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 the disks of the matrix and the transpose of the matrix? Let B1...Bn be the disks given by the Gerschgorin Circle Theorem for matrix A, and B~1...B~n be the disks given for the matrix A^T. Since the spectrum is the same for A and A^T we know that the spectrum is in: [ Union ( Bi ) ] Intersect [ Union (B~i) ] but can we instead look at the disks given for corresponding diagon elements (let's cl them a_11, a_22, ..., a_nn) and deduce that each pair of such disks must contain at least one eigenvue and that the eigenvue must be the same for both disks so that: Union [ Intersect ( Bi, B~i ) ] contains the spectrum. Effectively we are throwing away the bigger disk of each pair (Bi, B~i) to obtain a better estimate. I think it holds for l disk pairs (Bi, B~i) that are disjoint from the other disks but what if o or more of them intersect? ==== >What if we look at the disks of the matrix and the transpose of the >matrix? >Let B1...Bn be the disks given by the Gerschgorin Circle Theorem for >matrix A, and B~1...B~n be the disks given for the matrix A^T. Since >the spectrum is the same for A and A^T we know that the spectrum is >in: >[ Union ( Bi ) ] Intersect [ Union (B~i) ] True. >but can we instead look at the disks given for corresponding diagon >elements (let's cl them a_11, a_22, ..., a_nn) and deduce that each >pair of such disks must contain at least one eigenvue and that the >eigenvue must be the same for both disks so that: >Union [ Intersect ( Bi, B~i ) ] > contains the spectrum. No. There's nothing to say that Bi contains any eigenvue (unless it is disjoint from the other Bj's). Consider e.g. [ -1 -r 0 ] [ 1/r 0 1/r ] [ 0 -r 1 ] which has eigenvues 0, i, -i. The Gerschgorin disc about -1, 0 and 1 have radii, |r|, 2/|r| and |r| respectively, so if |r| is sml l the eigenvues are in the disc B2 and not in B1 or B3. But for the transpose, the eigenvues i and -i are in B~1 and B~3 but not in B~2. So 0 is the only eigenvue in (B1 intersect B~1) union (B2 intersect B~2) union (B3 intersect B~3). Department of Mathematics http://www.math.ubc.ca/~israel ==== While dx/dt at x has only one parameter x, the integr from x1 to x2 has o parameters x1 and x2. Iff x2 was ways equ to zero, then derivative and antiderivative would be more similar to each other. Mathematicians might feel this idea somewhat cheeky. However, it has a physic background. Eckard ==== > While dx/dt at x has only one parameter x, > the integr from x1 to x2 has o parameters x1 and x2. Iff x2 was ways equ to zero, then derivative and antiderivative > would be more similar to each other. Mathematicians might feel this idea somewhat cheeky. > However, it has a physic background. Eckard What about functions not defined at zero, such as f(x) = 1/x? The antiderivatives of this function are too important to ignore, but setting limits of integration to include zero is a no-no. ==== > While dx/dt at x has only one parameter x, > the integr from x1 to x2 has o parameters x1 and x2. Iff x2 was ways equ to zero, then derivative and antiderivative > would be more similar to each other. Mathematicians might feel this idea somewhat cheeky. > However, it has a physic background. Eckard But then if you had a function f such that f(0)=/=0, you could not say that f was an antiderivative of f', which would not make sense to do. It would work if you restricted yourself to functions f such that f^(n)(0)=0 for l n but this would exclude l of the nice functions. Have a tolerable existence. Eli ==== Incidently, see my error correction (17.38). By chance, I just posted a pertaining question in sci.physics: (0,inf) or [0,inf). Eckard >>While dx/dt at x has only one parameter x, >>the integr from x1 to x2 has o parameters x1 and x2. >>Iff x2 was ways equ to zero, then derivative and antiderivative >>would be more similar to each other. >>Mathematicians might feel this idea somewhat cheeky. >>However, it has a physic background. >>Eckard > But then if you had a function f such that f(0)=/=0, you could not say that > f was an antiderivative of f', which would not make sense to do. It would > work if you restricted yourself to functions f such that f^(n)(0)=0 for l > n but this would exclude l of the nice functions. Have a tolerable existence. Eli ==== > While dx/dt at x has only one parameter x, > the integr from x1 to x2 has o parameters x1 and x2. Iff x2 was ways equ to zero, then derivative and antiderivative > would be more similar to each other. Mathematicians might feel this idea somewhat cheeky. > However, it has a physic background. Actuly, the fundament theorem says that the former is the opposite of the latter when the first bound remains constant and is the point where the function takes the vue 0. I hope I've been clear. -- ==== > grava .88 la saucisse et au marteau: While dx/dt at x has only one parameter x, > the integr from x1 to x2 has o parameters x1 and x2. Iff x2 was ways equ to zero, then derivative and antiderivative > would be more similar to each other. Mathematicians might feel this idea somewhat cheeky. > However, it has a physic background. Actuly, the fundament theorem says that the former is the opposite > of the latter when the first bound remains constant and is the point > where the function takes the vue 0. Shouldn't that be a point instead of the point? ==== >> While dx/dt at x has only one parameter x, >> the integr from x1 to x2 has o parameters x1 and x2. Iff x2 was ways equ to zero, then derivative and antiderivative >> would be more similar to each other. Mathematicians might feel this idea somewhat cheeky. >> However, it has a physic background. >Actuly, the fundament theorem says that the former is the opposite > of the latter when the first bound remains constant and is the point > where the function takes the vue 0. I hope I've been clear. I do not even understand what fundament theorem you are referring to. Please get more specific. Antiderivative is the same like integr. Therefore one could expect some similarity. I blame R.8en.8e Descartes for the missing fix point of our coordinates. ==== grava .88 la saucisse et au marteau: > I do not even understand what fundament theorem you are referring to. > Please get more specific. The theorem I'm tking about is: If g: x-> Int(from a to x)f(t)dt, then g'(x) = f(x), where a is the vue such that g(a) = 0. But maybe it doesn't answer to your question. -- ==== > grava .88 la saucisse et au marteau: I do not even understand what fundament theorem you are referring to. > Please get more specific. The theorem I'm tking about is: If g: x-> Int(from a to x)f(t)dt, then g'(x) = f(x), where a is the > vue such that g(a) = 0. But maybe it doesn't answer to your question. IIRC, the explicit requirement that g(a) = 0 is redundant, since g(a) = Int(from a to a)f(t)dt = 0 in any case. ==== I have to apologize for confusing x and t. Read either dy/dx at x has only one parameter x or dx/dt at t has only one parameter t. The (in German) so cled integration constant (?) is arbitra. The border t=0 (or x=0) is natur to R+. > grava .88 la saucisse et au marteau: >> I do not even understand what fundament theorem you are referring to. >> Please get more specific. > The theorem I'm tking about is: If g: x-> Int(from a to x)f(t)dt, then g'(x) = f(x), where a is the > vue such that g(a) = 0. But maybe it doesn't answer to your question. > ==== > let A : T1 space show that derived set of A is closed set -------- if A is finite, it is trivi. but in the infinite, i 't know. advice ...please...thank you. In the proof I gave, I actuly used the fact that, in T1 spaces, x is a limit point of a set A if, and only if, eve neighborhood of x contains infinitely many points of A. I'm not sure if this is clear in my first post. Artur ==== Suppose we have a re function f(x) and we must find zero-crossing points. So this is equ to say that we have an equation in the form: f(x) = 0 and we want to find its solutions. If lim (x->t) f(x) / g(x) = 0 / 0 De L'Hopit says: lim (x->t) f(x) / g(x) = lim (x->t) f '(x) / g'(x) = lim (x->t) f ''(x) / g''(x) and so on until f(x) or g(x) become a constant vue. But if f(t)=0 (as hypotesis) then t must be a zero-crossing point, and so a solution of the equation f(x)=0. Now I must define a particular function g(x) that tends to 0 as x tends to t. I so need to eliminate the term t from inside the limit (and so from g(x) ) in order to cculate its vue. So a function that initily tends to 0 as x->t can be: g(x)= e^x - e^t Applying De L'Hopit: lim (x->t) f(x) / (e^x - e^t) = lim (x->t) f '(x) / (e^x) = lim (x->t) f ''(x) / (e^x) ....... In the first limit I can't cculate the vue of t because I have it so inside the limit. So I consider the 2nd and the 3rd limit: lim (x->t) f '(x) / (e^x) = lim (x->t) f ''(x) / (e^x) Applying the properties of limits: lim (x->t) (f '(x) - f ''(x) ) / (e^x) = 0 I define a new function as the argument of the limit: h(x) = (f '(x) - f ''(x) ) / (e^x) At this point suppose we have a limit in the form: lim (x->t) h(x) = 0 then if t exists in h(x), t = h^(-1) (0) (h^(-1) is the inverse function of h) where h^(-1) (0) is a solution of the equation f(x)=0. Anyone can help me find the mistakes? I tried it with Derive 5, but it won't work properly. ==== Suppose we have a re function f(x) and we must find zero-crossing points. > So this is equ to say that we have an equation in the form: f(x) = 0 and we want to find its solutions. If lim (x->t) f(x) / g(x) = 0 / 0 De L'Hopit says: That's putting it sloppily... you should rather say that the limits of f and g are zero, and lose the silly 0/0 notation. lim (x->t) f(x) / g(x) = lim (x->t) f '(x) / g'(x) = lim (x->t) f ''(x) / > g''(x) and so on until f(x) or g(x) become a constant vue. That's *not* what it says. The second equity in your expression holds *only* if both f'(x) and g'(x) have limit 0 at t. Not in gener. But if f(t)=0 (as hypotesis) then t must be a zero-crossing point, and so a > solution of the equation f(x)=0. Now I must define a particular function g(x) that tends to 0 as x tends to > t. > I so need to eliminate the term t from inside the limit (and so from > g(x) ) in order to cculate its vue. > So a function that initily tends to 0 as x->t can be: g(x)= e^x - e^t Applying De L'Hopit: lim (x->t) f(x) / (e^x - e^t) = lim (x->t) f '(x) / (e^x) = lim (x->t) f > ''(x) / (e^x) ....... The second equity does not hold, because the denominator (e^x) is not zero when x=t. ==== > lim (x->t) f(x) / g(x) = lim (x->t) f '(x) / g'(x) = lim (x->t) f ''(x) / > g''(x) and so on until f(x) or g(x) become a constant vue. That's *not* what it says. The second equity in your expression holds *only* if both > f'(x) and g'(x) have limit 0 at t. Not in gener. But if f '(x) and g '(x) are both 0 at t I have to re-apply De L'Hopit in order to cculate the limit... What's necessa is that the initi limit has f(t) and g(t) both = 0. > But if f(t)=0 (as hypotesis) then t must be a zero-crossing point, and so a > solution of the equation f(x)=0. > Now I must define a particular function g(x) that tends to 0 as x tends to > t. > I so need to eliminate the term t from inside the limit (and so from > g(x) ) in order to cculate its vue. > So a function that initily tends to 0 as x->t can be: > g(x)= e^x - e^t > Applying De L'Hopit: > lim (x->t) f(x) / (e^x - e^t) = lim (x->t) f '(x) / (e^x) = lim (x->t) f > ''(x) / (e^x) ....... The second equity does not hold, because the denominator (e^x) > is not zero when x=t. But it's the result of the iterative apply of the Hopit's Rule, and so the initi statement that f(t) = 0 and g(t) = 0 is still vid, I presume. ==== lim (x->t) f(x) / g(x) = lim (x->t) f '(x) / g'(x) = lim (x->t) f ''(x) > / > g''(x) and so on until f(x) or g(x) become a constant vue. > That's *not* what it says. > The second equity in your expression holds *only* if both > f'(x) and g'(x) have limit 0 at t. Not in gener. But if f '(x) and g '(x) are both 0 at t I have to re-apply De L'Hopit in > order to cculate the limit... Correct. But what if they are NOT both zero? Does a reapplication of the rule work in that case? Look again in your textbook if you think the answer is yes. > What's necessa is that the initi limit has f(t) and g(t) both = 0. But if f(t)=0 (as hypotesis) then t must be a zero-crossing point, and > so a > solution of the equation f(x)=0. > Now I must define a particular function g(x) that tends to 0 as x tends > to > t. > I so need to eliminate the term t from inside the limit (and so from > g(x) ) in order to cculate its vue. > So a function that initily tends to 0 as x->t can be: > g(x)= e^x - e^t > Applying De L'Hopit: > lim (x->t) f(x) / (e^x - e^t) = lim (x->t) f '(x) / (e^x) = lim (x->t) f > ''(x) / (e^x) ....... > The second equity does not hold, because the denominator (e^x) > is not zero when x=t. But it's the result of the iterative apply of the Hopit's Rule, and so the > initi > statement that f(t) = 0 and g(t) = 0 is still vid, I presume. Yes, but you can't *reapply* the rule unless f' and g' are *so* both zero at t. ==== Greetings! I know there are o ways to study a game in norm form: using pure strategies or using mixed ones. Consider, for instance, the battle of sexes: 2 +----------+-----------+ | F | C | +---+----------+-----------+ | F | (10, 5) | ( 0, 0) | 1 +---+----------+-----------+ | C | ( 0, 0) | ( 5, 10) | F = footbl +---+----------+-----------+ C = cinema The vues (2/3, 1/3) for player 1 and (1/3, 2/3) for player 2 are Nash Equilibrium (Mixed) Strategies for the game. How should I use this information? If I'm the player 1, should I choose strategy F since its probability 2/3 is greater than 1/3 (the probability of strategy C)? Or this means that if I play this game repeatedly, I should choose strategy F 2/3 of the time and strategy C 1/3 of the time? What is the practic interpretation of mixed strategies? ==== >Consider, for instance, the battle of sexes: > 2 > +----------+-----------+ > | F | C | > +---+----------+-----------+ > | F | (10, 5) | ( 0, 0) | > 1 +---+----------+-----------+ > | C | ( 0, 0) | ( 5, 10) | F = footbl > +---+----------+-----------+ C = cinema >The vues (2/3, 1/3) for player 1 and >(1/3, 2/3) for player 2 are Nash Equilibrium (Mixed) >Strategies for the game. >How should I use this information? If I'm the player 1, >should I choose strategy F since its probability 2/3 is greater >than 1/3 (the probability of strategy C)? >Or this means that if I play this game repeatedly, I should choose >strategy F 2/3 of the time and strategy C 1/3 of the time? If you want to use your mixed strategy from this Nash equilibrium, eve time you play you use some randomization device so that you have probability 2/3 for F and 1/3 for C. >What is the practic interpretation of mixed strategies? though this is a Nash equilibrium, it results in average payoffs of 5/3 for both playelower than the pure-strategy Nash equilibria (1,0),(1,0) and (0,1),(0,1). I 't know why anyone would want to use this Nash equilibrium. Note that though (according to the definition of Nash equilibrium) if one player follows her mixed strategy for the Nash equilibrium, the other has no incentive (in terms of expected payoff) to deviate from his mixed strategy for that equilibrium, in this case it's so true that there's no penty for such a deviation. So even if you know Player 2 is using her mixed strategy (1/3, 2/3), you may as well ways choose F. And then if Player 2 is ration (perhaps a doubtful assumption in Battle of the Sexes), once she figures out you are doing this she will so switch to F as this improves her average payoff. On the other hand, if Player 2 can communicate with you, she will just announce I'm choosing C, and you, if you are ration, will go ong with her choice. Department of Mathematics http://www.math.ubc.ca/~israel ==== This is a question about Fourier transform. The Fourier transform is a separation of a function into sinusoids of different frequency which sum to the origin waveform. I'm looking for a sofare tool whose input is a given function, and whose output is a row of sinusoids which sum to the origin. For example: 1. Lets say that the given function is: y=x^2; 2. I can brake that into (Fourier transform): y=(pi^2)/3-4*{cos(x)/(1^2) - cos(2*x)/(2^2) + cos(3*x)/(3^2) - cos(4*x)/(4^2) + cos(5*x)/(5^2) - cos(6*x)/(6^2)+cos(7*x)/(7^2)...}; Is there a sofare tool which if given (1.) will print (2.) ? Michael ==== >This is a question about Fourier transform. >The Fourier transform is a separation of a function into sinusoids >of different frequency which sum to the origin waveform. That's a ve, ve loose way of putting it. >I'm looking for a sofare tool whose input is a given function, and >whose output is a row of sinusoids which sum to the origin. >For example: >1. Lets say that the given function is: y=x^2; >2. I can brake that into (Fourier transform): >y=(pi^2)/3-4*{cos(x)/(1^2) - cos(2*x)/(2^2) + cos(3*x)/(3^2) - >cos(4*x)/(4^2) + cos(5*x)/(5^2) - cos(6*x)/(6^2)+cos(7*x)/(7^2)...}; I believe you're thinking of the Fourier series, not Fourier transform, of x^2 on the interv [-pi, pi]. The Fourier transform of x^2 (as a tempered distribution) on the re line is -sqrt(2 pi) times the second derivative of the Dirac delta distribution. >Is there a sofare tool which if given (1.) will print (2.) ? In Maple: FourierPartiSum:= proc(f::gebraic, R::name = range, N::nonnegint) # f should be an expression involving a variable x. # R should be of the form x = a .. b. # Returns the parti sum of the Fourier series for f on the interv # a..b up to the terms in sin and cos of 2 Pi N x/(b-a). loc n, x, a, b, res; x:= op(1,R); a:= op([2,1],R); b:= op([2,2],R); 1/(b-a)*int(f,R) + add(2/(b-a)*int(f*cos(2*Pi*n*x/(b-a)),R)*cos(2*Pi*n*x/(b-a)), n=1..N) + add(2/(b-a)*int(f*sin(2*Pi*n*x/(b-a)),R)*sin(2*Pi*n*x/(b-a)), n=1..N) + `...`; end; e.g.: > FourierPartiSum(x^2, x = -Pi .. Pi, 7); 2 Pi --- - 4 cos(x) + cos(2 x) - 4/9 cos(3 x) + 1/4 cos(4 x) 3 - 4/25 cos(5 x) + 1/9 cos(6 x) - 4/49 cos(7 x) + ... Department of Mathematics http://www.math.ubc.ca/~israel ==== Suppose R is a ring with s = s ^ 2 for each s in R. Why s + s = 0? ==== > Suppose R is a ring with s = s ^ 2 for each s in R. Why s + s = 0? What is (xy)^2, what is (x+y)^2? ==== >Suppose R is a ring with s = s ^ 2 for each s in R. Why s + s = 0? Does ring mean, to you, that R has a multiplicative identity 1? If it does, then there's a ve easy proof. ==== >>Suppose R is a ring with s = s ^ 2 for each s in R. Why s + s = 0? > Does ring mean, to you, that R has a multiplicative identity 1? > If it does, then there's a ve easy proof. Is it easier than: s + s = (s + s)^2 = s^2 + s^2 + s^2 + s^2 = s + s + s + s ? -- ==== >Suppose R is a ring with s = s ^ 2 for each s in R. Why s + s = 0? > Does ring mean, to you, that R has a multiplicative identity 1? >> If it does, then there's a ve easy proof. Is it easier than: s + s = (s + s)^2 > = s^2 + s^2 + s^2 + s^2 > = s + s + s + s ? Uh, erm, it's shorter by three instances of ^2. That's easier, right? ==== Suppose R is a ring with s = s ^ 2 for each s in R. Why s + s = 0? >> Does ring mean, to you, that R has a multiplicative identity 1? > If it does, then there's a ve easy proof. >>Is it easier than: >> s + s = (s + s)^2 >> = s^2 + s^2 + s^2 + s^2 >> = s + s + s + s Uh, erm, it's shorter by three instances of ^2. > That's easier, right? That's 2s = (2s)^2 = 4s versus s+1 = (s+1)^2 = 3s+1 Or milk more by homogenizing x^2 = x x + y = (x + y)^2 = x + xy + yx + y => xy + yx = 0 (homogenous in x,y) => 2x = 0 via y = 1 (or x) -Bill Dubuque ==== > Suppose R is a ring with s = s ^ 2 for each s in R. Why s + s = 0? What is (1 + s)^2 ? -- ==== -- Robin Chapman ìb.a6l«.97 > Suppose R is a ring with s = s ^ 2 for each s in R. Why s + s = 0? What is (1 + s)^2 ? -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > Needless to say, I had the last laugh. > an Partridge, _Bouncing Back_ (14 times) ==== Sor for that stupid question, but how is cled that function in english? I can't find this word in my dictiona. ==== > Sor for that stupid question, but how is cled that function in english? > I can't find this word in my dictiona. That isn't English. It's Mathematica. -- ==== >Sor for that stupid question, but how is cled that function in english? >I can't find this word in my dictiona. y = f(x) is read literly as y equs f of x. Or to be more expressive you could say y is a function of x given by y equs f of x. ==== What does the function do? Do you possibly mean the ceiling (floor) function? For example, f(x) = [x] --> f(3.4) = 4, an example of the ceiling function. Or, conversely, the floor function: f(3.4) = 3. Lurch > Sor for that stupid question, but how is cled that function in english? > I can't find this word in my dictiona. ==== I recently became aware of a book cled Probability Theo : The Logic of Science by E. T. Jaynes. The reviews of this book in Amazon were intriguing, particularly the one by Michael Hardy. I consider myself a pure mathematician to whom probability theo is a subset of measure theo, but I've so worked with statistics and inference models (in bioinformatics). I'm curious to know what people think about this book. Does it present a useful theo of statistic inference? How rigorous is it? an ==== >I recently became aware of a book cled Probability Theo : The >Logic of Science by E. T. Jaynes. The reviews of this book in Amazon >were intriguing, particularly the one by Michael Hardy. I consider >myself a pure mathematician to whom probability theo is a subset >of measure theo, but I've so worked with statistics and inference >models (in bioinformatics). This url will connect you with a thread of discussion on Sci.stat.math ng which was an interesting and lengthy debate in part beeen Hardy and me of one aspect Bayesian statistics. Based on your comment about Hardy's review--I got to Amazon to read his review for myself as fast as I could. http://home.rochester.rr.com/jbxroads/interests/sci.stat.math/ Rodriguez provided an interesting summa of the thread midway. (quoting) ===== Rubin: The people that use entropy or whatever other so cled neutr priors are using unjustified computation copouts. My position: 1) Hurray for Bailey! 2) Sure but they should know better. 3) I disagree with Rubin's position with l the energy in my reproductive system. (end of quote) Incidently, Chapter 11 of Jaynes book covers his views on the topic. A useful current reference: http://xyz.lanl.gov/abs/hep-ph/9512295 >I'm curious to know what people think about this book. Does it present >a useful theo of statistic inference? How rigorous is it? I will leave the judgement of rigor to mathematicians. (Quis justodiet ipsos justodes?) Judgments about usefulness are the perogative of engineers. As an engineer, I have found Jaynes and the works those building on his work to be enormously vuable. I am appled at the amount of relius fervor attending the criticism of Bayesian methods. To me they are logic and extremely useful. If they are flawed in rigor, perhaps a Laplace or Fourier to his Heavyside will emerge. ==== [..] >> However, in res, NO numbers are coprime, as for instance, 2(3/2) = >> 3, so 2 is a factor of 3. Again wrong, l but 0 are coprime. That is correct, and I was wrong, as in fact, since *l* numbers but >0 are units in res, it'd be the case that l would be considered >trivi factoso pushing the definition to the limit, 3 is coprime >to 6, though it's still a factor of 6. Pushing the definition to the limit? What nonsense is that? Nobody is pushing anything. We were simply APPLYING the definition YOU gave, explicitly and precisely as given, to the situation described. That's what one does with definitions. >It's starting to look like that word coprime is problematic, eh? Hmm... I think I finly understand. When you say problematic, you mean it confused me. So, when you say coprime is problematic, you mean you made a fool of yourself by insisting no less than 4 times (three of them while ting to gloat) that something was correct, when according to the definition you gave it was just plain fse. And, given l your confusion about gebraic integemaybe that's why you think there's a problem with the definition? Because you 't get it? No, there is nothing problematic about the word 'coprime', except perhaps that you are using a NONSTANDARD meaning of the word. But whether you use the standard meaning, or the meaning you gave it, so long as you STICK TO IT there are no problems. The only problem is you 't know what a definition is, and you 't know how to use them. That's what's problematic: your rampant incompetence and ignorance. [..] ====== ething; self-evident honesty of conviction, which does more; and a long purse, which does most of l. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine tgustus de Morgan ====== Welcome to sci.math/sci.logic, where loathsome mutants gyre > and gimble in good company. Cool. Way cool. But how do you tell them apart? (Assuming you want to.) If you 't know how yet, ask John Quincy Hutchings. ==== The task is to choose h(t) that maximizes the integr: int_a^b e^(-rt) * (p(t) - c(t)) * h(t) dt, subject to: int_a^b h(t) dt = W (some given constant) and where p(t)-c(t) is concave down (ie, upside-down-parabola-like). My book says that it is evident, due to the simple form of the integrand in the objective, that the required h(t) is given by: h(t) = W * delta (t - t_m), where t_m that time giving the maximum vue of e^(-rt) * (p(t) - c(t)). I 't dispute the truth of the claim, I just t see what's so evident about it - can anyone out there explain the obviousness of the Dirac delta as the solution to this problem? thanks a bunch for any insights, cdj ==== Random thoughts on creating a theo of sets prior to a theo of propositions and quantifiers: Let's start with the empty set, 0, and logic identity, =, then we can define T, for true, by T =def 0 = 0 Let's define ordered pair a la Kuratowski, then we can define conjunction by phi & psi =def

= (So formulae are sets--but why not? If our logic is to be infinita, with infinite conjunctions say, we'll want formulae to be sets anyway.) Now if-then is defined by phi -> psi =def (phi & psi) <-> phi (<-> is, of course, just another way of writing =; I told you we'd want formulae to be sets.) Throwing caution to the wind, we'll low ourselves to have a univers set, so that we can define the univers quantifier by (for l x)phi =def {x | phi} = {x | x = x} F (for fse) =def (for l p)p (p a formula without free variables.) ~phi =def phi -> F phi v psi =def (phi -> psi) -> psi (exists x)phi =def ~(for l x)~phi. Now, what theo of sets will yield what logic of propositions and quantifiers? -- G.C. ==== Are you saying that a set, empty or not, is identic with itself? How can something that is identic with itself be distinguished or counted as 'one' itself, if there are no means of distinguishing itself from what it is identic with? And if there is no means to distinguish or count, how do we present the case 0 = 0 as something being identic with itself? JJ > Random thoughts on creating a theo of sets prior to a theo of > propositions and quantifiers: Let's start with the empty set, 0, and logic identity, =, then we can > define T, for true, by T =def 0 = 0 Let's define ordered pair a la Kuratowski, then we can define > conjunction by phi & psi =def

= (So formulae are sets--but why not? If our logic is to be infinita, > with infinite conjunctions say, we'll want formulae to be sets anyway.) Now if-then is defined by phi -> psi =def (phi & psi) <-> phi (<-> is, of course, just another way of writing =; I told you we'd > want formulae to be sets.) > Throwing caution to the wind, we'll low ourselves to have a univers > set, so that we can define the univers quantifier by (for l x)phi =def {x | phi} = {x | x = x} F (for fse) =def (for l p)p (p a formula without free variables.) ~phi =def phi -> F > phi v psi =def (phi -> psi) -> psi > (exists x)phi =def ~(for l x)~phi. Now, what theo of sets will yield what logic of propositions and > quantifiers? -- ==== Could somebody up there help me out by plugging the following into Mathematica or equivent and (I m sure there is a result because the Mathworld 'Integrator' gives me one for the indefinite case - but it's a bit messy) Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 + 2*Cos[x]]],{x,0,Pi}] ex ==== > Could somebody up there help me out by plugging > the following into Mathematica or equivent and (I m sure there is a result because the Mathworld 'Integrator' > gives me one for the indefinite case - but it's a bit messy) Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 + 2*Cos[x]]],{x,0,Pi}] > ex What is D? ==== just a re constant ex flip escribi.97 en el mensaje > Could somebody up there help me out by plugging > the following into Mathematica or equivent and > (I m sure there is a result because the Mathworld 'Integrator' > gives me one for the indefinite case - but it's a bit messy) > Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 + 2*Cos[x]]],{x,0,Pi}] > ex What is D? ==== > Phil Holman grava .88 la saucisse et au marteau: f(p) = q > where f(140) = 15000 > and f'(140) = -100 R = pq > What is dR/dp (p=140) > I get (f'*p) + (f+f')*dp > = (-100*140) + (14900*1) > = -14000 + 14900 > = +900 R = pf(p) dR/dp = f(p) + pf'(p) So dR/dp |p =140 = f(140) + 140*f'(140) > = 15000 - 14000 > = 1000 If we apply this practicly. p=price, q=quantity sold and R=revenue. So if we increase the price from 140 to 141, we decrease the quantity sold from 15000 to 14900. This will increase the revenue by 900 and not 1000. I guess the problem here is with a changing derivative beeen a p=140 and p=141. This gave me a re problem and hence my hacked solution. Do you have any other comments about this. Phil Holman ==== >>In intege2 and 3 are coprime as are 12 and 13, as it simply means >>they 't share non-unit factors. >>However, in res, NO numbers are coprime, as for instance, 2(3/2) = >>3, so 2 is a factor of 3. But 2 is a unit, so your claim here is fse. Hmmm...now that's interesting, eve re but 0 is a unit, eh? Yup. Eve element of a field, except 0, is a unit. Fascinating perspective. Isn't it? So you have coprime with one clear meaning in a ring like integers, >but things are different in a field. Notice then that 3 and 6 are coprime, but 6 still has 3 as a factor, >so in fact, coprime simply loses any relevance. I think that's telling. I think so too. It tells us that you have little or no clue, and no desire to obtain one. Coprime, like divides and factor, are terms related to the divisibility structure of a ring. They are only as interesting as the divisibility structure of the ring is interesting. In a field, the divisibility structure is completely uninteresting: eve nonzero number divides eve number. Since the divisibility structure is not interesting, it so makes the notions of coprime and divides uninteresting. Just like the concept of odd or even: these are notions that are interesting in the integers. A number is even when it is a multiple of 2, and it is odd if it is NOT a multiple of o. The definition of odd and even becomes completely uninteresting in the ration numbers, where evething is a multiple of 2. That does not mean that the notion of odd or even is broken, or silly. It just means that it is useless and uninteresting in that context. So, yes, in a field, coprime loses its relevance; because coprime is only as relevant as the notion of divisibility is interesting, and the notion of divisibility is not interesting in a field. >Part of my point here is that the mathematics you've taken for >granted, with lots of definitions that seem ok, goes off into some >interesting places. The definitions do not seem ok. The definitions ARE. Just because a notion which is important in one context becomes unimportant in another does not mean that the notion is broken. It means that it is not interesting in l contexts. Few things are interesting in l contexts. For instance, the order in the integers: a> Actuly, in the re numbeANY O NONZERO NUMBERS ARE >> COPRIME. That's because, given any o nonzero re numbers x and y, >> any common divisor of x and y is a unit. Well, then eve re number but 0 is a unit follows from that >> position. As that is the definition of a field, it looks about right. So mathematicians take these positions, which not only go against >common sense, they 't make sense in gener, pushing definitions. Nobody is pushing definitions. We are USING the definitions. What is it that goes against common sense, anyway? You defined 'coprime' as not having any non-unit common factors. If l common factors are units, then they are coprime. 1 is coprime to EVE integer, is it not? And 1 DIVIDES eve integer, does it not? So, what is wrong with something both dividing and being coprime to something else? Nothing per se... Except that you screwed up, and rather than admit that you made a mistake, you t to blame it on evebody else. As usu. >So you have this broken word coprime which has no use at l if >you're in the field of re numbers. It has no use at l in a field, but that does not mean the word is broken; it means the concept it defines is not interesting in that context. >> I guess you could say that and it doesn't change things in any >> meaningful way. But it's *fascinating* that Arturo Magidin pushed that position! Are you contradicting it? Nope. Weasel. Of course you were ting to claim I was wrong. You went to great lengths to imply it, only to have it blow up in your face. So now you t to weasel out of your own errors by claiming that the problem is NOT that you screwed up (as is evidently the case), but that the definition is somehow broken. Same thing as with the gebraic integers. You screwed up, and rather than take responsibility, you blamed the definition. > I'm just highlighting how screwed up things get when >mathematicians are left to their own devices. There is nothing screwed up, except your notions and arguments. >> Well consider then, he's saying that 3 is not a factor of 6 in res >> because they're coprime! Hmmm...that's my statement. It looks off in retrospect, as instead >coprime is broken, so that what it means in integers isn't what it >means in res. Of course not. Like I TOLD YOU DOZENS OF TIMES: coprime is contextu. It is something that depends on what ring you are working in. Of COURSE it doesn't mean the same thing in the integers as it means in the res, just like divides does not mean the same thing, and factor does not mean the same thing. Duh. >So you have 3 is coprime to 6, 3 is still a factor of 6, but it's a >unit, or trivi factor, which in one sense is ok, In eve sense it is okay. >as eve re but 0 >is a factor of eve other re, but the word coprime is now a >liability. No, it is not a liability. It is simply a concept and notion that is useless in the context of re numbers. >> And you know what? I think the way mathematicians usuly go, he's >> right!!! And indeed. But apparently you have no idea about the mathematic >> meaning of coprime. For those who wonder, coprimeness is usuly defined to ignore >*trivi* factowhere unit factors are, of course, trivi. No, for those who wonder, coprimeness has many definitions; the most common one, in my experience, is that o elements are coprime when there is a linear combination that equs 1. The second most common in my experience is that there is no prime ide that contains the princip ides generated by the o elements. divisors other than units. This definition is equivent for the integers to the more usu ones, and so for the gebraic integers. It is not equivent in other settings. The definition is weaker in gener (that is, if o things are coprime in the sense of the previous paragraph, they are coprime in the way uses the work, but the converse does not ways hold). In addition, the definition uses does not work well when going to larger rings: if o things are coprime in one ring, they need not be coprime in a larger ring. And if o things are NOT coprime in a ring, they need not be non-coprime in a larger ring. By contrast, the definition in terms of linear combinations does satisfy one implication: if o things are coprime in R, then they are so coprime in any larger ring. However, it is possible for o things to be non-coprime in one ring, and yet be coprime in a larger one. So, for those who wonder: No, never take 's word for what is e usuly. He is a self-confessed ignoramus of what is 'usuly' e. >So 2 is coprime to 3 in integebut they both have 1 as a trivi >factor, of course. But notice how the word coprime gets broken when you end up where >*eve* factor is trivi! It does not get 'broken', it just becomes uninteresting. >Then you can say 3 is coprime to 6, in res. Yes. So what? >Most of you do a quick switch in your heads when you're operating in >the re world, so that you handle the problem, and operate in a >particular ring based on your particular needs at the time. Duh. Coprime depends on the particular ring you are working at. What ring you work in depends on your particular needs at the time. That's why you claim to have invented the object ring, after l: to fill the needs you think you have. >Mathematicians, well, they basicly do the same thing, but *claim* to >be more precise. We give the definition explicitly. We apply the definition precisely. You screw up, and you blame mathematicians for your errors. >> Fun stuff, eh? Yeah. HELL YEAH!!! Mathematicians have broken or screwed up terms l over >the place, but tend to make ad hoc rules to handle them. For instance, with coprime an ad hoc rule would be NOT to use the >word with a field!!! There is no such 'ad hoc' rule. You just pulled it out of a certain orifice, which is not the one that your dentist checks. >The math world is spectacularly illogic and quirky, and uses lots of >end-runs and ad hoc rules to cover up messes, Whereas you just use one constant rule: whenever you mess up, you blame someone or something else. I can see the advantages in it; certainly simpler. Too bad it brings you no closer to reity. ====== ething; self-evident honesty of conviction, which does more; and a long purse, which does most of l. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine tgustus de Morgan ====== ==== wow. I had noticed that he made a strange statement about primity (viz co-primity), a nonsequiter. do we have to define nonsequiter, now, in another endless regress? you've made your points, Magadin; it's time to let sleeping (braindead, whatever) dogs lie; is it not? he relly does have a problem with mathamaticians, more-so than any baccelaureate in physics I've heard of; one rely wonders if he was ways the smartest in his class, or just the smartest smart-ass. it's like the New York Legislature, supposedly nominated TR for President, just to get rid of him -- which was not a ve good idea! >So you have this broken word coprime which has no use at l if >you're in the field of re numbers. > No, it is not a liability. It is simply a concept and notion that is > useless in the context of re numbers. --ils duces d'Enron! http://tarpley.net/bush8.htm http://www.wlym.com/PDF-SpReps/SPRP13.pdf ==== [materi which has been repeated often ped] I think I begin to see what JSH is saying with this constant-term business. I suspect others have seen it as well, but below is my take on it. > 5. Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. Most posters seem ok with the steps up to the fin o. In what > follows it's important to understand the word coprime. In intege2 and 3 are coprime as are 12 and 13, as it simply means > they 't share non-unit factors. However, in res, NO numbers are coprime, as for instance, 2(3/2) = > 3, so 2 is a factor of 3. That's an important point to consider going forward. 6. However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. But for o of the factors of P(x), the constant terms is 7, which is > coprime to 22. Therefore, *none* of the constant terms can have 7 as > a factor. (By saying that 7 is coprime to 22, I'm making a choice as to where > the proof is going. Since I've been tking about gebraic integers, > where 7 is coprime to 22, it's natur to go with a choice where 7 is > coprime to 22.) > This is the heart of it. JSH says [1] 5*b3 + 22 has constant term 22, and that constant term is coprime to 7. Correct. Here, I believe, is how the thinking goes. If you have a polynomi with integer coefficients, say, F(x) = a*x^3 + b*x^2 + c*x + 22, then that *polynomi* is coprime to 7, because one of the coefficients itself is coprime to 7. That is true whether the coefficients are integers or gebraic integers. There is no possible common factor of F(x) with 7. The constant term, F(0), is 22, and that is l you need to know to say that the polynomi is coprime to 7. This statement seems to me to be correct. One says that a polynomi with integer coefficients is coprime to an integer m if *any one* of its coefficients is coprime to m. Here the constant coefficient happens to be coprime to 7, and that is enough. Of course in [1] above, b3 is not a polynomi. It is a function of x. It takes vues in the gebraic integers. But perhaps the same reasoning that applies for polynomis applies for b3(x). There are however o problems: 1. Even though F(x) *as a polynomi function* is coprime to 7, it may be the case that for particular vues of x, F(x) is NOT coprime to 7. For example, say F(x) = x^3 + 3*x + 22. Then F(0) = 22, which is certainly coprime to 7. However, if x = 4, F(x) = 64 + 12 + 22 = 98, which is 7*14. To which JSH may say: So what? It is still true that F(x) is coprime to 7 AS A POLYNOMI !!! To which I would say: Yes, but what you NEED is that, for *specific vues* of x, F(x) *as a number, not as a polynomi*, is coprime to 7. That is what you need in your argument. And you have no proof that it is true. Knowing that the constant F(0) of a polynomi is coprime to a number w does NOT tell that F(x) is coprime to w for l x. This is a slightly subtle point, and the crux is the phrase as a number, not as a polynomi for *specific vues* of x. 2. As noted above, b3(x) is not a polynomi function of x. In gener it cannot be written as a finite sum of powers of x. If it were written as an infinite power series, in gener the coefficients will not be integers or gebraic integers. In gener they will be elements of a field. In gener it makes no sense to consider whether the coefficients are divisible by, say, 7. They are elements of a field. For nonpolynomi functions in gener, one does not even consider whether the function is coprime to 7. Take sqrt(x + 1), for example. Of course sqrt(0 + 1) = 1, so the constant term is coprime to 7. If you expand g(x) = sqrt(x + 1) in powers of x, the coefficients are not integers; they are ration numbebut not integers. So the definition of coprime that one would use for polynomis does not apply. And again, for *specific vues of x*, sqrt(x + 1) is not coprime to 7: x = 48, for example. JSH may reply to the effect that I am ting to obfuscate or that I am lying, etc.. After l, I have said that for a polynomi F(x) with integer or even gebraic integer coefficients, if the constant term is coprime to 7, then the WHOLE POLYNOMI is coprime to 7. Therefore I am agreeing with what he says and that should be the end of it. Then I bring in this confusing side-business about having to consider F(x) for specific vues of x, and tking about F(x) AS A NUMBER rather than as a POLYNOMI. I giveth, and then, untrusorthy mathematician that I am, I t to taketh away. Not fair, right, JSH ? I am lying and obfuscating, right? I'm cheating. Right? No. It IS fair. I am NOT cheating. You DO have to consider F(x) AS A NUMBER, for specific vues of x. That in fact is at the core of the JSH step-by-step argument. This I think is the source of the disagreement beeen JSH and the rest of us this whole time. He is tking about coprimeness of a *function* to 7. We are tking about coprimeness of individu numbers to 7. Functions and evuations of functions are not the same kind of things. In the JSH argument, one needs *evuations*. Figuring out how JSH is thinking at any given point is not necessarily much of a reason to celebrate. Nora B. > Given that the constant terms are independent of x's vue, it must be > the case that dividing P(x) by 49 divides the o constant terms equ > to 7, by 7. 7. But to divide 7 from those constant terms requires dividing > through o of the factoso (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 from reverse use of the distributive property, which gives a constant > term coprime to 7, as required. For some odd reason I've STILL had mathematicians refusing to > acknowledge the truth. But, surprisingly, the proof here, from what I read in a link posted > further down, is tighter than what mathematicians usuly claim is a > proof, and represents perfection at a level most mathematicians, even > profession mathematicians, 't even attempt. To understand what I mean, you might want to see When is a proof? http://www.maa.org/devlin/devlin_06_03.html Here's an excerpt: (right-or-wrong, rule-of-law) definition is that a proof is a > logicly correct argument that establishes the truth of a given > statement. The left wing answer (fuzzy, democratic, and human > centered) is that a proof is an argument that convinces a typic > mathematician of the truth of a given statement. While vid in an ideistic sense, the right wing definition of a > proof has the problem that, except for trivi examples, it is not > clear that anyone has ever seen such a thing. > What I've shown you in my post is irrefutable proof, yet > mathematicians seem to think they can just ignore it, while they > themselves have far vaguer works, which they expect the world to > celebrate. They're cheating. http://mathforprofit.blogspot.com ==== [..] >> But for o of the factors of P(x), the constant terms is 7, which is >> coprime to 22. Therefore, *none* of the constant terms can have 7 as >> a factor. (By saying that 7 is coprime to 22, I'm making a choice as to where >> the proof is going. Since I've been tking about gebraic integers, >> where 7 is coprime to 22, it's natur to go with a choice where 7 is >> coprime to 22.) > This is the heart of it. JSH says [1] 5*b3 + 22 has constant term 22, and that constant term is coprime to 7. >Correct. Here, I believe, is how the thinking goes. If you have a >polynomi with integer coefficients, say, F(x) = a*x^3 + b*x^2 + c*x + 22, then that *polynomi* is coprime to 7, because one of the coefficients >itself is coprime to 7. That is true whether the coefficients are >integers or gebraic integers. There is no possible common factor >of F(x) with 7. The constant term, F(0), is 22, and that is l >you need to know to say that the polynomi is coprime to 7. This statement seems to me to be correct. One says that a polynomi >with integer coefficients is coprime to an integer m if *any one* of >its coefficients is coprime to m. Here the constant coefficient happens >to be coprime to 7, and that is enough. The statement is true IF by coprime you mean coprime in Z[x], and you mean have no common divisors other than units. However, it is not true if you mean coprime in Z^Z; the example of x^2+x and 2 comes to mind. Using Dot's proof that the vues of a polynomi with gebraic integer coefficients are ways divisible by an integer if and only if each coefficient is a multiple of that integer (in the gebraic integers) gives you that the statement is correct for polynomis in A[x]. > Of course in [1] above, b3 is not a polynomi. It is a function >of x. It takes vues in the gebraic integers. But perhaps the >same reasoning that applies for polynomis applies for b3(x). But it does not. [..] > Figuring out how JSH is thinking at any given point is not >necessarily much of a reason to celebrate. One possiblity that occurred to me yesterday is that has not yet reized that any function f(x) from A to C can be written as f(x) = g(x) + c where c is a constant, and g is a function that takes any specified vue at any specified point. That is, if a is in A and b is in C, then there ways exists a function g(x) from A to C and a constant c in A such that f(x) = g(x) + c, and g(a)=b. Just take c = f(a)-b, and let g(x) = f(x)-f(a)+b. So, given ANY f(x) function from A to C, there is a function g(x) such that g(0)=0 and f(x) = g(x)+c, c a constant. He has found ONE function that has the right c, namely, (5a_1(x)+7)/7; so he thinks this is the ONLY function that has the right c. Likewise, (5a_2(x)+7)/7 works, so it must be the only one that works; and (5b_3+22)/1 works, so it must be the only one that does. He does not reize that for ANY complex-vued functions w_1(x), w_2(x), w_3(x), such that w_1(0)=w_2(0)=7, w_3(0)=1, and w_1(x)*w_2(x)*w_3(x)=49, one can write (5a_1(x)+7)/w_1(x), (5a_2(x)+7)/w_2(x), and (5b_3(x)+22)/w_3(x) as a function which is 0 at 0, plus 1, a function which is 0 at 0, plus 1, and a function which is 0 at 0, plus 22. He found one possibility, he thinks there is only one possibility. And that could be the mistake. ====== ething; self-evident honesty of conviction, which does more; and a long purse, which does most of l. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine tgustus de Morgan ====== ==== For cing out loud. You should stop cutting and pasting incorrect stuff. In fact, you should stop cutting and pasting, period. Why not post a link to your origin post, instead, if l you are going to do is repeat it verbatim, ERRORS INCLUDED? [..] >Most posters seem ok with the steps up to the fin o. In what >follows it's important to understand the word coprime. In intege2 and 3 are coprime as are 12 and 13, as it simply means >they 't share non-unit factors. However, in res, NO numbers are coprime, as for instance, 2(3/2) = >3, so 2 is a factor of 3. That's an important point to consider going forward. It's such an important point that it is wrong. In the res, any o nonzero numbers are coprime by your definition (and by the standard definition). Because ANY common divisor of 2 and 3 is different from 0, and therefore a unit. Hence, any common divisor of 2 and 3 is a unit. So 2 and 3 do not share any non-unit factors (in the re numbers). Your statement that 2 and 3 are not coprime in the re numbers is just plain wrong. In the re numbeany nonzero number is a unit, and therefore any nonzero number is a (trivi) factor of eve number. ====== ething; self-evident honesty of conviction, which does more; and a long purse, which does most of l. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine tgustus de Morgan ====== ==== For cing out loud. You should stop cutting and pasting incorrect > stuff. In fact, you should stop cutting and pasting, period. Why not > post a link to your origin post, instead, if l you are going to do > is repeat it verbatim, ERRORS INCLUDED? [..] Why 't you off Arturo Magidin? You're such a pathetic piece of after l, why do you have to keep tagging ong like an unwanted brat? ==== For cing out loud. You should stop cutting and pasting incorrect >> stuff. In fact, you should stop cutting and pasting, period. Why not >> post a link to your origin post, instead, if l you are going to do >> is repeat it verbatim, ERRORS INCLUDED? [..] > Why 't you **** off Arturo Magidin? > You're such a pathetic piece of **** after l, why do you have to > keep tagging ong like an unwanted brat? Arturo hits a nerve, and as ways, drops even his *pretense* of deceny when he knows he's outclassed... ==== > For cing out loud. You should stop cutting and pasting incorrect > stuff. In fact, you should stop cutting and pasting, period. Why not > post a link to your origin post, instead, if l you are going to do > is repeat it verbatim, ERRORS INCLUDED? > [..] ==== For cing out loud. You should stop cutting and pasting incorrect >> stuff. In fact, you should stop cutting and pasting, period. Why not >> post a link to your origin post, instead, if l you are going to do >> is repeat it verbatim, ERRORS INCLUDED? [..] Why 't you off Arturo Magidin? Three times I offered to do so if you told me to stop posting replies. You declined to do so. Are you doing so now, in your oh-so-courteous way? Just say so: Do you want me to stop posting with comments about your statements? Yes or no? (And, note, that despite l your protestations to the contra, you have been unable to produce a SINGLE instance, barring the current disagreement, in which I made a statement, you made the opposite statement, and I was wrong and you were right; despite the fact that I have WAYS been correct in the past, nonetheless you think that you can accuse me of lying for years. What a pathetic little erp you are) >You're such a pathetic piece of after l, why do you have to >keep tagging ong like an unwanted brat? Did you, or did you not, cut and paste incorrect stuff? There is only one who keeps tagging ong, going to sci.math, sci.physics, even sci.chem in the vain hope to rly an army to march behind him while he imagines storming the strongholds of mathematics. The only brat is you. ====== ething; self-evident honesty of conviction, which does more; and a long purse, which does most of l. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine tgustus de Morgan ==== [..] > I guess you could say that and it doesn't change things in any > meaningful way. But it's *fascinating* that Arturo Magidin pushed that position! Are you contradicting it? > Well consider then, he's saying that 3 is not a factor of 6 in res > because they're coprime! And you know what? I think the way mathematicians usuly go, he's > right!!! And indeed. But apparently you have no idea about the mathematic >meaning of coprime. He slipped it past you, Dik. He's saying that 3 is not a factor of 6 in res because they're coprime! I never said that. I never even implied it. is either utterly confused about HIS OWN definition, or lying. ====== ething; self-evident honesty of conviction, which does more; and a long purse, which does most of l. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine tgustus de Morgan ====== ==== sci.physics ped > [deletia] >4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I >have >>P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) >>P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). >>Note that l you have e is add and subtract 3 to define b_3(x); >>that is, you are writing >>b_3(x) = (a_3(x) - 3) >>so >>5a_3(x) + 7 = 5(a_3(x)-3+3) + 7 >> = 5(a_3(x)-3) + 15 + 22 >> = 5b_3(x) + 22. >>The exact same process that you decried when I used it. You claimed >>that doing this ma[de] no sense mathematicly. Do you still make >>that claim? Just curious. That is a lie from Arturo Magidin as in fact he just subtracted and > added 3 on the same line. I'm focusing on constant terms, not ting > to hide a correct argument with meaningless operations like > subtracting and adding 3. No it's not, . You might want to check your facts before you accuse others of lying. Here by switching to b_3(x) I have 7,7, and 22, the three constant > term factors of the main constant term 1078 of P(x), shown so that > there's less room for confusion. I'm adding addition steps to handle people like Arturo Magidin who > have made it their business to lie for so long, and not surprisingly, > he's ting to find fault with the process. Why? Because you are not giving him credit for helping you? Posters like Arturo Magidin waste a lot of other people's time. >5. Now P(x) has a factor of 49 as >>P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 >>which means that >>(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) >>has a factor of 49. >>Most posters seem ok with the steps up to the fin o. In what >follows it's important to understand the word coprime. >>You need to understand that coprime is CONTEXTU, just as >>divides. Saying a and b are coprime per se has no meaning; one >>must say a and b are coprime IN SUCH AND SUCH A RING, unless the >>ring is understood from context. Like I said, posters like Arturo Magidin waste a LOT of people's time. > It's cled demanding precision. > >In intege2 and 3 are coprime as are 12 and 13, as it simply means >they 't share non-unit factors. >>However, in res, NO numbers are coprime, as for instance, 2(3/2) = >3, so 2 is a factor of 3. >>But 2 is a unit, so your claim here is fse. > Hmmm...now that's interesting, eve re but 0 is a unit, eh? Fascinating perspective. It follows quite naturly from the definitions. It's been suggested before, but you might want to take a class or o in abstract gebra before you continue your work. It will help you a great de. >>Actuly, in the re numbeANY O NONZERO NUMBERS ARE >>COPRIME. That's because, given any o nonzero re numbers x and y, >>any common divisor of x and y is a unit. > Well, then eve re number but 0 is a unit follows from that > position. Definition. I guess you could say that and it doesn't change things in any > meaningful way. But it's *fascinating* that Arturo Magidin pushed that position! Why? Because he's actuly concerned about accuracy? Well consider then, he's saying that 3 is not a factor of 6 in res > because they're coprime! No. This is a result of you not understanding the difference in the definitions. This type of statement is typic of what causes you problems. 3 is a factor of 6 in the res. 6 is a factor of 3 in the res. 6 and 3 are coprime in the res. l of these are true statements in the res. And you know what? I think the way mathematicians usuly go, he's > right!!! Please start using standard definitions. It would help. -- ==== > sci.physics ped > [deletia] > >4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I >have >>P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) >>P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). >>Note that l you have e is add and subtract 3 to define b_3(x); >>that is, you are writing >>b_3(x) = (a_3(x) - 3) >>so >>5a_3(x) + 7 = 5(a_3(x)-3+3) + 7 >> = 5(a_3(x)-3) + 15 + 22 >> = 5b_3(x) + 22. >>The exact same process that you decried when I used it. You claimed >>that doing this ma[de] no sense mathematicly. Do you still make >>that claim? Just curious. That is a lie from Arturo Magidin as in fact he just subtracted and > added 3 on the same line. I'm focusing on constant terms, not ting > to hide a correct argument with meaningless operations like > subtracting and adding 3. No it's not, . You might want to check your facts before you > accuse others of lying. You obviously are the one who didn't check facts as what I said IS correct. ==== You obviously are the one who didn't check facts as what I said IS correct. For seven yeaand more, JSH has been saying this and having later to recant. Why should today be any different? ==== >> sci.physics ped >[deletia] >> >>4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I >>have >>P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) >>P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). >>Note that l you have e is add and subtract 3 to define b_3(x); >that is, you are writing >>b_3(x) = (a_3(x) - 3) >>so >>5a_3(x) + 7 = 5(a_3(x)-3+3) + 7 > = 5(a_3(x)-3) + 15 + 22 > = 5b_3(x) + 22. >>The exact same process that you decried when I used it. You claimed >that doing this ma[de] no sense mathematicly. Do you still make >that claim? Just curious. That is a lie from Arturo Magidin as in fact he just subtracted and >> added 3 on the same line. I'm focusing on constant terms, not ting >> to hide a correct argument with meaningless operations like >> subtracting and adding 3. No it's not, . You might want to check your facts before you >> accuse others of lying. You obviously are the one who didn't check facts as what I said IS correct. Actuly, no. You said three things, and you implied a fourth. Of the three things you said, one is fse. Therefore, what you said is NOT correct. The three things you said are: (a) I lied; (b) I just subtracted and added 3 on the same line; (c) You are focusing on constant terms. The thing you implied was that I was ting to hide a correct argument with meaningless operations. Statements (b) and (c) are correct. Statement (a) is fse. Your definition of b_3 MEANS that you've added and subtracted 3 to go from 5a_3(x)+7 to 5b_3(x) + 22. Just because you did not say so explicitly You said three things. One of them is fse. Therefore, you claim that what you said IS correct is fse. ====== ething; self-evident honesty of conviction, which does more; and a long purse, which does most of l. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine tgustus de Morgan ====== ==== the *only* time that I ever made a mistake, was joining the JSH Show. maybe the matforprofit.blog is rely a sign, that he's been getting paid for this experiment, l-ong, per character of his would-be help-meets, including l the silly contextu quoting, which is supposed to be aviable in the thread. as, Usenet was too sml for the proof . >Note that l you have e is add and subtract 3 to define b_3(x); >that is, you are writing >>b_3(x) = (a_3(x) - 3) >>so >>5a_3(x) + 7 = 5(a_3(x)-3+3) + 7 > = 5(a_3(x)-3) + 15 + 22 > = 5b_3(x) + 22. >>The exact same process that you decried when I used it. You claimed >that doing this ma[de] no sense mathematicly. Do you still make >that claim? Just curious. That is a lie from Arturo Magidin as in fact he just subtracted and --ils duces d'Enron! http://tarpley.net/bush8.htm http://www.wlym.com/PDF-SpReps/SPRP13.pdf ==== > How do I integrate this? f(x) = sin(x)/x > Indefinite integr? That is not elementa. It is known as the sine integr function, written Si(x). http://www.math.ohio-state.edu/~edgar/ ==== T P says... How do I integrate this? f(x) = sin(x)/x There is a speci case that can be found by fourier transforms: Integr from x = -infinity to +infinity sin(x)/x dx First, note that sin(x)/x = integr from k= -1 to +1 of 1/2 cos(kx) dk Now, if we define F(k) according to F(k)=1/2 (for k beeen -1 and +1) =0 (otherwise) then we have sin(x)/x = integr from k= -infinity to +infinity of F(k) cos(kx) dk That's just a Fourier transform of the function F(k). We can take the inverse transform to find an integr representation of F(k): F(k) = 1/(2 pi) integr from x= -infinity to +infinity of sin(x)/x cos(kx) dx In the particular case k=0, we have (the cos(kx) drops out) F(0) = 1/(2 pi) integr from x= -infinity to +infinity of sin(x)/x dx Since we defined F so that F(0) = 1/2, we have integr from x= -infinity to +infinity of sin(x)/x dx = pi -- ==== Here are o geomet questions from a curious layperson. If they're unclear, please tell me. Ditto if they're addressed to the wrong forum. 1) Are there surfaces that can be bent to each other (the way a flat piece of paper is bent to a cylinder) but not to anything flat? 2) Besides planes and spheres, is there any other surface S such that a piece of S can be moved around adlibitum while each of its points remains in contact with S? I've wondered about things like this, off and on, for a long time; but I haven't e much studying about it, because the going has ways seemed too dense. If anybody knows a good place to make an entrance into this body of knowledge, I'd be grateful to know about it. ==== > Here are o geomet questions from a curious layperson. I think these are wonderful questions; such a curious layperson is welcome into my classes any time! Let met take them in opposite order. > 2) Besides planes and spheres, is there any other surface S such that > a piece of S can be moved around adlibitum while each of its points > remains in contact with S? What is meant by a surface is not ways clear, but let me assume that S is what would technicly be cled a smooth 2-dimension submanifold of R^3. For piece of S I will take an open neighborhood around a point P. If you zoom in ve near to P, you will see that this piece of S looks flat, that is, there is a well-defined tangent plane to S at the point P, which approximates the shape of S ve well near P. (The existence of such a plane is the definition of differentiability or smoothness.) For convenience, spin S around in space so that this plane is horizont. Now zoom out a little bit -- not too far! -- and you will see that the ren around P isn't rely flat. It will have hills and vleys near P, looking ve much like the graph of a function f defined on the horizont plane. (That's the Implicit Function Theorem at work.) If you see sever hills and complicated vleys, zoom in a bit more. In a sml enough ren around P, you can approximate the shape with the graph of a _quadratic polynomi_ Q instead of f itself. (That's Taylor's theorem -- the right Q can be computed if you know the parti derivatives of f at P.) With a bit of high school geomet you can rotate coordinates on the horizont plane so that Q is just a x^2 + b y^2 + c for some constants a, b, c. (This c is just the height of the point P off the horizont plane, so if this plane -- which I have never yet specified! -- is just the tangent plane mentioned earlier, then c = 0.) The constants a and b are pivot. They measure how markedly the graph of Q (and so of f, and so of S itself) ists up or down as you travel ong the axes. Moreover, these o directions are the extremes: ong any other line, Q changes more slowly (and if ab<0, there are even lines of travel where Q stays constant). The importance of a and b is that these are invariants of the shape of S at P. though they can be computed from various coordinates or other formulas, they are clearly geometric in nature and uniquely specify what is known as the _curvature_ of S at P. (Be careful of that word; it's so used for other quantities which summarize a and b , e.g. reporting only their sum or product.) So your request comes down to this: you want to know what are the surfaces which have the same curvature (tensor) at eve point. You have provided o important examples: on planes, we have a=b=0 at each point; on spheres we so have a=b (but nonzero, and constant) at each point. It turns out there are no other surfaces of this particular type: if at the point P we have a=b, then P is said to be an _umbilic point_ of S . If eve point is an umbilic point (with, a priori, possibly different vues of a=a(P) at each point) then it turns out that a is indeed constant, and the surface is part of a plane or sphere. You're forgetting another obvious example: on a cylinder we have a=0 at each point P, and a nonzero constant vue for b. (Unlike the previous case, here it IS now possible for eve point to be parabolic in this sense but to have b va with P, giving generized cylinders. But they 't have the property you asked for.) Any other examples of surfaces with the property you seek would have at each point a pair of distinct, nonzero vues of a and b which remain constant on the surface. In particular, their product would remain constant, which means your surface would be a surface of constant Gaussian curvature. Their sum is so constant, making the surface a constant mean curvature space. These are interesting categories of surfaces, each containing some beautiful examples. (Soap bubbles stretching across a wire frame form CMC surfaces.) But it turns out that you can't find any more examples having both Gaussian and mean curvatures be constant -- not among surfaces which are immersed in R^3. > 1) Are there surfaces that can be bent to each other (the way a flat > piece of paper is bent to a cylinder) but not to anything flat? You're asking for a one-parameter family of loc isometries. I 't know if you intended flat in the technic sense but you've got it right: flat means Gaussian curvature is zero, i.e. at each point P we have a=0 in the discussion above. (The Moebius strip is flat in this sense too, but maybe isn't what you had in mind as being flat. You can certainly bend a Moebius strip into many configurations.) Here's a nontrivi example, using what are cled _minim surfaces_. (These are surfaces of constant mean curvature equ to zero, that is, at each point we have a = -b.) One such surface is the helicoid, formed by taking a double helix wrapping around a vertic line and then joining points in one helix with a line segment stretching horizontly to the other helix. They sell such things as yard ornaments, packaged with l the horizont line segments lying in a plane, but the helicoid is not rely flat: the true surface would have some fabric stretching from one horizont line to the next, with more fabric near the helices, so that when you t to lay the thing flat there will be ruffles of fabric which are ve noticeable far from the centr vertic line. The other surface is the catenoid, formed by rotating a catena, e.g. the graph of y=(1/2)( exp(x)+exp(-x) ), around the horizont axis. This is rely a different surface from the helicoid. But you can make one from exactly one full turn of the helicoid by forming each of the helices into a circle; the centr vertic line of the helicoid then so forms a circle (smler than the other o circles). The horizont parlels have been bent into copies of the catena, and the fabric in the ruffles smoothes out to give the surface of the catenoid. You can perform this transformation in a such a way that at eve intermediate moment the fabric is fully stretched out, that is, this is a continuous bending. Cool, huh? ==== > Here are o geomet questions from a curious layperson. If they're unclear, > please tell me. Ditto if they're addressed to the wrong forum. 1) Are there surfaces that can be bent to each other (the way a flat piece of > paper is bent to a cylinder) but not to anything flat? I'm not sure I understand ve well... it depends on what kind of bending would be lowed. For instance, would a cube and a sphere be 'bendable' into each other? Spheres and eggs? If you think about it, I guess you will have a hard time explaining (and even defining) what you mean exactly with 'bending'. > 2) Besides planes and spheres, is there any other surface S such that a piece of > S can be moved around adlibitum while each of its points remains in contact with > S? This is a bit easier. As far as I can see, only a cylinder, but then you can only translate a piece ong the axis or perpendicular to it. You cannot rotate it like you *can* on a plane or sphere. > I've wondered about things like this, off and on, for a long time; but I > haven't e much studying about it, because the going has ways seemed too > dense. If anybody knows a good place to make an entrance into this body of > knowledge, I'd be grateful to know about it. Both questions can be handled by a re tough part of mathematics, cled differenti geomet. Keyword: radius of curvature. The second question might be partly treated in the more accessible part of 'anytic geomet of 3D-space', but only partly. But I'm afraid that both parts cover an area that is much and much larger than what you have in mind :-( So... good questions, but ve difficult to answer appropriately. ==== > Here are o geomet questions from a curious layperson. If they're unclear, >> please tell me. Ditto if they're addressed to the wrong forum. >> 1) Are there surfaces that can be bent to each other (the way a flat piece of >> paper is bent to a cylinder) but not to anything flat? I'm not sure I understand ve well... it depends on what >kind of bending would be lowed. For instance, would >a cube and a sphere be 'bendable' into each other? >Spheres and eggs? >If you think about it, I guess you will have a hard time >explaining (and even defining) what you mean exactly >with 'bending'. I assumed she meant bending without stretching: For example there's a natur map from part of the plane to part of a cylinder that preserves distances exactly (if distances are measured ong curves on the surface). >> 2) Besides planes and spheres, is there any other surface S such that a piece of >> S can be moved around adlibitum while each of its points remains in contact with >> S? This is a bit easier. As far as I can see, only a cylinder, >but then you can only translate a piece ong the axis >or perpendicular to it. You cannot rotate it like you >*can* on a plane or sphere. > I've wondered about things like this, off and on, for a long time; but I >> haven't e much studying about it, because the going has ways seemed too >> dense. If anybody knows a good place to make an entrance into this body of >> knowledge, I'd be grateful to know about it. Both questions can be handled by a re tough part >of mathematics, cled differenti geomet. >Keyword: radius of curvature. >The second question might be partly treated in the >more accessible part of 'anytic geomet of >3D-space', but only partly. >But I'm afraid that both parts cover an area that >is much and much larger than what you have in >mind :-( So... good questions, but ve difficult to answer >appropriately. ==== How can I prove, that the following inequation is ways true. (b+1/a)^(b+1) >= (1/a)*(1+b)^(b+1) The variables a and b are greater than zero. I have no plan, how i can do this? Any ideas? nice wishes from Austria Martin Ranzmaier ==== > How can I prove, that the following inequation is ways true. (b + (1/a))^(b+1) >= (1/a)*(1+b)^(b+1) The variables a and b are greater than zero. > I have no plan, how i can do this? Any ideas? How about fixing an arbitra b > 0? Then you want to show the above holds for l a > 0. Simplify by replacing 1/a with a. Then we want to show (b + a)^(b+1) >= a(1+b)^(b+1), for a > 0. Take logs like Arturo suggested, and define f(a) = (b+1)log(b+a) - [log(a) + (1+b)log(1+b)]. You want to show f(a) >= 0 for l a > 0. So you've got a cculus problem, namely, showing the minimum vue (if it exists) of f over (0,oo) is 0. ==== How can I prove, that the following inequation is ways true. (b+1/a)^(b+1) >= (1/a)*(1+b)^(b+1) You want to be a bit careful; I'm not sure if you mean (b + (1/a) )^(b+1) or ((b+1)/a)^(b+1) though I'm pretty sure you mean the latter, given what happens. >The variables a and b are greater than zero. >I have no plan, how i can do this? Any ideas? If both a and b are positive, then both numbers are positive, so you can apply a logarithm; this because for a,b>0, a>=b if and only if log(a) >= log(b). Using the logarithm would bring down the exponent: (b+1)*log(b+1/a) >= log(1/a) + (b+1)*log(b+1) if and only if (b+1)*(log (b+1/a)- log(b+1)) >= log(1/a) Now use the properties of logarithm and the fact that b>0, and it should be clear. ==== >How can I prove, that the following inequation is ways true. >(b+1/a)^(b+1) >= (1/a)*(1+b)^(b+1) You want to be a bit careful; I'm not sure if you mean (b + (1/a) )^(b+1) or ((b+1)/a)^(b+1) though I'm pretty sure you mean the latter, given what happens. The latter is fse: The factor (b+1)^(b+1) cancels and you're left with (1/a)^(b+1) >= (1/a), which fails for a > 1. ==== Sor for the double meaning in the inequation. I ment (b + (1/a) )^(b+1) Arturo Magidin schrieb im Newsbeitrag >How can I prove, that the following inequation is ways true. >(b+1/a)^(b+1) >= (1/a)*(1+b)^(b+1) You want to be a bit careful; I'm not sure if you mean (b + (1/a) )^(b+1) or ((b+1)/a)^(b+1) though I'm pretty sure you mean the latter, given what happens. >The variables a and b are greater than zero. >I have no plan, how i can do this? Any ideas? If both a and b are positive, then both numbers are positive, so you > can apply a logarithm; this because for a,b>0, a>=b if and only if log(a) >= log(b). Using the logarithm would bring down the exponent: (b+1)*log(b+1/a) >= log(1/a) + (b+1)*log(b+1) if and only if > (b+1)*(log (b+1/a)- log(b+1)) >= log(1/a) Now use the properties of logarithm and the fact that b>0, and it > should be clear. ====== ==== >Sor for the double meaning in the inequation. I ment (b + (1/a) )^(b+1) I still suggest taking logarithms and playing with those. ==== >Maybe i was a bit unclear as to what i was asking. >I know that we can apply the Descartes' Sign Rule to determine the >max number (upper limit) of positive and negative re roots of a >polynomi. But from this we cannot (ways) determine if a polynomi >of degree n has n re roots. >Now we can apply Sturms Theorem to determine the exact number of >distinct re roots of a polynomi on an interv(a,b). However we >must limit ourselves to an interv here. You can let the end points of the interv approach plus or minus infinity. The signs are constant far out and easily determined. Most textbooks give examples of this. so, it is easy to get an upper bound on the magnitude of the roots of an equation. The sum of the magnitudes of the other coefficients, assuming the leading coefficient is 1, is more than enough. >So is there any specific condition whereby given a polynomi of >degree n we can ways state that, given this specific condition >holds, l its roots are re? >Maybe what i am asking is rather silly/uneducated and i am missing >some basic points here, if so please tell :) >> >>could anyone tell me what conditions are neccessa in order for a >>polynomi to have re roots only? >> Look for Sturm sequences in an old Theo of equations >> book, or elsewhere. They tell you exactly how many >> re roots a re polynomi has. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University ==== >What are the conditions that guarantee, given a polynomi of degree >n, that the polynomi has n re zeros? For example given polynomi of degree 2, > ax^2 + bx + c, >The condition that, > b^2 - 4ac > 0 >guarantees that this polynomi shl have 2 re zeros. Does there exist such a condition that holds for polynomis of >arbita degree? There is a similar set of conditions which guarantee that the cubic ax^3 + bx^2 + cx + d has three re roots, expressed as conditions on signs of certain polynomis in a,b,c,d. There is a similar set of conditions which guarantee that a polynomi of degree 100 has 100 re roots. It is expressed in terms of the 101 coefficients. And so on. Now, what exactly are you expecting for a condition that holds for polynomis of arbitra degree? Do you expect to see a formula with infinitely many variables in it, which can be used equly well for quadratics (which have d=e=f=... = 0) and cubics (which have e=f=...=0) and l other degrees? How exactly would we write down such a thing? Once you fix the degree n, the family of l polynomis forms a vector space of dimension n+1. The polynomis with n distinct re roots form an open subset of this space. The bounda of this ren is contained in the variety which picks out the polynomis with multiple roots; this variety is defined by the vanishing of the discriminant, which is a polynomi in the n+1 coefficients. For polynomis of high degree, though, the open subset defined by the condition discriminant > 0 includes sever components, and not just the ren where l the roots are re. So you need multiple conditions on the coefficients to describe the ren you are interested in. Apart from l this, I should point out that you keep requesting conditions which guarantee the existence of n re roots, and not conditions which exactly describe the existence of n re roots. Is that what you rely want? If so, I can give you conditions like this: The cubic ax^3 + bx^2 + cx + d has three re roots if the point (a,b,c,d) is within [some distance] of (1,6,11,6). Obviously other cubics so have three re roots too! ==== > >What are the conditions that guarantee, given a polynomi of degree >n, that the polynomi has n re zeros? >For example given polynomi of degree 2, > ax^2 + bx + c, >The condition that, > b^2 - 4ac > 0 >guarantees that this polynomi shl have 2 re zeros. >Does there exist such a condition that holds for polynomis of >arbita degree? There is a similar set of conditions which guarantee that the > cubic ax^3 + bx^2 + cx + d has three re roots, expressed as > conditions on signs of certain polynomis in a,b,c,d. There is a similar set of conditions which guarantee that a polynomi > of degree 100 has 100 re roots. It is expressed in terms of the > 101 coefficients. And so on. Now, what exactly are you expecting for a condition that holds for > polynomis of arbitra degree? Do you expect to see a formula with > infinitely many variables in it, which can be used equly well for > quadratics (which have d=e=f=... = 0) and cubics (which have e=f=...=0) > and l other degrees? How exactly would we write down such a thing? Once you fix the degree n, the family of l polynomis forms a > vector space of dimension n+1. The polynomis with n distinct > re roots form an open subset of this space. The bounda of this > ren is contained in the variety which picks out the polynomis > with multiple roots; this variety is defined by the vanishing of the > discriminant, which is a polynomi in the n+1 coefficients. > For polynomis of high degree, though, the open subset defined by > the condition discriminant > 0 includes sever components, and > not just the ren where l the roots are re. So you need > multiple conditions on the coefficients to describe the ren you > are interested in. Apart from l this, I should point out that you keep requesting > conditions which guarantee the existence of n re roots, and > not conditions which exactly describe the existence of n re > roots. Is that what you rely want? If so, I can give you conditions > like this: The cubic ax^3 + bx^2 + cx + d has three re roots > if the point (a,b,c,d) is within [some distance] of (1,6,11,6). > Obviously other cubics so have three re roots too! Damn you sound brainy! I hear what your saying about being able express for eve polynomi some condition in terms of its coefficients that will guarantee you have re roots, just as with degree 2 we can express it as b^2-4ac>0. What it is precisely that i am looking for, im not quite sure! But i was just curious as to whether there was some gener condition that would hold, anagous to Descartes Sign method, i.e. something simple that even i could understand :)...But i suppose using the Sturm technique to show n re zeros exist or the Routh-Hurwitz Criterion is just like saying some condition must hold. Apologies if i am using incorrect/ambiguous/uneducated terminology, most likely due to me being uneducated in this area! Once again it becomes clear that guarantee is not what i was looking for and exactly describe sounds more like what i am. But i think i did say i was looking for a gener condition that must hold for polynomis of arbita degree. So your example just holds for this one specific example of a polynomi of one specific degree(yes i know you could just find a similar point for another polynomi and specify [some distance]), or maybe i took you up wrong and your example was to show where i was being incorrect in my terminology, which it did, thanks either way. pat ==== >> Now we can apply Sturms Theorem to determine the exact number of >> distinct re roots of a polynomi on an interv(a,b). However we >> must limit ourselves to an interv here. Yes, but one can effctively find a number M such that l zeroes >> of the polynomi satisfy |z| < M. Apply Sturm to (-M,M). Yes obviously we can set our range to (-infinity, +infinity). > But that is just a method for finding out the number of re zeros. What are the conditions that guarantee, given a polynomi of degree > n, that the polynomi has n re zeros? For example given polynomi of degree 2, > ax^2 + bx + c, > The condition that, > b^2 - 4ac > 0 > guarantees that this polynomi shl have 2 re zeros. Does there exist such a condition that holds for polynomis of > arbita degree? For a given polynomi you can build a particular matrix from its coefficients, such that you can read off the information about the numbers of (re) zeroes from the rank and signature. I have forgotten the details, it should be in any book on re gebra. Marc ==== >> Now we can apply Sturms Theorem to determine the exact number of >> distinct re roots of a polynomi on an interv(a,b). However we >> must limit ourselves to an interv here. Yes, but one can effctively find a number M such that l zeroes >> of the polynomi satisfy |z| < M. Apply Sturm to (-M,M). Yes obviously we can set our range to (-infinity, +infinity). > But that is just a method for finding out the number of re zeros. What are the conditions that guarantee, given a polynomi of degree > n, that the polynomi has n re zeros? For example given polynomi of degree 2, > ax^2 + bx + c, > The condition that, > b^2 - 4ac > 0 > guarantees that this polynomi shl have 2 re zeros. Does there exist such a condition that holds for polynomis of > arbita degree? For a given polynomi you can build a particular matrix from its > coefficients, such that you can read off the information about the > numbers of (re) zeroes from the rank and signature. I have forgotten the details, it should be in any book on re gebra. Marc Is this not the Routh-Hurwitz Criterion? ==== I would like to point out that if you have any odd-degree polynomi, you can be gurenteed that there will be at least one re zero. This is because the structure of x^3, x^5, x^7... will naturly cause it to cross the X-axis. I know this only answers that you will be certain of one re zero some of the time, but I hope this helps! Anthony > > could anyone tell me what conditions are neccessa in order for a > polynomi to have re roots only? > pat ==== I would like to point out that if you have any odd-degree polynomi, > you can be gurenteed that there will be at least one re zero. This > is because the structure of x^3, x^5, x^7... will naturly cause it > to cross the X-axis. I know this only answers that you will be certain > of one re zero some of the time, but I hope this helps! Anthony > could anyone tell me what conditions are neccessa in order for a > polynomi to have re roots only? > pat with l roots re, this is a nice piece of info, cheers! ==== While there is no simple rule to determine whether a polynomi of arbitra degree has only re zeros, from a stability perspective, it is ve unlikely. In addition, the coefficients must be big. In particular, I believe that the following is true for polynomis with positive coefficients : if the polynomi has the form p(z) = sum_{k=0}^n a_k z^k, where l a_k ge 0 and a_n=a_0=1, then it cannot have l re zeros if a_k < {n choose k} for any k --irascible since 1957 ==== >While there is no simple rule to determine whether a polynomi of arbitra >degree has only re zeros, from a stability perspective, it is ve unlikely. What does this mean, unlikely? There is the family of l polynomis (of a given degree, perhaps?) and within it the family of l polynomis l of whose roots are re. Do you have in mind a finite measure on the first space in which the measure of the subspace is zero? I am not familiar with such a result. Looking at the family of quadratics ax^2 + bx + c, for example, I see a 3-dimension vector space with an unbounded open subset of quadratics having o re roots. We could, for example, sce l polynomis down to the unit sphere in R^3 (i.e., assume a^2+b^2+c^2=1), in which case the ones with o re roots ( b^2-4ac > 0) lie in the portion of the sphere lying beeen the o sheets of the hyperbolic surface a^2 + c^2 + 4ac = 1. This certainly includes hf the sphere (the whole ren where a and c have opposite signs) and a little more. In this way I compute that the quadratics with o re roots constitute about 51.234% of l quadratics, using this measure. That's not unlikely, so I guess the claim is that this fraction decreases to zero as the degree rises. Is that true? Proof? Reference? he reminded me of a result of Polya which notes that if f has l roots re, so does t0 f + t1 f' for eve re t0 and t1. (That is, our l-roots-re ren includes whole planes passing through each of its points.) So this ren isn't l that sml. (Proof of Polya's result: apply Rolle's theorem to exp(t x) f(x).) ==== For example given polynomi of degree 2, > ax^2 + bx + c, > The condition that, > b^2 - 4ac > 0 > guarantees that this polynomi shl have 2 re zeros. Does there exist such a condition that holds for polynomis of > arbita degree? no So basicly for some low order polynomis we have conditions that low us state that a polynomi satisfying this condition has only re roots. But in gener, there is no such condition that guarantees polynomis of arbita size have only re roots, and the only way of finding out if a polynomi of degree n has n re roots (without actuly solving it) is to apply some technique like a Sturm function to it, and if it tells us the polynomi has n re roots, then, well we know it has n re roots :) Right? ==== bcc new advanced physics list Paul, I suspect the key CLASSIC argument against a LOC vacuum gravity stress-energy tensor different from Ruv itself in Einstein's vacuum field equation Ruv = 0 goes something like this 1. If ordina matter, radiation and near EM fields (l in Tuv) are locly present then Einstein's CLASSIC loc field equation Guv = Ruv - (1/2)Rguv = - (8piG/c^4)Tuv Can be written as tuv(vac) + Tuv = 0 Where tuv(vac) = (c^4/8piG)Guv = (String Tension)(Marble Curvature) i.e. in Einstein's colorful terms tuv(Marble) + Tuv(Wood) = 0 End of sto, simple solution. 2. Another aspect of the problem is to add new nonlinear terms like t'uv(vac) =Lp^2 (c^4/8piG)RuwlsRv^w^l^s to the field equation. I did not check B to see if that particular expression is a symmetric tensor. I am just ting to show the gener idea. 3. Quantum aspect of the problem is that random zero point vacuum fluctuations of l physic quantum fields contribute to the cosmologic term /zpfguv that must be added to Einstein's field equations to get Guv + /zpfguv = - (String Tension)^-1Tuv Note that Sakharov's Metric Elasticity = (String Tension)^-1 = Space-Time Stiffness The tighter the string tension the less elastic is the geomet. /zpf = exotic vacuum unified Dark Energy/Matter LOC field, which in large sce FRW regime is Einstein's cosmologic constant but on smler sce is a loc inhomogeneous dynamic spin 0 field that stabilizes the spatily-extended hidden variable lepto-quarks against their explosive self-electric charge. The lepto-quarks obey a micro-geon quasi Kerr-Newmann metric with addition topologic handles for the SU(2)xSU(3) charge generators in the extra space dimensions (i.e. Cabi-Yau spaces of M theo in engineer's language via black hole - meme duities). The lepto-quarks shrink from 10^-13 cm at low energy to 10^-18 cm for high energy large magnification Heisenberg microscope images because of the huge space warp from the strong G* ~ 10^40 G(Newton) zero point dark energy cores holding the explosive self-electric charge and the centrifug & Coriolis inerti forces in equilibrium. l this is IT in the Bohm Pilot BIT Wave theo. IT gets its marching orders from BIT (action) BIT is warped by IT (reaction) At a coarser purely IT level Matter emerges as micro-geons getting subsidia classic marching orders from geomet. The geomet is from the macro-quantum coherence, i.e., guv = (Minkowski)uv + (Metric Elasticity)(Modulation of Goldstone Vacuum Superfluid Phase)uv /zpf = 0 is classic non-gravitating non-exotic vacuum. /zpf > 0 is w = -1 anti-gravity exotic vacuum dark energy with negative pressure and positive zero point energy density. /zpf < 0 is w = -1 gravitating exotic vacuum dark matter with positive pressure and negative zero point energy density. Big clumps of w = -1 dark matter (e.g. dark gactic spheres) float in the Hubble flow like w ~ 0 for us distant observers. Lp*^2 = G*h/c^3 Univers Regge slope of hadronic resonances on micro-sce is G*/hc^5 in J = (G*/hc^5)E^2 + Jo = n/2 Blackett Effect seen in many rotating astrophysic objects with anomous magnetic dipole is reduced to G^1/2m = e (papers by Saul-Paul Sirag) at large sce where G* ---> G. Now Matt Visser's review paper has impractic perturbative quantum field theo formulae for both large-sce G(Newton) and cosmologic constant /. However my model is better because it is intrinsicly non-perturbative like the BCS theo is. I have NON-PERTURBATIVE Visser does not have that non-perturbative More is different (PW Anderson) Vacuum Coherence whose generized phase rigidity is precisely String Tension (in Blackhole-Meme Duity) = Sakharov's (Metric Elasticity)^-1. I so explain how LOC MACRO PHYSICS emerges from quantum field theo whose entangled Fock space states are nonloc. I so explain why post-inflationa early universe has low entropy from collapse of vacuum phase space volume hence how the cosmic irreversibility Arrow of Time is pointed. I so give a micro-QED dynamic instability for the large sce FRW chaotic inflation field of Linde et-, which previously was only a useful fudge factor with no fundament dynamics (according to Paul The M argument against Yilmaz is simply that one cannot make a loc classic vacuum pure gravity stress-energy density tensor from the torsion-free non tensor connection without violating the equivence principle. But we have to be ve careful about what we mean by the equivence principle. There is the so-cled EEP; there is Einstein's origin equivence principle; which can be viewed as an interpretation of the EEP; and there is the ternative interpretation of the EEP in terms of *inerti compensation* which rejects Einstein's fundament identification of gravitation and inerti fields. What you need to do is spell out carefully l of these different versions in form terms and in clear operation terms to see if there is rely any scientific difference of any importance. I do not understand what you mean by inerti compensation above. I think the gravity force, i.e. the fictitious or inerti g-force i.e. the metric torsion-free connection for parlel transport in the origin 1915 theo is LOCLY equivent to a gravity force. Whether or not that gravity force is re depends on whether or not there is a loc ti curvature differenti in the g-force as described by the geodesic deviation equation beeen o neighboring timelike geodesic LIF free float observers. This curvature tensor ti differenti force cannot be eliminated locly. l I mean effectively by EEP is the set of following form statements. 1. There exist loc tetrads ea^u(P) at point P such that nab = ea^u(P)eb^v(P)guv(P) 2. The loc symmetric torsion-free connection from first derivatives of the metric guv(P) can be set to zero locly at P. This defines the distinction beeen LIFs on timelike geodesics and LNIF's on timelike non-geodesics both intersecting at same P. Note timelike non-geodesics could not exist without electric charge. Neither presumably could light hence no light cones hence nothing - not even null geodesics. This suggests in itself that gravity is an emergent collective effect primarily from electromagnetism and Heisenberg quantum uncertainty as suggested by Sakharov's metric elasticity. Part of the confusion is that curvature = field strength in the modern fiber bundle picture of the spin 1 gauge forces. The connection for parlel transport in the intern dimensions beyond space-time (or extra space dimensions of Kuza-Klein et-) is theYang-Mills potenti like the EM 4-vector potenti Au(P) in the U(1) EM simplest case. On the other hand it is the metric, specificly goo(P) ~ (1 + 2U(Newton)/c^2) in weak field slow motion static sphly symmetric vacuum case outside M from source mass M leading to notion of gravity force as connection from Newton's 2nd law F = -m GradU(Newton) F is locly eliminitable on the LIF timelike geodesics in curved space-time. Ti inhomogeneous differentis F(P+dP) - F(P) are not. EEP, as I understand it, merely requires that the net gravitation-inerti forces ti effects can ways be locly neglected. This last statement needs clarification. It is strictly not true. It is only a weak field approximation that the ti effects are of order (L/r)^2 where r is the loc sce of radius of curvature. M make this approximate nature of the remark ve clear when I read them. You have misinterpreted them as making a much stronger statement and I think this is the root of the pseudo-problem you pose. If Einstein was a bit confused on this in the early days, I 't know, it is of no re consequence except in the histo of the evolution of the idea. This can be consistently interpreted in at least o different ways: (1) as the consequence of a (loc) *fundament physic identity* of gravitation and inerti fields (Einstein, Pauli); or (2) as merely the result of inerti compensation of the separately existing gravitation field (Eddington et .) in conjunction with the natur attenuation of ti effects within arbitrarily sml neighborhoods of spacetime. Either way , the theo remains generly covariant, and a weak correspondence relationship with SR holds -- locly measurable effects that locly discriminate beeen the predictions of SR (flat spacetime) from those of GR (curved spacetime) can be regarded as practicly neglile in a sufficiently sml neighborhood. Yes, in the weak field limit, but not at a black hole, which Puthoff wrongly claims does not exist. I would argue that a weak correspondence relation is l that one can reasonably demand of the re-interpreted theo. There is in reity no flat spacetime *anywhere*. Fine, that's how I read M anyway. so in my theo, globly flat space-time is intrinsicly dynamicly unstable from quantum electrodynamics. so, it is important to understand that none of this depends on the empiric correctness of Yilmaz's particular theo vis a vis GR. I find Yilmaz's idea completely obscure intuitively. For example, the ternative approach is taken in a number of bimetric theories that agree empiricly with GR. These employ a background metric that accounts for inerti effects in conjunction with a separate (though commensurable) gravitation metric that accounts for physic gravitation. Ve ugly. Less with more. In the ternative interpretation, I believe the way is open to consistently defining a true gravitation field stress-energy density, which is fully locizable, and an inertily-compensated *net effective* stress-energy density which corresponds to the force field that is actuly measured in an accelerated frame of reference. Show me the math otherwise the words have no meaning to me. This means that Mach's principle is effectively abaned -- ong with his rather loony d of strict empiricism (remember, Mach flatly rejected the atomic theo). Mach's idea reenters in the Holographic World where Lp* = Lp^2/3(c/H)^1/3 = 1 fermi now. But this may have a fat flaw since H is cosmic time dependent in the FRW regime. The connection for parlel transport locly vanishes in LIF's on timelike geodesics. Of course it does. But we can (at least locly) distinguish beeen the inerti and the gravitation *components* of the canonic connection field (distinct contributions to the net metric gradients). That comes right out of the differenti geomet (so, see e.g. Weinberg's Gravitation and Cosmology). I think I said that above. The changes observed in net gravitation-inerti forces under acceleration of one's frame of reference are then viewed not *merely* as an abstract mathematic property of the laws of GR under certain class of spacetime coordinate transformations, but so as a *re physic effect* arising from the interaction of ponderable matter with the physic vacuum (AKA the luminiferous aether). The Urstoff! Non-materi, yet physic... As I understand it, this is precisely the way the electromagnetic field is treated in GR when observed in accelerated frames -- as inertily compensated. There is in canonic GR no question of deriving EM forces from a geometrodynamic connection field, and so we are left (in standard GR) with inerti forces derived from an *inerti* connection field that can physicly compensate the EM forces in the EM anog of free-fl. Of course, we then no longer have gener relativity in Einstein's sense, since we do not have strict gravitation-inerti equivence (gravitation and inerti fields considered fundamently indistinguishable). Einstein's point may then be taken as this: the non-materi aether is so inextricably bound up with empty space, that the physic effects that result from interaction of ponderable matter with the vacuum might as well be viewed as arising from a certain kind of geomet. We then POV, we can ways decide as an ternative approach to investigate the properties of empty space as properties of a physic entity -- the vacuum -- which is to be distinguished from a void. The void of Einstein with its Riemannian geomet of spacetime manifolds then appears merely as the shadow cast by a deeper reity -- the physics of the *physic vacuum* (ghost of the departed aether). However, we do still have gener covariance of the laws. The strength of the gravitation field -- and, I would argue, its associated gravitation field stress-energy density -- is then no longer dependent on one's frame of reference, and the true gravitation stress-energy should itself then be represented by a tensor quantity (Diff(4) world-tensor). What *is* dependent on one's frame of reference in this ternative interpretation is the *inerti contribution* to the net forces experienced by massive bodies. Thus the inerti stress-energy will then obviously NOT be represented by a tensor. So we are tking about a (loc) decomposition of the toffel connection into its inerti (acceleration-dependent) and gravitation (acceleration-independent) parts. In this view, for a single gravitating mass, the true gravitation field is the force field that is observed in a frame at rest with respect to the source. In the Einstein approach, in contrast, *l* frames of reference are on an equ physic footing (gener relativity) and there is no true gravitation force field -- any more than there is in SR a true inerti frame in which clocks and rods assume their actu properties, or a true electric or magnetic field considered separately. If the connection is not zero at a point then one is in a LNIF at that point feeling weight or g-force on a timelike non-geodesic world line. Of course. None of this is disturbed by the shift of interpretation, nor by the (loc) decomposition of the connection field into its inerti and gravitation parts. We simply make a fundament physic distinction beeen inerti and gravitation forces, while fully recognizing the close mathematic anogy beeen them that was pointed out by Einstein (both are derivable from metric-tensor fields) . In the ternative paradigm, we look for the ground of this anogy at a deeper physic level, rather than basing it on some mystic Gilean-Platonistic faith in the reity of the forms presented to us at first glance by tensor anysis and differenti geomet. In the ternative (neo-Lorentzian) paradigm, gravitation is a physic effect that depends on the existence of gravitating matter and its time-dependent distribution in space. Inerti forces, on the other hand, are separate and distinct physic effects that do not depend on the existence of materi gravitation sources but only on the accelerative interaction of ponderable matter with the physic vacuum (and not, *contra* Mach, with other materi objects)-- as I would assume any workable field theo of quantum gravity would predict. While the curvature tensor ti force beeen o neighboring time-like LIF geodesic observers is sml in the weak field limit it is not locly zero in gener and cannot be eliminated exactly. But there is the awkward question of the multipole effects experienced Riemannian gravitation fields -- which I understand do not sce down with spacetime volume as do ti effects. And of course, there is so the closely related absolute character of spacetime curvature, by which I mean Riemannian curvature. It's either there or it isn't. Einstein was only able to save his gener relativity thesis by pretending that Riemann curvature has no direct empiric meaning -- at least locly -- thus fixing l attention on the connection field as physicly re. That now strikes me as a classic *ad hoc* stratagem in what Imre Lakatos might have cled a degenerating problemshift. These kinds of questions were thoroughly thrashed out beeen Einstein and von Laue in the early 1920s. I believe later Einstein quietly came to accept the fact that his concept of gener relativity had failed. Which raises fundament questions about the true character and status of speci relativity, since Einstein's theo of relativity program required both speci *and* gener relativity to succeed in order to be fully vindicated. One could t to make a loc classic vacuum stress-energy density tensor from contracted quadratic forms of the 4th rank curvature tensor, but then one has a field equation like Ruv - (1/2)Rguv + /zpfguv + Quadratic Function of Ruvwl = 8pi(G/c^4)Tuv(matter) The problem then is to conserve the loc currents. Wouldn't it be easier to look first at the definitions of gravitation and inerti stress-energy densities for flat fields? What is the stress-energy density of the pure inerti field (no sources)? What are its mathematic properties? What is the acceleration-independent gravitation field stress-energy density in the far field (asymptoticly flat spacetime)? What are its mathematic properties? Z. I will get back to you on last hf later. ==== >>Just as the >>answer to a mathematic question is either right or right, and a >>proof is either vid or invid, >Not where I went to school. A perfectly vid prrof would be marked >down if it was not elegant; quite properly, IMHO. I see no reason for this. This is not the way mathematics is e; first get the proof, then MAYBE look for elegance. so, it is quite common for elegant proofs to hide the concepts; using Liouville's Theorem to prove the Fundament Theorem of gebra is such. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University ==== Ì Herman Rubin ó.8d.98.87.8b.8c .97.99.95 .92.86.94.9d.92.87 >Not where I went to school. A perfectly vid prrof would be marked >down if it was not elegant; quite properly, IMHO. I see no reason for this. This is not the way mathematics is > e; first get the proof, then MAYBE look for elegance. Taking away the elegance from ANY proof, transforms the proof into a tedious, boring and mechanic procedure, which should interest no true mathematician. > so, > it is quite common for elegant proofs to hide the concepts; I'd rather have an elegant proof which hides some of the concepts yet gives the overl idea, than a mechanistic and non-interesting proofwhich boggles the mind with inane details which are of no interest to the reader. > using Liouville's Theorem to prove the Fundament Theorem of > gebra is such. > -- > This address is for information only. I do not claim that these views > are those of the Statistics Department or of Purdue University. > Herman Rubin, Department of Statistics, Purdue University -- http://users.forthnet.gr/ath/jg/ ==== > >> Well, that's ready been proved wrong. >> Since if you find a chemist that knows anything >> about science, you'll be a first. >Are you saying that chemists are not scientists, therefore brain >chemist does not influence intelligence? ZZBunker is a Markov text generator and can not be viewed as > saying anything. It would not pass the classic Turing test for > humanity. That is true. Since Markov being a probabity theorist didn't have have anything interesting to say about science. It's just the same old Gaussian test for low IQ. And we'll ignore the reference the term classic since a mathetoid wouldn't know the difference beeen classic and modern if Cantor rose from the dead and bit him in the ass. And thirdly given that the Turing test wasn't invented by Turing, but rather second rate Van Neumann-wannabee mathemaphilosophe it's irrelevent. ==== [added sci.edu, comp.edu] >> I Çt get it.IÇm a perfect 0 at math. Some people have no problems at l with it. > Am I too dumb for math? You're asking the wrong question -- you're not >dumb for math; you *might* have a brain/mind that >works in a way that is not compatible with the way >of thinking that you need for understanding maths >(emphasis on the *might* -- you claim to be dumb >for maths, but who knows, maybe you're much better >newsgroup and just 't know it, or maybe you're a >high-expectation kind of person, and then anything >below Newton, or Gauss, or Fourier's brains means >too dumb for math in your mind? ) I've come across various students who viewed that they were missing the mathematics gene (or programming gene, or whatever the particular subject happened to be). In those cases it was uniformly the case that their difficulty was emotion/attitudin, rather than cognitive. As you mention, one unhelpful attitude is perfectionism, especily in hard-edged subjects where some answers are clearly objectively *wrong* and thus the student has no wiggle room to avoid the conclusion that they made an error. >Anyway, this, plus many of the things that have >been ready said (mainly about math being genuinely >hard -- the more sophisticated level of math, the >harder, of course) Sever of the missing gene students had the unhelpful attitude that they expected maths to be easy, since they had found their schooling easy so far. ISTM that cognitive issues do kick in when deing with high levels of abstraction where there are no readily-accessible concrete models. For example, my brain hit the wl ting to visuise non-Hausdorff spaces, and my painful memo of the rest of that topology course is of generly mindless memorizing and proof cranking. >HTH, -- ---- | BBB b barbara minus knox at iname stop com | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ------ ==== I 't know how to quantify it, but it seems that a large subset of the people who get stuck in mathematics at some point actuly missed some key point earlier in their study of mathematics without reizing it. The way some of them get unstuck is by going back to the point that they didn't rely understand. One seemingly common version of this is the good student who is able for awhile to compensate for the weakness of their understanding of some group of concepts in math by memorizing more and learning more techniques. Ironicly a good ability to de with mathematics on the level of meaningless manipulations can betray these students later. They may be ting their best to succeed, as they imagine they should be, but it gets harder and harder. Unfortunately the way we teach students too often conduces to this strategy on their part. Again, backing up to what they didn't quite understand seems to be one of the best ways to de with it. Of course, this can be hard when you're on a fixed schedule and seem not to have the time to go back and learn things properly, but even so, tracking down a key concept and finly getting it can save a lot of wasted effort in ting to cope without quite understanding. This raises the issue of pacing. I once read a claim which I thought was interesting and somewhat plausible, that the best math students in school have a well-developed ability to adjust their pace in accord with how fast they need to go. This means first being aware when you actuly are understanding and when you aren't, and second being able to slow way down when you need to. Smart students are often accustomed to being able to read quickly and compose essays pages at a time, but in mathematics actuly digesting a few brief statements sometimes takes hours and hours. Of course there are parts that good math students can then breeze through rapidly too, especily once the concepts are clear. Best wishes to struggling students. It's important to exercise patience, especily not panicking when it seems not to be going well. Take a few slow deep breaths. Get other students and have a group study after working on the problems yourselves. Ask your teachers questions in and out of class. ==== I am writing a computer gorithm that cculates the APR for fixed-rate and adjustable-rate mortgages (ARM). The fixed-rate APR gorithm is working, and it is uses the actuari method that is defined in Title 12 Truth in Lending (Regulation Z Article J, section 226.36, pp. 43-46): http://www.federreserve.gov/regulations/title12/sec226/12cfr226_01.htm Does anyone know or know where to find the procedure/formula for cculating the APR of an ARM? While I cannot find a detailed description for the ARM APR procedure, I have pieced some of it together, but I'm not sure how accurate it is... An example 3/1 ARM: Loan Amt: $129,500 Initi Rate: 3.25% Years: 30 Unit Periods: 12 (i.e., monthly payments) Fees: $3,753.01 Index: 2.75% Margin: 1.13% Six-Step ARM APR Procedure (I 't think it is correct): 1. Cculate the monthly payment for the first three years based on the following inputs: Loan Amt: $129,500 Rate: 3.25 (initi rate) Years: 30 Unit Periods: 12 2. Cculate princip paid out over first three years based on the following inputs: Loan Amt: $129,500 Payment: (previously cculated payment) Rate: 3.25 (initi rate) Years: 3 Unit Periods: 12 3. Cculate the remaining bance: New bance: $129,500 less princip paid out over first three years 4. Cculate fully-indexed rate: Rate: 3.88 (index + margin) 5. Cculate the new monthly payment based on the new input vues: Loan Amt: (new bance) Rate: 3.88 (index + margin) Years: 27 (30 - 3) Unit Periods: 12 6. Cculate the APR: Loan Amt: (new bance) Payment: (previously cculated payment) Years: 27 Unit Periods: 12 Fees: $3,753.01 Estimated APR: 3.88 (input fully-indexed rate here as estimated APR -- the acturati method is a convergence gorithm) ==== Is it possible to determine what the uncountable product of l re > numbers in the interv [0.5, 1.5] is? Intuitively it seems like 1, > but is this concept studied in gener anywhere (e.g. the concept of > uncountably infinite products or sums of re numbers)? Lookup the term summable family. This is a concept which lows definition > of convergence for any numbers a_i with i in I where I is an arbitra > index set. By applying exp-log-formism (transform your product into a > sum) you get the anogue theo for products. > A necessa condition for such a family to converge is that its support (the > largest set J subset I such that a_j <> 0 forl j in J) is countable - > this means that your product cannot converge. > Because equly well than your reasoning I could argue that it converges to > any other re number <> 1. Cf. Riemann's rearrangement theorem. This is something that I have wondered about. Occasionly in previous threads the idea of summing an uncountable set of numbers has come up. It is usuly commented this is not possible unless l but a countable number are zero. However my reaction has ways been: is there any definition for infinite sums of any sort other than the common series. I wondered about summing the vues of a function f from a set to R or C. I wondered what sort of structure the set would have to have and what class of functions could be handled. Even when the set is countable, the answer is not obvious since the sequence can affect the result. I mentioned in a recent thread, the distinction beeen absolutely and conditionly convergent series. This is the first reference that I have noticed to a generisation of infinite sums. Unfortunately I could not find much on Summable families on the net. A search of the groups reveed little but this thread. A search of the web found more but many seemed to mention it only in passing or were in formats that I could not read. Mathworld did not seem to have anything. Can you point me somewhere or give a bit more detail? For non-countable sets, could measure theo be considered a generisation of these sums? J ==== |I guess I was thinking of an average, which is of course a sum not a |product. But how about this (which I'm sure is still wrong). What if |you consider the infinite product of the numbers in the interv |A=[2/3,3/2]. The product of the numbejust as a product of numbeis undefined as has been explained by others. I think you've moved to considering something which bears the same relationship to products as taking an integr does to sums. | Then is this product 1? I would attempt to prove this |as follows: | |Let x be in A. If x is 1 then it contributes nothing to the product, |otherwise assume 2/3 <= x < 1. Then 1 < 1/x <= 3/2. So 1/x is in A. |Likewise if 1 < x <= 3/2, then 1/x is in A. So for any number x in A, |1/x is in A. Hence the product of l of them is 1. | |Or is there some problem with this? For this sense of continuous product, it fails for the same reason that the following doesn't work: What's the average of x^2 from 0 to 1? Well, for each vue of x^2 > 1/2 there's a corresponding vue 1-x^2 at the point sqrt(1-x^2). So the vues of x^2 match up in pairs adding to 1, so the average vue is 1/2. This fails because the horizont spacing so counts. Transforming from x to sqrt(1-x^2) or 1/x stretches or compresses intervs on the x axis, which affects the result. ==== > The product of the numbejust as a product of numbeis > undefined as has been explained by others. I think you've > moved to considering something which bears the same > relationship to products as taking an integr does to sums. This is known as a product integr. in the case that the vues are positive re numbeit may be e by taking log, integrating, then exponentiating. It gets more interesting when the vues are in some space with non-commutative multiplication. Matrices, say. -- http://www.math.ohio-state.edu/~edgar/ ==== [..] >Natur numbers pose a speci problem since they may be viewed in o >different ways. Actuly, they can be defined in many, in many, many different ways, and viewed in many, many, many different ways. >We may view them either as something defined by >induction, or as the smlest gebraic structure which is closed >under the operations 0, 1 , +, and * (theese being subject to the >usu laws). As far as I can tell, {0,1} satisfies the hypothesis you give, through suitable definitions. Presumably, you mean the smlest subset of R/Q/Z which contains 0, 1 and is closed under + and *... But there are plenty of other ways to characterise the natur numbers; they are the smlest non-finite ordin; they are the first singular cardin; they are a set together with a function that satisfies the recursion theorem, and many others. ==== ..................... >> So it is not possible to tell from a description of a >> representation how the representation is to be interpreted. >This is a good example. However, I think that with the representation >of natur numbers I gave, unlike this one, there is no risk of >misunderstandings. >o questions whose answers I am seeking are: >1) What exactly might we mean by representation? >2) How do we define a mathematic object if we do not want to >identify it with some representation of it? This is e quite often. The Peano Postulates CHARACTERIZE the positive (or non-negative) integers. There are many other ways to characterize them. >Natur numbers pose a speci problem since they may be viewed in o >different ways. We may view them either as something defined by >induction, or as the smlest gebraic structure which is closed >under the operations 0, 1 , +, and * (theese being subject to the >usu laws). Only o? There is so the cardin version, which is still different. They are so the smlest subset of the re numbers (suitably defined) containing 1, which is defined as the re numbers form a field, and closed under addition. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University ==== >A natur number is a sequence of symbols in which only the symbol >'|' occurs. > That is ONE representation of a natur number. >>Indeed. But identifying mathematic objects with what are rely >>representations of them is ve common, and I was just doing what >>eveone else does, though maybe I should not have. It's crass formism, a major impediment to undertstanding mathematics. And not eveone else does it. >> This is an extremely poor way of attempting to understand >> mathematic objects. I agree. The first thing I do is ways to abstract away from > inessentis, which includes representations (or are there cases where > it is necessa to use representations?). I do not see how people > manage to do mathematics with so much garbage around. Some people like the garbage :-( >> Take, for example, the following: A positive integer is a non-empty finite string of symbols >> from a collection of 15 symbols, of which one of the >> symbols cannot be the first element of the string. Now what led me to produce this statement? You might >> consider that it came from the base-15 representation of >> integebut what led me to this was the Sumerian >> representation in base 60, in which a digit is either >> the zero symbol, or (to use our terminology) one of five >> symbols for multiples of ten, or one of nine symbols for >> multiples of one, or a symbol for a multiple of ten >> followed by one for a multiple of one. But a string >> without attention to the spacing can be interpreted in >> only one manner. So it is not possible to tell from a description of a >> representation how the representation is to be interpreted. This is a good example. However, I think that with the representation > of natur numbers I gave, unlike this one, there is no risk of > misunderstandings. You identification of natur numbers with a sequence of symbols is ready a msiunderstanding. Natur numbers pose a speci problem since they may be viewed in o > different ways. We may view them either as something defined by > induction, or as the smlest gebraic structure which is closed > under the operations 0, 1 , +, and * (theese being subject to the > usu laws). Isn't the field of o elements a smler gebraic structure which is closed under the operations 0, 1 , +, and * (theese (sic) being subject to the usu laws) than the natur numbers. -- ==== >A natur number is a sequence of symbols in which only the symbol >'|' occurs. That is ONE representation of a natur number. >Indeed. But identifying mathematic objects with what are rely >>representations of them is ve common, and I was just doing what >>eveone else does, though maybe I should not have. It's crass formism, a major impediment to undertstanding > mathematics. And not eveone else does it. > This is an extremely poor way of attempting to understand >> mathematic objects. I agree. The first thing I do is ways to abstract away from > inessentis, which includes representations (or are there cases where > it is necessa to use representations?). I do not see how people > manage to do mathematics with so much garbage around. Some people like the garbage :-( > >> Take, for example, the following: A positive integer is a non-empty finite string of symbols >> from a collection of 15 symbols, of which one of the >> symbols cannot be the first element of the string. Now what led me to produce this statement? You might >> consider that it came from the base-15 representation of >> integebut what led me to this was the Sumerian >> representation in base 60, in which a digit is either >> the zero symbol, or (to use our terminology) one of five >> symbols for multiples of ten, or one of nine symbols for >> multiples of one, or a symbol for a multiple of ten >> followed by one for a multiple of one. But a string >> without attention to the spacing can be interpreted in >> only one manner. So it is not possible to tell from a description of a >> representation how the representation is to be interpreted. This is a good example. However, I think that with the representation > of natur numbers I gave, unlike this one, there is no risk of > misunderstandings. You identification of natur numbers with a sequence of > symbols is ready a msiunderstanding. > Natur numbers pose a speci problem since they may be viewed in o > different ways. We may view them either as something defined by > induction, or as the smlest gebraic structure which is closed > under the operations 0, 1 , +, and * (theese being subject to the > usu laws). Isn't the field of o elements a smler gebraic structure which is > closed under the operations 0, 1 , +, and * (theese (sic) being subject to > the usu laws) than the natur numbers. You are right! Of course, Z_2 is the smlest such structure. One needs to add some axiom that forces there to be infinitely many elements. ==== In Set Theo, is an element of N so a proper subset of N? For example, I want to prove that a cardin k belonging to N is less than the cardin of N itself, eph null. If an element of the natur numbers so a proper subset of N, then I can show k < eph null by showing that a set A with cardin k is a proper subset of the natur numbers. Since a natur number n can't be mapped onto a proper subset of itself, I think I can say that eph null can't be mapped onto a proper subset of itself. ==== >In Set Theo, is an element of N so a proper subset of N? Depends on the definition, but under the usu one (e.g., the one in Hmos's _Naive Set Theo_), then yes. The natur numbers are an example of a transitive set, which means a set that satisfies that any element is so a subset. It is then not hard to show that in the case of N, they are in fact proper subsets. >For example, I want to prove that a cardin k belonging to N is less >than the cardin of N itself, eph null. That's true for any limit cardin; in particular for N. >If an element of the >natur numbers so a proper subset of N, then I can show k < eph >null by showing that a set A with cardin k is a proper subset of the >natur numbers. That's not enough; the natur numbers contain proper subsets which have the same cardinity as the full subset. In fact, that's one way to define infinite (at least, in the presence of the Axiom of Choice). > Since a natur number n can't be mapped onto a >proper subset of itself, I think I can say that eph null can't be >mapped onto a proper subset of itself. That would be fse, though. The natur numbers can be mapped onto the set of l even natur numbea proper subset of itself. ==== >>In Set Theo, is an element of N so a proper subset of N? Depends on the definition, but under the usu one (e.g., the one in >Hmos's _Naive Set Theo_), then yes. I'm using Jech's Inroduction to Set Theo, 3rd Ed. >That's true for any limit cardin; in particular for N. What's a *limit* cardin??? >That's not enough; the natur numbers contain proper subsets which >have the same cardinity as the full subset. In fact, that's one way >to define infinite (at least, in the presence of the Axiom of >Choice). You're right, Arturo . . . I forgot about that . . . And my professor won't cover the Axiom of Choice until late November I think. >> Since a natur number n can't be mapped onto a >>proper subset of itself, I think I can say that eph null can't be >>mapped onto a proper subset of itself. That would be fse, though. The natur numbers can be mapped onto >the set of l even natur numbea proper subset of itself. Now I see that a jump from a natur number to eph null would be too much. But how can I *show* k < eph null? It's like proving that 1 < infinity. It's obviously true, but how do you prove it? ==== >>In Set Theo, is an element of N so a proper subset of N? >>Depends on the definition, but under the usu one (e.g., the one in >>Hmos's _Naive Set Theo_), then yes. I'm using Jech's Inroduction to Set Theo, 3rd Ed. >That's true for any limit cardin; in particular for N. What's a *limit* cardin??? Oops. Looks like I screwed up, at least with respect to Jech's book. He only defines limit ordins, and of course, eve infinite cardin is a limit ordin when considered an ordin. I was ting to refer to cardins which were not successor cardins (like 0, eph_0, eph_omega, etc). But that is rather silly. The Axiom of regularity guarantees that if an element of a set A is so a subset, then it must be a PROPER subset. Otherwise, {A} would have no epsilon-minim element. >>That's not enough; the natur numbers contain proper subsets which >>have the same cardinity as the full subset. In fact, that's one way >>to define infinite (at least, in the presence of the Axiom of >>Choice). You're right, Arturo . . . I forgot about that . . . And my professor won't cover the Axiom of Choice until late November I >think. I only mentioned the axiom of choice because there are many different ways to define infinite and finite, and I'm not positive which one is usuly used when you are in ZF. I know that whichever it is, if we are in ZFC then the definition of infinite is equivent to there is a one-to-one bijective correspondence beeen A and a proper subset of A. > Since a natur number n can't be mapped onto a >proper subset of itself, I think I can say that eph null can't be >mapped onto a proper subset of itself. >>That would be fse, though. The natur numbers can be mapped onto >>the set of l even natur numbea proper subset of itself. Now I see that a jump from a natur number to eph null would be too >much. But how can I *show* k < eph null? It's like proving that 1 < >infinity. It's obviously true, but how do you prove it? Looking at my copy of Jech's Set Theo, 2nd edition, a set is defined to be finite if it has the same cardinity as some n in N. eph_0 is defined as the least infinite cardin; and the order of cardins/ordins is by inclusion. So if a is in N, then a is a subset of N, and thus a<=N. If you had a=N, then N would be an element of itself, which contradicts regularity, so a Let U and V be independent random variables each with Uniform(0,1) > distribution. Determine the joint density of X = maj(U,V) and Y= U+V. I get this: P(XThere are quite a few posters who reply in my threads. Now I've >noticed that I trounce one poster and then another pops up and starts >yapping. Later some poster who got his ass kicked is back trotting >out the *same* crap. That's your impression. In _fact_ you have never trounced anyone here. >It occurs to me that with l those posting in l those threads that >I create, there's lots of you who probably 't keep up, and 't >reize that I usuly win. You usuly win. Right. Except for the times you 't, which happen to be ways. It must be nice, living in a world where evething is the way you want it to be... >So I'm thinking of ways to nail these posters down, force them to have >to stand by their positions in clear sight, and when they're trashed, >make certain that they can't just wait for a while and come back later >with the same ol' crap. Now like the sneaky bastards they are, they t to act like I'm the >one who's keeping things confusing by being a moving target, but I'm >only moving towards simplification. That is, when I find that the argument is too complicated for most >readeand that other posters have it easy confusing people, I work >to simplify it, which is how I've gone from P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) to P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 and *still* I find posters ting to find ways to confuse, and as >usu, piling up posts in my threads!!! So, I'm looking for ways to end their games. Those who want a stable version of my argument, can just go to a blog >I've created, where you can read it, without distractions from dumb >posters!!! http://mathforprofit.blogspot.com/ Go, you know you want to go read it. Go ahead, no one will know. Just go and read it ready. Why are you still here reading? Go!!! http://mathforprofit.blogspot.com/ ==== > There are quite a few posters who reply in my threads. Now I've > noticed that I trounce one poster and then another pops up and starts > yapping. Later some poster who got his ass kicked is back trotting > out the *same* crap. It occurs to me that with l those posting in l those threads that > I create, there's lots of you who probably 't keep up, and 't > reize that I usuly win. So I'm thinking of ways to nail these posters down, force them to have > to stand by their positions in clear sight, and when they're trashed, > make certain that they can't just wait for a while and come back later > with the same ol' crap. Now like the sneaky bastards they are, they t to act like I'm the > one who's keeping things confusing by being a moving target, but I'm > only moving towards simplification. That is, when I find that the argument is too complicated for most > readeand that other posters have it easy confusing people, I work > to simplify it, which is how I've gone from P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) to P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 and *still* I find posters ting to find ways to confuse, and as > usu, piling up posts in my threads!!! So, I'm looking for ways to end their games. Those who want a stable version of my argument, can just go to a blog > I've created, where you can read it, without distractions from dumb > posters!!! http://mathforprofit.blogspot.com/ Go, you know you want to go read it. Go ahead, no one will know. Just go and read it ready. Why are you still here reading? Go!!! -- http://www.crbond.com ==== > There are quite a few posters who reply in my threads. Now I've > noticed that I trounce one poster and then another pops up and starts > yapping. Later some poster who got his ass kicked is back trotting > out the *same* crap. Seems you lost interest in prime counting gorithms since you got your ass kicked there... ==== >There are quite a few posters who reply in my threads. Now I've >noticed that I trounce one poster and then another pops up and starts >yapping. I've noticed that you never trounce anyone, except in your head. You usuly misunderstand what is said, engage in red herrings, strawmen, and ad hominems, and then declare unilater victo. > Later some poster who got his ass kicked is back trotting >out the *same* crap. Because you NEVER ADDRESS IT. Your idea of kicking ass seems to be to repeat your incorrect argument over and over, start new threads, claim victo, and lie about what has happened before. Here you go again. >It occurs to me that with l those posting in l those threads that >I create, there's lots of you who probably 't keep up, and 't >reize that I usuly win. You do not usuly win, except maybe in your imagination. You routinely fail to even comprehend what is written, let one answer it; in fact, you often fail to even read it. [..] ====== ething; self-evident honesty of conviction, which does more; and a long purse, which does most of l. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine tgustus de Morgan ====== ==== > There are quite a few posters who reply in my threads. Now I've > noticed that I trounce one poster and then another pops up and starts > yapping. Later some poster who got his ass kicked is back trotting > out the *same* crap. Probably imitating you. ==== Which number should replace the question mark so that each group of four numbers has the same logic mathematic property: 2240 3534 4451 8765 994? It is a quiz question and I just can't seem to work it out. I assume that a) it is a single digit that replaces the question mark and b) the numbers are not four digit numbebut are in fact groups of single digit numbers that are somehow related. Someone must have an idea about this! Lots of us have been ting to work it out but just can't get anything that makes sense. ==== from your address with no luck. I've been filtered! But I guess my message is suitable for gener viewing, so here goes...) I was thinking about these (and CF's in other number fields) and came across your posts on sci.cpt; it seems like the gener consensus was use the Euclidean gorithm for pha/1 and nearest neighbor in place of greatest integer. The nearest neighbor guarantees the magnitude of the remainer is less than one for Z[i] but not for gener number fields. Conjugates of pha have the same norm; is there ways a conjugate of pha with purely positive coefficients? If so, then one could take the gebraic integer nearest zero in the cell pha lies in. This guarantees that the coefficients remain positive, but has the disadvantage that the difference can be greater than 1. There is an upper bound on the magnitude of the remainder r, though; cl it b. Then we can take b/r and be guaranteed that it will be at least 1. Something similar happens with cf's over the res; if you subtract multiples of some integer n, then r < n; and rather than simply inverting, take n^2/r > n. Then the convergents p/q satisfy p^2 - pha^2 q^2 < (2 n pha) I haven't worked out yet whether there are similar bounds for the Gaussian integers given the integer nearest zero greatest integer function above. Did you find anything more after receiving those replies to your posts? ==== > .... I'm curious as to how Euler determined Euler's constant .... If you'd rely like to dig into the histo, here's a reference. published in Commentarii academiae Petropolitanae ad annum 1736, Tom. VIII, pp.14-16. Unfortunately Cajori doesn't give the title of the . ==== ==== > .... > According to http://members.aol.com/jeff570/mathword.html : The terms HYPERBOLIC GEOMET, ELLIPTIC GEOMET, and PARABOLIC > GEOMET were introduced by Felix Klein (1849-1925) in 1871 in .86ber > die sogenannte Nicht-Euklidische Geometrie (On so-cled > non-Euclidean geomet), reprinted in his Gesammelte mathematische > Abhandlungen I (1921) p. 246 ( and Smart, p. 301). I have not read that reference, but.... The reference has been translated into English, in John Stillwell, Sources of Hyperbolic Geomet, 1996, p.72 (3rd paragraph and footnote). I'd rather not type it l now, but anyway Stillwell's little book is well worth looking at. . ==== given factorization are roots of the given expression. And where the can applies is to the factorization as there are an > *infinity* of factorizations for P(x), but I'm pointing out that one > of them is the one that follows. So, there's a direct statement telling *exactly* what the a's are for > the factorization, which readers can see for themselves. > If you had bothered to read further, you would see that I agree > this is not rely a problem, or at least it is a problem that is > easily fixed. There's no problem there. Now when a poster makes mistakes I've gotten to where I just stop there, but since this one is pressing the issue, let's see what has gotten Nora Baron so excited. > But as I mentioned at the top there are o problems. The > second one is the re one. You will not be able to fix it. But you for some reason deleted off l reference to it. So here it is again. Evade it again if you want, but at least > quit pretending that you are willing to respond to l objections. ==== = = > Deleted part of my post: This problem is easy to fix. Just replace 5 by y or some other symbol. > You have then restored P(x) to being a polynomi in y. You would then > be factoring it in the form P(x) = (a1*y + 7)*(a2*y + 7)*(a3*y + 7), and indeed you can then conclude CORRECTLY that the a's > *must be* roots of [**]. Thus this is one place where you have a mistake, but it is easily fixed. > I will pretend that you have made the right fix and proceed. Did that make sense to anyone? >Notice it *appears* that the constant terms for the three factors are >l 7, which can't be right, as the constant term of P(x) is 1078, so >setting x=0, reves >P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) >as the cubic defining the a's with x=0 is >a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2 >for 0, so that leaves a_3 with a vue of 3 when x=0. >So let a_3 = b_3 + 3, where I keep indices matched. Then I have >P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7) >P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22) >and now my constant terms work out correctly. l this is basicly OK, condition on the fix described above. This poster Nora Baron is as usu, annoying. >But P(x) has 49 as a factor as eve term in >P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 >has 49 as a factor, so I can divide by 49, and dividing 1078 by 49 >gives me 22, as the new constant term. Yes. so true. No doubt about that. > >Well that means that >P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22) >is the only way that the constant terms keep matching. No. This last step is where your most serious error is. Well that's clearly wrong as it follows from the previous step. Now the poster Nora Baron has e this before as I chged the poster to point to an error in the argument, and the poster pointed at the conclusion!!! And here yet again, the poster points at the conclusion!!! However, it follows that since the constant term of P(x)/49 is 22 that the constant terms of the factors of P(x)/49 't have 7 as a factor because 22 doesn't have 7 as a factor. Notice how short and direct that is. I've connected the conclusion to what came before, and it's obvious enough, unless you wish to chge the assertion that 7 is not a factor of 22. ==== >> given factorization are roots of the given expression. And where the can applies is to the factorization as there are an >> *infinity* of factorizations for P(x), but I'm pointing out that one >> of them is the one that follows. So, there's a direct statement telling *exactly* what the a's are for >> the factorization, which readers can see for themselves. >> If you had bothered to read further, you would see that I agree >> this is not rely a problem, or at least it is a problem that is >> easily fixed. There's no problem there. Now when a poster makes mistakes I've gotten to where I just stop >there, but since this one is pressing the issue, let's see what has >gotten Nora Baron so excited. > >> But as I mentioned at the top there are o problems. The >> second one is the re one. You will not be able to fix it. But you for some reason deleted off l reference to it. So here it is again. Evade it again if you want, but at least >> quit pretending that you are willing to respond to l objections. ==== = = >> Deleted part of my post: This problem is easy to fix. Just replace 5 by y or some other symbol. >> You have then restored P(x) to being a polynomi in y. You would then >> be factoring it in the form P(x) = (a1*y + 7)*(a2*y + 7)*(a3*y + 7), and indeed you can then conclude CORRECTLY that the a's >> *must be* roots of [**]. Thus this is one place where you have a mistake, but it is easily fixed. >> I will pretend that you have made the right fix and proceed. Did that make sense to anyone? Yes. Did it make sense to you? >>Notice it *appears* that the constant terms for the three factors are >>l 7, which can't be right, as the constant term of P(x) is 1078, so >>setting x=0, reves >>P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) >>as the cubic defining the a's with x=0 is >>a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2 >>for 0, so that leaves a_3 with a vue of 3 when x=0. >>So let a_3 = b_3 + 3, where I keep indices matched. Then I have >>P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7) >>P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22) >>and now my constant terms work out correctly. l this is basicly OK, condition on the fix described above. This poster Nora Baron is as usu, annoying. And, as usu, correct. >>But P(x) has 49 as a factor as eve term in >>P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 >>has 49 as a factor, so I can divide by 49, and dividing 1078 by 49 >>gives me 22, as the new constant term. Yes. so true. No doubt about that. >> >>Well that means that >>P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22) >>is the only way that the constant terms keep matching. No. This last step is where your most serious error is. Well that's clearly wrong as it follows from the previous step. No, it does not follow from the previous steps. That's PRECISELY what we are objecting to. >Now the poster Nora Baron has e this before as I chged the >poster to point to an error in the argument, and the poster pointed at >the conclusion!!! No, she is pointing to the implicit claim that the last line follows from the previous ones. It does NOT follow from the previous ones. It is NOT true that it is the only way that the constant terms keep matching. I ready have given you many other ways, but here is the KEY points again: (1) Given ANY function f(x) from the gebraic integer to the complex numbeand given any gebraic a and any gebraic integer b, there exists a function g(x) from the gebraic integers to the complex numbeand a complex number c, such that f(x) = g(x)+c and g(a)=b. Set g(x) = f(x) - f(a) + b c = f(a) - b. (2) Let w_1(x), w_2(x), and w_3(x) be ANY functions from the gebraic integers to the complex numbers that satisfy the following conditions: (a) w_1(0)=7 (b) w_2(0)=7 (c) w_3(0)=1 (d) w_1(x)*w_2(x)*w_3(x) = 49 for eve gebraic integer x. Let h_1(x) = (5a_1(x)+7)/w_1(x) h_2(x) = (5a_2(x)+7)/w_2(x) h_3(x) = (5b_3(x)+22)/w_3(x). (3) Apply point (1), by setting a=b=0; that is, we write h_i(x) as h_i(x) = H_i(x) + c_i subject to the condition that H_i(0)=0. Then, according to the above, we have c_i = h_i(0)-0 = h_i(0). Since h_1(0) = 1, h_2(0) = 1, and h_3(0)=22, it follows that for ANY functions w_1(x), w_2(x), w_3(x) which satisfy (2a), (2b), (2c), and (2d), we will have P(x)/49 = (H_1(x) + 1)(H_2(x) + 1)(H_3(x) + 22) The function H_1(x) is equ to 5a_1(x)/7 if and only if w_1(x)=7 for l x; the function H_2(x) is equ to 5a_2(x)/7 if and only if w_2(x)=7 for l x; and the function H_3(x) is equ to 5b_3(x)+22 if and only if w_3(x)=1 for l x. This is clearly not the only way things can happen. In addition, we may ways choose w_1(x), w_2(x), w_3(x) to have vues which are GEBRAIC INTEGEand such that they divide (in the ring of l gebraic integers) 5a_1(x)+7, 5a_2(x)+7, and 5b_3(x)+22, for each x. These vues need not be constant (and in fact, cannot be constant). In any case, it should be clear that your claim that it is the only way is fse, since there are infinitely many other ways in which are not the only way you claim. >And here yet again, the poster points at the conclusion!!! No: she is pointing out that the conclusion does NOT follow from the previous statements. It is fse. >However, it follows that since the constant term of P(x)/49 is 22 that >the constant terms of the factors of P(x)/49 't have 7 as a factor >because 22 doesn't have 7 as a factor. No, the hypothesis does not imply the conclusion you claim they imply. >Notice how short and direct that is. Notice how wrong it is. >I've connected the conclusion to what came before, By a leap of logic. > and it's obvious >enough, that you have no idea what a logic argument is. > unless you wish to chge the assertion that 7 is not a >factor of 22. No, your assertion is NOT based merely on the fact that 7 is not a factor of 22 in the ring of l gebraic integers. It is based on the incorrect and unspoken assumption that there is only one possible choice for functions w_1(x), w_2(x), and w_3(x) that will low you to write the functions with constant terms 1, 1, and 22; you then assume that since you found one way, and there is only one way, your way is the only and therefore the correct way. The unspoken assumption that there is only one way is incorrect. It has nothing to do with 7 not dividing 22. ====== ething; self-evident honesty of conviction, which does more; and a long purse, which does most of l. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine tgustus de Morgan ====== ==== Methinks I finly disproved CH by finding something with cardinity of C, l think by looking in places noone else looks. amateur ==== > -----Origin Message----- > Conversation: How to prove that sqrt(2) exist? The fact that you 't find that convincing is a > good thing, from a purely mathematic point of view. :-) There are various ways to show that a re number > exists - the simplest is first to give a rigorous > definition of continuous function, prove the > theorem that if a < b, f is continuous, f(a) < 0 > and f(b) > 0 then f(x) = 0 for some x beeen > a and b, and then apply this theorem with > f(x) = x^2 - 2, a = 0, b = 2. I actuly thought about this (not in this much > detail, but what I mean is that I thought of the > continuity issue, but then I kind of dismissed it, > since I thought the ve notion of continuity > relies on the existence of numbers... So, the > function has a chance at being continuous because > the vues exist -- how could I use that to prove > that a vue exists such that f(x) = 0? What I find curious is that this continuity issue > seems to rely on the fact that beeen any o > distinct re numbethere is at least one other > re number -- but this fact is not a sufficient > condition for the existence of such number for > which f(x) = 0: it is so true for ration > numbers that beeen any o distinct ration > numbethere is at least one other ration > number. > The thing is that the res are complete (no gaps). It's the connectedness that makes it work. The rations satisfy the other properties of the res that 't rely on completeness. > (at this point, I'm guessing my doubt becomes > sort of philosophic? Something akin to how > do we prove that 0 exists? or how do we > prove that 1 is greater than 0?, or that sort > of thing?) Actuly none of it exists, but some notions are a little easier to swlow than others. ==== I.e., how to prove that there exists a re number x > such that x^2 = 2? I'm getting started with the rigourous side of maths > (as a hobby -- being an engineer, with sever years > of experience, you understand that I haven't rely > needed it), and rely enjoyed reading a proof that > sqrt(2) can not be a ration number. However, the justification they give is that from > Pythagoras' theorem, looking at a triangle with > sides 1 and 1, then the hypothenuse is such that > its square is 2. I 't find that convincing at > l (I mean, sure, intuitively yes -- I wouldn't > even need that evidence to convince me -- I have > ways accepted that sqrt(2) exists, just because, > well, it has to be there, somewhere beeen 1.4 > and 1.5 ... You know... engineers!! ) But anyway, they 't rely prove that sqrt(2) is > a re number: they prove that there is no ration > number p/q such that (p/q)^2 is 2. So, I'm curious: how can one prove that sqrt(2) > exists and it is a re number? > > -- I think this whole thing could be resolved if just bought, or borrowed, a book on Introducto Re ansis. Lurch ==== > I think this whole thing could be resolved if just bought, or > borrowed, a book on Introducto Re ansis. And what do you think prompted my question?!! I do have (just bough a month or so ago) Russell Gor's Re Anysis - A first course, and I'm enjoying reading it. It's precisely there where I saw the proof that sqrt(2) is not a ration number. But as I said, they just argue (in the intuitive sense, I think), that sqrt(2) exists, using the triangle as argument. Now that I think about it (after reading Kent's message) that maybe my skepticism was due to the fact that I 't have a clue on the rigorous side of geomet. It is true that the triangle is an actu construct that shows you that that length (which exists, because the triangle exists) is such that its square must be 2. But I guess my doubt came from the fact that I feel that whatever results in geomet (including Pythagoras) relies on existence and other properties of re numbers; so, using that as an argument to claim existence of a particular re number sounded odd to me... After reading l the posts that you guys have so kindly written, I see that the issue is indeed complex (well, I have more or less understood the arguments, but the whole anysis side of maths is new to me, so I guess it'll take me some time to fully digest the ideas). Anyway, I'm ve grateful for l the replies and the -- ==== Anysis, and higher math, can have ve subtle arguements. It sometimes takes a few readings to even get the idea of what is going on. They say there are stages to understanding proofs. Firstly, one understands the logic progression. Secondly, one understands the proof enough to reproduce it themselves. Thirdly, ???? I forget what the third one is. Anyways, it something like that. Lurch > I think this whole thing could be resolved if just bought, or > borrowed, a book on Introducto Re ansis. And what do you think prompted my question?!! I do have (just bough a month or so ago) Russell Gor's > Re Anysis - A first course, and I'm enjoying reading > it. It's precisely there where I saw the proof that sqrt(2) > is not a ration number. But as I said, they just > argue (in the intuitive sense, I think), that sqrt(2) > exists, using the triangle as argument. Now that I think about it (after reading Kent's > message) that maybe my skepticism was due to the fact > that I 't have a clue on the rigorous side of > geomet. It is true that the triangle is an actu > construct that shows you that that length (which > exists, because the triangle exists) is such that > its square must be 2. But I guess my doubt came from the fact that I feel > that whatever results in geomet (including Pythagoras) > relies on existence and other properties of re numbers; > so, using that as an argument to claim existence of a > particular re number sounded odd to me... After reading l the posts that you guys have so > kindly written, I see that the issue is indeed complex > (well, I have more or less understood the arguments, > but the whole anysis side of maths is new to me, > so I guess it'll take me some time to fully digest > the ideas). Anyway, I'm ve grateful for l the replies and the > > -- ==== In sci.math, Moreno >> Or one can go the slightly silly route: sqrt(2) exists because >> (sqrt(2)^2) = 2 You can ways define it that way (for instance, the way > you define the imagina unit). But that does not mean that > that number you just defined is in the set of re numbers. In particular, I could arbitrarily come up with such definition > for the ration numbers: I could simply say: sqrt(2) is the > ration number X such that X^2 is 2. I could claim that the > number exists because I defined it, and that means that it > exists. You can, however, prove that if such number exists, > it is not a ration number (and thus, not an integer or natur > number either). I can prove a contradiction from your definition, so your claim is bogus. However, X *is* re, for a properly rigorous definition thereof -- just not ration. But let's prove X isn't ration, as a diversion. Assume that it is: X = p/q, gcd(p,q) = 1, for some integers p and q. Now, since X^2 = 2, p^2/q^2 = 2, or p^2 = 2 * q^2. This means p is even. Write p = 2 * r, and we get p^2 / q^2 = 4 * r^2 / q^2 = 2. We flip this and note that q^2 / r^2 = 2, or q^2 = 2 * r^2, so that q is even as well. Oops...this means gcd(p,q) cannot be 1 and therefore the assumption that p and q exist at l leads us into a contradiction. Therefore X, if it exists at l, is not in Q. Obviously X doesn't necessarily exist until someone such as Dedekind or Cauchy step up and prove that it makes sense to treat such numbers with the usu arithmetic operations. The question remains -- when ting to be rigourous, at least > the way I'm understanding it, as a beginner in the rigourous > side of mathematics, you ask yourself the question: how do I > rely know that there is such number? How do I know that it > will not be the case that whatever number I can come up with, > its square will be either greater than 2, or less than 2? First, be ve clear on what you might mean by the term number. Peano, in particular, went the rather interesting route of starting with 1 and deriving l the other natur numbers therefrom, using 4 axioms and 1 axiom which I prefer to cl a meta-axiom, his weak induction axiom. We therefore get N, the set of natur numbers. If that's not rigorous enough, one can construct a sequence S_0 = empty, S_1 = S_0 union {S_0}, S_2 = S_1 union {S_1}, etc, and then prove that S_0 != S_1 != S_2 != ... by set theo, perhaps. Since S_{n+1} = S_n union {S_n} Peano's successor operation is well-defined here as well, and we get 0 as a bonus. Once one has the whole numbers one can construct the integers by assuming -a is such that (a + (-a)) = 0. Fractions are another extension: p/q, q != 0, gives us Q. Now it gets tricky. Can we extend Q in a rigorous fashion, defining re numbers with a consistent definition of +, -, *, and /? I would think that we can but would have to research the issue somewhat. However, I've no problem with pi existing, as one can easily approximate it by various sequences, l of which are Cauchy. (Most of them are derived from infinite series.) e is even easier, as the sequence e = 1/0! + 1/1! + 1/2! + 1/3! + ... + 1/n! + ... is absolutely convergent and generates a Cauchy sequence, namely e_n = 1/0! + 1/1! + 1/2! + 1/3! + ... + 1/n! One can even define sqrt(2) in such a manner. Note that the binomi expansion (1 + x)^(1/2) = 1 + (1/2)x + (1/2)(-1/2)/2!*x^2 + (1/2)(-1/2)(-3/2)/3!*x^3 + ... can be indefinitely extended for non-integr powers. Setting x= -1/2, one gets sqrt(1/2), which is sqrt(2)/2 -- if I've e this right. The crowning achievement, after one more extension (i=sqrt(-1)), is that our numbers are now complete -- we 't get any more by ting to solve cubics, quartics, quintics, etc., though we may have trouble with a precise formulation of the roots, resorting to num approximations instead for degree 5 and up. > >> Or one goes the engineering route, which you've stated you >> 't like: sqrt(2) = 1.4142; pi = 3.1416; e=2.71828. Well, yes. There's the practic side to it, which is what > I have l my life lived with -- it's not like now I want to > become an actu mathematician; but I'm enjoying that side > of maths, and am ting to adapt my mind to work in the > skeptic mode. It is a good question, and one that has bedeviled theorists for yeaif not centuries (it feels a bit like Zeno's Paradox, in a way). I'm not entirely certain we've fixed Cantor's rather famous diagon proof yet, though at least one webpage I've seen purports such. The main problem appears to be that Cantor's diagon is not purely constructive; one can fix that by force-selecting 0 or 1 in base 10, avoiding the .999... problem. Since the requirement is to show that an enumeration of l res isn't possible by constructing a new one, I 't see this as a big problem, but then, I'm not l that rigorous, either. Infinities, however, are like that: tricky. ==== The fact that you 't find that convincing is a > good thing, from a purely mathematic point of view. There are various ways to show that a re number > exists - the simplest is first to give a rigorous > definition of continuous function, prove the > theorem that if a < b, f is continuous, f(a) < 0 > and f(b) > 0 then f(x) = 0 for some x beeen > a and b, and then apply this theorem with > f(x) = x^2 - 2, a = 0, b = 2. I actuly thought about this (not in this much > detail, but what I mean is that I thought of the > continuity issue, but then I kind of dismissed it, > since I thought the ve notion of continuity > relies on the existence of numbers... So, the > function has a chance at being continuous because > the vues exist -- how could I use that to prove > that a vue exists such that f(x) = 0? What I find curious is that this continuity issue > seems to rely on the fact that beeen any o > distinct re numbethere is at least one other > re number -- but this fact is not a sufficient > condition for the existence of such number for > which f(x) = 0: it is so true for ration > numbers that beeen any o distinct ration > numbethere is at least one other ration > number. (at this point, I'm guessing my doubt becomes > sort of philosophic? Something akin to how > do we prove that 0 exists? or how do we > prove that 1 is greater than 0?, or that sort > of thing?) > > -- Many posters have touched on the re core of the issue, which is how to define a re number. As Ullrich says, you are right to distrust the approach that simply constructs a triangle one of whose sides has a length which squares to 2. There are a variety of ways to define re numbenone of which goes into terribly murky philosophic issues. We assume the existence of ration numbers together with their operations of addition and multiplication and the usu rules relating to these operations. It is so important to include the order properties so that we are clear on the meaning of statments like a > b when a and b are re numbers. Beyond that we have an intuition for what the fundament properties of re numbers are, and from this intuition we can construct a set that does indeed have these properties. Having e that we soon reach the conclusion that eve nonempty set of re numbers (whatever trivi to deduce a square root of 2 by applying to the set of re numbers whose squares are less than 2. l of the above evades the question of how to actuly construct the re numbers. The one that works best for me is the approach of using Cauchy sequences which was developed by Cantor and not by Cauchy. A sequence {a_0, a_1, ...) is cled Cauchy if for eve epsilon > 0, there is an N such that if m and n are both greater than N, then a_m and a_n are within epsilon of each other. Backtracking a llttle, a sequence is said to converge to a limit L if for eve epsilon > 0 there is an N such that if n > N, then a_N is within epsilon of L. It is easy to show that a sequence can converge to at most one limit, and that if it converges to a limit, then the sequence is a Cauchy sequence. At least it is easy after some experience. I recl having to wrestle pretty hard with these concepts when they were new to me. It is less clear, but one would hope, that eve Cauchy sequence converges to some limit. This is not true for the ration numbers. We intend to turn the collection of Cauchy sequences into the set of re numbers. It is obvious how to add, subtract, multiply, and divide Cauchy sequences, namely term by term. There is a minor, but solvable problem, with dividing Cauchy sequences when the dividend has zeros. We think intuitively of the Cauchy sequence as representing its limit. The ration number q is turned into the Cauchy sequence {q, q, q, q, q, ...}. The only remaining problem to resolve is that many Cauchy sequences may converge the same limit, if they are convergent, but if that is the case, the difference in these sequences must converge to 0. that is the key to the fin step. Re numbers are defined as equivence classes of Cauchy sequences, o sequences being equiventif and only if their difference converges to 0. It is then straightforward to extend the ordering to these re numbeto prove that l Cauchy sequences of re numbers (how about that, Cauchy sequences of Cauchy sequences) which can be defined because the order properties carried across, and then the least upper bound property. It is a fair amount of effort, the end result of which is that you drop l the baggage, think of the re numbers just the way you ways did, but are now confident that the re numbers have a variety of extremely useful properties such as least upper bounds, nested intervs, convergence of l Cauchy sequences, and others. A fin benefit is that you are in a ve good postion to grasp the nature of p-adic numbers without a great de of trouble. Dedekind chose another path. I never worked it out in great detail, but both the Dedekind cut approach and the Cauchy sequence approach can be found in a great many books with names like Re Anysis, or Introduction to Re Anysis, etc. Achava ==== > The fact that you 't find that convincing is a > good thing, from a purely mathematic point of view. > There are various ways to show that a re number > exists - the simplest is first to give a rigorous > definition of continuous function, prove the > theorem that if a < b, f is continuous, f(a) < 0 > and f(b) > 0 then f(x) = 0 for some x beeen > a and b, and then apply this theorem with > f(x) = x^2 - 2, a = 0, b = 2. I actuly thought about this (not in this much > detail, but what I mean is that I thought of the > continuity issue, but then I kind of dismissed it, > since I thought the ve notion of continuity > relies on the existence of numbers... So, the > function has a chance at being continuous because > the vues exist -- how could I use that to prove > that a vue exists such that f(x) = 0? What I find curious is that this continuity issue > seems to rely on the fact that beeen any o > distinct re numbethere is at least one other > re number -- but this fact is not a sufficient > condition for the existence of such number for > which f(x) = 0: it is so true for ration > numbers that beeen any o distinct ration > numbethere is at least one other ration > number. (at this point, I'm guessing my doubt becomes > sort of philosophic? Something akin to how > do we prove that 0 exists? or how do we > prove that 1 is greater than 0?, or that sort > of thing?) I 't think so. Of course, you can't prove evething. Some things you just have to assume. We cl such things axioms. Maybe you want to go back to the beginning and define what the natur numbethe integethe ration numbeand then the re numbers are. Until you know what they are, you can't rely prove anything about them. Now, basicly, l these different things are is stuff that satisfies certain axioms. Different people have different axioms, and they are l equivent. (Except the ones that aren't, which are cled non-standard, or intuitionist [I can buy non-standard, but intuitionism leaves me cold]. There may be others that I'm not aware of.) What equivent means is that the axioms of one system are either so axioms in another system, or are theorems in that system. (That means provable.) So, you need to go to your axioms for the res (whatever they are), and prove something cled the least upper bound theorem. Let's say you define a re number to be a pair of sets of rations, (L,R) such that a. Eve ration number is in exactly one of L and R and b. Eve member in L is less than eve member of R. This is cled a Dedekind cut. Now, if you can (in the language of your axioms) say x^2<2, then the set {x ration | x^2<2 or x < 0}will be your L, and {x ration | x^2 > 2 and x > 0} will be your R. Then the set (L,R) will be the square root of 2. Gee, this doesn't look like a re number that I'm familiar with, does it? Well, that's the price of formism. What you construct from simpler parts doesn't look like what you're familiar with. But what does sqrt(2) look like anyway? If you think of res as being infinite decims, then you probably want to use a Cauchy sequence construction (re numbers are equivence classes of Cauchy sequences of rations). But what do equvence classes of Cauchy sequences look like? Or you could just cut to the chase and define re numbers to be infinite decims (assuming you can define what an infinite decim is -- most people use the definition that an infinite decim is rely an infinite sum, which is just a particular type of [you guessed it] Cauchy sequence). Now, since you've got axioms, you might well ask Is there in fact anything that rely satisfies these axioms? Mathematicly, the answer to that is basicly We 't know. It turns out that the best we can hope for is to prove consistency or inconsistency. But even that isn't ways possible -- this is the content of Goedel's Incompleteness Theorem. However, we can prove that our axiomatization of the res is consistent with our axiomatization of the rations (this is cled relative consistency), which is in turn consistent with our axiomatization of the natur (counting) numbers. These seem intuitively to exist (well, we do count, 't we?), so the re concern (if there is any concern at l) would be that we haven't accurately axiomatized them. (Or that logic doesn't rely work at l.) But hey, you're an engineer, so you're comfortable with not being absolutely sure that l your assumptions are strictly true. I hope I haven't snowed you here. I've tried to keep l your options open, so you can use the approach you want (I'm strongly pro-choice [even up to axiomatization]), but so want to give you a vague idea of what the difficulties you might encounter are. If I've only confused you, ask again. Either I or someone else can rely help -- but a lot depends on where you are and what you want. ==== >>I.e., how to prove that there exists a re number x >>such that x^2 = 2? >>I'm getting started with the rigourous side of maths >>(as a hobby -- being an engineer, with sever years >>of experience, you understand that I haven't rely >>needed it), and rely enjoyed reading a proof that >>sqrt(2) can not be a ration number. >>However, the justification they give is that from >>Pythagoras' theorem, looking at a triangle with >>sides 1 and 1, then the hypothenuse is such that >>its square is 2. I 't find that convincing at >>l ... > If we low that re numbers are in one-to-one > correspondence with points on the x-axis, then > we can mark off sqrt(2) on this re number line, > using compass and straightedge. Voila. There it is. That's not a solid convincing argument -- it's like saying it must exist because I can picture it in my mind. That may be a good way to understand something, but definitely not to prove it. You could have thought the same about it with ration numbers. In fact, I can most definitely picture in my mind a fraction whose square is o (not picture as in telling you the numerator and denominator, but picturing as in visuizing that there is one) -- if it weren't because I know, and I can prove, that there can not be a fraction whose square is o. After l, you could say well, any point in there is a fraction of the length of a unit interv; so there, I put a mark in there, and that must be a fraction... And yet, one woudl be wrong. There are plenty of arguments that would otherwise convince me that there is a ration number whose square is 2, such as beeen any o distinct ration numbethere is at least one other ration number; with this, you could reason that whatever addition decims you put, there's ways a fraction that represents that, and thus, no matter how far we go, there will ways be a fraction, and thus, there is a fraction that is equ to sqrt(2)... Voila, it's right there, one might say. And no, intuition fails in that particular case. I know, it feels strange ting to argue that your attempt at convincing me of something that not only I'm convinced, but that I know for a fact is true, is not convincing enough... (when I say I know for a fact, I mean that any mathematician or mathematics book would tell you so, and would tell you that it can be proven). Still, if we play the game of being skeptic and proving evething, then let's be truly skeptic. ==== >I.e., how to prove that there exists a re number x >such that x^2 = 2? >I'm getting started with the rigourous side of maths >(as a hobby -- being an engineer, with sever years >of experience, you understand that I haven't rely >needed it), and rely enjoyed reading a proof that >sqrt(2) can not be a ration number. >>However, the justification they give is that from >Pythagoras' theorem, looking at a triangle with >sides 1 and 1, then the hypothenuse is such that >its square is 2. I 't find that convincing at >l ... Why not? The segment has a length, and the square with sides of that length has ice the area of a square with sides of length 1. What would that length be, if not the square root of 2? There is a good (and inexpensive, in paperback) book by L. Blumenth, A Modern View of Geomet, that has a quite rigorous development of co-ordinate systems and gebraic structures from purely geometric ideas. ==== >I.e., how to prove that there exists a re number x >>such that x^2 = 2? > >>However, the justification they give is that from >>Pythagoras' theorem, looking at a triangle with >>sides 1 and 1, then the hypothenuse is such that >>its square is 2. I 't find that convincing at >>l ... >Why not? The segment has a length, and the square >with sides of that length has ice the area of a >square with sides of length 1. What would that >length be, if not the square root of 2? I think the issue was, is it re? I might know that there exists a number z such that z = Sqrt(-1), but is it a re number? And if you define Sqrt(2) like that, what happens in non-euclidean geometries? Does the vue of Sqrt(2) change? ==== Toni Lassila grava .88 la saucisse et au marteau: > I think the issue was, is it re? I might know that there exists a > number z such that z = Sqrt(-1), but is it a re number? And if you No, there is _not_ any number z such that z = sqrt(-1). There is z such that z^2 = -1, which is _not_ the same thing. Otherwise, I could write: 1 = (-1)(-1), so 1 = sqrt(1) = sqrt[(-1)(-1)] = sqrt(-1)sqrt(-1) = i*i = -1 Eve number, positive or negative, has o square roots. For positive res, we define THE square root by the positive one. On negative res, as there is no order on C, we can't pick one to be THE square root. Therefore, we decide that there are o square roots, i and -i. -- ==== > I think the issue was, is it re? I might know that there exists a >> number z such that z = Sqrt(-1), but is it a re number? And if you No, there is _not_ any number z such that z = sqrt(-1). There is z such >that z^2 = -1, which is _not_ the same thing. No, but it's still a vid definition. http://mathworld.wolfram.com/ImaginaUnit.html ==== >> I think the issue was, is it re? I might know that there exists a >> number z such that z = Sqrt(-1), but is it a re number? And if you >No, there is _not_ any number z such that z = sqrt(-1). There is z such >that z^2 = -1, which is _not_ the same thing. No, but it's still a vid definition. Not so! The square root of -1 does not exist! A square root does, in fact, if any square rootof -1 exists then o of them do, with no way of distinguishing beeen them. What we sometimes label i is picked arbitrarily to be one of them and -i the other. ==== > No, but it's still a vid definition. http://mathworld.wolfram.com/ImaginaUnit.html I've ready seen mistakes on MathWrold and I would not take it as a reference. But if you have more serious references, I may change my mind. -- ==== >Toni Lassila grava .88 la saucisse et au marteau: > No, but it's still a vid definition. >> http://mathworld.wolfram.com/ImaginaUnit.html I've ready seen mistakes on MathWrold and I would not take it as a >reference. But if you have more serious references, I may change my >mind. I'm not equipped with a lot of serious complex anysis books, but at least some physics texts (Brehm-Mullin: Introduction to the Structure of Matter, for one) define i like that. Most mathematic approaches take i as the ordered pair of res (0, 1) and work from there. ==== > If we low that re numbers are in one-to-one > correspondence with points on the x-axis, then > we can mark off sqrt(2) on this re number line, > using compass and straightedge. Voila. There it is. > That's not a solid convincing argument -- it's like > saying it must exist because I can picture it in my > mind. Maybe we disagree on what proof is. I see a proof as being a convincing argument that theoreticly can be made rigorous. > That may be a good way to understand something, but > definitely not to prove it.... ==== > If we low that re numbers are in one-to-one >> correspondence with points on the x-axis, then >> we can mark off sqrt(2) on this re number line, >> using compass and straightedge. Voila. There it is. > That's not a solid convincing argument -- it's like >> saying it must exist because I can picture it in my >> mind. Maybe we disagree on what proof is. >I see a proof as being a convincing argument >that theoreticly can be made rigorous. But as far as I can see making it rigorous involves a lot more work than with the other approaches that have been suggested (the _simplest_ way to make it rigorous seems to be to _include_ one of those other approaches...) >> That may be a good way to understand something, but >> definitely not to prove it.... ==== >> If we low that re numbers are in one-to-one >> correspondence with points on the x-axis, then >> we can mark off sqrt(2) on this re number line, >> using compass and straightedge. Voila. There it is. >> Or do you think this line has holes or gaps? >> That's not a solid convincing argument -- it's like >> saying it must exist because I can picture it in my >> mind. >> That may be a good way to understand something, but >> definitely not to prove it.... >Maybe we disagree on what proof is. >I see a proof as being a convincing argument >that theoreticly can be made rigorous. > But as far as I can see making it rigorous involves > a lot more work than with the other approaches that > have been suggested (the _simplest_ way to make > it rigorous seems to be to _include_ one of those > other approaches...) asked: How to prove that sqrt(2) exists? Right? I assumed he had an idea what re numbers are; i.e., maybe a completion of the rations (that's not p-adic) or maybe a continuous extension, formly constructed by taking limits of Cauchy sequences of equivence classes of rations or by the cuts of Eudoxius and Dedekind, which both underlie the one-to-one correspondence. I that case, one would just have to show that sqrt(2) fls within his understanding of the res. However, he has reveed that he knows little or no anysis, even at the level of baby Rudin, etc. And his question is being answered by posters defining the res. l-in-l, IMHO, his question was not well-posed. ==== Hey l When we say a function f(t) is smooth, does this mean that f has infinite differentis with respect to t? Or any other form definition on this? Fred ==== > Hey l When we say a function f(t) is smooth, does this mean that > f has infinite differentis with respect to t? Or any other form definition on this? The number of continuous derivatives that a function possesses is a measure of its smoothness. Bob Kolker ==== >> Hey l When we say a function f(t) is smooth, does this mean that >> f has infinite differentis with respect to t? Or any other form definition on this? The number of continuous derivatives that a function possesses is a >measure of its smoothness. Well, yes. But more directly responsively to Fred: yes, in many areas of mathematics, When we say a function f(t) is smooth and 't specify anything further, we often mean that f is infinitely differentiable. This is particularly common among differenti topologists. I have a feeling that re anysts, for instance, will tend to be more precise here; but maybe not. ==== > I am ting to cculate the probability that a gambler with capit C > and who uses a Martinge betting strategy will be wiped out in m > turns at the game. This would happen with a run of n consecutive > losses and I want to cculate the probability of this. The textbooks treat this problem at an advanced level, invoking > generating functions or difference equations, but I wonder if a > solution satisfacto for the purpose could be arrived at in a simpler > way. l I need is the probability that there would be AT LEAST one > run of length AT LEAST n. If in n independent tris the probability of a loss at a single tri > is q, the probability of l losses in n tris would be q^n. A string > of n losses could begin at any tri from the first up to the m - n + > 1 th, so would the probability be (m - n + 1) * q^n? As far as I can see, if q is the probabilty of a loss and P(n,m) is the probability of at least 1 run of at least length n in m tris then P(n,m) = 0 if m < n, = q^n.(1 + (1-q).( (m-n) - sum(i=n..m-n-1, P(n,i) ) ). Having read the materi in this thread, I assume I've made a simple mistake or that a similar formula is given in Feller but not considered a practic method of cculation in 1968. Admittedly, you wouldn't want to use this definition for hand cculations, but it's pretty easy to program up. A spreadsheet is fine if the vues for m and n are not too big (as was suggested by the mention of a number of gambling games). Ian Smith P.S. I have tried with sanity checks to make sure the formula isn't ludicrous so I'd appreciate an indication of where the following logic is faulty or an example which shows up the error rather than 't be stupid or see Feller (I haven't got access to a copy). The logic is simple. Either the first n tris are losses or the first tri is a success and then we have n losses or we have no runs of n or more in i tris followed by a success and then n losses (for i = 1.. m-n-1). I think this enumerates l the possibile combinations without duplication. ==== >There have been a number of threads on this topic in sci.math, e.g. >Consecutive runs in Bernoulli tris in June 1997. |> In Bernoulli tris, an event A occurs with probability p, and the |> probability |> of A occurring exactly M times in N tris is: |> |> (N!/(M!*(N-M)!)) * p^M * q^(N-M), |> |> where q = 1-p is the probability of A not occurring. |> What I would like to know is: what is the probability of A occurring at |> least |> M *consecutive* times in N tris (with p being the probability of A |> occurring |> at any given tri)? I am assuming that the tris are independent. Let f(n) be the probability that the first run of M consecutive successes occurs at tri n (i.e. that tris n-M+1, n-M+2, ..., n are successes, and there is no previous run of M consecutive successes). The generating function of this is F(s) = sum_{n=r}^infinity f(n) s^n = p^M s^M (1-ps)/(1-s+q p^M s^(M+1)) = p^M s^M/(1 - qs - qps^2 - ... - q p^(M-1) s^M) This can be expanded in parti fractions: F(s) = -p/q + sum_r a_r/(s-r) where the sum runs over the roots of the denominator (this must be modified in the case of multiple roots). The probability of no run of M consecutive successes in N tris is sum_{n=N+1}^infinity f(n) = sum_r -a_r(1/r^(N+2) + 1/r^(N+3) + ...) = sum_r -a_r/((r^2-r) r^N). When M is not too sml, the smlest root in absolute vue, which is the unique positive root, gives the main contribution, and the resulting approximation is quite good. See Feller, An Introduction to Probability Theo and Its Applications, Vol. 1, sec. XIII.7. Comment: This is one of sever standard approaches and I am fishing for an origin one or one that culminates in a set of formulas that can be put into a reference book and be useful to non-mathematicians. Supposedly this problem was first propounded and solved by Abraham DeMoivre and could well be the deepest problem in l of Probability. ==== Since making the first post I made a post to t.sci.math.probability and received a request for references, which I am including here. >That will be an upper bound on the probability you are wishing to compute. Can you please tell me which textbook did you refer to and how does it >propose to solve the problem. Supposedly this problem was first studied by Abraham DeMoivre, and a collection of his works might be a good place to start for anyone who is interested. References may be found at http://mathworld.wolfram.com/Run.html I so have looked at Burnside, William The Theo of Probability (1928) Dover Publications, 1959 Chapter III An approximate solution is arrived at by means of a difference equation. Uspensky, J. V. Introduction to Mathematic Probability Chapter V, Section 3 A difference equation and a generating function are discussed and an approximation using a Lagrange series is derived. Burnside gives a lower bound and his derivation is much simpler than Uspensky's. A collection of probabilities of runs is a distribution of runs and sometimes these are used in tests for randomness.