1(m) * g_2(m), with g_1(m) and g_2(m) functions from A to A. Then: there exist s and t in A such that s*t=f; and for l m in A, s divides g_1(m) (in A) and t divides g_2(m) (in A). Core Error Proof: Consider the function P(m,f,x) = mfx^2 +2fx + f^2 (m,f,x in A) [Among other possibilities we can consider this as a polynomi in m, with parameters f and x, or a polynomi in x with parameters m and f. The latter only matters if we want to motivate the following] Consider the quadratic t^2 + 2t + mf (*) Let a_1(m,f) = 1 - sqrt(1-mf) and a_2(m,f) = 1 + sqrt(1-mf) be the negatives of the roots of (*). Some simple gebra gives P(m,f,x) = ( a_1(m,f) x + f ) ( a_2(m,f) x + f ) or (with a minor abuse of notation) letting P(m) = P(m,f,x), g_1(m) = a_1(m,f) x + f and g_2(m) = a_2(m,f) x + f P(m) = g_1(m) * g_2(m) Noting that P(m) can be considered a polynomi in m with coefficients in A, l coefficients divisible by f we apply lemma 1 and obtain: There exists s and t, such that s*t=f; and for l m, s divides g_1(m) = a_1(m,f) x + f t divides g_2(m) = a_2(m,f) x + f Let m = 0. g_1(0) = f and g_2(0) = 2x + f. The only choice for s and t (up to units) is s=f and t=1. thus f divides (a_1(m,f) x + f) thus f divides a_1(m,f) Now speciize by letting m=1, f =3 3 divides a_1(1,3) = 1-sqrt(1-(1)(3)) = 1 - i sqrt(2) but (1- i sqrt(2))/3 is a root of (3x^2 - 2x + 1) At this point the mathematic universe collapses Or does it, The problem is with Lemma 1. While this holds for integer polynomis [? does this hold for gebraic integer polynomis] it is not true in gener. Counterexamples are easy to construct if g_1 and g_2 are not continuous. We get a more interesting example by letting x=1 and f = 3 above. Let P(m) = 3m + 15, this is divisible by 3. By simple gebra P(m) = 3m+15 = (4-sqrt(1-3m)) (4+sqrt(1-3m)) so let g_1(m) = (4-sqrt(1-3m)) and g_2(m) = (4+sqrt(1-3m)) P( -1) = 12, g_1( -1) = 2, g_2( -1) = 6 P( -5) = 0, g_1( -5) = 0. g_2( -5) = 8 P( -8) = -9, g_1( -8) = -1. g_2( -8) = 9 P(-16) = -33, g_1(-16) = -3. g_2(-16) = 11 We note that while P(m) is ways divisible by 3, the factor of 3 switches back and forth beeen g_1 and g_2. In gener l we have is Lemma 1': Let: P(m) be a polynomi with coefficients in A, the gebraic integers. Let each coefficient be divisible (in A) by the gebraic integer f. Let P(m) = g_1(m) * g_2(m), with g_1(m) and g_2(m) functions from A to A. Then: for l m in A there exist s(m) and t(m) in A such that s(m)*t(m)=f;, s(m) divides g_1(m) and t(m) divides g_2(m). So l we can conclude above is that s(0)=f and t(0)=1. We cannot conclude that s(1)=f, so we cannot conclude that f divides a_1(1,f), so we cannot conclude that 3 divides a_1(1,3), so mathematics does not collapse in ruins. ==== > Let me start by saying that I didn't mean to offend or annoy you in > case you're offended or annoyed. Next, let me just say, for the > record, that as a probabilist, in particular, one who works with > stochastic ordina and parti differenti equations, I'm ve > familiar with white noise and the techniques for working with it in a > mathematicly rigorous fashion. I am, however, a student of pure > mathematics, so anything I say will be from that perspective. I'll t > to answer your questions below. I guess I did fly off the handle a bit fast there. I will certainly keep your perspective in mind. Indeed it is what I expected of this newsgroup. And I am thankful that a person of your quifications is ting to answer my questions - I appreciate that. I apologize for jumping to the wrong conclusions about your motives. > >> [..] >from a mathematic perspective, your question doesn't even make >sense. You state, let X(t) be a zero-mean IID process with variance >of sigma^2. By this, I assume you mean that X(t) is a mean zero >stationa process such that >>(i) X(t) and X(s) are independent whenever t < s, and >>Er, last I checked IID includes dependent, so you have simply >>repeated my conditions. >(ii) E|X(t)|^2 = sigma^2 < infty. >>Do you rely have to be that an? Actuly, you're not being an >>enough - if the covariance was not finite, then I would've stated that >>it does not exist. > It's true that, here, I am simply repeating your conditions. I'm > ting to use the terminology you will be more likely to see in the > pure math literature. For example, when it comes to a process of a > re parameter, t, the term IID is less common. The reason is > ofold: (1) postulating independence can lead to difficulties > regarding the existence of the process, as I mention below; and (2) > processes of a re parameter which are identicly distributed are > more commonly referred to as stationa. Point taken. I am still getting comfortable with the language and conventions. > The < infty is there to emphasize that this is the important part of > the assumption. If the variance is lowed to be infinite, then such a > process will have a measurable version, though it won't be continuous. >Well, there is no reasonable process that satisfies these conditions. >(In this case, such a process would not even be measurable. If you >drop (ii), such a process would not be continuous.) >>And how did you get to these conclusions? > See the first page of Chapter III of Stochastic Differenti > Equations by Bernt Oksen and the references therein. > (Incidently, white noise is *not* the non-continuous process > constructed by dropping assumption (ii). See below.) >On the applied side, there is a lot of intuition about white noise. >Translating that into rigorous mathematic definitions is no sml >task. The white noise mathematicians work with is a completely >different kind of object than you're probably used to. (In fact, it's >not even a process in the tradition sense.) >>Please, enlighten me, or point me to the subjects and/or books >>that will enlighten me. > An ordina process is a function X(t,omega) of o variables. For > each omega, we have a function of t, and for each t, we have a random > variable. Even though Papoulis and others state this, I somehow got this wrong long ago and I'd appreciate it if you could confirm it: Are the omegas the elements of the sample space, so that for any specific t, say, t0, X(t0, omega) is the mapping from S (the sample space) to R (the res) that we normly speak of when considering a random variable? > It is possible to regard white noise as a process in the > sense that for each t we have an object cled a generized random > variable or a generized Wiener function. See Stochastic Parti > Differenti Equations: A Modeling White Noise Approach by Oksen, > et , and the references therein for an introduction to this > approach. More commonly, though, white noise is regarded as a random > variable taking vues in the space of distributions: for each omega, > we get a distribution rather than a function of t. To be precise, > white noise would be the distribution derivative of Brownian motion. Do you mean distributions in the sense that the Dirac delta function is a distribution? So X(t, omega) for each omega gives a distribution (kinda like a pathologic function) instead of a norm function? > In the former approach, white noise does not have poinise vues for > each omega, in the latter it does not have poinise vues for each > t. Hmmm. >I know this doesn't answer your question, but I hope it at least >partily explains why you haven't received more responses. In some >sense, you're posting to the wrong group. >>I am absolutely not posting to the wrong group. If the answer to this >>this question involves delving into pure math, so be it. >If you post your question in >an engineering group, they may be able to give you an answer based on >their own intuition and jargon, which may be what you're looking for. >>No, that's not what I'm looking for. And I HAVE posted it to an engineering >>group (comp.dsp) SEVER times now over the last o or three years. It >>is not resolvable by the folks there. >If you want a mathematic answer, then the question must be posed in >a more form way. >>Hey Jason, bend a little - give me a clue as to what is lacking in formity >>here. > Okay, I'll give it a t. If someone said to me, consider an IID > process whose covariance function is EX(s)X(t) = delta(t-s), I would > understand that to mean the white noise I mentioned earlier, which I > would think of as the distribution derivative of Brownian motion. > The connection beeen the mathematics and the heuristics is this: if > X_r(t) = [B(t+r)-B(t)]/r, where B(t) is Brownian motion, then > EX_r(s)X_r(t) = f_r(t-s) for some functions f_r which coverge in the > sense of distributions to the delta distribution. Now, I think you want to consider an IID process whose covariance > function is EX(s)X(t) = sigma^2 if s=t, 0 otherwise. Yes, in the one sense that's exactly what I want. Heuristicly speaking, it seems like it should be reasonable to construct a process which is stationa and independent at any time difference tau (i.e., X(t) and X(t+tau) are independent random variables when tau is not zero), so that the autocorrelation function is zero when tau is not zero. Yet each random variable in this process's family of random variables could have the same distribution and that distribution could have a finite variance - for example, a simple N(0,1) distribution - so that the autocorrelation function is the finite vue sigma^2 at tau = 0 and zero otherwise. Yes, that's precisely the anim I want to construct. > How should I > interpret this in a rigorous sense? Is it the distribution > derivative of some ordina process? If so, which one? Yeah, yeah! > Okay, once we establish what the process is, we come to next part, > which is that it has a white PSD. As I understand it, the PSD is > simply the Fourier transform of the covariance function. You keep on saying the covariance function, but most engineering texts I've seen say autocorrelation function (the probabilistic autocorrelation, i.e., Rxx(tau) = E[X(t)X(t+tau)], NOT the one which does time-domain correlation, i.e., the convolution of the time-domain function with a time-reversed version of itself). > But o > functions which differ only a set of measure zero have the same > Fourier transform. This process has a covariance function which is 0 > most evewhere. Therefore, its Fourier transform is the zero > function. Right. That's the problem. > Notice that we 't have this problem when taking the > Fourier transform of the delta function, even though the delta > function is zero evewhere except the origin. This is because the > delta function is not a function at l, but a distribution, so we > must take the Fourier transform in the sense of distributions. So, in summa, the process you've introduced does not exist as an > ordina process. Perhaps it exists as a random variable taking vues > in the space of Schwartz distributions. If so, it's not clear exactly > what that random variable should be, so you need to explicitly define > it in the language of distribution theo. Once e, you need to be > more explicit about the white PSD part, since, as it stands, the > Fourier transform of your covariance function is the zero function, > which is not white. A good place to start reading about distribution > theo (if you're not ready familiar with it) is Introduction to > the Theo of Distributions by F. G. Friedlander. Excellent - I've been looking for a reference to this exact topic - thanks! > Anyway, I hope this helps. I enjoy discussing this stuff because I > enjoy stochastic anysis. But I 't want to annoy or offend you and > I 't want to engage in any kind of emotionly charged discussion > over this stuff. Good luck in your investigations. Please let me know > if you come upon the form way to describe your process. I'd be ve > interested in seeing it. help. The key seems to be in understanding why a continuous-time independent process isn't a norm process. I have the Oksen book on order. -- http://home.earthlink.net/~yatescr ==== > An ordina process is a function X(t,omega) of o variables. For > each omega, we have a function of t, and for each t, we have a random > variable. Even though Papoulis and others state this, I somehow got this wrong > long ago and I'd appreciate it if you could confirm it: Are the omegas > the elements of the sample space, so that for any specific t, say, t0, > X(t0, omega) is the mapping from S (the sample space) to R (the res) > that we normly speak of when considering a random variable? Yes, that's right. It is possible to regard white noise as a process in the > sense that for each t we have an object cled a generized random > variable or a generized Wiener function. See Stochastic Parti > Differenti Equations: A Modeling White Noise Approach by Oksen, > et , and the references therein for an introduction to this > approach. More commonly, though, white noise is regarded as a random > variable taking vues in the space of distributions: for each omega, > we get a distribution rather than a function of t. To be precise, > white noise would be the distribution derivative of Brownian motion. Do you mean distributions in the sense that the Dirac delta function > is a distribution? So X(t, omega) for each omega gives a distribution > (kinda like a pathologic function) instead of a norm function? Right again. In the former approach, white noise does not have poinise vues for > each omega, in the latter it does not have poinise vues for each > t. Hmmm. > ... > Okay, I'll give it a t. If someone said to me, consider an IID > process whose covariance function is EX(s)X(t) = delta(t-s), I would > understand that to mean the white noise I mentioned earlier, which I > would think of as the distribution derivative of Brownian motion. > The connection beeen the mathematics and the heuristics is this: if > X_r(t) = [B(t+r)-B(t)]/r, where B(t) is Brownian motion, then > EX_r(s)X_r(t) = f_r(t-s) for some functions f_r which coverge in the > sense of distributions to the delta distribution. Now, I think you want to consider an IID process whose covariance > function is EX(s)X(t) = sigma^2 if s=t, 0 otherwise. Yes, in the one sense that's exactly what I want. Heuristicly > speaking, it seems like it should be reasonable to construct > a process which is stationa and independent at any time difference > tau (i.e., X(t) and X(t+tau) are independent random variables when tau > is not zero), so that the autocorrelation function is zero when tau is not > zero. Yet each random variable in this process's family of random > variables could have the same distribution and that distribution could > have a finite variance - for example, a simple N(0,1) distribution - so > that the autocorrelation function is the finite vue sigma^2 at tau = 0 > and zero otherwise. Yes, that's precisely the anim I want to construct. Is this anim that you want to construct something that others on the applied side are working with? I mean, is it well known in the outside world, or is it just something you thought of at some point and were curious about? I've thought some more about this. Let's let sigma=1. For sml r, the process X_r(t)=[B(t+r)-B(t)]/sqrt{r} is pretty close to what you're looking for. The question is, do these processes, as r-->0, coverge in any reasonable sense? I 't see it, off hand. For fixed omega, they converge to zero as Schwartz distributions. For fixed t, they converge to zero as generized random variables. In fact, if we write EX_r(t+tau)X_r(t)=f_r(tau), then f_r converges to 0 most evewhere and in L^1. Of course, for fixed t, the X_r(t)'s coverge in law (in fact they're l identic in law) to a norm(0,1) random variable. I 't think this helps though, because you want some kind of convergence that des with l t simultaneously. There's a lot of interesting behavior in these processes X_r relating to the law of the iterated logarithm and its variants. Chapter 2, section 2.9, subsection E of Brownian Motion and Stochastic Cculus by Karatzas and Shreve contains the law of the iterated logarithm. The paragraph after the proof of it discusses this process X_r. It contains references to o previous remarks and to comments in the notes at the end of the chapter. Part of remark 9.15 is worth posting here: It is quite possible that for each fixed t>=0, a certain property holds most surely, but then it fails to hold for l t>=0 simultaneously on an event whose probability is one (or even positive!). As an extreme and rather trivi example, consider that P[B(t)!=1]=1 holds for eve 0<=t interpret this in a rigorous sense? Is it the distribution > derivative of some ordina process? If so, which one? Yeah, yeah! Okay, once we establish what the process is, we come to next part, > which is that it has a white PSD. As I understand it, the PSD is > simply the Fourier transform of the covariance function. You keep on saying the covariance function, but most engineering > texts I've seen say autocorrelation function (the probabilistic > autocorrelation, i.e., Rxx(tau) = E[X(t)X(t+tau)], NOT the one > which does time-domain correlation, i.e., the convolution of > the time-domain function with a time-reversed version of itself). For a process X(t), the covariance function is the function of o variables given by p(s,t)=EX(s)X(t). If X(t) is stationa, then p depends only on the difference t-s, so that we may write EX(s)X(t)=f(t-s) for some function f. This f is the autocorrelation function. I've been being sloppy. though I've been saying covariance function, I've meant autocorrelation function. But o > functions which differ only a set of measure zero have the same > Fourier transform. This process has a covariance function which is 0 > most evewhere. Therefore, its Fourier transform is the zero > function. Right. That's the problem. Notice that we 't have this problem when taking the > Fourier transform of the delta function, even though the delta > function is zero evewhere except the origin. This is because the > delta function is not a function at l, but a distribution, so we > must take the Fourier transform in the sense of distributions. So, in summa, the process you've introduced does not exist as an > ordina process. Perhaps it exists as a random variable taking vues > in the space of Schwartz distributions. If so, it's not clear exactly > what that random variable should be, so you need to explicitly define > it in the language of distribution theo. Once e, you need to be > more explicit about the white PSD part, since, as it stands, the > Fourier transform of your covariance function is the zero function, > which is not white. A good place to start reading about distribution > theo (if you're not ready familiar with it) is Introduction to > the Theo of Distributions by F. G. Friedlander. Excellent - I've been looking for a reference to this exact topic - thanks! Anyway, I hope this helps. I enjoy discussing this stuff because I > enjoy stochastic anysis. But I 't want to annoy or offend you and > I 't want to engage in any kind of emotionly charged discussion > over this stuff. Good luck in your investigations. Please let me know > if you come upon the form way to describe your process. I'd be ve > interested in seeing it. help. The key seems to be in understanding why a continuous-time independent > process isn't a norm process. I have the Oksen book on order. ==== > The key seems to be in understanding why a continuous-time independent > process isn't a norm process. The problem with a continuous-time iid process X(t) is not that it doesn't exist---Kolmogorov's theorem guarantees existence---but that as a function X(t,w) of time and sample point it behaves rather pathologicly. For instance, it cannot be a measurable function of (t,w), so various things one would like to do are problematic. For example, an integr like int_u^v X(s) ds need not exist, let one be a random variable. The process you may be looking for can be described as follows. Let H be the class of square-integrable re-vued functions on the re line R. The white noise W = {W(h) : h in H} is a collection of random variables defined on some probability space with joint norm distributions; each W(h) is mean zero, and for g,h in H E[W(g)W(h)] = int_R g(t)h(t) dt. In particular, if g and h are orthogon, then W(g) and W(h) are independent. Notice that W is stationa in the sense that the joint distributions of the W(h) are unchanged if, for fixed s in R, each h is replaced by the translate h(.+s). ==== > about picking up a higher level book so I can read some more about it. It > rely intrigues me that this one is not elementa, because it seems it > should be. Integr of e^x is elementa, e^-(x^2) is elementa, e^x^2 is > not. I need to read up some more about this. I did find an approximation >> The indefinite integr of e^(-x^2) is no more elementa than that of > e^x^2. > I was working this integr out last night (a definite integr) and power > series didn't seem to work. What am I overlooking? It's obvious we are not tking about the same integr. What makes you think you are overlooking anything? -- Seaman Judge Yohn's mistakes reveed in Mumia Abu-Jam ruling. ==== >> thinking > about picking up a higher level book so I can read some more about it. It > rely intrigues me that this one is not elementa, because it seems it > should be. Integr of e^x is elementa, e^-(x^2) is elementa, e^x^2 is > not. I need to read up some more about this. I did find an approximation >> The indefinite integr of e^(-x^2) is no more elementa than that of > e^x^2. > I was working this integr out last night (a definite integr) and power > series didn't seem to work. What am I overlooking? It's obvious we are not tking about the same integr. What makes you > think you are overlooking anything? -- > Seaman > Judge Yohn's mistakes reveed in Mumia Abu-Jam ruling. > I was evuating the integr of e^(x^2) from 2 to 3 and got something around 12, which I know cannot be right (and I verified it on my TI-83). Moran ==== >> thinking >> about picking up a higher level book so I can read some more about it. > It >> rely intrigues me that this one is not elementa, because it seems it >> should be. Integr of e^x is elementa, e^-(x^2) is elementa, e^x^2 > is >> not. I need to read up some more about this. I did find an > approximation >> The indefinite integr of e^(-x^2) is no more elementa than that of >> e^x^2. >> I was working this integr out last night (a definite integr) and > power >> series didn't seem to work. What am I overlooking? >> It's obvious we are not tking about the same integr. What makes you >> think you are overlooking anything? > I was evuating the integr of e^(x^2) from 2 to 3 and got something > around 12, which I know cannot be right (and I verified it on my TI-83). I 't see what this has to do with what I said about the indefinite integbut since you ask, it depends on what series you were integrating and how many terms you used. For example, if we ask Mathematica to integrate the MacLaurin series, we can get something like In[1]:= Integrate[Norm[Series[Exp[x^2],{x,0,20}]],x] 3 5 7 9 11 13 15 17 19 x x x x x x x x x Out[1]= x + -- + -- + -- + --- + ---- + ---- + ----- + ------ + ------- + 3 10 42 216 1320 9360 75600 685440 6894720 21 x > -------- 76204800 where, if you substitute x = 3, you find that the x^21 term still evuates to around 137.3 and therefore lots more terms are needed for convergence. I found that going to x^100 seems to work. But you probably need more significant digits than you can get on a cculator to add up l those numbers without losing accuracy. On the other hand, if we use a Taylor series centered at the midpoint of the interv of integration, we get In[2]:= Integrate[Norm[Series[Exp[x^2],{x,5/2,10}]],x] 25/4 3 5 Out[2]= (E (-408240 (5 - 2 x) - 177093 (5 - 2 x) + 5 7 5 8 5 9 > 5615280 (-(-) + x) + 4789350 (-(-) + x) + 3752610 (-(-) + x) + 2 2 2 5 11 20393423 (-(-) + x) 5 10 2 > 2727745 (-(-) + x) + --------------------- - 8346240 x + 2 11 6 2 4 373275 (-5 + 2 x) > 1814400 x + 292950 (-5 + 2 x) + ------------------)) / 725760 4 and this converges much more quickly, since the powers of (-5/2 + x) go quickly to zero when evuated at x=2 or x=3. -- Seaman Judge Yohn's mistakes reveed in Mumia Abu-Jam ruling. ==== there's abundant evidence to show taht Newton is a hoax of the Venetian Party. to say that publishers are biased is a gross understatement; > I was a bit surprised to read this. I looked in a few Aman > cculus texts on my shelf, and found Newton & Leibniz with equ > billing for this. --les ducs d'Enron! http://www.wlym.com/antidummies/part17.html ==== that's a great little book! > Getting back to Newton's mathematics, Huygens & Barrow, Newton & Hooke > by V.I. Arnol'd (Birkhauser 1990) is definitely worth reading. --les ducs d'Enron! http://www.wlym.com/antidummies/part17.html ==== r.e.s. ~ you (and many others out there) are l above my head. However, it is nice to know you are out there making sure the information given is correct. I am rely in awe. Keep us l on track!! Kavon > I second that. r.e.s. > Kavon >> and is perfectly consistent (and no, I'm NOT tking about > Non-Standard >> anysis here). >If d^2 = 0, and the anysis is *not* non-standard, > ... which has nothing to do with anything that was said that you're > replying to... > http://search.yahoo.com/search?p=How To Search The Web > http://search.yahoo.com/search?p=Non-Standard Anysis > http://search.yahoo.com/search?p=Wikipedia >then how can d differ from 0? > *Sigh.* > http://search.yahoo.com/search?p=Associative Linear gebras > http://search.yahoo.com/search?p=Vector Spaces > http://search.yahoo.com/search?p=Nilpotent gebras > http://search.yahoo.com/search?p=Finitely Presented gebras > http://search.yahoo.com/search?p=Ides+Infinitesims > http://search.yahoo.com/search?p=Hypercomplex Numbers+Infinitesims > ==== I got redirected to this newgroup for this problem: 5 customers (A thru E) requesting a load of supply: 25, 10, 3, 6 and 7 > tons respectively. Delive-cars (1 thru 6), capacity 8, 16, 24, 32, 32 > and 32 tons resp. go each to 1 customer, go back and stay. In matrix are the haulage-costs: Cust: A B C D E > Car: > 1 6 7 8 9 10 > 2 2 6 9 11 14 > 3 6 7 10 11 12 > 4 6 8 10 12 13 > 5 8 10 11 13 13 > 6 7 8 10 12 13 Request:25 10 3 6 7 tons How do I solve this using Linear Programming (with objective function > and constraints) so that tot haulagecosts are minim? I'm required to > use Maple, tho in that newsgroup they say it can hardly be e that way. > I have difficulties with this since it's an un-equ matrix. There's no requirement in Linear Programming that the number of rows and columns in the constraints be equ. By the way, I learned about LPs before interior point methods were shown to be good ways to solve them. My program (see sig) for solving them is dated, in that sense. I'm not sure if I fully understand what your LP should be. But to find it, here's how I would proceed: 1. Define the decisions variables. (Some of these will be zero in the fin solution.) Must your decision variables be non-negative? 2. Express tot haulage costs in terms of the decision variables. I think the matrix of the haulage-costs will be used here. This is the objective function you want to minimize. 3. Express constraints. It seems to me that satisfying the request of a customer will yield a constraint. I think the capacities of the delive cars will provide coefficients for the constraint. The size of the requested supply will provide the right-hand-side, I think. And you should not get an equity for the constraint. What gorithm have you been taught to solve LPs? If this is not sufficient help, say so. -- T http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html ==== > How do you integrate the function f(x) = x/(tanx) ? > please send substitutes, etc. thanks There is a tricky way. Let S be the definite integr. S=int{0 to pi/2} t cos t / sin t = int{0 to pi/4} t cos t / sin t + int{pi/4 to pi/2} t cos t / sin t = int{0 to pi/4} t cos t / sin t + int{0 to pi/4} (pi/2 - t) sin t / cos t = int{0 to pi/4} t [ cos t / sin t - sin t / cos t ] - pi/2 log cos pi/4 = int{0 to pi/4} t [ cos 2t / sin 2t ] - pi/2 log 1/sqrt(2) = S/2 + pi/4 log 2 and therefore S = pi/2 log 2. Another way: pi cot pi.t = 1/t + 2t/(tt-1) + 2t/(tt-4) + 2t/(tt-9) + ... Now multiply by t and integrate termwise. It's quite cute. T it. ==== > >> How do you integrate the function f(x) = x/(tanx) ? >> please send substitutes, etc. thanks > There is a tricky way. Let S be the definite integr. > S=int{0 to pi/2} t cos t / sin t > = int{0 to pi/4} t cos t / sin t + int{pi/4 to pi/2} t cos t / sin t > = int{0 to pi/4} t cos t / sin t + int{0 to pi/4} (pi/2 - t) sin t / cos > t = int{0 to pi/4} t [ cos t / sin t - sin t / cos t ] - pi/2 log cos > pi/4 = int{0 to pi/4} t [ cos 2t / sin 2t ] - pi/2 log 1/sqrt(2) > = S/2 + pi/4 log 2 > and therefore > S = pi/2 log 2. > Another way: > pi cot pi.t = 1/t + 2t/(tt-1) + 2t/(tt-4) + 2t/(tt-9) + ... > Now multiply by t and integrate termwise. It's quite cute. T it. > -- ==== > >If a sequence of res {s_n} converges, then the corresponding Limsup >equs its Liminf. These 2, however, appear to be independent of the >particular ordering of {s_n}. >>This would be true, by looking at their definitions and considering that >>for any epsilon, the s_i that cause a problem can be excluded finitely >>far into the sequence, regardless of the order. >> Is it accurate to state, then, that >Limsup=Liminf does not necessarily imply the convergence of {s_n}? >Any help/explanation appreciated. >>{s_n} converges iff Limsup=Liminf. Think about epsilon/N proofs. > for a given relationship (n,s_n). For an arbitra e, there exists an > N such that n>N ----> |s_n - A| (k,s_k)it is entirely possible that the implication does not hold. > Obiously I'm confused here. Is there some simple proof that, no > matter what the relationship, the limit of the resulting sequence is > the same? Again, many thanks. Since there are finitely many natur numbers m <=N, for the rearrangement (and possible subset), there must be a K for which l the s_n's <=N are in the s_k's <=K. You compute the new bounda K from the old bounda N. The new bounda may be higher, but it does exist. -- ==== > I wouldn't put in a lot of technic things, nor would I tk about > other news readers or means of posting besides Google Groups because I > use it l the time and I think it has transformed Usenet. Rely... how? Besides the archiving part, I mean. ==== >> I wouldn't put in a lot of technic things, nor would I tk about >> other news readers or means of posting besides Google Groups because I >> use it l the time and I think it has transformed Usenet. > Rely... how? > Besides the archiving part, I mean. Well, it's made it easier to get access to USENET for people who either 't want, or are too clueless, to learn to operate a proper newsreader. himself is a prime example of the latter class. Sometime I miss the days when getting to USENET required configuring UUCP, compiling, instling and configuring B News or C News, finding and negotiating with an admin somewhere for a news feed... or, if you were a student, getting an account from your university news admin. would be excluded from both methods: too technicly illiterate to do it himself, and too afraid of education to have a student account. ==== I wouldn't put in a lot of technic things, nor would I tk about > other news readers or means of posting besides Google Groups because I > use it l the time and I think it has transformed Usenet. Rely... how? Besides the archiving part, I mean. Google Groups makes Usenet easily accessible over the Internet. Some newsgroups, like sci.math, are mirrored to the Internet, so that you have what I now cl Usenet-Internet. That vastly increases the potenti audience. Usenet is the one sure-fire way to say something, and probably get a response--if you know what you're doing--as you're read around the world. Sure, you can put up a webpage, and watch as no one comes. And it takes more effort to start and *maintain* a webpage, as remember, it's not just about throwing up a webpage, as you have to keep it up as well. And that applies to blogs as well. Now posting, you're competing with a lot of other postebut you have the possibility of standing out based on your content and style, so there's a greater potenti I think to step out and have an impact. My feeling is that Usenet-Internet will explode into an ever more important medium of exchange as people see its important features. If you want to see the power of knowing what you're doing, do a Google search on core error and see what person has number 1 and 2. Or you can do a search on my name and math, yes, it's a lot of hate stuff, but my point remains, as notice how many *different* LANGUAGES you'll bump into if you dig, where for some reason, my work is forward enough that Google highlights it, so you end up directed to it. http://lostincomment.blogspot.com/ ==== > If you want to see the power of knowing what you're doing, do a Google > search on core error and see what person has number 1 and 2. Repeating the same phrase over and over despite the fact that it has no re mathematic meaning at l is proof of knowing what you're doing? > Or you can do a search on my name and math, yes, it's a lot of hate > stuff, but my point remains, as notice how many *different* LANGUAGES > you'll bump into if you dig, where for some reason, my work is forward > enough that Google highlights it, so you end up directed to it. Google highlights based on how forward your mathematics is? I tell ya, I've learned l about this Usenet-Internet thing today. -- ==== > I wouldn't put in a lot of technic things, nor would I tk about > other news readers or means of posting besides Google Groups because I > use it l the time and I think it has transformed Usenet. Rely... how? Besides the archiving part, I mean. Google Groups makes Usenet easily accessible over the Internet. Some > newsgroups, like sci.math, are mirrored to the Internet, so that you > have what I now cl Usenet-Internet. That vastly increases the potenti audience. ** You're basicly singing about how good usenet is. We l know that. anyone to post, but how does that affect usenet itself? > My feeling is that Usenet-Internet will explode into an ever more > important medium of exchange as people see its important features. You know, if google searches ways turned up a mush of usenet-internet then we'd have the kind of noise problem that our weblogs are creating now. I 't want this post to turn up when someone types google groups into the google web search. The sign to noise ratio on usenet is terrible, and I'm expecting weblogs to go that way--useful and intellectu for a while, but eventuly degenerating into flames, flames about flames, histories of flames being flamed, etc. My point is that just because something turns out #1 or #2 for google's results for a core error search doesn't mean anyone has to take those results seriously. Plus that mirroring would happen without Google Groups too. Google groups have not transformed Usenet. ==== Crossposts to writing newsgroups left in because, for once, they're relevant. > I wouldn't put in a lot of technic things, nor would I tk about > other news readers or means of posting besides Google Groups because I > use it l the time and I think it has transformed Usenet. Rely... how? Besides the archiving part, I mean. Google Groups makes Usenet easily accessible over the Internet. Some > newsgroups, like sci.math, are mirrored to the Internet, so that you > have what I now cl Usenet-Internet. Quibble: Usenet was available over the internet before there was widespread web access. I think you mean Google Groups makes Usenet accessible over the Web. The Internet is both much older and much larger than the Web. archive. Google extended the capabilities of the search engine and made some improvements in the interface. > That vastly increases the potenti audience. Usenet is the one sure-fire way to say something, and probably get a > response--if you know what you're doing--as you're read around the > world. provide the same heard around the world capability. ==== I wouldn't put in a lot of technic things, nor would I tk about >> other news readers or means of posting besides Google Groups because I >> use it l the time and I think it has transformed Usenet. Rely... how? Besides the archiving part, I mean. Google Groups makes Usenet easily accessible over the Internet. As opposed to the days before Google, when usenet was easily accessible over the internet? (I think you mean easily accessible over the web...) >Some >newsgroups, like sci.math, are mirrored to the Internet, so that you >have what I now cl Usenet-Internet. Um, usenet is _part_ of the internet - ways has been. Again, you mean the web, not the internet. You should definitely write that guide to posting. Just what we need, authoritative information from someone who thinks that the internet means the web. >That vastly increases the potenti audience. Usenet is the one sure-fire way to say something, and probably get a >response--if you know what you're doing--as you're read around the >world. Sure, you can put up a webpage, and watch as no one comes. And it >takes more effort to start and *maintain* a webpage, as remember, it's >not just about throwing up a webpage, as you have to keep it up as >well. And that applies to blogs as well. Now posting, you're competing with a lot of other postebut you >have the possibility of standing out based on your content and style, >so there's a greater potenti I think to step out and have an impact. My feeling is that Usenet-Internet will explode into an ever more >important medium of exchange as people see its important features. If you want to see the power of knowing what you're doing, do a Google >search on core error and see what person has number 1 and 2. How does this show that usenet is important? l it shows is that anyone can post any nonsense he wants. You've derived some sort of power from being able to make a fool of yourself in front of a glob audience? Good for you. >Or you can do a search on my name and math, yes, it's a lot of hate >stuff, but my point remains, as notice how many *different* LANGUAGES >you'll bump into if you dig, where for some reason, my work is forward >enough that Google highlights it, so you end up directed to it. http://lostincomment.blogspot.com/ ==== ... >Google Groups makes Usenet easily accessible over the Internet. As opposed to the days before Google, when usenet was easily > accessible over the internet? (I think you mean easily accessible > over the web...) Indeed, is confused. >Some >newsgroups, like sci.math, are mirrored to the Internet, so that you >have what I now cl Usenet-Internet. Um, usenet is _part_ of the internet - ways has been. No, usenet has various ways of transmission. Internet is one of them, but it can be e (and has been e) with UUCP through di-up lines. -- home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== >... >Google Groups makes Usenet easily accessible over the Internet. As opposed to the days before Google, when usenet was easily > accessible over the internet? (I think you mean easily accessible > over the web...) Indeed, is confused. >Some >newsgroups, like sci.math, are mirrored to the Internet, so that you >have what I now cl Usenet-Internet. Um, usenet is _part_ of the internet - ways has been. No, usenet has various ways of transmission. Oh. I'd ways assumed that internet was the same as described in RFC's. I guess not. >Internet is one of them, >but it can be e (and has been e) with UUCP through di-up lines. ==== >Google Groups makes Usenet easily accessible over the Internet. As opposed to the days before Google, when usenet was easily > accessible over the internet? (I think you mean easily accessible > over the web...) >Some >newsgroups, like sci.math, are mirrored to the Internet, so that you >have what I now cl Usenet-Internet. Um, usenet is _part_ of the internet - ways has been. > >No, usenet has various ways of transmission. Oh. I'd ways assumed that internet was the same as > described in RFC's. I guess not. transport mechanisms than the Internet. Di-up lines with UUCP connection, FIDO, BITNET, EARN. The RFC's for Usenet when transmitted through the Internet. -- home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > Given my *years* of posting on sever newsgroups, and my current > funding situation (as in, few funds) I'm considering writing up a > guide to posting on newsgroups. The way I figure it, in keeping with > the open source mentity, I'd give it away, with people having the > option to contribute, if they found it useful. > Do a search first, it appears that this has ready been e. > I wouldn't put in a lot of technic things, nor would I tk about > other news readers or means of posting besides Google Groups because I > use it l the time and I think it has transformed Usenet. > You are limiting your audience if you do this. It might be worth while to look at a few other newsreaders. -- ==== Springer books, > i have a mini-guide on reparing books at geocities.com/math_libra/links.html -- ==== not sure about those i would have to go through it l again. but you have > to remember that we are determining the potenti in the plane of z=0 Lets > do it here.... we have Laplacian(V) = -n(r)*delta(z)*U(a-r) U=heaviside step function, n=charge density, delta=dirac delta. the > Laplacian is taken in cylindric coords where we assume the potenti is > axiy symmetric about the z-axis. (dropping l dependance on theta). Take the Hankel transform of both sides (or in reity, multiply by > J(kr):=zero order bessel function of first kind, integrate from 0 to oo over > r). i.e. integrate(0,oo) d/dr(r*dV/dr)*J(kr)dr + d^2U/dz^2 = -delta(z) integrate(0,a) > r*n(r)*J(kr) where U(k,z):= transformed V over r . integrate the first term by parts ice - then use the well known recurrance > relations of the bessel functions and we get -k^2*U + d^2U/dz^2 = -delta(z) * N(k) first solve -k^2*U + d^2U/dz^2 = 0 and we get U+/- = A(k)*exp(+/-kz) for z>0 and z<0 respectively. (plus and > minus on U is a reminder of which one we are looking at....we have to have > it like this as the potenti must go to zero as z->+/-oo) where N(k) is the transform of n(r) with limited support. Integrate the > above just over the plane of z=0 i.e. integrate(-e,e), then let e->0 and use > continuity to show the first term vanishes, and we are left with dU-/dz - dU+/dz = N(k) substitute in for U obtained above and take the strict limit as z->0 as we > get A(k) = 1/2 * N(k)/k thus U = 1/2 N(k)exp(-k|z|) / k inverting this using the inverse hankel transform we have V = 1/2 Integrate(0,oo) exp(-k|z|)*J(rk)*J(r'k) dk Integrate(0,a) r'*n(r') > dr' let n(r') = constant =1. and lets evuate this in the plane i.e. z=0 V(r,0) = 1/2 Integrate(0,oo) J(rk)*J(r'k) dk Integrate(0,a) r'dr' the integr over the bessel functions (i think) is V(r,0) = 1/(pi*r) * Integrate(0,a) E((r'/r)^2) r' dr' > Up to here, you are correct (as far as I can tell). The integr formula however is only correct if r'r Sor, for beeing so late with this remark, but I only just found time for it. so forget about my comment about E(-4 r/...), it was just a stupid error on my part. ==== I have a 2n*2n matrix over a field of characteristic 2 of the following form ( I A) ( B I ) where A and B are both n*n matrices such that A_ij = (-1)^(i+j) det (B_j,i) here j and i mean omit the j'th row and the i'th column. The same holds the other way around: B_ij = (-1)^(i+j) det (A_j,i) Moreover, A and B both have determinant 0. My question: is this a well-known matrix? And do these matrices exist if the entire matrix has to have determinant 1. I know that for n=2, these matrices do exist, but need something for bigger n (n=4 is my go). ==== I have a 2n*2n matrix over a field of characteristic 2 of the following form > ( I A) > ( B I ) > where A and B are both n*n matrices such that > A_ij = (-1)^(i+j) det (B_j,i) > here j and i mean omit the j'th row and the i'th column. The same holds > the other way around: > B_ij = (-1)^(i+j) det (A_j,i) Moreover, A and B both have determinant 0. My question: is this a well-known matrix? And do these matrices exist if the > entire matrix has to have determinant 1. I know that for n=2, these matrices > do exist, but need something for bigger n (n=4 is my go). > The flip answer is that the matrices exit for l n. Just take A = B =0. Then the determinant of the block matrix is 1 in any field. Here's a Maple proof that for n = 4 over GF(2) this is the only solution. (Note that, of course, A and B are the (classic) adjoints of each other.) I first find l such matrices and then see which lead to determinant 1 mod 2: > restart: > with(Lineargebra): > A:=Matrix(4,4,(i,j)->x[i,j]): > B:=Adjoint(A): > C:=Adjoint(B): > S:={}: > for i from 1 to 4 do > for j from 1 to 4 do > S:=S union {A[i,j]=C[i,j]}: > od; > od; > sol:=[msolve(S,2)]: > nops(sol); 5 > for i from 1 to 5 do W[i]:=map(x->subs(sol[i],x),A); od; [0 0 0 0] [ ] [0 0 0 0] W[1] := [ ] [0 0 0 0] [ ] [0 0 0 0] [0 x[1, 2] 1 x[1, 4]] [ ] [0 1 0 0 ] W[2] := [ ] [0 0 0 1 ] [ ] [1 0 1 0 ] [x[1, 1] x[1, 2] 1 x[1, 4]] [ ] [ 0 1 0 0 ] W[3] := [ ] [ 0 0 0 1 ] [ ] [ 1 0 0 x[4, 4]] [1 x[1, 2] 0 x[1, 4]] [ ] [0 1 0 0 ] W[4] := [ ] [0 0 0 1 ] [ ] [1 0 1 0 ] [1 x[1, 2] x[1, 3] x[1, 4]] [ ] [0 1 0 0 ] W[5] := [ ] [0 x[3, 2] 0 1 ] [ ] [0 x[4, 2] 1 x[4, 4]] > Id:=IdentityMatrix(4): > for i from 1 to 5 do > M:=Matrix([[Id,W[i]],[Adjoint(W[i]),Id]]): > if Determinant(%) mod 2 = 1 then print(M) fi; > od: [1 0 0 0 0 0 0 0] [ ] [0 1 0 0 0 0 0 0] [ ] [0 0 1 0 0 0 0 0] [ ] [0 0 0 1 0 0 0 0] [ ] [0 0 0 0 1 0 0 0] [ ] [0 0 0 0 0 1 0 0] [ ] [0 0 0 0 0 0 1 0] [ ] [0 0 0 0 0 0 0 1] --Edwin ==== >> I have a 2n*2n matrix over a field of characteristic 2 of the following >form >> ( I A) >> ( B I ) >> where A and B are both n*n matrices such that >> A_ij = (-1)^(i+j) det (B_j,i) >> here j and i mean omit the j'th row and the i'th column. The same >holds >> the other way around: >> B_ij = (-1)^(i+j) det (A_j,i) >> Moreover, A and B both have determinant 0. >The flip answer is that the matrices exit for l n. Just take A = B =0. >Then the determinant of the block matrix is 1 in any field. >Here's a Maple proof that for n = 4 over GF(2) this is the only solution. >(Note that, of course, A and B are the (classic) adjoints of each other.) Wait a minute. For n x n matrices, if I'm not mistaken, Adjoint(Adjoint(A)) = (det A)^(n-2) A. This is an identity of polynomis over Z, so it's vid over any field. So for n > 2, if B = Adjoint(A) and A = Adjoint(B) and det(A) = 0 then A = B = 0. Department of Mathematics http://www.math.ubc.ca/~israel ==== Adj(Adj(A))=det(A)^(n-2) A I can see easily that this holds for invertible A. But for non-invertible A it is a different sto actuly... I am able to proof this so holds for non-invertible A but I would like to prove it in a more direct and simpler way. Am I missing some (simple) thinking step and does it exist? My proof would be: { X | det(X) ne 0 } is an open set in the Zarinski topology. Eve open set is dense in the Zarinski Topology. Since Adj(Adj(X))=det(X)^{n-2}X is a continuous function, which holds for l matrices A, with det(A) ne 0, then it holds for l A, thus so for A s.t. det(A)=0 (since the former matrices are dense). I would think there is a directer proof. But I fail to see it. ideas anybody? Or references if it is ready proven by someone (which will probably be the case, though I cannot find it). thanks! Robert Israel schreef in bht >> I have a 2n*2n matrix over a field of characteristic 2 of the following >form >> ( I A) >> ( B I ) >> where A and B are both n*n matrices such that >> A_ij = (-1)^(i+j) det (B_j,i) >> here j and i mean omit the j'th row and the i'th column. The same >holds >> the other way around: >> B_ij = (-1)^(i+j) det (A_j,i) >> Moreover, A and B both have determinant 0. >The flip answer is that the matrices exit for l n. Just take A = B =0. >Then the determinant of the block matrix is 1 in any field. >Here's a Maple proof that for n = 4 over GF(2) this is the only solution. >(Note that, of course, A and B are the (classic) adjoints of each other.) Wait a minute. For n x n matrices, if I'm not mistaken, > Adjoint(Adjoint(A)) = (det A)^(n-2) A. This is an identity of > polynomis over Z, so it's vid over any field. So for n > 2, if > B = Adjoint(A) and A = Adjoint(B) and det(A) = 0 then A = B = 0. > Department of Mathematics http://www.math.ubc.ca/~israel ==== > frac{parti I(x,y,t)}{parti > t}=left{begin{array}{cc}Phi_{0}(I)&(x,y)in > Omega_{0}Phi_{1}(I)&(x,y)in Omega_{1}end{array}right. > where Omega=[0,M] times [0,N]=Omega_{0} cup Omega_{1}. FYI. The OP is l ASCII. ==== Given a countable set of letters A={L_1,L_2,L_3,...}, consider the >> set of finite strings S(A) which can be formed using letters in A. Is >> S(A) a countable set? yes Yes, but it's not well-orderable is the correct answer. How could you have a countable set that's not well-orderable? How could you have a ration discussion with ZZBunker? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html ==== |At one time I seriously considered ting to find someone else |interested and spend enough time and energy to t to master his |methods and see what the result was. Unfortunately I suppose, I |never did that. Yeah, that's the way things often seem to work, isn't it? I was interested enough to copy key excerpts from one paper on how to do it. It was one of many little projects I have sitting around here somewhere. Since then I've moved, so I suppose my notes are in a box somewhere here. ==== To reprise ' proof, or at least the first o steps (I've included '*' so that GP/Pari works): [begin excerpt] 1. Let P(x) = 14706125 * x^3 - 900375 * x^2 - 17640 * x + 1078, where x is in the ring of gebraic integenotice that P(x) has a constant term that is 1078. 2. It can be shown that P(x)= 7^2*(2401*x^3 - 147*x^2 + 3*x)*(5^3) - 3*(-1 + 49*x)*(5)*(7^2)+7^3 where the *same* polynomi has been put in a form which lows a factorization into non-polynomi factors so that I have P(x) = (5*a_1(x) + 7)(5*a_2(x)+ 7)(5*a_3(x) + 7) where the a's are roots of Q(a) = a^3 + 3*(-1 + 49*x)*a^2 - 49*(2401*x^3 - 147*x^2 + 3*x). [end excerpt] Now, here's where I have a problem. How did Mr. Harris get to this point? It's clear that Q(a) has a binding problem as written; it should be written Q(x,a). I've attempted to compute this polynomi explicitly by assuming three roots in P(x) and getting an ungodly mess, but I do wonder how Q(a) can be derived from P(x), and if so, whether the derivation is correct. To further that end, I'll reexamine the problem, hopefully with more rigor than in my other post (which I 't have the ID for, sor). First, what we want is a polynomi Q(x,y), where Q(x,y) = (y - a_1(x)) * (y - a_2(x)) * (y - a_3(x)). a_i(x) is the actu 'a' root, and I've switched variables for clarity -- at least, I hope it's clear. We know P(x) = K * (x - r_1) * (x - r_2) * (x - r_3), for some K and r_i. Mr. Harris so purports that P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7). This is fine, and actuly works in my favor. K of course is 14706125 = 5^3 * 7^6. Therefore, P(x) = 7^6*(5*x - 5*r_1) * (5*x - 5*r_2) * (5*x - 5*r_3) = (5*7^2*x - 5*7^2*r_1 - 7 + 7) * (5*7^2*x - 5*7^2*r_2 - 7 + 7) * (5*7^2*x - 5*7^2*r_3 - 7 + 7) = (5*(7^2*x - 5*7^2*r_1 - 7/5) + 7) * (5*(7^2*x - 5*7^2*r_2 - 7/5) + 7) * (5*(7^2*x - 5*7^2*r_3 - 7/5) + 7) now I've actuly derived the a_i(x) I want. If I define the function a(r) = 5*(7^2*x - 5*7^2*r - 7/5), I now wonder: what is (y - a(r_1) ) * (y - a(r_2) ) * (y - a(r_3)) ? That's the cubic I want in step 2. Recl that P(x)/14706125 = (x - r_1) * (x - r_2) * (x - r_3) = x^3 - (r_1+r_2+r_3)*x^2 + (r_1*r_2+r_1*r_3+r_2*r_3)*x - r_1*r_2*r_3 therefore r_1 + r_2 + r_3 = 900375/14706125 = 3/49 r_1*r_2 + r_1*r_3 + r_2*r_3 = -17640/14706125 = -72/60025 r_1*r_2*r_3 = -1078/14706125 = -22/300125 And now it's time for the grinding. Apologies for the long line lengths. If you have a monospaced font you might verify that I've matched up terms in the first four equations (the first is the raw output from GP/Pari); I've tried to do this ve carefully. Q(x,y) = (y - a(r_1) ) * (y - a(r_2) ) * (y - a(r_3)) = -14706125*x^3 + (180075*y + (73530625*r_1 + (73530625*r_2 + (73530625*r_3 + 1260525))))*x^2 + (-735*y^2 + (-600250*r_1 + (-600250*r_2 + (-600250*r_3 - 10290)))*y + ((-367653125*r_2 + (-367653125*r_3 - 4201750))*r_1 + ((-367653125*r_3 - 4201750)*r_2 + (-4201750*r_3 - 36015))))*x + (y^3 + (1225*r_1 + (1225*r_2 + (1225*r_3 + 21)))*y^2 + ((1500625*r_2 + (1500625*r_3 + 17150))*r_1 + ((1500625*r_3 + 17150)*r_2 + (17150*r_3 + 147)))*y + (((1838265625*r_3 + 10504375)*r_2 + (10504375*r_3 + 60025))*r_1 + ((10504375*r_3 + 60025)*r_2 + (60025*r_3 + 343)))) = -14706125*x^3 + (180075*y + (73530625*r_1 + 73530625*r_2 + 73530625*r_3 + 1260525))*x^2 + (-735*y^2 + (-600250*r_1 - 600250*r_2 - 600250*r_3 - 10290)*y - 367653125*r_1*r_2 -367653125*r_1*r_3 - 4201750*r_1 - 367653125*r_2*r_3 - 4201750*r_2 - 4201750*r_3 - 36015)*x + (y^3 + (1225*r_1 + 1225*r_2 + 1225*r_3 + 21)*y^2 + (1500625*r_1*r_2 + 1500625*r_1*r_3 + 17150*r_1 + 1500625*r_2*r_3 + 17150*r_2 + 17150*r_3 + 147)*y + (1838265625*r_1*r_2*r_3 + 10504375*r_1*r_2 + 10504375*r_1*r_3 + 60025*r_1 + 10504375*r_2*r_3 + 60025*r_2 + 60025*r_3 + 343)) = -14706125*x^3 + (180075*y + (73530625*(r_1 + r_2 + r_3) + 1260525))*x^2 + (-735*y^2 + (-600250*(r_1 + r_2 + r_3) - 10290)*y - 367653125*(r_1*r_2+r_1*r_3+r_2*r_3) - 4201750*(r_1+r_2+r_3) - 36015)*x + (y^3 + (1225*(r_1+r_2+r_3) + 21)*y^2 + (1500625*(r_1*r_2+r_1*r_3+r_2*r_3) + 17150*(r_1+r_2+r_3) + 147)*y + (1838265625*r_1*r_2*r_3 + 10504375*(r_1*r_2+r_1*r_3+r_2*r_3) + 60025*(r_1+r_2+r_3) + 343)) = -14706125*x^3 + (180075*y + (73530625*(3/49 ) + 1260525))*x^2 + (-735*y^2 + (-600250*(3/49 ) - 10290)*y - 367653125*(-72/60025 ) - 4201750*(3/49 ) - 36015)*x + (y^3 + (1225*(3/49 ) + 21)*y^2 + (1500625*(-72/60025 ) + 17150*(3/49 ) + 147)*y + (1838265625*(-22/300125)+ 10504375*(-72/60025 ) + 60025*(3/49 ) + 343)) = -14706125*x^3 + (180075*y + (73530625*(3/49) + 1260525))*x^2 + (-735*y^2 + (-600250*(3/49) - 10290)*y - 367653125*(-72/60025) - 4201750*(3/49) - 36015)*x + (y^3 + (1225*(3/49) + 21)*y^2 + (1500625*(-72/60025) + 17150*(3/49) + 147)*y + (1838265625*(-22/300125)+ 10504375*(-72/60025) + 60025*(3/49) + 343)) = -14706125*x^3 + (180075*y + 5762400)*x^2 + (-735*y^2 - 47040*y + 147735)*x + (y^3 + 96*y^2 - 603*y - 143332) = y^3 + (-735*x + 96)*y^2 + (180075*x^2 - 47040*x - 603)*y+(-14706125*x^3 + 5762400*x^2 + 147735*x - 143332) Since this does not match Mr. Harris' cubic I must conclude that one of us has made an gebraic or logic error somewhere; I hope it wasn't me but these equations are ugly; much of the work was e by GP/Pari but it isn't smart enough to substitute the symmetric root forms I detailed above. Please check my work here (if you dare! ); I think that I have found the form that the y's (or a's) must satisfy, dependent on x. But it's not pretty -- and it doesn't match '. But at least now it doesn't have fractions in it. ==== > To reprise ' proof, or at least the first o steps > (I've included '*' so that GP/Pari works): [begin excerpt] 1. Let P(x) = 14706125 * x^3 - 900375 * x^2 - 17640 * x + 1078, where x is > in the ring of gebraic integenotice that P(x) has a constant > term that is 1078. 2. It can be shown that P(x)= 7^2*(2401*x^3 - 147*x^2 + 3*x)*(5^3) - 3*(-1 + 49*x)*(5)*(7^2)+7^3 where the *same* polynomi has been put in a form which lows a > factorization into non-polynomi factors so that I have P(x) = (5*a_1(x) + 7)(5*a_2(x)+ 7)(5*a_3(x) + 7) where the a's are roots of Q(a) = a^3 + 3*(-1 + 49*x)*a^2 - 49*(2401*x^3 - 147*x^2 + 3*x). [end excerpt] Now, here's where I have a problem. How did Mr. Harris get to > this point? Actuly it is quite simple. Q(a) is one of the many possible polynomis. And it works the other way around, when you have a1(x), a2(x) and a3(x) roots of Q(a), then P(x) = (5 a1(x)+7)(5 a2(x)+7)(5 a3(x)+7). Q(a) gives the vues for (a1+a2+a3), (a1 a2+a1 a3+a2 a3) and (a1 a2 a3). Writing out the factorisation of P(x) as a single expression and filling in these three vues gives the origin form of P(x). -- home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > I've cled the factoring of polynomi into non-polynomi factors > advanced for a reason, understanding requires that you have the basics > down, are VE well-founded in mathematic logic, and reize that > you will have to make some serious effort. If you honestly believe they are advanced, you know most nothing about math. Unfortunately, I have people who seem to think that replying to me is > an easy exercise, or just some kind of a lark, which has no meaning or > consequence. Easy? No. Frustrating is more like it, especily when you fail to understand the objections. However, if you THINK you're going to be a mathematician, or think you > ready are a mathematician, what you say in reply to me, is a part of > your work as a mathematician. And thankfully, as long as Google is > around, it looks like it'll be part of the historic record till you > die, and beyond. Hmm... I best know where I'll stand, then. Now then, I've come across a nice example to give some perspective on > advanced concepts in factoring polynomis, which has the nice feature > of, I hope, being easy to understand. The polynomi is F(x) = x^2 + x + 2, and the factorization relies on > something you probably know about, but didn't pursue, which is that > x^2 + x is ways even if x is an integer, so I can factor as F(x) = 2(x(x+1)/2 + 1) Where 2 and (x(x+1)/2+1) are the factoI presume. for a factorization vid in integebut not in gener vid in > gebraic integers. This is not a vid factorization in the integethough both factors are integers for any integer x. When you say it's a factorization in the integeyou are *normly* referring to the ring of coefficients being integers. Here you have 1/2 for o coefficients, which is an gebraic number. To move beyond factoring polynomis into polynomi factoyou need > to have a *thorough* understanding of coprimeness, ring operations, > and logic argument. Then you are unquified, unless you are now able to correctly define coprime and understand that Z[1/2] is *not* the res. What I've found is a puzzling lack of a thorough foundation in these > areas from LOTS of people in the math world, as I'll include Bar > Mazur, Andrew Granville, and Rph McKenzie, in with the posters who > have continuly posted in reply to me on this forum. We agree on our terminology. This suggests that the problem lies elsewhere. Basicly, quite a few of you have demonstrated a failure to > understand mathematics at a level I'd cl competent, as the basics > necessa are what I'd *think* any mathematician would have. No, we disagree with you on how *you* think mathematics works. We've studied the subject, you haven't. Why should *you* be the authority? But here, in the 21st centu, mathematicians are displaying a > woeful lack of the basics, besides often displaying childish > petulance, arrogance, and a stubborn refus to acknowledge the math. Who is the one who cls people liars for disagreeing? Who is the one who resorts to cursing? Who is being childish? http://mathforprofit.blogspot.com/ Have you made any profit yet? -- ==== you're just bullting, since you've demonstrated for years that you 't have a *thorough* understanding of coprimeness etc., through mathematic logic. is that presumed to be ordina predicate logic, the symbolic notation of the Aristotelian stuff? does nonpolynomi factors mean monomi ones? > I've cled the factoring of polynomi into non-polynomi factors > advanced for a reason, understanding requires that you have the basics > down, are VE well-founded in mathematic logic, and reize that > you will have to make some serious effort. > your work as a mathematician. And thankfully, as long as Google is > around, it looks like it'll be part of the historic record till you > The polynomi is F(x) = x^2 + x + 2, and the factorization relies on > something you probably know about, but didn't pursue, which is that > x^2 + x is ways even if x is an integer, so I can factor as F(x) = 2(x(x+1)/2 + 1) for a factorization vid in integebut not in gener vid in > gebraic integers. To move beyond factoring polynomis into polynomi factoyou need > to have a *thorough* understanding of coprimeness, ring operations, > and logic argument. > http://mathforprofit.blogspot.com/ --les Dicks d'Enron! http://www.wlym.com/antidummies/part36.html ==== Cantor was mistaken. This is clear to anyone who considers an arbitra even prime >2. though some parts of Cantor have truth, I believe that Cantor is completely bereft of truth. (Actuly, that last sentance was inconsistent and needs correction, but we can still adhere to it without loss of generity.) I have obtained empiric evidence that Cantor's diagon number does not exist, through a long thought experiment. There is no doubt in my mind that Cantor is full of CONTRADICTIONS. Einstien, Hawkins, and Feynmann were l well aware of this. In fact, quantum physics subtly relies on Cantor's work, and it too is therefor fundamently misguided. I have attended prestius universities, so I know what I'm tking about. I have been working on this theo ever since I was 10 years old. Before you t to ste any of the things I will write below, know am now willing to offer a $1000 reward to anyone who finds fault in my theo. A simple curso glance at the Cantoresque Numbers will ready make it clear the something isn't quite right. I'm not good at math, but my theo is conceptuly right, so l I need is for someone to express it in terms of equations. We must remember that the current axiomatic models are only theories, nothing more. though ZFC can be used to accurately predict the outcome of typing things into a cculator, it hardly explains *why* the cculator gives the answers it gives. Like Einstein, I have a speci insight into these matters. My work is on the cutting edge of a paradigm shift. Before we proceed, I'd like to mention that I am ve displeased with the crackpot index located at http://math.ucr.edu/home/baez/crackpot.html, and have written the author to complain that it suppresses origin thinkers (the author, by the way, can't even spell Einstein correctly). When you have read my theo I am sure you will agree it is of Nobel prize cibre. Like Newton, I know I will be preserved in memo as a great pioneer of the sciences. We know from C.S. Lewis ready that Cantor is flawed, I am merely making his defeat more rigorous. In the past, my theories were ridiculed but I was ways able to carefully defend them, leaving my assailants floudering and discredited. They were hidebound reactionaries and self-appointed defenders of the orthodoxy. It is a little known fact that in their correspondence, Dedekind relentlessly opposed Cantor's work, and the only reason he did not make his objections public was out of friendship to Cantor. In his later yeaEinstein himself was groping toward a physic disproof of Cantor's nonsensic works. I am certain that if there are any other civilizations in this universe, more advanced than our own, surely they are aware of the countability of l things. I would have brought this to light sooner, but was constantly being resisted by a damn psychologist (what do psychologists know about these things??) Those who disagree with my theories and support Cantor are no better than nazis, stormtroopers or brownshirts. As Harris has proven time and again, the scientific community is actively engaged in an outright conspiracy to prevent theories like mine from surfacing. At times, amid l this persecution, I truly believe I know how Gileo felt. But only for a time. When my theories break through, modern day mathematics and science will be seen as the laughingstock that they are. I look forward to the day my opponents are forced to recant their objections, or be ostracized from the community. As I have no shown, my theo is ve revolutiona. It is only a matter of time now, for knowledge cannot be suppressed forever. ==== <> For anyone who hadn't noticed yet, this post contains L the 35 items of the crackpot index IN ORDER. -- Wim Benthem ==== < > For anyone who hadn't noticed yet, this post contains L the 35 items of > the crackpot index IN ORDER. > And where is this index? / ==== > >> <> >> For anyone who hadn't noticed yet, this post contains L the 35 items of >> the crackpot index IN ORDER. >> And where is this index? http://www.math.ucr.edu/home/baez/crackpot.html ==== http://www.math.ucr.edu/home/baez/crackpot.html ==== I'm interested in how you have remained at age 11 for the past 5 years? One merely needs to look up Nathan the Great (or Nathaniel Deeth) and Age 11 on Google Groups to see that you have been claiming this age for some time now. You were even accused of this back then: http://mathforum.org/discuss/sci.math/a/m/219236/219315 I can only guess that you are speaking about the equivelent age of your thinking and spelling. . ==== >I have obtained empiric evidence that Cantor's diagon number >does not exist, through a long thought experiment. The diagon number isn't just one number. For any infinite listing of re numbethere is a corresponding diagon number. For example, consider the following infinite list r_0 = 0.5 r_1 = 0.05 r_2 = 0.005 r_3 = 0.0005 etc. Let's take the diagon by the following rule: add 1 to any diagon digit that is less than 5, and subtract 1 from any diagon digit that is greater than 5. This gives us diagon = 0.6666... = 2/3 2/3 definitely exists. -- ==== >I have obtained empiric evidence that Cantor's diagon number >does not exist, through a long thought experiment. The diagon number isn't just one number. For any infinite > listing of re numbethere is a corresponding diagon number. > For example, consider the following infinite list r_0 = 0.5 > r_1 = 0.05 > r_2 = 0.005 > r_3 = 0.0005 > etc. Let's take the diagon by the following rule: add 1 to any > diagon digit that is less than 5, and subtract 1 from any > diagon digit that is greater than 5. You have the right idea, but a slight error in execution: Your rule as stated fails to give your stated result, as a 5 on the diagon is not changed. To get your result, the rule should say something like add 1 to any diagon digit that is less than 6, and subtract 1 from any diagon digit that is greater than 5. This will now modify any diagon digit, including 5, as your rule does not, and will give your stated result. This gives us diagon = 0.6666... = 2/3 2/3 definitely exists. -- ==== Virgil says... >> Let's take the diagon by the following rule: add 1 to any >> diagon digit that is less than 5, and subtract 1 from any >> diagon digit that is greater than 5. You have the right idea, but a slight error in execution: >Your rule as stated fails to give your stated result, as a 5 on >the diagon is not changed. -- ==== Unfortunately Nathan, you fl into the same logic trap that Cantor did. You tacitely assume that it is possible to construct a _countably_ infinite set. I would be interested in seeing your proof, so that I could point out it's flaws. You have f in with those who would suppress the work of luminaries such as Harris when you take such a view. The truth cannot be hidden forever. On my webpage I have posted a proof of the 1-1 mapping, N -> R. However, as a corrola to my proof, I demonstrate that N itself is uncountable. It's there for anyone to view, and unless someone can find a flaw in my proof, we will have to agree that it is correct. Until someone posts a FLAW in this proof I insist that no one be critic of me or the proof, since clearly in mathematics we base things on *PROOF* and not character assasination. I am not especily concerned, as I have shown the proof to my chemist teacher and he assured me it was correct. This certainly quifies me for a Fields met. The proof is right there on my homepage, for anyone with internet access to see. Notice that I do not withhold proof from the newsgroup, so sure am I that it is correct. ==== I am of the firm belief that you and the author of this thread are one and the same (and possibility just an ias of JSH, which would explain his immaturity). Both of you sign your name in the same way and right your age afterwards. Both of you refer to proofs but do not provide links to them (Charlie, neither you nor Justin have websites with your name). You both you are wrong. You cannot state that the natur numbers are uncountable. By definition, a countable set is that of the same cardinity as the naturs. By the way, Nathan how have you been through University at age 11, and Charlie, how have you finished law school by age 9? n > Unfortunately Nathan, you fl into the same logic trap that Cantor did. > You tacitely assume that it is possible to construct a _countably_ infinite > set. I would be interested in seeing your proof, so that I could point out > it's flaws. You have f in with those who would suppress the work of > luminaries such as Harris when you take such a view. The truth cannot be > hidden forever. On my webpage I have posted a proof of the 1-1 mapping, N -> R. However, as > a corrola to my proof, I demonstrate that N itself is uncountable. It's > there for anyone to view, and unless someone can find a flaw in my proof, we > will have to agree that it is correct. Until someone posts a FLAW in this proof I insist that no one be critic of > me or the proof, since clearly in mathematics we base things on *PROOF* and > not character assasination. I am not especily concerned, as I have shown > the proof to my chemist teacher and he assured me it was correct. This > certainly quifies me for a Fields met. The proof is right there on my > homepage, for anyone with internet access to see. Notice that I do not > withhold proof from the newsgroup, so sure am I that it is correct. Charlie Van Winkle, esquire, > Age 9 ==== >On my webpage I have posted a proof of the 1-1 mapping, N -> R. However, as >a corrola to my proof, I demonstrate that N itself is uncountable. It's >there for anyone to view, and unless someone can find a flaw in my proof, we >will have to agree that it is correct. Bah, amateurs. I've proof that the finite set of integers [1, 10^10^10] is uncountable. Just t and count them, one by one, out loud. Report back to me when ready. ==== > Einstien, Hawkins, and Feynmann were l well > aware of this. Is that Jim Hawkins? > Before we proceed, I'd like to mention that I am ve displeased > with the crackpot index located at > http://math.ucr.edu/home/baez/crackpot.html, and have written the > author to complain that it suppresses origin thinkers (the author, > by the way, can't even spell Einstein correctly). So no one who spells Einstein incorrectly is worthy of notice? -- ==== OK, so backtracking a few lines: Take Gamma(1-s)(-2*pi*n)^s-1. Summing > over l integers n other than n=0 and using 3) -Zeta(s)-Sigma(Gamma > 1-s/2*pi*i)Int_|x+-2*pi*n|=Epsilon ((-x)^s)/((e^x)-1) * dx/x=0 then > gives Zeta(s)=Sigma_n=1^infinity(Gamma(1-s)[(-2*pi*n)^s-1 + (2*pi*n)^s-1]. I > 't see how 3) is implemented to give this result. Sor ahead of > time for any Ughs. Well can I present you with an ugh for each asterisk above Let's translate this. Does Gamma(1-s)(-2*pi*n)^s-1 mean Gamma(1-s)(-2pi n)^s-1 or Gamma(1-s)(-2pi n)^{s-1}. The latter seems to make more sense in this context. I can't make head nor tail of what you write for 3). Is n the summation variable? what has happened to the first integr. Is the equs sign before the epsilon rely meant to be there? I presume you are using the contour integr argument for continuing Gamma(s)zeta(s). One introduces a countour C_e in three parts: imagina axis from -infinity to -e, circle radius e about origin, imagina axis from -e to -infinity and take the integr of z^{s-1} e^z on C_e (with ch cut on negative re axis). Thse integr f(s) is independent of e by Cauchy's theorem. It is so an entire function of s: convegence is nice since e^t -> 0 rapidly as t -> -infinity. The integr of z^{s-1} e^z on the first part of the contour is integr_e^infinity t^{s-1} exp(-pi i(s-1)) e^{-t} dt = - integr_e^infinity t^{s-1} exp(-pi is) e^{-t} dt. Similarly on the third part of the contour it is integr_e^infinity t^{s-1} exp(pi is) e^{-t} dt. These add to 2i integr_e^infinity t^{s-1} sin(pi s) e^{-t} dt. If Re(s) > 0 the integr over the circle of radius e is O(e^Re(s)). Letting e -> 0 we get that f(s) = 2i sin(pi s) Gamma(s) for Re(s) > 0. Now we consider g(s) = integr_{C_e} z^{s-1} e^z/(1-e^z) dz where we insist e < 2pi (so that e^z =/= 1 for 0 < |z| < e ). For Re(s) > 1, as with f(s), we can take e -> 0 so that g(s) = 2i sin(pi s) integr_0^infinity t^{s-1} e^{-t}/(1-e^{-t}) dt = 2i sin(pi s) Gamma(s)zeta(s). Consider G_N(s) = integr_{C_{(2N+1)pi}} z^{s-1} e^z/(1-e^z) dz. By Cauchy's theorem the difference G_N(s) - g_e(s) is 2pi i times the sum of the residues of the poles of the integrand at +-2pi i, +- 4pi i, ..., +- 2Npi i, that is 2pi i sum_{n=1}^N [-(2pi ni)^{s-1} - (-2pi ni)^{s-1}] = 2pi i (2pi)^{1-s} sum_{n=1} (-2) cos(pi(s-1)/2)/n^{1-s} = something nasty times sum_{n=1}^N 1/n^{1-s} If Re(s) < 0, as N -> infinity, G_N(s) -> 0. This is because |e^z/(1-e^z)| = 1/(1-e^{-z}) and on the contours C_{(2N+1)pi} e^{-z} is bounded away from 1. Thus for Re(s) < 0 Gamma(s)zeta(s) = something nasty times zeta(1-s). -- ==== OK, so backtracking a few lines: Take Gamma(1-s)(-2*pi*n)^s-1. Summing > over l integers n other than n=0 and using 3) -Zeta(s)-Sigma(Gamma > 1-s/2*pi*i)Int_|x+-2*pi*n|=Epsilon ((-x)^s)/((e^x)-1) * dx/x=0 then > gives Zeta(s)=Sigma_n=1^infinity(Gamma(1-s)[(-2*pi*n)^s-1 + (2*pi*n)^s-1]. I > 't see how 3) is implemented to give this result. Sor ahead of > time for any Ughs. Well can I present you with an ugh for each asterisk above Let's translate this. Does > Gamma(1-s)(-2*pi*n)^s-1 > mean > Gamma(1-s)(-2pi n)^s-1 > or > Gamma(1-s)(-2pi n)^{s-1}. > The latter seems to make more sense in this context. You are correct > I can't make head nor tail of what you write for 3). > Is n the summation variable? what has happened to the first integr. > Is the equs sign before the epsilon rely meant to be there? Yes. I lifted this verbatim from p 13. I presume you are using the contour integr argument for continuing > Gamma(s)zeta(s). One introduces a countour C_e in three parts: > imagina axis from -infinity to -e, circle radius e about origin, > imagina axis from -e to -infinity and take the integr > of z^{s-1} e^z on C_e (with ch cut on negative re axis). > Thse integr f(s) is independent of e by Cauchy's theorem. > It is so an entire function of s: convegence is nice since e^t -> 0 > rapidly as t -> -infinity. The integr of z^{s-1} e^z on the first part of the contour is > integr_e^infinity t^{s-1} exp(-pi i(s-1)) e^{-t} dt > = - integr_e^infinity t^{s-1} exp(-pi is) e^{-t} dt. > Similarly on the third part of the contour it is > integr_e^infinity t^{s-1} exp(pi is) e^{-t} dt. > These add to > 2i integr_e^infinity t^{s-1} sin(pi s) e^{-t} dt. > If Re(s) > 0 the integr over the circle of radius e is > O(e^Re(s)). Letting e -> 0 we get that > f(s) = 2i sin(pi s) Gamma(s) > for Re(s) > 0. Now we consider > g(s) = integr_{C_e} z^{s-1} e^z/(1-e^z) dz > where we insist e < 2pi (so that e^z =/= 1 for 0 < |z| < e ). > For Re(s) > 1, as with f(s), we can take e -> 0 so that > g(s) = 2i sin(pi s) integr_0^infinity t^{s-1} e^{-t}/(1-e^{-t}) dt > = 2i sin(pi s) Gamma(s)zeta(s). Consider > G_N(s) = integr_{C_{(2N+1)pi}} z^{s-1} e^z/(1-e^z) dz. > By Cauchy's theorem the difference G_N(s) - g_e(s) is 2pi i times > the sum of the residues of the poles of the integrand > at +-2pi i, +- 4pi i, ..., +- 2Npi i, that is > 2pi i sum_{n=1}^N [-(2pi ni)^{s-1} - (-2pi ni)^{s-1}] > = 2pi i (2pi)^{1-s} sum_{n=1} (-2) cos(pi(s-1)/2)/n^{1-s} > = something nasty times sum_{n=1}^N 1/n^{1-s} If Re(s) < 0, as N -> infinity, G_N(s) -> 0. This is because > |e^z/(1-e^z)| = 1/(1-e^{-z}) and on the contours C_{(2N+1)pi} > e^{-z} is bounded away from 1. Thus for Re(s) < 0 > Gamma(s)zeta(s) = something nasty times zeta(1-s). and residues more before I can follow this. Plus read other books on RZF to get other perspectives. I'll continue to study this. ==== > >> I can't make head nor tail of what you write for 3). >> Is n the summation variable? what has happened to the first integr. >> Is the equs sign before the epsilon rely meant to be there? Yes. I lifted this verbatim from p 13. I fear not :-( Edwards uses standard notation. -- ==== > >I can't make head nor tail of what you write for 3). > Is n the summation variable? what has happened to the first integr. > Is the equs sign before the epsilon rely meant to be there? Yes. I lifted this verbatim from p 13. I fear not :-( Edwards uses standard notation. > Yes, this is directly from the book, or as close as you can get in ASCII. I have it in front of me now. The summation variable is missing and the equs sign is before the epsilon. Edwards so uses capit Pi for the 'factori' function instead of the Gamma function. He attributes this usage to Gauss and Riemann (footnote p8). The OP has silently changed this. In this chapter Edwards is explaining and expanding upon Riemann's origin paper which is ve terse. This limits him to following Riemann's developement and notation. I find nport, Multiplicative Number Theo chapter 8 clearer as an introduction to the function equation. Jack Fearnley ==== I HATE sports, I hate people who PLAY sports, and I hate the coach! =[ d message > quantify Basebl to other sports; Optim Strategy for > Basebl > Archimedes Plutonium whole entire Universe is just one big atom where dots of > the electron-dot-cloud are gaxies > sci.logic, soc.histo, sci.math > A friend asked me to watch this years World Series to render my > opinion. > I sketchedly watched and here is my opinion, as I usuly do not have > time for such recreation. I watched only parts of game 2 & 3 where the Florida Marlins were > being > overpowered, and missed game 1. And from game 3 I decided it was a > waste > of time to watch games 4 or 5 in that the New York Yanks would > probably > overpower the Marlins and so only watched a few innings of pitching. I > watched nearly the full game of 6. I had seen some clips of Basebl sluggers such as B. Bonds and S. > Sosa > (excuse me if name is incorrect spelling) from the sports section of > the > loc TV news when getting the weather report so I have some awareness > of the best hitters of Basebl. I am not interested in sports only to the extent in which my > anytic > mind can > recast the sport as to Optim Strategies. Basebl Optim Strategy: after watching this last game of World > Series for Basebl > which includes these threads. (1) The pitcher in Basebl, unlike many other sports, is the dominate > feature of basebl when you consider that of 9 players, the pitcher > has > a say in eve offensive action. (2) We can math quantify the dominance of Basebl pitcher with other > sports such as Footbl. Where the quarterback has a big control of > the > offense but not as much of control of the overl game as the Basebl > pitcher for defense. The quarterback if he ran eve play would have > as > much control as the Basebl pitcher. But he does not and has a > handoff > to a running back or a receiver. So we can say mathwise that the > Basebl pitcher has 100% control of defense Say-of-Action whereas in > Footbl the quarterback at most has 25% to 33% Say-of-Action concept. > Because in each offensive play in Footbl, the quarterback either > runs > himself or hands to a runningback or throws to a receiver. But in > Basebl, eve offensive play has a Say by the pitcher with his > pitch. (3) So, unlike eve other sport, the pitcher in Basebl is the > dominate feature because the Say-of-Action is 100% the pitcher. So, the OS of Basebl in order to assemble a team that will win the > World Series for that year, is key in having as many pitchers that are > ace pitchers such as Beckett and Pavano of Marlins. Beckett pitching > was > superior to that of Pettite and Pavano to that of Clemens. Beckett, they > would > have the highest chances of reaching the World Series. By the way, I did not see the pitchers batting, so I guess the game of > Basebl has made some progressive rule change where it is option > for > the pitchers to go to bat; and in the old days I remember the pitcher > was usuly the 9th spot hitter and usuly an easy out. I guess the > new > rule is that when the option is picked that the team rotates the 8 > hitters and leaves out the pitchers as hitters. Getting back to this idea that pitching is the key to basebl > winning. > What does it matter for a team to have great hitting such as a Bonds > or > Sosa if they choke on pitching in that the other team with average > hitters but with great pitching. By the way, did Beckett pitch to > Bonds Now suppose the Yankees had both Bonds and Sosa in their lineup. > Facing the Marlins > had 3 pitchers of the quity of Beckett then I would guess that the > Marlins would have still won the game. I did see clips of the Boston game versus the Yankees and the Boston > ace > pitcher of Martinez (forgive the spelling). So I am guessing that > teams > that have at least one or o good pitchers make it to the playoffs. > Pitching is number one key. So, these concepts would then ask for a Basebl historian to look at > the winning teams and to see whether eve World Series champ had 2 > excellent pitchers such as Beckett and Pavano for Florida So, the above should guide l club owners who aspire to win the World > Series that if your team has at least 2 excellent pitcheis the > basic > prerequisite. And to concentrate the effort more in getting great > pitching than in getting great batting. > The batting of the Marlins was often frustrating and it seemed as > though > homeruns were a rarity for the Marlins so it goes to show that teams > that focus on hitting are not rely the Optim Strategy thread. And finly a discussion of Pitching itself. I would hate to be a > pitcher personly because throwing a bl at 95 mph for many hours > has > a toll on the arm and is a job that does not last too long. I think > there is a Optim-Strategy-Subset for pitching itself in basebl. > The > key is to get the batter out in fast quick time. That means it is no > good to strike out a batter rather than to get him to fly out. If a > pitcher is so good at devising a pitch that is popped up for a fly > out, > then that is better than going for 3 strikes because if you pitch a > bl > with a great spin on it such that the probability when hit popps up > then > conceivably 3 pitches for the inning can retire the side whereas > strikeouts require at least 9 pitches. By the way is there any > statistic > where a pitcher retires a complete side with only 3 pitches in l??? So, the pitcher that can devise a pitch that is prone to pop up is > superior to the pitcher that relies on fastbls and strikes. I am > guessing that some pitch has such a spin on it that the batter is > likely > to pop up or ground out. The perfect pitched Basebl game would have 9 innings and only 9 X 3 > = > 27 > pitches where l batters swung and hit the first pitch and popped out > or grounded out. Not the game where the pitcher made 9 X 9 = 81 > pitches > and l strikes and l strikeouts. whole entire Universe is just one big atom where dots > of the electron-dot-cloud are gaxies ==== > I know the construction of cardins less than eph_0, but this > construction suppose the existence of the concept of set and > implicitly the theo supposses that there are only integer (and > positive) cardins. > The construction could be the most elaborated theo as you want, but > the implicit ideas of the theo, the ideas that have l the > mathematicions when (we or) they think about it, are simpler: the > concept of the (finite) set have to be integer cardin. > Respecting ve much John Conway, and from my knowless, I think that > it's only a way to construct numbebut not the way to contruct > sets, because if we want to extend the idea of cardin, I think > that we have to extend the idea of set. And the problem is doing this > without perversion-circle problem. > Xan. These are actuly games, not sets. For example if you have 3 games of -1/3 > and one game of 1, it is equivent to the game of zero (a second player > win). Likewise, if you a large finite amount of negative infitisms and > one game of one, it is going to be a game where right wins. I 't think > it is possible to have non whole numbered cardins like you want. Well, thanks anyway, n. I will continue think about this problem. If you know anything about Xan. ==== may yo use more details please? ==== given E = sum_{i=1}^{N} sum_{j=1}^{N} w_{i,j} | u_i v_j - d_{i,j}|^2 = sum_{i=1}^{N} sum_{j=1}^{N} w_{i,j} e_{i,j}^2 the paper said, Grad E = 2 sum_{j=1}^N w_{m,j} e_{m,j} v_j for m = 1..N why not treat | u_i v_j - d_{i,j}|^2 as e_{i,j} cdot e_{i,j}^*, if so should we treat u_i^* independent of u_i, leading to Grad E = sum_{j=1}^N w_{m,j} e_{m,j}^* v_j for m = 1..N Lin. S. ==== >>Deck of 52 cards. You start out with k dollars. You place a bet on the >first hand. The game is that you guess whether the next card will be black >or red. The deer flips over a card and if you're right, you receive >whatever you bet, and if you lose, your bet is gone. >My question is, how much would you pay to play this game? In other words, >what is the optim strategy for this game and what is your expected >winnings using this strategy? () >>A review of a collection of various papers on the Kelly Criterion, aka >>Proportion Betting leads to the conclusion that the Kelly approach >>provide a basis for a solution. Kelly betting prescribes betting on >>each turn a proportion of a players holdings as a function of the >>players edge: 2p -1, where p is the probability of a win on that turn. Unless I've made a mistake, Kelly's criterion will lead to exactly the >same expected vue. It does indeed! It took some eaking but I now have an Excel simulation which demonstrates the behavior of the game, using both the pure strategy of waiting for a sure thing bet and the proportion strategy of Kelly betting. The waiting strategy gives exponentily large payoffs with inverse exponenti rarity, averaging to 8.0133 while the Kelly criterion approach gives consistent payoffs at vues near the theoretic vue. Said another way: using sure thing bets, you come up short most of the time while with Kelly betting, the players reizes the vue of the game most of the time. Note that most of the time is not equivent to on average. >Maybe it will help to look at some simple situations. ==== >> I 't feel anything, but l of these were covered in the first >> o years of my undergraduate degree. Well it certainly isn't a fact that they should be covered at the > undergraduate level, hence it's a matter of opinion. Either you feel > 1) that they should be covered at the graduate level, 2) that they > should be covered at the undergraduate level, or 3) you have no > opinion or can't decide If 3) were the case, you would not respond with such a snotty remark > as snotty? that's abuse. That's graduate gebra? because you would have no opinion. Thus one can only conclude that you either feel 1) or you feel 2). In your opinion, what topics SHOULD be covered in a first year > graduate gebra course? And what textbook is suitable for such a > course? 't be asinine. I was merely expressing my surprise > that such elementa topics should be considered as graduate. > And I did suggest a book. Perhaps the use of snotty is different in the UK than it is here in the US. Chez nous, snotty is often used by the person who feels abused. It is used to indicate that one's counterpart is elevated by airs of superiority, unable to maintain a tone of equity. Among us, it would be cled a put-down, rather than abuse. ==== > What's a good book for a first year graduate gebra course? >> Something with ot of emphasis on factorization, polynomi rings, >> fields, PIDs, Gois Theo, and of course l the more basic topics >> of gebra as well like groups, ides, integr domains, etc. >> That's graduate gebra? >> There's a recent book by Joe Rotman Advanced Modern gebra > which does l that, and more. Uhh... Yeah. I mean do you feel like the topics I mentioned should >> be covered at the undergraduate level and not at the graduate level? >> If so, then why do both Lang and Hungerford treat l of the above >> topics in detail in their GTM books? Granted, of course groups, >> ides, and integr domains should be covered at the undergraduate >> level, but should free groups, finitely generated abelian groups, and >> gois theo? Maybe if you go to Harvard or MIT. l the listed topics can (and should) be taught in any good first course >> in abstract gebra. Anyone lacking such fundament gebraic knowledge >> would be poorly prepared for graduate-level studies. -Bill Dubuque Are you serious? I'm sure he is. > If so, maybe I'm just a complete idiot. You said it. > But are you > telling me that Gois theo can and should be taught in a first > course in abstract gebra? We wren't tking about what goes in a first course, but what is graduate and by inference undergraduate level. Gois theo is certainly undergraduate level. -- ==== > >Are you serious? If so, maybe I'm just a complete idiot. But are you >telling me that Gois theo can and should be taught in a first >course in abstract gebra? Granted, l the topics I mentioned >(groups, rings, ides, integr domains, fields, PIDs, factorization) >should be mentioned in a first course on abstract gebra, but how >much detail can you possibly go into in such a short amount of time? >Should free abelian groups, sylow theo, and solvable groups be >included in this introducto course? How about Splitting Fields and >Gois groups? If you say yes then you've lost your noodle, however >these still fl under the catego of group theo or field >theo. I said in my origin post that I'll have finished >hungerford by the time I start reading the new book, so when I refer >to group theo obviously I'm not tking about the definition of a >group, or a homomorphism, though many books will probably mention >that anyway. ==== >> I 't feel anything, but l of these were covered in the first >> o years of my undergraduate degree. Well it certainly isn't a fact that they should be covered at the > undergraduate level, hence it's a matter of opinion. Either you feel > 1) that they should be covered at the graduate level, 2) that they > should be covered at the undergraduate level, or 3) you have no > opinion or can't decide If 3) were the case, you would not respond with such a snotty remark > as snotty? that's abuse. That's graduate gebra? because you would have no opinion. Thus one can only conclude that you either feel 1) or you feel 2). In your opinion, what topics SHOULD be covered in a first year > graduate gebra course? And what textbook is suitable for such a > course? 't be asinine. I was merely expressing my surprise > that such elementa topics should be considered as graduate. > And I did suggest a book. You still haven't said what topics in gebra you consider advanced enough to belong in a graduate gebra course. I'd be curious to see the syllabus for an undergraduate introduction to abstract gebra course which covered: Groups Subgroups Homomorphisms Norm Subgroups / Isomorphism Theorems Lagrange's Theorem ternating & Symmetric Groups Free Groups Finitely Generated Abelian Groups Sylow Theorems Solvable and Nilpotent Groups Rings Factorization in Polynomi Rings (Maxim) Ides Field Extensions Splitting Fields Gois Groups And as I've pointed out, you've avoided at least 3 times the question of what materi DOES belong in a graduate gebra course. ==== > You still haven't said what topics in gebra you consider advanced > enough to belong in a graduate gebra course. I'd be curious to see > the syllabus for an undergraduate introduction to abstract gebra > course which covered: The thing here is that you seem to be equating undergraduate gebra with introducto gebra. I would say that for *most* universities, these o are not the same. The university I went to for undergrad (after 3 courses in linear gebra) have SIX upper-year undergradute courses in abstract gebra (granted, o of these are not intended for pure/applied math majobut still the other four courses cover l the topics you listed earlier.) I copy their course descriptions below: PMATH 345 LEC 0.50 Course ID: 007667 Polynomis, Rings and Finite Fields Elementa properties of rings, polynomi rings, Gaussian intege integr domains and fields of fractions, homomorphisms and ides, Basis theorem, Gauss' lemma, Eisenstein's criterion, unique factorization, computation aspects of polynomis, construction of finite fields with applications, primitive roots and polynomis, addition topics. [Offered: F,S] Prereq: MATH 235/245; PMATH 346 LEC 0.50 Course ID: 007668 Group Theo Elementa properties of groups, cyclic groups, permutation groups, Lagrange's theorem, norm subgroups, homomorphisms, isomorphism theorems and automorphisms, Cayley's theorem and generizations, class equation, combinatori applications, p-groups, Sylow theorems, groups of sml order, simplicity of the ternating groups, direct product, fundament structure theorem for finitely generated Abelian groups. [Offered: W] Prereq: MATH 235/245; PMATH 442 LEC 0.50 Course ID: 007692 Fields and Gois Theo Norm series, elementa properties of solvable groups and simple groups, gebraic and transcendent extensions of fields, adjoining roots, splitting fields, geometric constructions, separability, norm extensions, Gois groups, fundament theorem of Gois theo, solvability by radics, Gois groups of equations, cyclotomic and Kummer extensions. [Offered: F] Prereq: PMATH 345, 346; PMATH 444 LEC 0.50 Course ID: 007694 Non-Commutative gebra Jacobson structure theo, density theorem, Jacobson radic, Maschke's theorem. Artinian rings, Artin-Wedderburn theorem, modules over semi-simple Artinian rings. Division rings. Representations of finite groups. [Note: Offered in the Winter of odd years.] Prereq: PMATH 345; Coreq: PMATH 346 As well as a 400-level course in gebraic number theo and a 400-level course in gebraic graph theo, which use the 345/346 courses as prereqs. The other o abstract gebra courses that are not commonly taken by (pure)math majors: PMATH 334 LEC 0.50 Course ID: 007662 Introduction to Rings and Fields with Applications Rings, ides, factor rings, homomorphisms, finite and infinite fields, polynomis and roots, field extensions, gebraic numbeand applications, for example, to Latin squares, finite geometries, geometric constructions, error-correcting codes. [Note: PMATH 345 may be substituted for PMATH 334 whenever the latter is a requirement in an Honours plan. Offered: F,S] Prereq: MATH 235/245; Not open to Gener Mathematics students PMATH 336 LEC 0.50 Course ID: 007663 Introduction to Group Theo with Applications Groups, permutation groups, subgroups, homomorphisms, symmet groups in 2 and 3 dimensions, direct products, Polya-Burnside enumeration. [Note: PMATH 346 may be substituted for PMATH 336 whenever the latter is a requirement in an Honours plan. Offered: W,S] Prereq: MATH 235/245; Not open to Gener Mathematics students ==== <> : You still haven't said what topics in gebra you consider advanced : enough to belong in a graduate gebra course. I'd be curious to see : the syllabus for an undergraduate introduction to abstract gebra : course which covered: : Groups : Subgroups : Homomorphisms : Norm Subgroups / Isomorphism Theorems : Lagrange's Theorem : ternating & Symmetric Groups : Free Groups : Finitely Generated Abelian Groups : Sylow Theorems : Solvable and Nilpotent Groups : Rings : Factorization in Polynomi Rings : (Maxim) Ides : Field Extensions : Splitting Fields : Gois Groups ==== You still haven't said what topics in gebra you consider advanced >enough to belong in a graduate gebra course. I'd be curious to see >the syllabus for an undergraduate introduction to abstract gebra >course which covered: Nobody said that l of the topics below were suitable for an undergraduate introduction to abstract gebra. The claim was rather that they are l topics that can readily be taught at undergraduate level. In the UK, undergraduate mathematics courses take 3 or 4 yeaand the topics below would be covered in various courses during that period. At my university, Warwick, things like groups, subgroups, homomorphisms, Lagrange's theorem, An and Sn, factorization in polynomi rings over fields would be covered in core (compulso) courses during the first year. Splitting fields, rings, ides, f.g. abelian groups, possibly Sylow's Theorem might be covered in the second year, and most of that would still be core. The other topics, like group presentations, solvable groups, Gois theo would be in option courses in the third or fourth year, but would be studied in some depth. >Groups >Subgroups >Homomorphisms >Norm Subgroups / Isomorphism Theorems >Lagrange's Theorem >ternating & Symmetric Groups >Free Groups >Finitely Generated Abelian Groups >Sylow Theorems >Solvable and Nilpotent Groups >Rings >Factorization in Polynomi Rings >(Maxim) Ides >Field Extensions >Splitting Fields >Gois Groups And as I've pointed out, you've avoided at least 3 times the question >of what materi DOES belong in a graduate gebra course. Where I am, you would not have a graduate level course cled gebra. The graduate courses are much more speciized, and are generly given by people doing (or planning to do) research in that area, who are ting to attract research students. Examples of such topics are homologic gebra, Kac-Moody Lie gebras, simple groups of Lie type, reflection and Coxeter groups, non-commutative rings, ... Having said that, there has been some discussion in recent years about whether we should be giving a collection of gener graduate level courses in topics like gebra, (gebraic) topology, manifolds, ... which would remain more or less stable from year to year. Particularly in gebra, we would have enormous difficulty in agreeing on a syllabus. Derek Holt. ==== Groups > Subgroups > Homomorphisms > Norm Subgroups / Isomorphism Theorems > Lagrange's Theorem > ternating & Symmetric Groups > Free Groups > Finitely Generated Abelian Groups > Sylow Theorems > Solvable and Nilpotent Groups > Rings > Factorization in Polynomi Rings > (Maxim) Ides > Field Extensions > Splitting Fields > Gois Groups l undergraduate level topics. For postgraduate level: commutative gebra modules (projective/injective) homologic gebra simple groups (Lie type and sporadics) centr simple gebras Lie gebras group representations gebraic number theo Hopf gebras etc. though some of these are borderline undergraduate topics. -- ==== > You still haven't said what topics in gebra you consider advanced > enough to belong in a graduate gebra course. I'd be curious to see > the syllabus for an undergraduate introduction to abstract gebra > course which covered: Groups > Subgroups > Homomorphisms > Norm Subgroups / Isomorphism Theorems > Lagrange's Theorem > ternating & Symmetric Groups > Free Groups > Finitely Generated Abelian Groups > Sylow Theorems > Solvable and Nilpotent Groups > Rings > Factorization in Polynomi Rings > (Maxim) Ides > Field Extensions > Splitting Fields > Gois Groups And as I've pointed out, you've avoided at least 3 times the question > of what materi DOES belong in a graduate gebra course. I 't have a syllabus, but below is the course catog description for the undergrad gebra course at Harvard. When I took it, most of the students were sophomores and we used Michael Artin's textbook gebra, which is usuly marketed as an undergrad book. We covered evething in your list, plus did some stuff with representation theo. Math 122. Abstract gebra I: Theo of Groups and Vector Spaces gebra is the language of modern mathematics. Provides an introduction to this language, through the study of groups and group actions, vector spaces and their linear transformations, and some gener theo of rings and fields. Math 123. Abstract gebra II: Theo of Rings and Fields Rings, ides, and modules; unique factorization domains, princip ide domains and Euclidean domains and factorization of ides in each; structure theorems for modules; fields, field extensions. Automorphism groups of fields are studied through the fundament theorems of Gois theo. The descriptions of the graduate gebra courses are: Course introduces ubiquitous gebraic structures and discusses some of Gois theo; the Brauer theo of centr simple gebras; representation theo of finite groups; introduction to gebraic number theo.Prerequisite: Mathematics 123 or equivent. Continuation of Mathematics 250a. Some basic commutative gebra. Loc and glob fields. Study of ide class groups. Lang is often used as a text for 250a, and 250b usuly uses Atiyah and Macd. ==== > I 't feel anything, but l of these were covered in the first > o years of my undergraduate degree. That's Europe. In Ama, undergraduates study liber arts, and > squeeze in only a little of their specity in the first o years. In fact, I believe the standard is that an applicant to a graduate program should have seen the standard topics. You aren't expected to *understand* them until you take your quifiers (first set of exams). In other words, the first year of US grad school catches us up to European BA/BS standards. It's the price we pay for not closing anyone out of higher education. ==== ==== > El Moritz grava .88 la saucisse et au marteau: Question: > o coins were flipped and at least one is a head. What are the > chances for o heads? If you understand French, I suggest you go see > http://www.eleves.ens.fr:8080/home/madore/math/proba.html which des > about l the traps in probability. A problem similar to yours is treated. I'll t to translate it into > English: I visit the Martin's Family. I know that the parents have o children. > I ring at the door and a girl answers. What is the probability for the > other child to be a kid? Your problem is similar, but it adds a lot of conjecture. Conjecture changes the question; sometime it changes the answer. Consider our question, o coins were flipped and at least one is a tail. What are the chances for o tails? Then consider that heads represents boys, and tails represents girls, or let tails represent boys, and heads represents girls. Then, we have similar questions. Then suppose that our question was computer generated. The computer read the coin flip, color coded the outcome and we saw o different colored lights. We selected one and the computer generated our question. That is our question, without conjecture. Thanx for the input. I suppose that if you're arguing this question in France, there are so people over there who believe as I do. I'm in the minority here, it's a sml percentage, but it's a large number. The people over here who disagree with me, plead ignorance. They 't seem to be able to understand my argument. I 't know if they can't, or won't. Your friend from Texas, El Moritz > The natur answer is 1/2. > The sophisticated one is No, because there are four likewise solutions > for the composition of the family (G/G, G/B, B/B, B/G) and the fact that > a girl answered excludes the third possibility, which makes the > probability for the other child to be a boy be 2/3 and 1/3 for a girl. Explication: The problem is too vague to determine which of the solutions is correct. > If we say the older child answers to the door and it's a girl or any > of the child (with the same probability) opens the door and it's a > girl, then the first proposed solution is correct. If, on the other > hand, one says boys never answer to the door and I've been answered to > the door or boys answer to the door only if there is no girl in the > house, and a girl has answered, then the second solution is correct. I tend to think the first solution is correct, but without any further > information, we can't answer to this question. To sum up: > Among families with o children, one of which is a girl, 2/3 have a boy > for other child. But among families with o children, the oldest being > a girl, 1/2 have a boy for second child. ==== > Your problem is similar, but it adds a lot of conjecture. Conjecture > changes the question; sometime it changes the answer. Consider our question, o coins were flipped and at least one is a > tail. What are the chances for o tails? Ok, I'll tell you my answer (I 't even recl who was arguing for which answer, so the other one, feel free to yell): I guess the answer is 1/2 The one we saw has a chance 1/2 to be the forst coin and 1/2 to be the second. Then P(TT|one is a tail) = P(we saw the first)P(TT|the first is tail) + P(we saw the second)P(TT|the second is tail) = 1/2 * 1/2 + 1/2 * 1/2 = 1/2 One other way to see it is that there are four possible double-throws: H H T H H T T T Assuming we saw a tail, the first one must be removed. It remains: T H H T T T But those three throws have not the same probability to have been seen. P(it was the first throw|we see a tail) = P(it was the first throw and we see a tail)/P(we see a tail) = (1/6)/(4/6) = 1/4 The 1/6 comes that there are 6 possible coins to be seen and only one tail in the first throw (by throw, I mean line, not column). The 4/6 comes from the fact that there are 4 tails over the 6 coins. The same probability holds for the second throw. For the third one, we have: P(it was the second throw|we see a tail) = P(it was the third throw and we see a tail)/P(we see a tail) = (2/6)/(4/6) = 1/2 Therefore, given that the first coin we see is a tail, the probability that the o coins are tails is 1/2 . -- ==== Is it correct that for any matrix a Gerschgorin circle disjoint with l the other Gerschgorin circles contains exactly one eigenvue (or the same eigenvue multiple times), and subsequently if the matrix is re then this eigenvue must be re? I think this is what Brauer's Theorem says, but need confirmation. ==== >Is it correct that for any matrix a Gerschgorin circle disjoint with >l the other Gerschgorin circles contains exactly one eigenvue (or >the same eigenvue multiple times), and subsequently if the matrix is >re then this eigenvue must be re? Exactly one eigenvue, with multiplicity one (i.e. only one time), according to the Gerschgorin Circle Theorem. [ Of course you're tking about the closed discs {x: |x-a_{ii}| <= r_i}, not just the circles ] Non-re eigenvues of a re matrix come in complex-conjugate paiand a disc with centre on the re line that contains one member of the pair would contain both. So yes, if the matrix is re this eigenvue must be re. Department of Mathematics http://www.math.ubc.ca/~israel ==== >>Is it correct that for any matrix a Gerschgorin circle disjoint with >>l the other Gerschgorin circles contains exactly one eigenvue (or >>the same eigenvue multiple times), and subsequently if the matrix is >>re then this eigenvue must be re? Exactly one eigenvue, with multiplicity one (i.e. only one time), >according to the Gerschgorin Circle Theorem. [ Of course you're >tking about the closed discs {x: |x-a_{ii}| <= r_i}, not just the >circles ] What if we look at