mm-118 is primitive in GF(4),> then how tofind a^4+a^2, why it is equal to 1 ?> a^4+a^3, why it is equal to a^2 ? Anybody can tell me what is the trick?> -Walalaalso, in GF(4), or any GF(q),is all the -a equal to +a, thats to say, there is no minus sign inGF(q), and once we see - in computation, we just the following question: suppose a is primitive in GF(4),>then how tofind> a^4+a^2, why it is equal to 1 ?>a^4+a^3, why it is equal 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 also has to be a^2. So a^2 = a + 1. Then a^3 = a(a + 1) = a^2 + a = (a + 1) + a = 1. Etc. > also, in GF(4), or any GF(q),is all the -a equal to +a, thats 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, its spelled Galois.-- following question: suppose a is primitive in GF(4),>> then how tofind>a^4+a^2, why it is equal to 1 ?>> a^4+a^3, why it is equal 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 also has to be a^2.----------------------------------->>why The fourth has to be a + 1, but it also has to be a^2?> So a^2 = a + 1.> Then a^3 = a(a + 1) = a^2 + a = (a + 1) + a = 1.> Etc. > also, in GF(4), or any GF(q),> is all the -a equal to +a, thats to say, there is no minus signin>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 very much!-Walala 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 nite elds cardinality has to be a power of a prime.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =[cut]>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 also has to be a^2.-----------------------------------> why The fourth has to be a + 1, but it also 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 equal 0 or 1 etc.Consider a + 1. It is certainly in GF(4).Could it be equal to one of the accounted three elements so far?No, for a + 1 = 0 says a = 1; a + 1 = 1 says a = 0; anda + 1 = a says 1 = 0. Thus, a + 1 must be the fourth unaccounted element.As an aside, you can approach your original 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 alsothe way that a primitive element is used for other GF(q), q not 4.You want tofind 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 = 1The second and third are easy to rule outsince they imply the false statements a^2 = 0 and a = 0.The rst is also easy to rule outsince you get 1 + a = 0 by dividing both sides by a.But, 1 + a = 0 implies the false 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 expressionyou want to evaluate to the four possible choicesand eliminate the three that lead to false statementsby simple algebraic manipulations.-- Bill Hale =one can make a dystinction between a)omniprescience, b)broad predictions, c)valuable forecasts, d)promoting **** that youre 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 hadnt happened yet counter the assertion> that the future cant be known in advance. QED.Whining or ridiculing about it is irrelevant. The original> assertion (by Popper) is proven wrong. Period.--les Dicks dEnron! Ive called the factoring of polynomial into non-polynomial factors> advanced for a reason....> To move beyond factoring polynomials into polynomial factors, you need> to have a *thorough* understanding of coprimeness, ring operations,> and logical argument.LOL.www.crank.net/harris.html =Does anyone know how to convert .JMP data to comma delimited format (orbetter yet, MINITAB format)? I really, really need to convert this databecause I dont own JMP--only MINITAB.. any points to conversion utilities(or if someone would be so kind as to convert it), I would be very grateful.Alex Chavezalex@divide0.netData (2 les):http://www-stat.wharton.upenn.edu/~lzhao/stat431/ DataFiles/Call%20Center%20Arrivals.JMPhttp:// www-stat.wharton.upenn.edu/~lzhao/stat431/DataFiles/ Philadelphia%20(Suburban).JMP =I found a program that does .JMP data to comma delimited format (or> better yet, MINITAB format)? I really, really need to convert this data> because I dont own JMP--only MINITAB.. any points to conversion utilities> (or if someone would be so kind as to convert it), I would be verygrateful. Alex Chavez> alex@divide0.net Data (2 les):http://www-stat.wharton.upenn.edu/~lzhao/stat431/ DataFiles/Call%20Center%20Arrivals.JMPhttp:// www-stat.wharton.upenn.edu/~lzhao/stat431/DataFiles/ Philadelphia%20(Suburban).JMP =what a beautiful evocation of rationality, that old report! a lot more than I can say for Jacks hearsayabout Roswell, which had a couple of interesting associationsaround WW2, which Art Bell (e.g.) never mentions (althoughhe must know about one of them). what Arts promoted is the thesisof the late Lt.Col. Corso, thatthis little pile of **** was parcelled-outto the big companies, and we reverse-engineered ber-optics,the transistor, RADAR ... you name it; I forgot. (fortunately,Corsos co-author exposed him, although I dont knowif its in the book; he did in the promotional tour !-) >The Religious Experience of Philip K. Dick by R. Crumb> http://www.philipkdick.com/weirdo.htm >Yes! There is no religion, no Way of God, no Way of Divine>Realization, no Way of Enlightenment, and no Way of>.... Therefore, Reality (Itself) Is Truth,>and Reality (Itself) Is the Only Truth.> -- Adi Da Samraj, a.k.a. Bubba Free John,> a.k.a. Da Free John, a.k.a. Da Kalki,> a.k.a. the Ruchira Avatar, Adi Da Love-Ananda Samraj,> a.k.a. Franklin 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. Condon, Scientic Director [1968]> C O N T E N T S> http://ncas.sawco.com/condon/text/contents.htmSection IConclusions and RecommendationsEdward U. Condon> http://ncas.sawco.com/condon/text/sec-i.htm > This formulation carries with it the corollary that we do not > think that at this time the federal government ought to set up > a major new agency, as some have suggested, for the scientic > study of UFOs. This conclusion may not be true for all time. > If, by the progress of research based on new ideas in this eld, > it then appears worthwhile to create such an agency, the decision> to do so may be taken at that time. Wefind that there are important areas of atmospheric optics, > including radio wave propagation, and of atmospheric 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 also of fundamental scientic interest, > and they are relevant to practical problems related to the> improvement of safety of military and civilian ying. > As the reader of this report will readily judge, we have focussed > attention almost entirely on the physical 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 reality of UFOs as > craft from remote galactic or intergalactic civilizations. We believe > that the rigorous study of the beliefs--unsupported by valid > evidence--held by individuals and even by some groups might prove of > scientic value to the social and behavioral sciences. There is no > implication here that individual or group psychopathology is a > principal area of study. Reports of UFOs offer interesting challenges > to the student of cognitive processes as they are affected by > individual and social variables. By this connection, we conclude that > a content-analysis of press and television coverage of UFO reports > might yield data of value both to the social scientist and the > communications specialist. 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 physical sciences. On the contrary, we > conclude that the same specicity in proposed research in these areas > is as desirable as it is in the physical sciences. > It has been contended that the subject has been shrouded in > ofcial secrecy. We conclude otherwise. We have no evidence > of secrecy concerning UFO reports. What has been miscalled > 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 small number of individuals who have given sensationalized > 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 whofind 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 critical analysis of arguments for fantastic > propositions that are being supported by appeals to fallacious > reasoning or false data. We hope that the results of our study will prove useful to > scientists and those responsible for the formation of public > policy generally in dealing with this problem which has now > been with us for 21 years. The Condon Report - 1968 CE> http://ncas.sawco.com/condon/index.html> http://www.csicop.org/klassles/posner_klass.html >Parallel Universes [...] http://www.sciam.com/issue.cfm>Not just a staple of science ction, other universes>are a direct implication of cosmological observations>By Max Tegmark--les ducs dEnron! grava ? la saucisse et au marteau:>By the way, what Le Roux calls an application is usually called a >Oops, sorry. Thats because there is a difference between fonction and>application in French, and I thought there was the same difference in>English.>>Thats interesting. Whats the difference? I have not seen >>application used this way in English, but I am not a professional >>mathematician.>>Could it be the same distinction as between function and map?Marc>translate to the English map. However, I always thought that map and function were synonymous.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu >[...]>>Oops, sorry. Thats because there is a difference between fonction and>>application in French, and I thought there was the same difference in>>English.>>Thats interesting. Whats the difference? I have not seen >application used this way in English, but I am not a professional >mathematician.>Could it be the same distinction as between function and map?>>Marc>> translate to the English map. However, I always thought that map > and function were synonymous.It could be just a hard dying habit (cf. Funktion and Abbildung),from Analysis, where maps with the reals as codomain are more equal than the restMarc 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 rst 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)>>Ben Scott>And thank *you* for using my preferred plural of formula!-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu =Many of you may be confused by the use by James Harrisof an overly complicated cubic polynomial and unneededvariables. Here is a somewhat simpler version.Lemma 1: Let: P(m) be a polynomial with coefcients inA, the algebraic integers. Let each coefcient bedivisible (in A) by the algebraic 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 Asuch that s*t=f; and for all m in A,s divides g_1(m) (in A) and t divides g_2(m) (in A).Core Error Proof: Consider the functionP(m,f,x) = mfx^2 +2fx + f^2 (m,f,x in A)[Among other possibilities we can consider this as a polynomial in m, with parameters f and x, or a polynomial in x with parameters m and f. The latter only matters if we want to motivate the following]Consider the quadratict^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 algebra givesP(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 + fand g_2(m) = a_2(m,f) x + fP(m) = g_1(m) * g_2(m)Noting that P(m) can be considered a polynomial in mwith coefcients in A, all coefcients divisible by fwe apply lemma 1 and obtain:There exists s and t, such that s*t=f; and for all m, s divides g_1(m) = a_1(m,f) x + ft divides g_2(m) = a_2(m,f) x + fLet 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 specialize by letting m=1, f =33 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 mathematical universe collapsesOr does it, The problem is with Lemma 1.While this holds for integer polynomials[? does this hold for algebraic integer polynomials]it is not true in general. Counterexamples areeasy to construct if g_1 and g_2 are not continuous.We get a more interesting example by lettingx=1 and f = 3 above.Let P(m) = 3m + 15, this is divisible by 3.By simple algebraP(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) = 6P( -5) = 0, g_1( -5) = 0. g_2( -5) = 8P( -8) = -9, g_1( -8) = -1. g_2( -8) = 9P(-16) = -33, g_1(-16) = -3. g_2(-16) = 11We note that while P(m) is always divisible by3, the factor of 3 switches back and forthbetween g_1 and g_2. In general all we have isLemma 1: Let: P(m) be a polynomial with coefcients inA, the algebraic integers. Let each coefcient bedivisible (in A) by the algebraic 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 all m in A there exist s(m) and t(m) in Asuch that s(m)*t(m)=f;,s(m) divides g_1(m) and t(m) divides g_2(m).So all we can conclude above is that s(0)=fand 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. William Hughes Let me start by saying that I didnt mean to offend or annoy you in> case youre offended or annoyed. Next, let me just say, for the> record, that as a probabilist, in particular, one who works with> stochastic ordinary and partial differential equations, Im very> familiar with white noise and the techniques for working with it in a> mathematically rigorous fashion. I am, however, a student of pure> mathematics, so anything I say will be from that perspective. Ill try> to answer your questions below.I guess I did y off the handle a bit fast there. I will certainlykeep your perspective in mind. Indeed it is what I expected of thisnewsgroup. And I am thankful that a person of your qualications istrying to answer my questions - I appreciate that. I apologize forjumping to the wrong conclusions about your motives.>>[..]>from a mathematical perspective, your question doesnt 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>stationary 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 really have to be that anal? Actually, youre not being anal>>enough - if the covariance was not nite, then I wouldve stated that>>it does not exist.> Its true that, here, I am simply repeating your conditions. Im> trying 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> real parameter, t, the term IID is less common. The reason is> twofold: (1) postulating independence can lead to difculties> regarding the existence of the process, as I mention below; and (2)> processes of a real parameter which are identically distributed are> more commonly referred to as stationary.Point taken. I am still getting comfortable with the language andconventions.> The < infty is there to emphasize that this is the important part of> the assumption. If the variance is allowed to be innite, then such a> process will have a measurable version, though it wont be continuous.>Well, there is no reasonable process that satises 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 rst page of Chapter III of Stochastic Differential> Equations by Bernt Oksendal and the references therein.> (Incidentally, 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 mathematical denitions is no small>task. The white noise mathematicians work with is a completely>different kind of object than youre probably used to. (In fact, its>not even a process in the traditional sense.)>>Please, enlighten me, or point me to the subjects and/or books>>that will enlighten me.> An ordinary process is a function X(t,omega) of two 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 wronglong ago and Id appreciate it if you could conrm it: Are the omegasthe elements of the sample space, so that for any specic t, say, t0,X(t0, omega) is the mapping from S (the sample space) to R (the reals)that we normally 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 called a generalized random> variable or a generalized Wiener functional. See Stochastic Partial> Differential Equations: A Modeling White Noise Approach by Oksendal,> et al, and the references therein for an introduction to this> approach. More commonly, though, white noise is regarded as a random> variable taking values 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 distributional derivative of Brownian motion.Do you mean distributions in the sense that the Dirac delta functionis a distribution? So X(t, omega) for each omega gives a distribution(kinda like a pathological function) instead of a normal function?> In the former approach, white noise does not have pointwise values for> each omega, in the latter it does not have pointwise values for each> t.Hmmm.>I know this doesnt answer your question, but I hope it at least>partially explains why you havent received more responses. In some>sense, youre 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 youre looking for.>>No, thats not what Im looking for. And I HAVE posted it to an engineering>>group (comp.dsp) SEVERAL times now over the last two or three years. It>>is not resolvable by the folks there.>If you want a mathematical answer, then the question must be posed in>a more formal way.>>Hey Jason, bend a little - give me a clue as to what is lacking in formality>>here.> Okay, Ill give it a try. 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 distributional derivative of Brownian motion.> The connection between 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 thats exactly what I want. Heuristicallyspeaking, it seems like it should be reasonable to constructa process which is stationary and independent at any time differencetau (i.e., X(t) and X(t+tau) are independent random variables when tauis not zero), so that the autocorrelation function is zero when tau is notzero. Yet each random variable in this processs family of randomvariables could have the same distribution and that distribution couldhave a nite variance - for example, a simple N(0,1) distribution - sothat the autocorrelation function is the nite value sigma^2 at tau = 0and zero otherwise. Yes, thats precisely the animal I want to construct.> How should I> interpret this in a rigorous sense? Is it the distributional> derivative of some ordinary 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 engineeringtexts Ive seen say autocorrelation function (the probabilisticautocorrelation, i.e., Rxx(tau) = E[X(t)X(t+tau)], NOT the onewhich does time-domain correlation, i.e., the convolution ofthe time-domain function with a time-reversed version of itself).> But two> functions which differ only a set of measure zero have the same> Fourier transform. This process has a covariance function which is 0> almost everywhere. Therefore, its Fourier transform is the zero> function. Right. Thats the problem. > Notice that we dont have this problem when taking the> Fourier transform of the delta function, even though the delta> function is zero everywhere except the origin. This is because the> delta function is not a function at all, but a distribution, so we> must take the Fourier transform in the sense of distributions.So, in summary, the process youve introduced does not exist as an> ordinary process. Perhaps it exists as a random variable taking values> in the space of Schwartz distributions. If so, its not clear exactly> what that random variable should be, so you need to explicitly dene> it in the language of distribution theory. Once done, 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> theory (if youre not already familiar with it) is Introduction to> the Theory of Distributions by F. G. Friedlander.Excellent - Ive been looking for a reference to enjoy discussing this stuff because I> enjoy stochastic analysis. But I dont want to annoy or offend you and> I dont want to engage in any kind of emotionally charged discussion> over this stuff. Good luck in your investigations. Please let me know> if you come upon the formal way to describe your process. Id be very> interested in seeing it.help.The key seems to be in understanding why a continuous-time independentprocess isnt a normal process. I have the Oksendal book on order.-- % Randy Yates % ...the answer lies within your soul%% Fuquay-Varina, NC % cause no one knows which side%%% 919-577-9882 % the coin will fall.%%%% % Big Wheels, *Out of the Blue*, ELOhttp://home.earthlink.net/~yatescr An ordinary process is a function X(t,omega) of two 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 Id appreciate it if you could conrm it: Are the omegas> the elements of the sample space, so that for any specic t, say, t0,> X(t0, omega) is the mapping from S (the sample space) to R (the reals)> that we normally speak of when considering a random variable?Yes, thats right.>It is possible to regard white noise as a process in the>sense that for each t we have an object called a generalized random>variable or a generalized Wiener functional. See Stochastic Partial>Differential Equations: A Modeling White Noise Approach by Oksendal,>et al, and the references therein for an introduction to this>approach. More commonly, though, white noise is regarded as a random>variable taking values 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 distributional 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 pathological function) instead of a normal function?Right again.>In the former approach, white noise does not have pointwise values for>each omega, in the latter it does not have pointwise values for each>t.Hmmm.> ...>Okay, Ill give it a try. 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 distributional derivative of Brownian motion.>The connection between 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 thats exactly what I want. Heuristically> speaking, it seems like it should be reasonable to construct> a process which is stationary 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 processs family of random> variables could have the same distribution and that distribution could> have a nite variance - for example, a simple N(0,1) distribution - so> that the autocorrelation function is the nite value sigma^2 at tau = 0> and zero otherwise. Yes, thats precisely the animal I want to construct.Is this animal that you want to construct something that others on theapplied side are working with? I mean, is it well known in theoutside world, or is it just something you thought of at some pointand were curious about?Ive thought some more about this. Lets let sigma=1. For small r, theprocess X_r(t)=[B(t+r)-B(t)]/sqrt{r} is pretty close to what yourelooking for. The question is, do these processes, as r-->0, coverge inany reasonable sense? I dont see it, off hand. For xed omega, theyconverge to zero as Schwartz distributions. For xed t, they convergeto zero as generalized random variables. In fact, if we writeEX_r(t+tau)X_r(t)=f_r(tau), then f_r converges to 0 almost everywhereand in L^1. Of course, for xed t, the X_r(t)s coverge in law (infact theyre all identical in law) to a normal(0,1) random variable. Idont think this helps though, because you want some kind ofconvergence that deals with all t simultaneously. Theres a lot onteresting behavior in these processes X_r relating to the law of theiterated logarithm and its variants. Chapter 2, section 2.9,subsection E of Brownian Motion and Stochastic Calculus by Karatzasand Shreve contains the law of the iterated logarithm. The paragraphafter the proof of it discusses this process X_r. It containsreferences to two previous remarks and to comments in the notes at theend of the chapter. Part of remark 9.15 is worth posting here:It is quite possible that for each xed t>=0, a certain propertyholds almost surely, but then it fails to hold for all t>=0simultaneously on an event whose probability is one (or evenpositive!). As an extreme and rather trivial example, consider thatP[B(t)!=1]=1 holds for every 0<=tHow should I>interpret this in a rigorous sense? Is it the distributional>derivative of some ordinary 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 Ive 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 twovariables given by p(s,t)=EX(s)X(t). If X(t) is stationary, then pdepends only on the difference t-s, so that we may writeEX(s)X(t)=f(t-s) for some function f. This f is the autocorrelationfunction. Ive been being sloppy. Although Ive been saying covariancefunction, Ive meant autocorrelation function.>But two>functions which differ only a set of measure zero have the same>Fourier transform. This process has a covariance function which is 0>almost everywhere. Therefore, its Fourier transform is the zero>function. Right. Thats the problem.Notice that we dont have this problem when taking the>Fourier transform of the delta function, even though the delta>function is zero everywhere except the origin. This is because the>delta function is not a function at all, but a distribution, so we>must take the Fourier transform in the sense of distributions.>>So, in summary, the process youve introduced does not exist as an>ordinary process. Perhaps it exists as a random variable taking values>in the space of Schwartz distributions. If so, its not clear exactly>what that random variable should be, so you need to explicitly dene>it in the language of distribution theory. Once done, 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>theory (if youre not already familiar with it) is Introduction to>the Theory of Distributions by F. G. Friedlander.Excellent - Ive been >Anyway, I hope this helps. I enjoy discussing this stuff because I>enjoy stochastic analysis. But I dont want to annoy or offend you and>I dont want to engage in any kind of emotionally charged discussion>over this stuff. Good luck in your investigations. Please let me know>if you come upon the formal way to describe your process. Id be very>interested in seeing it.help.The key seems to be in understanding why a continuous-time independent> process isnt a normal process. I have the Oksendal book on order. The key seems to be in understanding why a continuous-time independent> process isnt a normal process. The problem with a continuous-time iid process X(t) is not that it doesnt exist---Kolmogorovs theorem guarantees existence---but that as a function X(t,w) of time and sample point it behaves rather pathologically. For instance, it cannot be a measurable function of (t,w), so various things one would like to do are problematic. For example, an integral like int_u^v X(s) ds need not exist, let alone be a random variable.The process you may be looking for can be described as follows. Let H be the class of square-integrable real-valued functions on the real line R. The white noise W = {W(h) : h in H} is a collection of random variables dened on some probability space with joint normal distributions; each W(h) is mean zero, and for g,h in HE[W(g)W(h)] = int_R g(t)h(t) dt.In particular, if g and h are orthogonal, then W(g) and W(h) are independent. Notice that W is stationary in the sense that the joint distributions of the W(h) are unchanged if, for xed s in R, each h is replaced by the translate h(.+s).-- A. about picking up a higher level book so I can read some more about it. It> really intrigues me that this one is not elementary, because it seems it> should be. Integral of e^x is elementary, e^-(x^2) is elementary, e^x^2 is> not. I need to read up some more about this. I didfind an approximation>> The indenite integral of e^(-x^2) is no more elementary than that of> e^x^2.> I was working this integral out last night (a denite integral) and power> series didnt seem to work. What am I overlooking?Its obvious we are not talking about the same integral. What makes youthink you are overlooking anything?-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. = >>thinking>about picking up a higher level book so I can read some more about it.It>really intrigues me that this one is not elementary, because it seems it>should be. Integral of e^x is elementary, e^-(x^2) is elementary, e^x^2is>not. I need to read up some more about this. I didfind anapproximation> The indenite integral of e^(-x^2) is no more elementary than that of>e^x^2. > I was working this integral out last night (a denite integral) andpower>series didnt seem to work. What am I overlooking? Its obvious we are not talking about the same integral. What makes you> think you are overlooking anything? -- > Dave Seaman> Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.> I was evaluating the integral of e^(x^2) from 2 to 3 and got somethingaround 12, which I know cannot be right (and I veried it on my TI-83).David Moran >thinking>>about picking up a higher level book so I can read some more about it.> It>>really intrigues me that this one is not elementary, because it seems it>>should be. Integral of e^x is elementary, e^-(x^2) is elementary, e^x^2> is>>not. I need to read up some more about this. I didfind an> approximation> The indenite integral of e^(-x^2) is no more elementary than that of>>e^x^2.>>I was working this integral out last night (a denite integral) and> power>>series didnt seem to work. What am I overlooking?>> Its obvious we are not talking about the same integral. What makes you>> think you are overlooking anything?> I was evaluating the integral of e^(x^2) from 2 to 3 and got something> around 12, which I know cannot be right (and I veried it on my TI-83).I dont see what this has to do with what I said about the indeniteintegrals, but since you ask, it depends on what series you wereintegrating and how many terms you used. For example, if we askMathematica to integrate the MacLaurin series, we can get something likeIn[1]:= Integrate[Normal[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 xOut[1]= x + -- + -- + -- + --- + ---- + ---- + ----- + ------ + ------- + 3 10 42 216 1320 9360 75600 685440 6894720 21 x> -------- 76204800where, if you substitute x = 3, youfind that the x^21 term stillevaluates to around 137.3 and therefore lots more terms are needed forconvergence. I found that going to x^100 seems to work. But youprobably need more signicant digits than you can get on a calculator toadd up all those numbers without losing accuracy.On the other hand, if we use a Taylor series centered at the midpointof the interval of integration, we getIn[2]:= Integrate[Normal[Series[Exp[x^2],{x,5/2,10}]],x] 25/4 3 5Out[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 4and this converges much more quickly, since the powers of (-5/2 + x)go quickly to zero when evaluated at x=2 or x=3.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. =theres abundant evidence to show tahtNewton 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 American> calculus texts on my shelf, and found Newton & Leibniz with equal> billing for this.--les ducs dEnron!http://www.wlym.com/antidummies/part17.html =thats a great little book! > Getting back to Newtons mathematics, Huygens & Barrow, Newton & Hooke> by V.I. Arnold (Birkhauser 1990) is denitely worth reading.--les ducs dEnron!http://www.wlym.com/antidummies/part17.html =r.e.s. ~ you (and many others out there) are all above my head. However,it is nice to know you are out there making sure the information given iscorrect. I am really in awe. Keep us all on track!!Kavon> I second that. r.e.s. > Kavon> and is perfectly consistent (and no, Im NOT talking about> Non-Standard>> analysis here).>>If d^2 = 0, and the analysis is *not* non-standard,>... which has nothing to do with anything that was said that youre>> replying to...http://search.yahoo.com/search?p=How To Search The Web>> http://search.yahoo.com/search?p=Non-Standard Analysis>> http://search.yahoo.com/search?p=Wikipedia>then how can d differ from 0?>*Sigh.*http://search.yahoo.com/search?p=Associative Linear Algebras>> http://search.yahoo.com/search?p=Vector Spaces>> http://search.yahoo.com/search?p=Nilpotent Algebras>> http://search.yahoo.com/search?p=Finitely Presented Algebras>> http://search.yahoo.com/search?p=Ideals+Innitesimals>> http://search.yahoo.com/search?p=Hypercomplex Numbers+Innitesimals> =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. Delivery-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 13Request:25 10 3 6 7 tonsHow do I solve this using Linear Programming (with objective function > and constraints) so that total haulagecosts are minimal? Im required to > use Maple, tho in that newsgroup they say it can hardly be done that way.> I have difculties with this since its an un-equal matrix.Theres no requirement in Linear Programming that the number of rowsand columns in the constraints be equal. By the way, I learned aboutLPs before interior point methods were shown to be good ways to solvethem. My program (see sig) for solving them is dated, in that sense.Im not sure if I fully understand what your LP should be. But to ndit, heres how I would proceed:1. Dene the decisions variables. (Some of these will be zero in thenal solution.) Must your decision variables be non-negative?2. Express total haulage costs in terms of the decision variables. Ithink the matrix of the haulage-costs will be used here. Thisis the objective function you want to minimize.3. Express constraints. It seems to me that satisfying the request of acustomer will yield a constraint. I think the capacities of thedelivery cars will provide coefcients for the constraint. The sizeof the requested supply will provide the right-hand-side, I think. Andyou should not get an equality for the constraint.What algorithm have you been taught to solve LPs?If this is not sufcient help, say so.-- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.htmlr c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question t perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau =Georg Goerg> How do you integrate the function f(x) = x/(tanx) ?> please the denite integral.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 2and thereforeS = 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. Its quite cute. Try it.Larry Georg Goerg>> How do you integrate the function f(x) = x/(tanx) ?>> please send denite integral.> 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. Its quite cute. Try it.> LarryTHANK YOU very much,georg-- Georg Goerg If a sequence of reals {s_n} converges, then the corresponding Limsup>equals its Liminf. These 2, however, appear to be independent of the>particular ordering of {s_n}. >>This would be true, by looking at their denitions and considering that >>for any epsilon, the s_i that cause a problem can be excluded nitely >>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 arbitrary 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 Im confused here. Is there some simple proof that, no> matter what the relationship, the limit of the there are nitely many natural numbers m <=N, for the rearrangement (and possible subset), there must be a K for which all the s_ns <=N are in the s_ks <=K. You compute the new boundary K from the old boundary N. The new boundary may be higher, but it does exist.-- Will Twentyman I wouldnt put in a lot of technical things, nor would I talk about> other news readers or means of posting besides Google Groups because I> use it all the time and I think it has transformed Usenet.Really... how?Besides the archiving part, I mean. > I wouldnt put in a lot of technical things, nor would I talk about>> other news readers or means of posting besides Google Groups because I>> use it all the time and I think it has transformed Usenet.> Really... how?> Besides the archiving part, I mean.Well, its made it easier to get access to USENET for people who eitherdont want, or are too clueless, to learn to operate a proper newsreader.James himself is a prime example of the latter class.Sometime I miss the days when getting to USENET required conguringUUCP, compiling, installing and conguring B News or C News,findingand negotiating with an admin somewhere for a news feed... or, if youwere a student, getting an account from your university news admin.James would be excluded from both methods: too technically illiterateto do it himself, and too afraid of education to have a student account.-- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvisefwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock I wouldnt put in a lot of technical things, nor would I talk about>other news readers or means of posting besides Google Groups because I>use it all the time and I think it has transformed Usenet.Really... how?Besides the archiving part, I mean.Google Groups makes Usenet easily accessible over the Internet. Somenewsgroups, like sci.math, are mirrored to the Internet, so that youhave what I now call Usenet-Internet.That vastly increases the potential audience.Usenet is the one sure-re way to say something, and probably get aresponse--if you know what youre doing--as youre read around theworld.Sure, you can put up a webpage, and watch as no one comes. And ittakes more effort to start and *maintain* a webpage, as remember, itsnot just about throwing up a webpage, as you have to keep it up aswell. And that applies to blogs as well.Now posting, youre competing with a lot of other posters, but youhave the possibility of standing out based on your content and style,so theres a greater potential I think to step out and have an impact.My feeling is that Usenet-Internet will explode into an ever moreimportant medium of exchange as people see its important features.If you want to see the power of knowing what youre doing, do a Googlesearch on core error and see what person has number 1 and 2.Or you can do a search on my name and math, yes, its a lot of hatestuff, but my point remains, as notice how many *different* LANGUAGESyoull bump into if you dig, where for some reason, my work is forwardenough that Google highlights it, so you end up directed to it.James Harrishttp://lostincomment.blogspot.com/ If you want to see the power of knowing what youre 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 hasno real mathematical meaning at all is proof of knowing what youredoing?> Or you can do a search on my name and math, yes, its a lot of hate> stuff, but my point remains, as notice how many *different* LANGUAGES> youll 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, Ive learned all about this Usenet-Internet thing today.-- Jesse F. HughesWhat you call reasonable is suspect since youve proven yourself tobe an enemy of mathematics. -- James S. Harris defends the cause. I wouldnt put in a lot of technical things, nor would I talk about>> other news readers or means of posting besides Google Groups because I>> use it all the time and I think it has transformed Usenet.>>Really... 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 call Usenet-Internet.That vastly increases the potential audience.*snip*Youre basically singing about how good usenet is. We all 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 searches always turned up a mush ofusenet-internet then wed have the kind of noise problem that ourweblogs are creating now. I dont want this post to turn up whensomeone types ratio on usenet is terrible, and Im expectingweblogs to go that way--useful and intellectual for a while, buteventually degenerating into ames, ames about ames, histories ofames being amed, etc.My point is that just because something turns doesnt mean anyone has totake those results seriously. Plus that mirroring would happen withoutGoogle Groups too. Google groups have not transformed Usenet. =Crossposts to writing newsgroups left in because, for once,theyre relevant.> I wouldnt put in a lot of technical things, nor would I talk about>> other news readers or means of posting besides Google Groups because I>> use it all the time and I think it has transformed Usenet.>>Really... 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 call Usenet-Internet.Quibble: Usenet was available over the internet before therewas widespread web access. I think you mean Google Groupsmakes Usenet accessible over the Web. The Internet isboth much older and much larger than the Web.archive. Google extended the capabilities of the searchengine and made some improvements in the interface.> That vastly increases the potential audience.Usenet is the one sure-re way to say something, and probably get a> response--if you know what youre doing--as youre read around the> world.provide the same heard around the world capability. - Randy >>I wouldnt put in a lot of technical things, nor would I talk about>>other news readers or means of posting besides Google Groups because I>>use it all the time and I think it has transformed Usenet.> Really... 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 easilyaccessible over the internet? (I think you mean easily accessibleover the web...)>Some>newsgroups, like sci.math, are mirrored to the Internet, so that you>have what I now call Usenet-Internet.Um, usenet is _part_ of the internet - always has been. Again,you mean the web, not the internet.You should denitely write that guide to posting. Just what weneed, authoritative information from someone who thinks thatthe internet means the web.>That vastly increases the potential audience.Usenet is the one sure-re way to say something, and probably get a>response--if you know what youre doing--as youre 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, its>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, youre competing with a lot of other posters, but you>have the possibility of standing out based on your content and style,>so theres a greater potential 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 youre 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? All it shows is thatanyone can post any nonsense he wants. Youve derived somesort of power from being able to make a fool of yourself in frontof a global audience? Good for you.>Or you can do a search on my name and math, yes, its a lot of hate>stuff, but my point remains, as notice how many *different* LANGUAGES>youll 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.>James Harris>http://lostincomment.blogspot.com/******************** ****David C. Ullrich =... >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, James is confused. >Some >newsgroups, like sci.math, are mirrored to the Internet, so that you >have what I now call Usenet-Internet. > Um, usenet is _part_ of the internet - always has been.No, usenet has various ways of transmission. Internet is one of them,but it can be done (and has been done) with UUCP through dial-up lines.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: 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, James is confused. Some>>newsgroups, like sci.math, are mirrored to the Internet, so that you>>have what I now call Usenet-Internet.>>Um, usenet is _part_ of the internet - always has been.No, usenet has various ways of transmission. Oh. Id always assumed that internet was the same asdescribed in RFCs. I guess not.>Internet is one of them,>but it can be done (and has been done) with UUCP through dial-up lines.************************David C. Ullrich = >... >>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, James is confused. >>Some >>newsgroups, like sci.math, are mirrored to the Internet, so that you >>have what I now call Usenet-Internet. > Um, usenet is _part_ of the internet - always has been. >No, usenet has various ways of transmission. > Oh. Id always assumed that internet was the same as > described in RFCs. I guess not.transport mechanisms than the Internet. Dial-up lines withUUCP connection, FIDO, BITNET, EARN. The RFCs for Usenetwhen transmitted through the Internet.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Given my *years* of posting on several newsgroups, and my current> funding situation (as in, few funds) Im considering writing up a> guide to posting on newsgroups. The way I gure it, in keeping with> the open source mentality, Id give it away, with people having the> option to contribute, if they found it useful.> Do a search rst, it appears that this has already been done.> I wouldnt put in a lot of technical things, nor would I talk about> other news readers or means of posting besides Google Groups because I> use it all 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.-- Will Twentyman =Springer books,> i have a mini-guide on reparing books at geocities.com/math_library/links.html-- Karl M. Bunday Christ has set us free. Galatians 5:1Learn in Freedom (TM) http://learninfreedom.org/ would have to go through it all again. but you have> to remember that we are determining the potential in the plane of z=0 Lets> do it here....we haveLaplacian(V) = -n(r)*delta(z)*U(a-r)U=heaviside step function, n=charge density, delta=dirac delta. the> Laplacian is taken in cylindrical coords where we assume the potential is> axialy symmetric about the z-axis. (dropping all dependance on theta).Take the Hankel transform of both sides (or in reality, multiply by> J(kr):=zero order bessel function of rst 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 rst term by parts twice - 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)rst solve-k^2*U + d^2U/dz^2 = 0and 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 potential 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 rst term vanishes, and we are left withdU-/dz - dU+/dz = N(k)substitute in for U obtained above and take the strict limit as z->0 as we> getA(k) = 1/2 * N(k)/kthusU = 1/2 N(k)exp(-k|z|) / kinverting this using the inverse hankel transform we haveV = 1/2 Integrate(0,oo) exp(-k|z|)*J(rk)*J(rk) dk Integrate(0,a) r*n(r)> drlet n(r) = constant =1. and lets evaluate this in the plane i.e. z=0V(r,0) = 1/2 Integrate(0,oo) J(rk)*J(rk) dk Integrate(0,a) rdrthe integral over the bessel functions (i think) isV(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 integralformula however is only correct if rrSorry, for beeing so late with this remark, but I only just foundtime for it. Also forget about my comment about E(-4 r/...), itwas just a stupid error on my part.HTH,Michael.-- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&Dr. Michael UlmFB Mathematik, Universitaet Rostockmichael.ulm@mathematik.uni-rostock.de have a 2n*2n matrix over a eld of characteristic 2 of the following form( I A)( B I )where A and B are both n*n matrices such thatA_ij = (-1)^(i+j) det (B_j,i)here j and i mean omit the jth row and the ith column. The same holdsthe 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 theentire matrix has to have determinant 1. I know that for n=2, these matricesdo exist, but need something for bigger n (n=4 is my goal).chris eld of characteristic 2 of the followingform> ( 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 jth row and the ith column. The sameholds> 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 ifthe> entire matrix has to have determinant 1. I know that for n=2, thesematrices> do exist, but need something for bigger n (n=4 is my goal).> chris>The ip answer is that the matrices exit for all n. Just take A = B =0.Then the determinant of the block matrix is 1 in any eld.Heres a Maple proof that for n = 4 over GF(2) this is the only solution.(Note that, of course, A and B are the (classical) adjoints of each other.)I rstfind all such matrices and then see which lead to determinant 1 mod2:> restart:> with(LinearAlgebra):> 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) ;> 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 eld 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 jth row and the ith 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 ip answer is that the matrices exit for all n. Just take A = B =0.>Then the determinant of the block matrix is 1 in any eld.>Heres a Maple proof that for n = 4 over GF(2) this is the only solution.>(Note that, of course, A and B are the (classical) adjoints of each other.)Wait a minute. For n x n matrices, if Im not mistaken, Adjoint(Adjoint(A)) = (det A)^(n-2) A. This is an identity of polynomials over Z, so its valid over any eld. So for n > 2, if B = Adjoint(A) and A = Adjoint(B) and det(A) = 0 then A = B = 0.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =Adj(Adj(A))=det(A)^(n-2) AI can see easily that this holds for invertible A. But for non-invertible Ait is a different story actually... I am able to proof this also holds fornon-invertible A but I would like to prove it in a more direct and simplerway. 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. Every openset is dense in the Zarinski Topology. Since Adj(Adj(X))=det(X)^{n-2}X is acontinuous function, which holds for all matrices A, with det(A) ne 0, thenit holds for all A, thus also for A s.t. det(A)=0 (since the former matricesare dense).I would think there is a directer proof. But I fail to see it.ideas anybody? Or references if it is already proven by someone (which willprobably be the schreef in bericht> I have a 2n*2n matrix over a eld of characteristic 2 of the followingform> ( 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 jth row and the ith 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 ip answer is that the matrices exit for all n. Just take A = B =0.>Then the determinant of the block matrix is 1 in any eld. >Heres a Maple proof that for n = 4 over GF(2) this is the only solution.>(Note that, of course, A and B are the (classical) adjoints of eachother.) Wait a minute. For n x n matrices, if Im not mistaken,> Adjoint(Adjoint(A)) = (det A)^(n-2) A. This is an identity of> polynomials over Z, so its valid over any eld. So for n > 2, if> B = Adjoint(A) and A = Adjoint(B) and det(A) = 0 then A = B = 0. Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2 =|| > negentropic collapse of phase space of the unstable Dirac electron| > vacuum to a much smaller volume of phase space that ismacroscopically| > occupied by bound lepto-quark pairs.|It is a good laugh reading it though. How does a much smaller volumeof phase space t with macroscopically occupied? The weird thingposted recently. Jack, what happened? Did those aliens form theUFOs get to you to try to confuse us. Hehe.FrediFizzx >negentropic collapse of phase space of the unstable Dirac electron >>vacuum to a much smaller volume of phase space that is macroscopically >>occupied by bound lepto-quark pairs.Nope. Not even close. = Titan Point: >> negentropic collapse of phase space of the unstable Dirac electron >> vacuum to a much smaller volume of phase space that is macroscopically >> occupied by bound lepto-quark pairs. > A few of the buzzwords would if he hadnt used them in a sentence,but even then, only a few. In other words, put on hip boots and graba large shovel. = > frac{partial I(x,y,t)}{partial>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 all ASCII. Given a countable set of letters A={L_1,L_2,L_3,...}, consider the> set of nite strings S(A) which can be formed using letters in A. Is> S(A) a countable set?>>yes> Yes, but its not well-orderable is the correct answer.How could you have a countable set thats not well-orderable? How could you have a rational discussion with ZZBunker? :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) =|At one time I seriously considered trying tofind someone else|interested and spend enough time and energy to try to master his|methods and see what the result was. Unfortunately I suppose, I|never did that.Yeah, thats the way things often seem to work, isnt it?I was interested enough to copy key excerpts from one paperon how to do it. It was one of many little projects I have sittingaround here somewhere. Since then Ive moved, so I supposemy notes are in a box somewhere here.Keith Ramsay =To reprise James Harris proof, or at least the rst two steps(Ive included * so that GP/Pari works):[begin excerpt]1. Let P(x) = 14706125 * x^3 - 900375 * x^2 - 17640 * x + 1078, where x isin the ring of algebraic integers, notice that P(x) has a constantterm that is 1078.2. It can be shown thatP(x)= 7^2*(2401*x^3 - 147*x^2 + 3*x)*(5^3) - 3*(-1 + 49*x)*(5)*(7^2)+7^3where the *same* polynomial has been put in a form which allows afactorization into non-polynomial factors so that I haveP(x) = (5*a_1(x) + 7)(5*a_2(x)+ 7)(5*a_3(x) + 7)where the as are roots ofQ(a) = a^3 + 3*(-1 + 49*x)*a^2 - 49*(2401*x^3 - 147*x^2 + 3*x).[end excerpt]Now, heres where I have a problem. How did Mr. Harris get tothis point?Its clear that Q(a) has a binding problem as written; it shouldbe written Q(x,a).Ive attempted to compute this polynomial explicitly by assumingthree roots in P(x) and getting an ungodly mess, but I dowonder how Q(a) can be derived from P(x), and if so, whetherthe derivation is correct.To further that end, Ill reexamine the problem, hopefully withmore rigor than in my other post (which I dont have theID for, sorry).First, what we want is a polynomial Q(x,y), whereQ(x,y) = (y - a_1(x)) * (y - a_2(x)) * (y - a_3(x)).a_i(x) is the actual a root, and Ive switchedvariables for clarity -- at least, I hope its clear.We know P(x) = K * (x - r_1) * (x - r_2) * (x - r_3), for some K and r_i.Mr. Harris also purports thatP(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7).This is ne, and actually 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 Ive actually derived the a_i(x) I want.If I dene the function a(r) = 5*(7^2*x - 5*7^2*r - 7/5), Inow wonder: what is (y - a(r_1) ) * (y - a(r_2) ) * (y - a(r_3)) ?Thats the cubic I want in step 2.Recall 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_3therefore r_1 + r_2 + r_3 = 900375/14706125 = 3/49r_1*r_2 + r_1*r_3 + r_2*r_3 = -17640/14706125 = -72/60025r_1*r_2*r_3 = -1078/14706125 = -22/300125And now its time for the grinding. Apologies for thelong line lengths. If you have a monospaced font youmight verify that Ive matched up terms in the rstfour equations (the rst is the raw output from GP/Pari);Ive tried to do this very 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 mustconclude that one of us has made an algebraic or logicerror somewhere; I hope it wasnt me but these equationsare ugly; much of the work was done by GP/Pari but itisnt smart enough to substitute the symmetric root formsI detailed above.Please check my work here (if you dare! :-) ); I think thatI have found the form that the ys (or as) must satisfy,dependent on x. But its not pretty -- and it doesnt match James.But at least now it doesnt have fractions in it. :-)-- #191, ewill3@earthlink.net -- insert random ugly math hereIts still legal to go .sigless. = > To reprise James Harris proof, or at least the rst two steps > (Ive 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 algebraic integers, notice 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* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have > P(x) = (5*a_1(x) + 7)(5*a_2(x)+ 7)(5*a_3(x) + 7) > where the as 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, heres where I have a problem. How did Mr. Harris get to > this point?Actually it is quite simple. Q(a) is one of the many possible polynomials.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 values 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 llingin these three values gives the original form of P(x).-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Ive called the factoring of polynomial into non-polynomial factors> advanced for a reason, understanding requires that you have the basics> down, are VERY well-founded in mathematical logic, and realize that> you will have to make some serious effort.If you honestly believe they are advanced, you know almost 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, especially when you fail to understand the objections.However, if you THINK youre going to be a mathematician, or think you> already 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 itll be part of the historical record till you> die, and beyond.Hmm... I best know where Ill stand, then.Now then, Ive come across a nice example to give some perspective on> advanced concepts in factoring polynomials, which has the nice feature> of, I hope, being easy to understand.The polynomial is F(x) = x^2 + x + 2, and the factorization relies on> something you probably know about, but didnt pursue, which is that> x^2 + x is always even if x is an integer, so I can factor asF(x) = 2(x(x+1)/2 + 1)Where 2 and (x(x+1)/2+1) are the factors, I presume.for a factorization valid in integers, but not in general valid in> algebraic integers.This is not a valid factorization in the integers, although both factors are integers for any integer x.When you say its a factorization in the integers, you are *normally* referring to the ring of coefcients being integers. Here you have 1/2 for two coefcients, which is an algebraic number.To move beyond factoring polynomials into polynomial factors, you need> to have a *thorough* understanding of coprimeness, ring operations,> and logical argument.Then you are unqualied, unless you are now able to correctly dene coprime and understand that Z[1/2] is *not* the reals.What Ive found is a puzzling lack of a thorough foundation in these> areas from LOTS of people in the math world, as Ill include Barry> Mazur, Andrew Granville, and Ralph McKenzie, in with the posters who> have continually posted in reply to me on this forum.We agree on our terminology. This suggests that the problem lies elsewhere.Basically, quite a few of you have demonstrated a failure to> understand mathematics at a level Id call competent, as the basics> necessary are what Id *think* any mathematician would have.No, we disagree with you on how *you* think mathematics works. Weve studied the subject, you havent. Why should *you* be the authority?But here, in the 21st century, mathematicians are displaying a> woeful lack of the basics, besides often displaying childish> petulance, arrogance, and a stubborn refusal to acknowledge the math.Who is the one who calls people liars for disagreeing? Who is the one who resorts to cursing? Who is being childish?> James Harris> http://mathforprot.blogspot.com/Have you made any prot yet?-- Will Twentyman =youre just bullting, sinceyouve demonstrated for years thatyou dont have a *thorough* understanding of coprimeness etc.,through mathematical logic. is that presumedto be ordinary predicate logic, the symbolic notationof the Aristotelian stuff? does nonpolynomial factors mean monomial ones?> Ive called the factoring of polynomial into non-polynomial factors> advanced for a reason, understanding requires that you have the basics> down, are VERY well-founded in mathematical logic, and realize 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 itll be part of the historical record till you > The polynomial is F(x) = x^2 + x + 2, and the factorization relies on> something you probably know about, but didnt pursue, which is that> x^2 + x is always even if x is an integer, so I can factor asF(x) = 2(x(x+1)/2 + 1)for a factorization valid in integers, but not in general valid in> algebraic integers.To move beyond factoring polynomials into polynomial factors, you need> to have a *thorough* understanding of coprimeness, ring operations,> and logical argument. > http://mathforprot.blogspot.com/--les Dicks dEnron!http://www.wlym.com/antidummies/part36.html =Cantor was mistaken. This is clear to anyone who considers anarbitrary even prime >2. Although some parts of Cantor have truth, Ibelieve that Cantor is completely bereft of truth. (Actually, thatlast sentance was inconsistent and needs correction, but we can stilladhere to it without loss of generality.) I have obtained empiricalevidence that Cantors diagonal number does not exist, through a longthought experiment. There is no doubt in my mind that Cantor is fullof CONTRADICTIONS. Einstien, Hawkins, and Feynmann were all wellaware of this. In fact, quantum physics subtly relies on Cantorswork, and it too is therefor fundamentally misguided. I have attendedprestigious universities, so I know what Im talking about. I have been working on this theory ever since I was 10 years old. Before you try to steal any of the things I will write below, knowam now willing to offer a $1000 reward to anyone whofinds fault in mytheory. A simple cursory glance at the Cantoresque Numbers will alreadymake it clear the something isnt quite right. Im not good at math,but my theory is conceptually right, so all I need is for someone toexpress it in terms of equations. We must remember that the currentaxiomatic models are only theories, nothing more. Although ZFC can beused to accurately predict the outcome of typing things into acalculator, it hardly explains *why* the calculator gives the answersit gives. Like Einstein, I have a special insight into these matters. My work is on the cutting edge of a paradigm shift. Before we proceed, Id like to mention that I am very displeasedwith the crackpot index located athttp://math.ucr.edu/home/baez/crackpot.html, and have written theauthor to complain that it suppresses original thinkers (the author,by the way, cant even spell Einstein correctly). When you have read my theory I am sure you will agree it is ofNobel prize calibre. Like Newton, I know I will be preserved inmemory as a great pioneer of the sciences. We know from C.S. Lewisalready that Cantor is awed, I am merely making his defeat morerigorous. In the past, my theories were ridiculed but I was alwaysable to carefully defend them, leaving my assailants oudering anddiscredited. They were hidebound reactionaries and self-appointeddefenders of the orthodoxy. It is a little known fact that in their correspondence, Dedekindrelentlessly opposed Cantors work, and the only reason he did notmake his objections public was out of friendship to Cantor. In hislater years, Einstein himself was groping toward a physical disproofof Cantors nonsensical works. I am certain that if there are anyother civilizations in this universe, more advanced than our own,surely they are aware of the countability of all things. I would havebrought this to light sooner, but was constantly being resisted by adamn psychologist (what do psychologists know about these things??) Those who disagree with my theories and support Cantor are no betterthan nazis, stormtroopers or brownshirts. As Harris has proven timeand again, the scientic community is actively engaged in an outrightconspiracy to prevent theories like mine from surfacing. At times,amid all this persecution, I truly believe I know how Galileo felt. But only for a time. When my theories break through, modern daymathematics and science will be seen as the laughingstock that theyare. I look forward to the day my opponents are forced to recanttheir objections, or be ostracized from the community. As I have no shown, my theory is very revolutionary. It is onlya matter of time now, for knowledge cannot be suppressed forever. Your correspondent, Nathan Deeth Age 11 =For anyone who hadnt noticed yet, this post contains ALL the 35 items ofthe crackpot index IN ORDER.-- Wim Benthem = For anyone who hadnt noticed yet, this post contains ALL the 35 items of> the crackpot index IN ORDER.> And where is this index?/David > > For anyone who hadnt noticed yet, this post contains ALL the 35 items of>> the crackpot index IN ORDER.>> And where is this index?http://www.math.ucr.edu/home/baez/crackpot.html = == =Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu =http://www.math.ucr.edu/ home/baez/crackpot.html> LOL!/David Your correspondent,> Nathan Deeth> Age 11Im interested in how you have remained at age 11 for the past 5years? One merely needs to look up Nathan the Great (or NathanielDeeth) and Age 11 on Google Groups to see that you have beenclaiming this age for some time now. You were even accused of thisback then: http://mathforum.org/discuss/sci.math/a/m/219236/219315I can only guess that you are speaking about the equivelent age ofyour thinking and spelling.Jonathan Hoyle =Mike Deeth says...>I have obtained empirical evidence that Cantors diagonal number>does not exist, through a long thought experiment.The diagonal number isnt just one number. For any innitelisting of real numbers, there is a corresponding diagonal number.For example, consider the following innite list r_0 = 0.5 r_1 = 0.05 r_2 = 0.005 r_3 = 0.0005 etc.Lets take the diagonal by the following rule: add 1 to anydiagonal digit that is less than 5, and subtract 1 from anydiagonal digit that is greater than 5. This gives us diagonal = 0.6666... = 2/32/3 denitely exists.--Daryl McCulloughIthaca, NY Mike Deeth says...>I have obtained empirical evidence that Cantors diagonal number>does not exist, through a long thought experiment.The diagonal number isnt just one number. For any innite> listing of real numbers, there is a corresponding diagonal number.> For example, consider the following innite list r_0 = 0.5> r_1 = 0.05> r_2 = 0.005> r_3 = 0.0005> etc.Lets take the diagonal by the following rule: add 1 to any> diagonal digit that is less than 5, and subtract 1 from any> diagonal 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 diagonal is not changed. To get your result, the rule should say something like add 1 to any diagonal digit that is less than 6, and subtract 1 from any diagonal digit that is greater than 5. This will now modify any diagonal digit, including 5, as your rule does not, and will give your stated result.This gives us diagonal = 0.6666... = 2/32/3 denitely exists.--> Daryl McCullough> Ithaca, NY> =Virgil says...>> Lets take the diagonal by the following rule: add 1 to any>> diagonal digit that is less than 5, and subtract 1 from any>> diagonal 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 diagonal is not changed.--Daryl =Unfortunately Nathan, you fall into the same logical trap that Cantor did.You tacitely assume that it is possible to construct a _countably_ inniteset. I would be interested in seeing your proof, so that I could point outits aws. You have fallen in with those who would suppress the work ouminaries such as Harris when you take such a view. The truth cannot behidden forever.On my webpage I have posted a proof of the 1-1 mapping, N -> R. However, asa corrolary to my proof, I demonstrate that N itself is uncountable. Itsthere for anyone to view, and unless someone canfind a aw in my proof, wewill have to agree that it is correct.Until someone posts a FLAW in this proof I insist that no one be critical ofme or the proof, since clearly in mathematics we base things on *PROOF* andnot character assasination. I am not especially concerned, as I have shownthe proof to my chemistry teacher and he assured me it was correct. Thiscertainly qualies me for a Fields metal. The proof is right there on myhomepage, for anyone with internet access to see. Notice that I do notwithhold proof from the newsgroup, so sure am I that it is correct.Charlie Van Winkle, esquire,Age 9 =I am of the rm belief that you and the author of this thread are one andthe same (and possibility just an alias of JSH, which would explain hisimmaturity). Both of you sign your name in the same way and right your ageafterwards. Both of you refer to proofs but do not provide links to them(Charlie, neither you nor Justin have websites with your name). You bothyou are wrong. You cannot state that the natural numbers are uncountable.By denition, a countable set is that of the same cardinality as thenaturals. By the way, Nathan how have you been through University at age11, and Charlie, how have you nished law school by age 9?Steven> Unfortunately Nathan, you fall into the same logical trap that Cantor did.> You tacitely assume that it is possible to construct a _countably_innite> set. I would be interested in seeing your proof, so that I could pointout> its aws. You have fallen 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 corrolary to my proof, I demonstrate that N itself is uncountable. Its> there for anyone to view, and unless someone canfind a aw 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 criticalof> me or the proof, since clearly in mathematics we base things on *PROOF*and> not character assasination. I am not especially concerned, as I haveshown> the proof to my chemistry teacher and he assured me it was correct. This> certainly qualies me for a Fields metal. 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 corrolary to my proof, I demonstrate that N itself is uncountable. Its>there for anyone to view, and unless someone canfind a aw in my proof, we>will have to agree that it is correct.Bah, amateurs. Ive proof that the nite set of integers [1,10^10^10] is uncountable.Just try and count them, one by one, out loud. Report back to me whenready. Einstien, Hawkins, and Feynmann were all well> aware of this.Is that Jim Hawkins?> Before we proceed, Id like to mention that I am very 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 original thinkers (the author,> by the way, cant even spell Einstein correctly).So no one who spells Einstein incorrectly is worthy of notice?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) =OK, so backtracking a few lines: Take Gamma(1-s)(-2*pi*n)^s-1. Summing> over all 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> givesZeta(s)=Sigma_n=1^innity(Gamma(1-s)[(-2*pi*n)^s-1 + (2*pi*n)^s-1]. I> dont see how 3) is implemented to give this result. Sorry ahead of> time for any Ughs.Well can I present you with an ugh for each asterisk above :-)Lets translate this. DoesGamma(1-s)(-2*pi*n)^s-1meanGamma(1-s)(-2pi n)^s-1orGamma(1-s)(-2pi n)^{s-1}.The latter seems to make more sense in this context.I cant make head nor tail of what you write for 3).Is n the summation variable? what has happened to the rst integral.Is the equals sign before the epsilon really meant to be there?I presume you are using the contour integral argument for continuingGamma(s)zeta(s). One introduces a countour C_e in three parts:imaginary axis from -innity to -e, circle radius e about origin,imaginary axis from -e to -innity and take the integralof z^{s-1} e^z on C_e (with branch cut on negative real axis).Thse integral f(s) is independent of e by Cauchys theorem.It is also an entire function of s: convegence is nice since e^t -> 0rapidly as t -> -innity.The integral of z^{s-1} e^z on the rst part of the contour isintegral_e^innity t^{s-1} exp(-pi i(s-1)) e^{-t} dt= - integral_e^innity t^{s-1} exp(-pi is) e^{-t} dt.Similarly on the third part of the contour it isintegral_e^innity t^{s-1} exp(pi is) e^{-t} dt.These add to2i integral_e^innity t^{s-1} sin(pi s) e^{-t} dt.If Re(s) > 0 the integral over the circle of radius e isO(e^Re(s)). Letting e -> 0 we get thatf(s) = 2i sin(pi s) Gamma(s)for Re(s) > 0.Now we considerg(s) = integral_{C_e} z^{s-1} e^z/(1-e^z) dzwhere 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 thatg(s) = 2i sin(pi s) integral_0^innity t^{s-1} e^{-t}/(1-e^{-t}) dt = 2i sin(pi s) Gamma(s)zeta(s).ConsiderG_N(s) = integral_{C_{(2N+1)pi}} z^{s-1} e^z/(1-e^z) dz.By Cauchys theorem the difference G_N(s) - g_e(s) is 2pi i timesthe sum of the residues of the poles of the integrandat +-2pi i, +- 4pi i, ..., +- 2Npi i, that is2pi 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 -> innity, 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) < 0Gamma(s)zeta(s) = something nasty times zeta(1-s).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) >OK, so backtracking a few lines: Take Gamma(1-s)(-2*pi*n)^s-1. Summing>over all 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^innity(Gamma(1-s)[(-2*pi*n)^s- 1 + (2*pi*n)^s-1]. I>dont see how 3) is implemented to give this result. Sorry ahead of>time for any Ughs.Well can I present you with an ugh for each asterisk above :-)Lets 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 cant make head nor tail of what you write for 3).> Is n the summation variable? what has happened to the rst integral.> Is the equals sign before the epsilon really meant to be there?Yes. I lifted this verbatim from p 13.I presume you are using the contour integral argument for continuing> Gamma(s)zeta(s). One introduces a countour C_e in three parts:> imaginary axis from -innity to -e, circle radius e about origin,> imaginary axis from -e to -innity and take the integral> of z^{s-1} e^z on C_e (with branch cut on negative real axis).> Thse integral f(s) is independent of e by Cauchys theorem.> It is also an entire function of s: convegence is nice since e^t -> 0> rapidly as t -> -innity.The integral of z^{s-1} e^z on the rst part of the contour is> integral_e^innity t^{s-1} exp(-pi i(s-1)) e^{-t} dt> = - integral_e^innity t^{s-1} exp(-pi is) e^{-t} dt.> Similarly on the third part of the contour it is> integral_e^innity t^{s-1} exp(pi is) e^{-t} dt.> These add to> 2i integral_e^innity t^{s-1} sin(pi s) e^{-t} dt.> If Re(s) > 0 the integral 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) = integral_{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) integral_0^innity t^{s-1} e^{-t}/(1-e^{-t}) dt> = 2i sin(pi s) Gamma(s)zeta(s).Consider> G_N(s) = integral_{C_{(2N+1)pi}} z^{s-1} e^z/(1-e^z) dz.> By Cauchys 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 -> innity, 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 RZFto get other perspectives. Ill continue to study this. > I cant make head nor tail of what you write for 3).>> Is n the summation variable? what has happened to the rst integral.>> Is the equals sign before the epsilon really meant to be there?Yes. I lifted this verbatim from p 13.I fear not :-( Edwards uses standard notation. :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) I cant make head nor tail of what you write for 3).> Is n the summation variable? what has happened to the rst integral.> Is the equals sign before the epsilon really 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. Ihave it in front of me now. The summation variable is missing and theequals sign is before the epsilon. Edwards also uses capital Pi for thefactorial function instead of the Gamma function. He attributes thisusage to Gauss and Riemann (footnote p8). The OP has silently changed this.In this chapter Edwards is explaining and expanding upon Riemanns originalpaper which is very terse. This limits him to following Riemannsdevelopement and notation.Ifind Davenport, Multiplicative Number Theory chapter 8 clearer as anintroduction to the functional equation. Jack Fearnley =I HATE sports, I hate people who PLAY sports, and I hate thecoach!=[ dmessage> quantify Baseball to other sports; OptimalStrategy for> Baseball> Archimedes Plutonium whole entire Universe is just one big atomwhere dots of> the electron-dot-cloud are galaxies> sci.logic, soc.history, sci.math> A friend asked me to watch this years World Series torender my> opinion.> I sketchedly watched and here is my opinion, as I usuallydo not have> time for such recreation. I watched only parts of game 2 & 3 where the FloridaMarlins were> being> overpowered, and missed game 1. And from game 3 I decidedit was a> waste> of time to watch games 4 or 5 in that the New York Yankswould> probably> overpower the Marlins and so only watched a few innings ofpitching. I> watched nearly the full game of 6. I had seen some clips of Baseball sluggers such as B.Bonds and S.> Sosa> (excuse me if name is incorrect spelling) from the sportssection of> the> local TV news when getting the weather report so I havesome awareness> of the best hitters of Baseball. I am not interested in sports only to the extent in whichmy> analytical> mind can> recast the sport as to Optimal Strategies. Baseball Optimal Strategy: after watching this last gameof World> Seriesfor Baseball> which includes these threads. (1) The pitcher in Baseball, unlike many other sports, isthe dominate> feature of baseball when you consider that of 9 players,the pitcher> has> a say in every offensive action. (2) We can math quantify the dominance of Baseball pitcherwith other> sports such as Football. Where the quarterback has a bigcontrol of> the> offense but not as much of control of the overall game asthe Baseball> pitcher for defense. The quarterback if he ran every playwould have> as> much control as the Baseball pitcher. But he does not andhas a> handoff> to a running back or a receiver. So we can say mathwisethat the> Baseball pitcher has 100% control of defense Say-of-Actionwhereas in> Football the quarterback at most has 25% to 33%Say-of-Action concept.> Because in each offensive play in Football, thequarterback either> runs> himself or hands to a runningback or throws to a receiver.But in> Baseball, every offensive play has a Say by the pitcherwith his> pitch. (3) So, unlike every other sport, the pitcher in Baseballis the> dominate feature because the Say-of-Action is 100% thepitcher. So, the OS of Baseball in order to assemble a team thatwill win the> World Series for that year, is key in having as manypitchers that are> ace pitchers such as Beckett and Pavano of Marlins.Beckett pitching> was> superior to that had 2 pitchers of the quality ofBeckett, they> would> have the highest chances of reaching the World Series. By the way, I did not see the pitchers batting, so I guessthe game of> Baseball has made some progressive rule change where it isoptional> for> the pitchers to go to bat; and in the old days I rememberthe pitcher> was usually the 9th spot hitter and usually an easy out. Iguess the> new> rule is that when the option is picked that the teamrotates the 8> hitters and leaves out the pitchers as hitters. Getting back to this idea that pitching is the key tobaseball> winning.> What does it matter for a team to have great hitting suchas a Bonds> or> Sosa if they choke on pitching in that the other team withaverage> hitters but with great pitching. By the way, did Beckettpitch to> Bonds Now suppose the Yankees had both Bonds and Sosa in theirlineup.> Facingthe Marlins> had 3 pitchers of the quality of Beckett then I wouldguess that the> Marlins would have still won the game. I did see clips of the Boston game versus the Yankees andthe Boston> ace> pitcher of Martinez (forgive the spelling). So I amguessing that> teams> that have at least one or two good pitchers make it to theplayoffs.> Pitching is number one key. So, these concepts would then ask for a Baseball historianto look at> the winning teams and to see whether every World Serieschamp had 2> excellent pitchers such as Beckett and Pavano for Florida So, the above should guide all club owners who aspire towin the World> Series that if your team has at least 2 excellentpitchers, is the> basic> prerequisite. And to concentrate the effort more ingetting great> pitching than in getting great batting.> The batting of the Marlins was often frustrating and itseemed as> though> homeruns were a rarity for the Marlins so it goes to showthat teams> that focus on hitting are not really the Optimal Strategythread. And nally a discussion of Pitching itself. I would hateto be a> pitcher personally because throwing a ball at 95 mph formany hours> has> a toll on the arm and is a job that does not last toolong. I think> there is a Optimal-Strategy-Subset for pitching itself inbaseball.> The> key is to get the batter out in fast quick time. Thatmeans it is no> good to strike out a batter rather than to get him to yout. If a> pitcher is so good at devising a pitch that is popped upfor a y> out,> then that is better than going for 3 strikes because ifyou pitch a> ball> with a great spin on it such that the probability when hitpopps up> then> conceivably 3 pitches for the inning can retire the sidewhereas> strikeouts require at least 9 pitches. By the way is thereany> statistic> where a pitcher retires a complete side with only 3pitches in all??? So, the pitcher that can devise a pitch that is prone topop up is> superior to the pitcher that relies on fastballs andstrikes. I am> guessing that some pitch has such a spin on it that thebatter is> likely> to pop up or ground out. The perfect pitched Baseball game would have 9 innings andonly 9 X 3> 27> pitches where all batters swung and hit the rst pitchand popped out> or grounded out. Not the game where the pitcher made 9 X 9= 81> pitches> and all strikes and all strikeouts. whole entire Universe is just one big atom where dots> of the electron-dot-cloud are galaxies I know the construction of cardinals less than aleph_0, but this>construction suppose the existence of the concept of set and>implicitly the theory supposses that there are only integer (and>positive) cardinals.> The construction could be the most elaborated theory as you want, but>the implicit ideas of the theory, the ideas that have all the>mathematicions when (we or) they think about it, are simpler: the>concept of the (nite) set have to be integer cardinal.> Respecting very much John Conway, and from my knowless, I think that>its only a way to construct numbers, but not the way to contruct>sets, because if we want to extend the idea of cardinal, I think>that we have to extend the idea of set. And the problem is doing this>without perversion-circle problem.> Xan.These are actually games, not sets. For example if you have 3 games of -1/3> and one game of 1, it is equivalent to the game of zero (a second player> win). Likewise, if you a large nite amount of negative intismals and> one game of one, it is going to be a game where right wins. I dont think> it is possible to have non Steven.I will continue think about this problem. If you know anything aboutXan. =may yo use more details please? =givenE = 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}^2the paper said,Grad E = 2 sum_{j=1}^N w_{m,j} e_{m,j} v_j for m = 1..Nwhy not treat | u_i v_j - d_{i,j}|^2 as e_{i,j} cdot e_{i,j}^*, if soshould we treat u_i^* independent of u_i, leading toGrad E = sum_{j=1}^N w_{m,j} e_{m,j}^* v_j for m = 1..NLin. S. >Deck of 52 cards. You start out with k dollars. You place a bet on the>rst hand. The game is that you guess whether the next card will be black>or red. The dealer ips over a card and if youre right, you receive>whatever you bet, and if you lose, your bet is gone.>>snip>My question is, how much would you pay to play this game? In other words,>what is the optimal strategy for this game and what is your expected>winnings using this strategy?(snip)>>A review of a collection of various papers on the Kelly Criterion, aka>>Proportional 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 Ive made a mistake, Kellys criterion will lead to exactly the>same expected value.It does indeed! It took some tweaking but I now have an Excelsimulation which demonstrates the behavior of the game, using both thepure strategy of waiting for a sure thing bet and the proportionalstrategy of Kelly betting. The waiting strategy gives exponentiallylarge payoffs with inverse exponential rarity, averaging to 8.0133while the Kelly criterion approach gives consistent payoffs at valuesnear the theoretical value. Said another way: using sure thing bets,you come up short most of the time while with Kelly betting, theplayers realizes the value of the game most of the time. Note thatmost of the time is not equivalent to on average. >Maybe it will help to look at some simple situations.John Bailey = > I dont feel anything, but all of these were covered in the rst> two years of my undergraduate degree.>>Well it certainly isnt a fact that they should be covered at the>undergraduate level, hence its 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 cant decide>>If 3) were the case, you would not respond with such a snotty remark>assnotty? thats abuse.>Thats graduate algebra?>>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 rst year>graduate algebra course? And what textbook is suitable for such a>course?Dont be asinine. I was merely expressing my surprise> that such elementary 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 inthe US. Chez nous, snotty is often used by the person who feelsabused. It is used to indicate that ones counterpart is elevated byairs of superiority, unable to maintain a tone of equality. Among us,it would be called a put-down, rather than abuse.David Ames =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 ones counterpart is elevated by> airs of superiority, unable to maintain a tone of equality. Among us,> it would be called a put-down, rather than abuse.But an unwarranted use of a perjorative term is surely abusein anyones lexicon?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) > Whats a good book for a rst year graduate algebra course?>> Something with alot of emphasis on factorization, polynomial rings,>> elds, PIDs, Galois Theory, and of course all the more basic topics>> of algebra as well like groups, ideals, integral domains, etc.>> Thats graduate algebra?> Theres a recent book by Joe Rotman Advanced Modern Algebra>> which does all 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 all of the above>>topics in detail in their GTM books? Granted, of course groups,>>ideals, and integral domains should be covered at the undergraduate>>level, but should free groups, nitely generated abelian groups, and>>galois theory? Maybe if you go to Harvard or MIT.> All the listed topics can (and should) be taught in any good rst course>> in abstract algebra. Anyone lacking such fundamental algebraic knowledge>> would be poorly prepared for graduate-level studies.> -Bill DubuqueAre you serious? Im sure he is.> If so, maybe Im just a complete idiot. You said it.> But are you> telling me that Galois theory can and should be taught in a rst> course in abstract algebra? We wrent talking about what goes in a rst course, but what is graduateand by inference undergraduate level. Galois theory is certainlyundergraduate level.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 But are you>> telling me that Galois theory can and should be taught in a rst>> course in abstract algebra? We wrent talking about what goes in a rst course, but what is graduate> and by inference undergraduate level. Galois theory is certainly> undergraduate level.In Nobodys defense, Ill point out that I never saw Galois theorybefore my graduate course. And look how I turned out.Wait. Never mind.-- And yes, for those who think that just maybe I didfind a short proofof Fermats Last Theorem, and THE prime counting function, if Isucceed at what Im working on now world economy as you know it willbe gone. -- James Harris branches out. =|We wrent talking about what goes in a rst course, but what is graduate|and by inference undergraduate level. Galois theory is certainly|undergraduate level.In the U.S. what counts as graduate level varies enormouslyfrom one school to the next. I met some masters students oncewho were learning the concept of compactness. Certainly thefundamental theorem of Galois theory has been taught to rst-yeargraduate students at reasonable schools. One hopes that its byway of review... but its hard to be sure.Keith Ramsay [...]>Are you serious? If so, maybe Im just a complete idiot. But are you>telling me that Galois theory can and should be taught in a rst>course in abstract algebra? Granted, all the topics I mentioned>(groups, rings, ideals, integral domains, elds, PIDs, factorization)>should be mentioned in a rst course on abstract algebra, but how>much detail can you possibly go into in such a short amount of time? >Should free abelian groups, sylow theory, and solvable groups be>included in this introductory course? How about Splitting Fields and>Galois groups? If you say yes then youve lost your noodle, however>these still fall under the category of group theory or eld>theory. I said in my original post that Ill have nished>hungerford by the time I start reading the new book, so when I refer>to group theory obviously Im not talking about the denition of a>group, or a homomorphism, although many books will probably mention>that anyway.>My second semester undergraduate algebra course (at Wesleyan in 1978-79) covered Galois theory; the culmination of the rst semester was Sylows theorem. Basically, the two semseters went through Herstein, Topics in Algebra, more or less. However, we also used Stewarts Galois Theory.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu = > I dont feel anything, but all of these were covered in the rst> two years of my undergraduate degree.>>Well it certainly isnt a fact that they should be covered at the>undergraduate level, hence its 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 cant decide>>If 3) were the case, you would not respond with such a snotty remark>assnotty? thats abuse.>Thats graduate algebra?>>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 rst year>graduate algebra course? And what textbook is suitable for such a>course?Dont be asinine. I was merely expressing my surprise> that such elementary topics should be considered as graduate.> And I did suggest a book.You still havent said what topics in algebra you consider advancedenough to belong in a graduate algebra course. Id be curious to seethe syllabus for an undergraduate introduction to abstract algebracourse which covered:GroupsSubgroupsHomomorphismsNormal Subgroups / Isomorphism TheoremsLagranges TheoremAlternating & Symmetric GroupsFree GroupsFinitely Generated Abelian GroupsSylow TheoremsSolvable and Nilpotent GroupsRingsFactorization in Polynomial Rings(Maximal) IdealsField ExtensionsSplitting FieldsGalois GroupsAnd as Ive pointed out, youve avoided at least 3 times the questionof what material DOES belong in a graduate algebra course. You still havent said what topics in algebra you consider advanced> enough to belong in a graduate algebra course. Id be curious to see> the syllabus for an undergraduate introduction to abstract algebra> course which covered: The thing here is that you seem to be equating undergraduate algebra with introductory algebra. I would say that for *most* universities,these two are not the same. The university I went to for undergrad (after3 courses in linear algebra) have SIX upper-year undergradute courses inabstract algebra (granted, two of these are not intended for pure/appliedmath majors, but still the other four courses cover all the topics youlisted earlier.)I copy their course descriptions below:PMATH 345 LEC 0.50 Course ID: 007667 Polynomials, Rings and Finite Fields Elementary properties of rings, polynomial rings, Gaussian integers, integral domains and elds of fractions, homomorphisms and ideals, Basis theorem, Gauss lemma, Eisensteins criterion, unique factorization, computational aspects of polynomials, construction of nite elds with applications, primitive roots and polynomials, additional topics. [Offered: F,S] Prereq: MATH 235/245;PMATH 346 LEC 0.50 Course ID: 007668 Group Theory Elementary properties of groups, cyclic groups, permutation groups, Lagranges theorem, normal subgroups, homomorphisms, isomorphism theorems and automorphisms, Cayleys theorem and generalizations, class equation, combinatorial applications, p-groups, Sylow theorems, groups of small order, simplicity of the alternating groups, direct product, fundamental structure theorem for nitely generated Abelian groups. [Offered: W] Prereq: MATH 235/245;PMATH 442 LEC 0.50 Course ID: 007692 Fields and Galois Theory Normal series, elementary properties of solvable groups and simple groups, algebraic and transcendental extensions of elds, adjoining roots, splitting elds, geometric constructions, separability, normal extensions, Galois groups, fundamental theorem of Galois theory, solvability by radicals, Galois groups of equations, cyclotomic and Kummer extensions. [Offered: F] Prereq: PMATH 345, 346; PMATH 444 LEC 0.50 Course ID: 007694 Non-Commutative Algebra Jacobson structure theory, density theorem, Jacobson radical, Maschkes theorem. Artinian rings, Artin-Wedderburn theorem, modules over semi-simple Artinian rings. Division rings. Representations of nite groups. [Note: Offered in the Winter of odd years.] Prereq: PMATH 345;Coreq: PMATH 346 As well as a 400-level course in Algebraic number theory and a 400-level course in algebraic graph theory, which use the 345/346 courses as prereqs. The other two abstract algebra 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, ideals, factor rings, homomorphisms, nite and innite elds, polynomials and roots, eld extensions, algebraic numbers, and applications, for example, to Latin squares, nite geometries, geometrical 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 General Mathematics students PMATH 336 LEC 0.50 Course ID: 007663 Introduction to Group Theory with Applications Groups, permutation groups, subgroups, homomorphisms, symmetry 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 General Mathematics students =: You still havent said what topics in algebra you consider advanced: enough to belong in a graduate algebra course. Id be curious to see: the syllabus for an undergraduate introduction to abstract algebra: course which covered:: Groups: Subgroups: Homomorphisms: Normal Subgroups / Isomorphism Theorems: Lagranges Theorem: Alternating & Symmetric Groups: Free Groups: Finitely Generated Abelian Groups: Sylow Theorems: Solvable and Nilpotent Groups: Rings: Factorization in Polynomial Rings: (Maximal) Ideals: Field Extensions: Splitting Fields: Galois Groups =You still havent said what topics in algebra you consider advanced>enough to belong in a graduate algebra course. Id be curious to see>the syllabus for an undergraduate introduction to abstract algebra>course which covered:Nobody said that all of the topics below were suitable for an undergraduateintroduction to abstract algebra. The claim was rather that they are alltopics that can readily be taught at undergraduate level. In the UK,undergraduate mathematics courses take 3 or 4 years, and the topics belowwould be covered in various courses during that period.At my university, Warwick, things like groups, subgroups, homomorphisms,Lagranges theorem, An and Sn, factorization in polynomial rings over eldswould be covered in core (compulsory) courses during the rst year.Splitting elds, rings, ideals, f.g. abelian groups, possibly SylowsTheorem might be covered in the second year, and most of that would still becore. The other topics, like group presentations, solvable groups, Galoistheory would be in optional courses in the third or fourth year, but wouldbe studied in some depth.>Groups>Subgroups>Homomorphisms>Normal Subgroups / Isomorphism Theorems>Lagranges Theorem>Alternating & Symmetric Groups>Free Groups>Finitely Generated Abelian Groups>Sylow Theorems>Solvable and Nilpotent Groups>Rings>Factorization in Polynomial Rings>(Maximal) Ideals>Field Extensions>Splitting Fields>Galois GroupsAnd as Ive pointed out, youve avoided at least 3 times the question>of what material DOES belong in a graduate algebra course.Where I am, you would not have a graduate level course called algebra. Thegraduate courses are much more specialized, and are generally given bypeople doing (or planning to do) research in that area, who are trying toattract research students. Examples of such topics are homological algebra,Kac-Moody Lie Algebras, simple groups of Lie type, reection and Coxetergroups, non-commutative rings, ...Having said that, there has been some discussion in recent years aboutwhether we should be giving a collection of general graduate level courses intopics like algebra, (algebraic) topology, manifolds, ... which would remainmore or less stable from year to year. Particularly in algebra, we would haveenormous difculty in agreeing on a syllabus.Derek Holt. =Groups> Subgroups> Homomorphisms> Normal Subgroups / Isomorphism Theorems> Lagranges Theorem> Alternating & Symmetric Groups> Free Groups> Finitely Generated Abelian Groups> Sylow Theorems> Solvable and Nilpotent Groups> Rings> Factorization in Polynomial Rings> (Maximal) Ideals> Field Extensions> Splitting Fields> Galois GroupsAll undergraduate level topics.For postgraduate level:commutative algebramodules (projective/injective)homological algebrasimple groups (Lie type and sporadics)central simple algebrasLie algebrasgroup representationsalgebraic number theoryHopf algebrasetc.although some of these are borderline undergraduate topics.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) You still havent said what topics in algebra you consider advanced> enough to belong in a graduate algebra course. Id be curious to see> the syllabus for an undergraduate introduction to abstract algebra> course which covered:Groups> Subgroups> Homomorphisms> Normal Subgroups / Isomorphism Theorems> Lagranges Theorem> Alternating & Symmetric Groups> Free Groups> Finitely Generated Abelian Groups> Sylow Theorems> Solvable and Nilpotent Groups> Rings> Factorization in Polynomial Rings> (Maximal) Ideals> Field Extensions> Splitting Fields> Galois GroupsAnd as Ive pointed out, youve avoided at least 3 times the question> of what material DOES belong in a graduate algebra course.I dont have a syllabus, but below is the course catalog description for the undergrad algebra course at Harvard. When I took it, most of the students were sophomores and we used Michael Artins textbook Algebra, which is usually marketed as an undergrad book. We covered everything in your list, plus did some stuff with representation theory.Math 122. Abstract Algebra I: Theory of Groups and Vector SpacesAlgebra 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 general theory of rings and elds.Math 123. Abstract Algebra II: Theory of Rings and FieldsRings, ideals, and modules; unique factorization domains, principal ideal domains and Euclidean domains and factorization of ideals in each; structure theorems for modules; elds, eld extensions. Automorphism groups of elds are studied through the fundamental theorems of Galois theory.The descriptions of the graduate algebra courses are:Course introduces ubiquitous algebraic structures and discusses some of Galois theory; the Brauer theory of central simple algebras; representation theory of nite groups; introduction to algebraic number theory.Prerequisite: Mathematics 123 or equivalent.Continuation of Mathematics 250a. Some basic commutative algebra. Local and global elds. Study of ideal class groups.Lang is often used as a text for 250a, and 250b usually uses Atiyah and MacDonald. = > I dont feel anything, but all of these were covered in the rst>two years of my undergraduate degree. Thats Europe. In America, undergraduates study liberal arts, and> squeeze in only a little of their specialty in the rst two years.In fact, I believe the standard is that an applicant to a graduate programshould have seen the standard topics. You arent expected to *understand*them until you take your qualiers (rst set of exams).In other words, the rst year of US grad school catches us up to EuropeanBA/BS standards.Its the price we pay for not closing anyone out of higher education.Jon Miller It really has nothing to do with what environment one was actually> raised in. What truly matters is how one ts into a given> environment. Certainly it is a *lot* easier if one grew up> in the targeted environment.Consider Pygmalion, the Kennedy family, and others who actually> achieved that much sought after vertical mobility promised by> the American dream.How many people die in America due to economic pressure on its citizens for every American that achieves wealth? Minimum wealth being dened to sustain a yearly cost of $40KUS/yr for thirty years in ones retirement. Approximately $2.7MUSis needed as a base. A game in the U.S. is to achieve capital growth over a lifetime. A $60K salary over 40 years is $2.6M. Less cost of living, taxes etc. Kinda hard to earn the capital needed for retirement. People in the US should be OUTRAGED. The social/economic/political system is skewed for the wealthy to continue their wealth at the expense of the majority in this system. Just like Republicrats, The US can provide an intangible;access of opportunity, not a functional economic system that in fact accomplishes the goal of wealth achievable for its citizenry. We should all weep for joy at American oppotunities supposed opportunities that will provide each citizen with a capital base needed when each citizen desires to live a different chapter in their life upon retirement. Let us cheer for an educated citizenry and their image of wealth.wboer Is there anything good about California? Anyone? Id like to hear>something positive that I can hold onto, living here.>I was going to say the weathers nice. So nice that the entire state>goes up in ames when someone drops a cigarette. So nice that when it does manage to rain, the mountain sides go sliding into the>valleys. So I guess youre right, theres nothing positive about>living in California.Well, it could be worse...>My friends that live in NY say things like, Well, it could be>worse, we could be living in California!oops.It was all foreseen in utter detail, even down to the words.>No matter how bad things are, take heart, it could be worse:you could be in California.... in reference to ...>The best test of predictions is to simply wait and see them>emerge right before your very eyes. [...]>California will go bust by the end of the decade.... and to the times following it in the early 21st century.In reference to the sudden power shift in October>[...] receding back to male-majority right now (California in>2002 [...])>Yes, Virginia. There is a Santa Claus. It is all really that>simple, after all. Humans dont drive history. Demographics>do. Its all just a numbers game; and humans (and their>movements) taking false credit for the inevitable consequences>of the natural historical evolutionary processes engendered>by numbers games.Pun intended.my own reason. I hope that the economy will get better. There are somany places to visit around here; it costs though especially since Icannot stay at cheap motel/hotel for fear of mold. Message-id: anything good about California? Anyone? Id like to hearsomething positive that I can hold onto, living here.> I was going to say the weathers nice. So nice that the entire state> goes up in ames when someone drops a cigarette. So nice that > when it does manage to rain, the mountain sides go sliding into the> valleys. So I guess youre right, theres nothing positive about> living in California.> Well, I live in Illinois, so it was a jealosy thing. I once spent six weeks in> San Diego during October-November and never wanted to go back home to the> rainy, cold weather in Chicago. I have to have the sour grapes attitude that> its a nice place to visit, but you wouldnt want to live there in order to> justify why I continue to put up with humid continental climate. I tell myself> that at least there are no earthquakes or hurricanes. We got tornadoes, but> Ive lived here my entire life and I have never seen a tornado. Contrast that> with the fact that although I have only visited the East coast a dozen times, I> have seen a hurricane.On that San Diego trip, one day it started thundering and someone yelled Hey> everybody, ITS RAINING!. Everybody in the factory ran outside to watch. I> thought havent these people ever seen a thunderstorm?. Apparently not, the> next day, the headlines in the paper were TORNADO FEARED. Wow, they got more> worked up over a potential tornado than we do back home over a real one.If you still feel down about living in California, try visiting Chicago for six> weeks in, say, February-March and youll really appreciate what youve got.I do appreciate the climate in Sacramento as I had to move out ofHouston due to the effects from its humidity & heat combination andPolltuion on my so called Non-Allergic Rhinistis condition. I went toget a dignosis only after things got out of control. My stay inHouston were the most unproductive years in my life from low energy ondaily basis.I know how Chicago is. I visited there (in summer) when I was in goingto grad school in IL (not in Chicago) for 2 1/2 year. I got winterblues and it was depressing for not getting enough sunshine. The beefwas tasty though.I guess I miss Houston, the familiar place after moving out ofIllinois, and also cheap housing for much better quality. I wouldntmind to bear the climate if it were not affecting me health-wise.With CAs poor economy, energy crisis, and expensive phone rate swithphone companies not giving the option for distinctive rings on a phoneline so that people would have to get a second line for fax (but nowthere are fax machines which can detects whether the incoming call isa fax or a phone call), I miss the deals in Houston. Restaurants aregood deals too. I am taking about ne seafood restaurants.But, I am thinking positively about CA, i.e the climate. At least, Idont get sinus headache like I did in Houston in the summer. I havemore energy.some family members. >Message-id: anything good about California? Anyone? Id like to hearsomething positive that I can hold onto, living here.>I was going to say the weathers nice. So nice that the entire state> goes up in ames when someone drops a cigarette. So nice that > when it does manage to rain, the mountain sides go sliding into the> valleys. So I guess youre right, theres nothing positive about> living in California.Well, I live in Illinois, so it was a jealosy thing. I once spent six weeks in>San Diego during October-November and never wanted to go back home to the>rainy, cold weather in Chicago. I have to have the sour grapes attitude that>its a nice place to visit, but you wouldnt want to live there in order to>justify why I continue to put up with humid continental climate. I tell myself>that at least there are no earthquakes or hurricanes. We got tornadoes, but>Ive lived here my entire life and I have never seen a tornado. Contrast that>with the fact that although I have only visited the East coast a dozen times, I>have seen a hurricane.>>On that San Diego trip, one day it started thundering and someone yelled Hey>everybody, ITS RAINING!. Everybody in the factory ran outside to watch. I>thought havent these people ever seen a thunderstorm?. Apparently not, the>next day, the headlines in the paper were TORNADO FEARED. Wow, they got more>worked up over a potential tornado than we do back home over a real one.>>If you still feel down about living in California, try visiting Chicago for six>weeks in, say, February-March and youll really appreciate what youve got.> I do appreciate the climate in Sacramento as I had to move out of> Houston due to the effects from its humidity & heat combination and> Polltuion on my so called Non-Allergic Rhinistis condition. I went to> get a dignosis only after things got out of control. My stay in> Houston were the most unproductive years in my life from low energy on> daily basis.I know how Chicago is. I visited there (in summer) when I was in going> to grad school in IL (not in Chicago) for 2 1/2 year. I got winter> blues and it was depressing for not getting enough sunshine. The beef> was tasty though.I guess I miss Houston, the familiar place after moving out of> Illinois, and also cheap housing for much better quality. I wouldnt> mind to bear the climate if it were not affecting me health-wise.With CAs poor economy, energy crisis, and expensive phone rate swith> phone companies not giving the option for distinctive rings on a phone> line so that people would have to get a second line for fax (but now> there are fax machines which can detects whether the incoming call is> a fax or a phone call), I miss the deals in Houston. Restaurants are> good deals too. I am taking about ne seafood restaurants.But, I am thinking positively about CA, i.e the climate. At least, I> dont get sinus headache like I did in Houston in the summer. I have> more energy.some family members.history was at New Madrid.I replied that nobody believes that that will happen again any time soon.Thats why southern Illinois is called Little Egypt......because they live in denial. Eldon Moritz grava .88 la saucisse et au marteau:Question:> Two coins were ipped and at least one is a head. What are the> chances for two heads?If you understand French, I suggest you go see> http://www.eleves.ens.fr:8080/home/madore/math/proba.html which deals> about all the traps in probability.A problem similar to yours is treated. Ill try to translate it into> English:I visit the Martins Family. I know that the parents have two 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. Conjecturechanges the question; sometime it changes the answer.Consider our question, Two coins were ipped and at least one is atail. What are the chances for two tails?Then consider that heads represents boys, and tails represents girls,or let tails represent boys, and heads represents girls. Then, we havesimilar questions.Then suppose that our question was computer generated. The computerread the coin ip, color coded the outcome and we saw two differentcolored lights. We selected one and the computer generated ourquestion.That is our question, without conjecture.Thanx for the input. I suppose that if youre arguing this question inFrance, there are also people over there who believe as I do. Im inthe minority here, its a small percentage, but its a large number.The people over here who disagree with me, plead ignorance. They dontseem to be able to understand my argument.I dont know if they cant, or wont.Your friend from Texas,Eldon Moritz > The natural 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 its a girl or any> of the child (with the same probability) opens the door and its a> girl, then the rst proposed solution is correct. If, on the other> hand, one says boys never answer to the door and Ive 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 rst solution is correct, but without any further> information, we cant answer to this question.To sum up:> Among families with two children, one of which is a girl, 2/3 have a boy> for other child. But among families with two children, the oldest being> a girl, 1/2 have a boy for second child.hth =Eldon Moritz grava .88 la saucisse et au marteau:> Your problem is similar, but it adds a lot of conjecture. Conjecture> changes the question; sometime it changes the answer.> Consider our question, Two coins were ipped and at least one is a> tail. What are the chances for two tails?Ok, Ill tell you my answer (I dont even recall who was arguing forwhich answer, so the other one, feel free to yell):I guess the answer is 1/2The one we saw has a chance 1/2 to be the forst coin and 1/2 to be thesecond. ThenP(TT|one is a tail) = P(we saw the rst)P(TT|the rst is tail) + P(wesaw the second)P(TT|the second is tail) = 1/2 * 1/2 + 1/2 * 1/2 = 1/2One other way to see it is that there are four possible double-throws:H HT HH TT TAssuming we saw a tail, the rst one must be removed. It remains:T HH TT TBut those three throws have not the same probability to have been seen.P(it was the rst throw|we see a tail) = P(it was the rst throw and wesee a tail)/P(we see a tail) = (1/6)/(4/6) = 1/4The 1/6 comes that there are 6 possible coins to be seen and only onetail in the rst throw (by throw, I mean line, not column). The 4/6comes 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 andwe see a tail)/P(we see a tail) = (2/6)/(4/6) = 1/2Therefore, given that the rst coin we see is a tail, the probabilitythat the two coins are tails is 1/2 .-- Nicolas =Is it correct that for any matrix a Gerschgorin circle disjoint withall the other Gerschgorin circles contains exactly one eigenvalue (orthe same eigenvalue multiple times), and subsequently if the matrix isreal then this eigenvalue must be real?I think this is what Brauers Theorem says, but need conrmation. Is it correct that for any matrix a Gerschgorin circle disjoint with>all the other Gerschgorin circles contains exactly one eigenvalue (or>the same eigenvalue multiple times), and subsequently if the matrix is>real then this eigenvalue must be real?Exactly one eigenvalue, with multiplicity one (i.e. only one time),according to the Gerschgorin Circle Theorem. [ Of course youretalking about the closed discs {x: |x-a_{ii}| <= r_i}, not just the circles ]Non-real eigenvalues of a real matrix come in complex-conjugate pairs, and a disc with centre on the real line that contains one member of the pair would contain both. So yes, if the matrix is realthis eigenvalue must be real.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 >Is it correct that for any matrix a Gerschgorin circle disjoint with>>all the other Gerschgorin circles contains exactly one eigenvalue (or>>the same eigenvalue multiple times), and subsequently if the matrix is>>real then this eigenvalue must be real?Exactly one eigenvalue, with multiplicity one (i.e. only one time),>according to the Gerschgorin Circle Theorem. [ Of course youre>talking 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 thematrix? Let B1...Bn be the disks given by the Gerschgorin Circle Theorem formatrix A, and B~1...B~n be the disks given for the matrix A^T. Sincethe spectrum is the same for A and A^T we know that the spectrum isin:[ Union ( Bi ) ] Intersect [ Union (B~i) ]but can we instead look at the disks given for corresponding diagonalelements (lets call them a_11, a_22, ..., a_nn) and deduce that eachpair of such disks must contain at least one eigenvalue and that theeigenvalue must be the same for both disks so that:Union [ Intersect ( Bi, B~i ) ]contains the spectrum. Effectively we are throwing away the biggerdisk of each pair (Bi, B~i) to obtain a better estimate. I think itholds for all disk pairs (Bi, B~i) that are disjoint from the otherdisks but what if two 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 diagonal>elements (lets call them a_11, a_22, ..., a_nn) and deduce that each>pair of such disks must contain at least one eigenvalue and that the>eigenvalue must be the same for both disks so that:>Union [ Intersect ( Bi, B~i ) ]> contains the spectrum. No. Theres nothing to say that Bi contains any eigenvalue (unless it is disjoint from the other Bjs). Consider e.g. [ -1 -r 0 ][ 1/r 0 1/r ][ 0 -r 1 ]which has eigenvalues 0, i, -i.The Gerschgorin disc about -1, 0 and 1 have radii, |r|, 2/|r| and |r|respectively, so if |r| is small all the eigenvalues are in the disc B2and not in B1 or B3. But for the transpose, the eigenvalues i and -iare in B~1 and B~3 but not in B~2. So 0 is the only eigenvalue in(B1 intersect B~1) union (B2 intersect B~2) union (B3 intersect B~3).Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =While dx/dt at x has only one parameter x,the integral from x1 to x2 has two parameters x1 and x2.Iff x2 was always equal to zero, then derivative and antiderivative would be more similar to each other.Mathematicians might feel this idea somewhat cheeky.However, it has a physical background.Eckard While dx/dt at x has only one parameter x,> the integral from x1 to x2 has two parameters x1 and x2.Iff x2 was always equal to zero, then derivative and antiderivative > would be more similar to each other.Mathematicians might feel this idea somewhat cheeky.> However, it has a physical background.EckardWhat about functions not dened 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 integral from x1 to x2 has two parameters x1 and x2.Iff x2 was always equal to zero, then derivative and antiderivative> would be more similar to each other.Mathematicians might feel this idea somewhat cheeky.> However, it has a physical background.EckardBut 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 all n but this would exclude all of the nice functions.Have a tolerable existence. Eli =Incidentally, 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 integral from x1 to x2 has two parameters x1 and x2.>>Iff x2 was always equal to zero, then derivative and antiderivative>>would be more similar to each other.>>Mathematicians might feel this idea somewhat cheeky.>>However, it has a physical 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 all > n but this would exclude all of the nice functions.Have a tolerable existence. Eli = grava .88 la saucisse et au marteau:> While dx/dt at x has only one parameter x,> the integral from x1 to x2 has two parameters x1 and x2.> Iff x2 was always equal to zero, then derivative and antiderivative > would be more similar to each other.> Mathematicians might feel this idea somewhat cheeky.> However, it has a physical background.Actually, the fundamental theorem says that the former is the oppositeof the latter when the rst bound remains constant and is the pointwhere the function takes the value 0.I hope Ive been clear.-- Nicolas grava .88 la saucisse et au marteau:While dx/dt at x has only one parameter x,> the integral from x1 to x2 has two parameters x1 and x2.Iff x2 was always equal to zero, then derivative and antiderivative > would be more similar to each other.Mathematicians might feel this idea somewhat cheeky.> However, it has a physical background.Actually, the fundamental theorem says that the former is the opposite> of the latter when the rst bound remains constant and is the point> where the function takes the value 0.Shouldnt that be a point instead of the point? > While dx/dt at x has only one parameter x,>> the integral from x1 to x2 has two parameters x1 and x2.> Iff x2 was always equal to zero, then derivative and antiderivative >> would be more similar to each other.> Mathematicians might feel this idea somewhat cheeky.>> However, it has a physical background.>Actually, the fundamental theorem says that the former is the opposite> of the latter when the rst bound remains constant and is the point> where the function takes the value 0.I hope Ive been clear.I do not even understand what fundamental theorem you are referring to. Please get more specic.Antiderivative is the same like integral. Therefore one could expect some similarity.I blame R.8en.8e Descartes for the missing x point of our coordinates. = grava .88 la saucisse et au marteau:> I do not even understand what fundamental theorem you are referring to. > Please get more specic.The theorem Im talking about is:If g: x-> Int(from a to x)f(t)dt, then g(x) = f(x), where a is thevalue such that g(a) = 0.But maybe it doesnt answer to your question.-- Nicolas grava .88 la saucisse et au marteau:I do not even understand what fundamental theorem you are referring to. > Please get more specic.The theorem Im talking about is:If g: x-> Int(from a to x)f(t)dt, then g(x) = f(x), where a is the> value such that g(a) = 0.But maybe it doesnt 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 xor dx/dt at t has only one parameter t.The (in German) so called integration constant (?)is arbitrary. The border t=0 (or x=0) is natural to R+.> grava .88 la saucisse et au marteau:>> I do not even understand what fundamental theorem you are referring to. >> Please get more specic.> The theorem Im talking about is:If g: x-> Int(from a to x)f(t)dt, then g(x) = f(x), where a is the> value such that g(a) = 0.But maybe it doesnt answer to your question.> ----------------------------> let A : T1 spaceshow that derived set of A is closed set-------------------------------if A is nite, it is trivial.but in the innite, i dont know.advice ...please...thank you.In the proof I gave, I actually used the fact that, in T1 spaces, x isa limit point of a set A if, and only if, every neighborhood of xcontains innitely many points of A. Im not sure if this is clear inmy rst post.Artur =Suppose we have a real function f(x) and we mustfind zero-crossing points.So this is equal to say that we have an equation in the form: f(x) = 0and we want tofind its solutions.If lim (x->t) f(x) / g(x) = 0 / 0 De LHopital 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 value.But if f(t)=0 (as hypotesis) then t must be a zero-crossing point, and so asolution of the equation f(x)=0.Now I must dene a particular function g(x) that tends to 0 as x tends tot.I also need to eliminate the term t from inside the limit (and so fromg(x) ) in order to calculate its value.So a function that initially tends to 0 as x->t can be: g(x)= e^x - e^tApplying De LHopital: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 rst limit I cant calculate the value of t because I have it alsoinside 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) = 0I dene 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) = 0then 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 mefind the mistakes?I tried it with Derive 5, but it wont work properly. =Suppose we have a real function f(x) and we must nd zero-crossing points.> So this is equal to say that we have an equation in the form: f(x) = 0and we want tofind its solutions.If lim (x->t) f(x) / g(x) = 0 / 0 De LHopital says:Thats putting it sloppily... you should rather say that thelimits 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 value.Thats *not* what it says.The second equality in your expression holds *only* if bothf(x) and g(x) have limit 0 at t. Not in general.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 dene a particular function g(x) that tends to 0 as x tends to> t.> I also need to eliminate the term t from inside the limit (and so from> g(x) ) in order to calculate its value.> So a function that initially tends to 0 as x->t can be: g(x)= e^x - e^tApplying De LHopital: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 equality 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 value. Thats *not* what it says. The second equality in your expression holds *only* if both> f(x) and g(x) have limit 0 at t. Not in general.>But if f (x) and g (x) are both 0 at t I have to re-apply De LHopital inorder to calculate the limit...Whats necessary is that the initial limit has f(t) and g(t) both = 0.> But if f(t)=0 (as hypotesis) then t must be a zero-crossing point, andso a>solution of the equation f(x)=0.> Now I must dene a particular function g(x) that tends to 0 as x tendsto>t.>I also need to eliminate the term t from inside the limit (and so from>g(x) ) in order to calculate its value.>So a function that initially tends to 0 as x->t can be:> g(x)= e^x - e^t> Applying De LHopital:> 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 equality does not hold, because the denominator (e^x)> is not zero when x=t.But its the result of the iterative apply of the Hopitals Rule, and so theinitialstatement that f(t) = 0 and g(t) = 0 is still valid, 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 value.> Thats *not* what it says.> The second equality in your expression holds *only* if both>f(x) and g(x) have limit 0 at t. Not in general.> But if f (x) and g (x) are both 0 at t I have to re-apply De LHopital in> order to calculate the limit...Correct. But what if they are NOT both zero? Does a reapplicationof the rule work in that case?Look again in your textbook if you think the answer is yes.> Whats necessary is that the initial 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 dene a particular function g(x) that tends to 0 as x tends> to>> t.>> I also need to eliminate the term t from inside the limit (and so from>> g(x) ) in order to calculate its value.>> So a function that initially tends to 0 as x->t can be:> g(x)= e^x - e^t>Applying De LHopital:>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 equality does not hold, because the denominator (e^x)>is not zero when x=t.But its the result of the iterative apply of the Hopitals Rule, and so the> initial> statement that f(t) = 0 and g(t) = 0 is still valid, I presume.Yes, but you cant *reapply* the rule unless f and g are *also*both zero at t. know there are two ways to study a game in normal 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 = football +---+----------+-----------+ C = cinemaThe values (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 Im the player 1,should I choose strategy F since its probability 2/3 is greaterthan 1/3 (the probability of strategy C)?Or this means that if I play this game repeatedly, I should choosestrategy F 2/3 of the time and strategy C 1/3 of the time?What is the practical 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 = football> +---+----------+-----------+ C = cinema>The values (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 Im 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, every 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 practical interpretation of mixed strategies?Although this is a Nash equilibrium, it results in average payoffs of 5/3for both players, lower than the pure-strategy Nash equilibria(1,0),(1,0) and (0,1),(0,1). I dont know why anyone would want touse this Nash equilibrium. Note that although (according to the denition 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 its also true that theres no penalty for such a deviation. So even if you know Player 2 is using her mixedstrategy (1/3, 2/3), you may as well always choose F. And then ifPlayer 2 is rational (perhaps a doubtful assumption in Battle of theSexes), once she gures out you are doing this she will also switch toF as this improves her average payoff.On the other hand, if Player 2 can communicate with you, she will just announce Im choosing C, and you, if you are rational, will go along with her choice.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =ashwin scribbled the following:#include int main(void) { if (0!=1) printf(Its true!n); return 0;}-- /-- Joona Palaste (palaste@cc.helsinki.) ------------- Finland ---------- http://www.helsinki./~palaste --------------------- rules! --------/B-but Angus! Youre a dragon! - Mickey Mouse =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 original waveform. Im looking for a software tool whose input is a given function, andwhose output is a row of sinusoids which sum to the original.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 software 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 original waveform. Thats a very, very loose way of putting it.>Im looking for a software tool whose input is a given function, and>whose output is a row of sinusoids which sum to the original.>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 youre thinking of the Fourier series, not Fourier transform,of x^2 on the interval [-pi, pi]. The Fourier transform of x^2 (as a tempered distribution) on the real line is -sqrt(2 pi) times the second derivative of the Dirac delta distribution.>Is there a software tool which if given (1.) will print (2.) ?In Maple:FourierPartialSum:= proc(f::algebraic, 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 partial sum of the Fourier series for f on the interval # a..b up to the terms in sin and cos of 2 Pi N x/(b-a).local 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.:> FourierPartialSum(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) + ...Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =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 theres a very easy proof.Lee Rudolph >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 theres a very easy proof.Is it easier than: s + s = (s + s)^2 = s^2 + s^2 + s^2 + s^2 = s + s + s + s?-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. 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 theres a very 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, its shorter by three instances of ^2. Thatseasier, right?Lee Rudolph > 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 theres a very easy proof.>>Is it easier than:>> s + s = (s + s)^2>> = s^2 + s^2 + s^2 + s^2>> = s + s + s + sUh, erm, its shorter by three instances of ^2. > Thats easier, right?Thats 2s = (2s)^2 = 4sversus s+1 = (s+1)^2 = 3s+1Or 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, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) =-- Robin Chapman 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.> Alan Partridge, _Bouncing Back_ (14 times) =Sorry for that stupid question, but how is called that function in english?I cantfind this word in my dictionary. Sorry for that stupid question, but how is called that function in english?> I cantfind this word in my dictionary.That isnt English. Its Mathematica.-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling. Sorry for that stupid question, but how is called that function in english?>I cantfind this word in my dictionary.>y = f(x) is read literally as y equals f of x. Or to be moreexpressive you could say y is a function of x given by y equals f ofx.--Lynn =What does the function do? Do you possibly mean the ceiling (oor)function? For example,f(x) = [x] --> f(3.4) = 4, an example of the ceiling function. Or,conversely, the oor function:f(3.4) = 3.Lurch> Sorry for that stupid question, but how is called that function inenglish?> I cantfind this word in my dictionary. =I recently became aware of a book called Probability Theory : TheLogic of Science by E. T. Jaynes. The reviews of this book in Amazonwere intriguing, particularly the one by Michael Hardy. I considermyself a pure mathematician to whom probability theory is a subsetof measure theory, but Ive also worked with statistics and inferencemodels (in bioinformatics).Im curious to know what people think about this book. Does it presenta useful theory of statistical inference? How rigorous is it?Alan I recently became aware of a book called Probability Theory : 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 theory is a subset>of measure theory, but Ive also worked with statistics and inference>models (in bioinformatics).This url will connect you with a thread of discussion on Sci.stat.mathng which was an interesting and lengthy debate in part between Hardyand me of one aspect Bayesian statistics. Based on your comment aboutHardys review--I got to Amazon to read his review for myself as fastas I could. http://home.rochester.rr.com/jbxroads/interests/sci.stat.math /Carlos Rodriguez provided an interesting summary of the thread midway.(quoting) ==Rubin: The people that use entropy or whatever other so calledneutral priors are using unjustied computational copouts.My position: 1) Hurray for Bailey!2) Sure Mike but they should know better.3) I disagree with Rubins position with all the energy in myreproductive 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>Im curious to know what people think about this book. Does it present>a useful theory of statistical inference? How rigorous is it?I will leave the judgement of rigor to mathematicians. (Quis justodietipsos justodes?) Judgments about usefulness are the perogative ofengineers. As an engineer, I have found Jaynes and the works thosebuilding on his work to be enormously valuable. I am appalled at theamount of religious fervor attending the criticism of Bayesianmethods. To me they are logical and extremely useful. If they areawed in rigor, perhaps a Laplace or Fourier to his Heavyside willemerge.John Bailey = [.snip.]>>However, in reals, NO numbers are coprime, as for instance, 2(3/2) >3, so 2 is a factor of 3.> Again wrong, all but 0 are coprime.That is correct, and I was wrong, as in fact, since *all* numbers but>0 are units in reals, itd be the case that all would be considered>trivial factors, so pushing the denition to the limit, 3 is coprime>to 6, though its still a factor of 6.Pushing the denition to the limit?What nonsense is that? Nobody is pushing anything. We were simplyAPPLYING the denition YOU gave, explicitly and precisely as given,to the situation described.Thats what one does with denitions.>Its starting to look like that word coprime is problematic, eh?Hmm... I think I nally understand. When you say problematic, youmean it confused me. So, when you say coprime is problematic, youmean you made a fool of yourself by insisting no less than 4 times(three of them while trying to gloat) that something was correct, whenaccording to the denition you gave it was just plain false. And, given all your confusion about algebraic integers, maybe thatswhy you think theres a problem with the denition? Because youdont get it?No, there is nothing problematic about the word coprime, exceptperhaps that you are using a NONSTANDARD meaning of the word. Butwhether you use the standard meaning, or the meaning you gave it, solong as you STICK TO IT there are no problems. The only problem is youdont know what a denition is, and you dont know how to usethem. Thats whats problematic: your rampant incompetence andignorance. [.snip.] == ==Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidinmagidin@math.berkeley.eduHere one may shill for whatsoever and whosoever one pleases-->including that puke, Lyndon LaRouche--as long as one>res off a few at JSH.>> I dont read everything on sci.math; but my choices are informed in> part by morbid curiosity, so I am surprised to have missed the plugs> for Lyndon LaRouche. /~~~ / | /~~| `=` |/~~ /_/ | `` | | `` |/~~ / | | | | `` | { <| | | | | | ==. . == . == | | | | / . . ` / | . . / . / . / / / | | | |--les ducs dEnron!>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 dont know how yet, ask John Quincy Hutchings. =The task is to choose h(t) that maximizes the integral: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 theintegrand 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 value of e^(-rt) * (p(t) -c(t)).I dont dispute the truth of the claim, I just dont see whats soevident about it - can anyone out there explain the obviousness of theDirac delta as the solution to this =Random thoughts on creating a theory of sets prior to a theory ofpropositions and quantiers:Lets start with the empty set, 0, and logical identity, =, then we candene T, for true, by T =def 0 = 0Lets dene ordered pair a la Kuratowski, then we can deneconjunction by phi & psi =def = (So formulae are sets--but why not? If our logic is to be innitary,with innite conjunctions say, well want formulae to be sets anyway.) Now if-then is dened by phi -> psi =def (phi & psi) <-> phi(<-> is, of course, just another way of writing =; I told you wedwant formulae to be sets.)Throwing caution to the wind, well allow ourselves to have a universalset, so that we can dene the universal quantier by (for all x)phi =def {x | phi} = {x | x = x} F (for false) =def (for all p)p (p a formula without free variables.) ~phi =def phi -> F phi v psi =def (phi -> psi) -> psi (exists x)phi =def ~(for all x)~phi.Now, what theory of sets will yield what logic of propositions andquantiers?-- G.C. =Are you saying that a set, empty or not, is identical with itself?How can something that is identical with itself be distinguished or countedas one itself, if there are no means of distinguishing itself from what itis identical with? And if there is no means to distinguish or count, how dowe present the case 0 = 0 as something being identical with itself?JJ> Random thoughts on creating a theory of sets prior to a theory of> propositions and quantiers: Lets start with the empty set, 0, and logical identity, =, then we can> dene T, for true, by T =def 0 = 0 Lets dene ordered pair a la Kuratowski, then we can dene> conjunction by phi & psi =def = (So formulae are sets--but why not? If our logic is to be innitary,> with innite conjunctions say, well want formulae to be sets anyway.) Now if-then is dened by phi -> psi =def (phi & psi) <-> phi (<-> is, of course, just another way of writing =; I told you wed> want formulae to be sets.)> Throwing caution to the wind, well allow ourselves to have a universal> set, so that we can dene the universal quantier by (for all x)phi =def {x | phi} = {x | x = x} F (for false) =def (for all p)p (p a formula without free variables.) ~phi =def phi -> F> phi v psi =def (phi -> psi) -> psi> (exists x)phi =def ~(for all x)~phi. Now, what theory of sets will yield what logic of propositions and> quantiers? --> G.C. =Could somebody up there help me out by pluggingthe following into Mathematica or equivalent and(I m sure there is a result because the Mathworld Integratorgives me one for the indenite case - but its a bit messy)Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 + 2*Cos[x]]],{x,0,Pi}]Alex Could somebody up there help me out by plugging> the following into Mathematica or equivalent and (I m sure there is a result because the Mathworld Integrator> gives me one for the indenite case - but its a bit messy) Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 + 2*Cos[x]]],{x,0,Pi}]> AlexWhat is D? =just a real constantAlexip escribi.97 en el following into Mathematica or equivalent and> (I m sure there is a result because the Mathworld Integrator>gives me one for the indenite case - but its a bit messy)> Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 + 2*Cos[x]]],{x,0,Pi}]> Alex What is D? Phil Holman grava .88 la saucisse et au marteau:f(p) = q> where f(140) = 15000> and f(140) = -100R = pq> What is dR/dp (p=140)> I get (f*p) + (f+f)*dp>= (-100*140) + (14900*1)>= -14000 + 14900>= +900R = pf(p)dR/dp = f(p) + pf(p)So dR/dp |p =140 = f(140) + 140*f(140)> = 15000 - 14000> = 1000If we apply this practically. p=price, q=quantity sold and R=revenue.So if we increase the price from 140 to 141, we decrease the quantitysold from 15000 to 14900. This will increase the revenue by 900 andnot 1000. I guess the problem here is with a changing derivativebetween a p=140 and p=141. This gave me a real problem and hence myhacked solution. Do you have any other comments about this.Phil Holman > ...>In integers, 2 and 3 are coprime as are 12 and 13, as it simply means>they dont share non-unit factors.>>However, in reals, 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 false.> Hmmm...now thats interesting, every real but 0 is a unit, eh?> Yup. Every element of a eld, except 0, is a unit.> Fascinating perspective.> Isnt it?So you have coprime with one clear meaning in a ring like integers,>but things are different in a eld.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 thats telling.I think so too. It tells us that you have little or no clue, and nodesire to obtain one.Coprime, like divides and factor, are terms related to thedivisibility structure of a ring. They are only as interesting as the divisibility structure of the ringis interesting.In a eld, the divisibility structure is completely uninteresting:every nonzero number divides every number. Since the divisibilitystructure is not interesting, it also makes the notions of coprimeand divides uninteresting.Just like the concept of odd or even: these are notions that areinteresting in the integers. A number is even when it is a multiple of2, and it is odd if it is NOT a multiple of two. The denition of oddand even becomes completely uninteresting in the rational numbers,where everything is a multiple of 2. That does not mean that thenotion of odd or even is broken, or silly. It just means thatit is useless and uninteresting in that context.So, yes, in a eld, coprime loses its relevance; because coprimeis only as relevant as the notion of divisibility is interesting, andthe notion of divisibility is not interesting in a eld.>Part of my point here is that the mathematics youve taken for>granted, with lots of denitions that seem ok, goes off into some>interesting places.The denitions do not seem ok. The denitions ARE. Just because anotion which is important in one context becomes unimportant inanother does not mean that the notion is broken. It means that it isnot interesting in all contexts. Few things are interesting in allcontexts.For instance, the order in the integers: a>Actually, in the real numbers, ANY TWO NONZERO NUMBERS ARE>>COPRIME. Thats because, given any two nonzero real numbers x and y,>>any common divisor of x and y is a unit.> Well, then every real number but 0 is a unit follows from that>> position.> As that is the denition of a eld, it looks about right.So mathematicians take these positions, which not only go against>common sense, they dont make sense in general, pushing denitions.Nobody is pushing denitions. We are USING the denitions. What isit that goes against common sense, anyway? You dened coprime asnot having any non-unit common factors. If all common factors areunits, then they are coprime. 1 is coprime to EVERY integer, is itnot? And 1 DIVIDES every integer, does it not? So, what is wrong withsomething both dividing and being coprime to something else? Nothingper se...Except that you screwed up, and rather than admit that you made amistake, you try to blame it on everybody else.As usual.>So you have this broken word coprime which has no use at all if>youre in the eld of real numbers.It has no use at all in a eld, but that does not mean the word isbroken; it means the concept it denes is not interesting in thatcontext. >> I guess you could say that and it doesnt change things in any>> meaningful way.> But its *fascinating* that Arturo Magidin pushed that position!> Are you contradicting it?Nope. Weasel. Of course you were trying to claim I was wrong. You went togreat lengths to imply it, only to have it blow up in your face. Sonow you try to weasel out of your own errors by claiming that theproblem is NOT that you screwed up (as is evidently the case), butthat the denition is somehow broken.Same thing as with the algebraic integers. You screwed up, andrather than take responsibility, you blamed the denition.> Im 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, hes saying that 3 is not a factor of 6 in reals>> because theyre coprime!Hmmm...thats my statement. It looks off in retrospect, as instead>coprime is broken, so that what it means in integers isnt what it>means in reals.Of course not. Like I TOLD YOU DOZENS OF TIMES: coprime iscontextual. It is something that depends on what ring you are workingin. Of COURSE it doesnt mean the same thing in the integers as itmeans in the reals, 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 its a>unit, or trivial factor, which in one sense is ok, In every sense it is okay.>as every real but 0>is a factor of every other real, but the word coprime is now a>liability.No, it is not a liability. It is simply a concept and notion that isuseless in the context of real numbers.>> And you know what? I think the way mathematicians usually go, hes>> right!!!> And indeed. But apparently you have no idea about the mathematical>> meaning of coprime.For those who wonder, coprimeness is usually dened to ignore>*trivial* factors, where unit factors are, of course, trivial.No, for those who wonder, coprimeness has many denitions; the mostcommon one, in my experience, is that two elements are coprime whenthere is a linear combination that equals 1. The second most common inmy experience is that there is no prime ideal that contains the principalideals generated by the two having no commondivisors other than units. This denition is equivalent for theintegers to the more usual ones, and also for the algebraicintegers. It is not equivalent in other settings. The denition isweaker in general (that is, if two things are coprime in the sense ofthe previous paragraph, they are coprime in the way James uses thework, but the converse does not always hold). In addition, the denition James uses does not work well when goingto larger rings: if two things are coprime in one ring, they neednot be coprime in a larger ring. And if two things are NOT coprimein a ring, they need not be non-coprime in a larger ring.By contrast, the denition in terms of linear combinations doessatisfy one implication: if two things are coprime in R, then they arealso coprime in any larger ring. However, it is possible for twothings to be non-coprime in one ring, and yet be coprime in a largerone.So, for those who wonder: No, never take Jamess word for what is doneusually. He is a self-confessed ignoramus of what is usually done.>So 2 is coprime to 3 in integers, but they both have 1 as a trivial>factor, of course.But notice how the word coprime gets broken when you end up where>*every* factor is trivial!It does not get broken, it just becomes uninteresting. >Then you can say 3 is coprime to 6, in reals.Yes. So what?>Most of you do a quick switch in your heads when youre operating in>the real 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. Whatring you work in depends on your particular needs at the time. Thatswhy you claim to have invented the object ring, after all: to llthe needs you think you have.>Mathematicians, well, they basically do the same thing, but *claim* to>be more precise.We give the denition explicitly. We apply the denitionprecisely. You screw up, and you blame mathematicians for your errors.>> Fun stuff, eh?> Yeah.HELL YEAH!!! Mathematicians have broken or screwed up terms all 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 eld!!!There is no such ad hoc rule. You just pulled it out of a certainorice, which is not the one that your dentist checks.>The math world is spectacularly illogical 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, youblame someone or something else.I can see the advantages in it; certainly simpler. Too bad it bringsyou no closer to reality. == ==Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidinmagidin@math.berkeley.edu =wow. I had noticed that he made a strange statementabout primality (viz co-primality), a nonsequiter. do we have to dene nonsequiter, now,in another endless regress? youve made your points, Magadin;its 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 Ive heard of;one really wonders if he was always the smartest in his class,or just the smartest smart-ass. its like the New York Legislature,supposedly nominated TR for President, just to get rid of him-- which was not a very good idea! >So you have this broken word coprime which has no use at all ifyoure in the eld of real numbers. > No, it is not a liability. It is simply a concept and notion that is> useless in the context of real numbers.--ils duces dEnron!http://tarpley.net/bush8.htmhttp://www.wlym.com/ PDF-SpReps/SPRP13.pdf =[material which has been repeated often snipped] I think I begin to see what JSH is saying with thisconstant-term business. I suspect others have seen itas well, but below is my take on it.> 5. Now P(x) has a factor of 49 asP(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22which 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 nal two. In what> follows its important to understand the word coprime.In integers, 2 and 3 are coprime as are 12 and 13, as it simply means> they dont share non-unit factors.However, in reals, NO numbers are coprime, as for instance, 2(3/2) 3, so 2 is a factor of 3.Thats an important point to consider going forward.6. However, the constant term of P(x)/49 is 22, which is veried by> again setting x=0, which gives P(0)/49 = 22.But for two 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, Im making a choice as to where> the proof is going. Since Ive been talking about algebraic integers,> where 7 is coprime to 22, its natural to go with a choice where 7 is> coprime to 22.)> This is the heart of it. JSH says[1] 5*b3 + 22has constant term 22, and that constant term is coprime to 7.Correct. Here, I believe, is how the thinking goes. If you have apolynomial with integer coefcients, say, F(x) = a*x^3 + b*x^2 + c*x + 22,then that *polynomial* is coprime to 7, because one of the coefcientsitself is coprime to 7. That is true whether the coefcients areintegers or algebraic integers. There is no possible common factorof F(x) with 7. The constant term, F(0), is 22, and that is all you need to know to say that the polynomial is coprime to 7. This statement seems to me to be correct. One says that a polynomialwith integer coefcients is coprime to an integer m if *any one* ots coefcients is coprime to m. Here the constant coefcient happensto be coprime to 7, and that is enough. Of course in [1] above, b3 is not a polynomial. It is a functionof x. It takes values in the algebraic integers. But perhaps the same reasoning that applies for polynomials applies for b3(x). There are however two problems: 1. Even though F(x) *as a polynomial function* is coprime to 7, it may be the case that for particular values 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 POLYNOMIAL !!! To which I would say: Yes, but what you NEED is that, for *specic values* of x, F(x) *as a number, not as a polynomial*, 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 polynomial is coprime to a number w does NOT tell that F(x) is coprime to w for all x. This is a slightly subtle point, and the crux is the phrase as a number, not as a polynomial for *specic values* of x. 2. As noted above, b3(x) is not a polynomial function of x. In general it cannot be written as a nite sum of powers of x. If it were written as an innite power series, in general the coefcients will not be integers or algebraic integers. In general they will be elements of a eld. In general it makes no sense to consider whether the coefcients are divisible by, say, 7. They are elements of a eld. For nonpolynomial functions in general, 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 coefcients are not integers; they are rational numbers, but not integers. So the denition of coprime that one would use for polynomials does not apply. And again, for *specic values of x*, sqrt(x + 1) is not coprime to 7: x = 48, for example. JSH may reply to the effect that I am trying to obfuscate orthat I am lying, etc.. After all, I have said that for a polynomial F(x) with integer or even algebraic integer coefcients, if the constantterm is coprime to 7, then the WHOLE POLYNOMIAL is coprime to 7.Therefore I am agreeing with what he says and that should be theend of it. Then I bring in this confusing side-business abouthaving to consider F(x) for specic values of x, and talking about F(x) AS A NUMBER rather than as a POLYNOMIAL. I giveth,and then, untrustworthy mathematician that I am, I try to takethaway. Not fair, right, JSH ? I am lying and obfuscating, right?Im cheating. Right? No. It IS fair. I am NOT cheating. You DO have to consider F(x) AS A NUMBER, for specic values 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 between JSH and the rest of us this whole time.He is talking about coprimeness of a *function* to 7. We are talkingabout coprimeness of individual numbers to 7. Functions and evaluations of functions are not the same kind of things. In the JSHargument, one needs *evaluations*. Figuring out how JSH is thinking at any given point is notnecessarily much of a reason to celebrate. Nora B.> Given that the constant terms are independent of xs value, it must be> the case that dividing P(x) by 49 divides the two constant terms equal> to 7, by 7.7. But to divide 7 from those constant terms requires dividing> through two of the factors, so(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 + 22from reverse use of the distributive property, which gives a constant> term coprime to 7, as required.For some odd reason Ive 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 usually claim is a> proof, and represents perfection at a level most mathematicians, even> professional mathematicians, dont even attempt.To understand what I mean, you might want to seeWhen is a proof? http://www.maa.org/devlin/devlin_06_03.htmlHeres an excerpt: (right-or-wrong, rule-of-law) denition is that a proof is a> logically 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 typical> mathematician of the truth of a given statement.While valid in an idealistic sense, the right wing denition of a> proof has the problem that, except for trivial examples, it is not> clear that anyone has ever seen such a thing.> What Ive 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.Theyre cheating.> James Harris> http://mathforprot.blogspot.com = [.snip.]>> But for two 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, Im making a choice as to where>> the proof is going. Since Ive been talking about algebraic integers,>> where 7 is coprime to 22, its natural to go with a choice where 7 is>> coprime to 22.)>> This is the heart of it. JSH says[1] 5*b3 + 22has constant term 22, and that constant term is coprime to 7.>Correct. Here, I believe, is how the thinking goes. If you have a>polynomial with integer coefcients, say, F(x) = a*x^3 + b*x^2 + c*x + 22,then that *polynomial* is coprime to 7, because one of the coefcients>itself is coprime to 7. That is true whether the coefcients are>integers or algebraic integers. There is no possible common factor>of F(x) with 7. The constant term, F(0), is 22, and that is all >you need to know to say that the polynomial is coprime to 7. This statement seems to me to be correct. One says that a polynomial>with integer coefcients is coprime to an integer m if *any one* of>its coefcients is coprime to m. Here the constant coefcient happens>to be coprime to 7, and that is enough.The statement is true IF by coprime you mean coprime in Z[x], andyou mean have no common divisors other than units. However, it isnot true if you mean coprime in Z^Z; the example of x^2+x and 2comes to mind.Using Dots proof that the values of a polynomial with algebraicinteger coefcients are always divisible by an integer if and only ifeach coefcient is a multiple of that integer (in the algebraicintegers) gives you that the statement is correct for polynomials inA[x]. > Of course in [1] above, b3 is not a polynomial. It is a function>of x. It takes values in the algebraic integers. But perhaps the >same reasoning that applies for polynomials applies for b3(x).But it does not. [.snip.]> 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 James has not yetrealized that any function f(x) from A to C can be written asf(x) = g(x) + cwhere c is a constant, and g is a function that takes any speciedvalue at any specied point. That is, if a is in A and b is in C,then there always exists a function g(x) from A to C and a constant cin 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) suchthat 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 theright c. Likewise, (5a_2(x)+7)/7 works, so it must be the only onethat works; and (5b_3+22)/1 works, so it must be the only one thatdoes. He does not realize that for ANY complex-valued functionsw_1(x), w_2(x), w_3(x), such that w_1(0)=w_2(0)=7, w_3(0)=1, andw_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 0at 0, plus 1, a function which is 0 at 0, plus 1, and a functionwhich is 0 at 0, plus 22. He found one possibility, he thinks thereis only one possibility. And that could be the mistake. == ==Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidinmagidin@math.berkeley.eduFor crying out loud. You should stop cutting and pasting incorrectstuff. In fact, you should stop cutting and pasting, period. Why notpost a link to your original post, instead, if all you are going to dois repeat it verbatim, ERRORS INCLUDED? [.snip.]>Most posters seem ok with the steps up to the nal two. In what>follows its important to understand the word coprime.In integers, 2 and 3 are coprime as are 12 and 13, as it simply means>they dont share non-unit factors.However, in reals, NO numbers are coprime, as for instance, 2(3/2) 3, so 2 is a factor of 3.Thats an important point to consider going forward.Its such an important point that it is wrong.In the reals, any two nonzero numbers are coprime by your denition(and by the standard denition).Because ANY common divisor of 2 and 3 is different from 0, andtherefore a unit. Hence, any common divisor of 2 and 3 is a unit. So 2and 3 do not share any non-unit factors (in the real numbers).Your statement that 2 and 3 are not coprime in the real numbers isjust plain wrong. In the real numbers, any nonzero number is a unit, and therefore anynonzero number is a (trivial) factor of every number. =Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidinmagidin@math.berkeley.eduFor crying 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 original post, instead, if all you are going to do> is repeat it verbatim, ERRORS INCLUDED? [.snip.]Why dont you off Arturo Magidin?Youre such a pathetic piece of after all, why do you have tokeep tagging along like an unwanted brat?James Harris >> For crying 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 original post, instead, if all you are going to do>> is repeat it verbatim, ERRORS INCLUDED?> [.snip.]> Why dont you **** off Arturo Magidin?> Youre such a pathetic piece of **** after all, why do you have to> keep tagging along like an unwanted brat?Arturo hits a nerve, and as always, James drops even his *pretense*of deceny when he knows hes outclassed...-- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvisefwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock For crying 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 original post, instead, if all you are going to do>is repeat it verbatim, ERRORS INCLUDED?> [.snip.] Why dont you off Arturo Magidin? Youre such a pathetic piece of after all, why do you have to> keep tagging along like an unwanted brat?> James HarrisWhats wrong James, has the truth knocked your nonsense down yet again?You are such a dishonest and rude person who is only in search if glory andwill create lies, insults and totally wrong conclusions to get there ... thetruth is truly pathetic! >> For crying 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 original post, instead, if all you are going to do>> is repeat it verbatim, ERRORS INCLUDED?> [.snip.]Why dont you off Arturo Magidin?Three times I offered to do so if you told me to stop postingreplies. 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 yourstatements? Yes or no?(And, note, that despite all your protestations to the contrary, youhave been unable to produce a SINGLE instance, barring the currentdisagreement, in which I made a statement, you made the oppositestatement, and I was wrong and you were right; despite the fact that Ihave ALWAYS been correct in the past, nonetheless you think that youcan accuse me of lying for years. What a pathetic little twerp you are)>Youre such a pathetic piece of after all, why do you have to>keep tagging along like an unwanted brat?Did you, or did you not, cut and paste incorrect stuff?There is only one who keeps tagging along, going to sci.math,sci.physics, even sci.chem in the vain hope to rally an army to marchbehind him while he imagines storming the strongholds of mathematics.The only brat is you. == ==Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidinmagidin@math.berkeley.edu [.snip.]>I guess you could say that and it doesnt change things in any>meaningful way.>>But its *fascinating* that Arturo Magidin pushed that position!Are you contradicting it? > Well consider then, hes saying that 3 is not a factor of 6 in reals>because theyre coprime!>>And you know what? I think the way mathematicians usually go, hes>right!!!And indeed. But apparently you have no idea about the mathematical>meaning of coprime.He slipped it past you, Dik.Hes saying that 3 is not a factor of 6 in reals because theyrecoprime!I never said that. I never even implied it. James is either utterlyconfused about HIS OWN denition, or lying. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidinmagidin@math.berkeley.edu =sci.physics snipped>[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 all you have done is add and subtract 3 to dene 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 mathematically. 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. Im focusing on constant terms, not trying> to hide a correct argument with meaningless operations like> subtracting and adding 3.No its not, James. 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> theres less room for confusion.Im adding additional steps to handle people like Arturo Magidin who> have made it their business to lie for so long, and not surprisingly,> hes trying tofind fault with the process.Why? Because you are not giving him credit for helping you?Posters like Arturo Magidin waste a lot of other peoples 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 nal two. In what>follows its important to understand the word coprime.>>You need to understand that coprime is CONTEXTUAL, 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 peoples time.> Its called demanding precision.>In integers, 2 and 3 are coprime as are 12 and 13, as it simply means>they dont share non-unit factors.>>However, in reals, 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 false.> Hmmm...now thats interesting, every real but 0 is a unit, eh?Fascinating perspective.It follows quite naturally from the denitions.Its been suggested before, but you might want to take a class or two in abstract algebra before you continue your work. It will help you a great deal.>>Actually, in the real numbers, ANY TWO NONZERO NUMBERS ARE>>COPRIME. Thats because, given any two nonzero real numbers x and y,>>any common divisor of x and y is a unit.> Well, then every real number but 0 is a unit follows from that> position.Denition.I guess you could say that and it doesnt change things in any> meaningful way.But its *fascinating* that Arturo Magidin pushed that position!Why? Because hes actually concerned about accuracy?Well consider then, hes saying that 3 is not a factor of 6 in reals> because theyre coprime!No. This is a result of you not understanding the difference in the denitions. This type of statement is typical of what causes you problems.3 is a factor of 6 in the reals. 6 is a factor of 3 in the reals. 6 and 3 are coprime in the reals. All of these are true statements in the reals.And you know what? I think the way mathematicians usually go, hes> right!!!Please start using standard denitions. It would help.-- Will Twentyman sci.physics snipped>[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 all you have done is add and subtract 3 to dene 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 mathematically. 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. Im focusing on constant terms, not trying>to hide a correct argument with meaningless operations like>subtracting and adding 3.No its not, James. You might want to check your facts before you > accuse others of lying.You obviously are the one who didnt check facts as what I said IS correct.James Harris =You obviously are the one who didnt check facts as what I said IS correct.>For seven years, and more, JSH has been saying this and having later to recant. Why should today be any different? James Harris > sci.physics snipped> [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 all you have done is add and subtract 3 to dene 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 mathematically. 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. Im focusing on constant terms, not trying>>to hide a correct argument with meaningless operations like>>subtracting and adding 3.> No its not, James. You might want to check your facts before you >> accuse others of lying.You obviously are the one who didnt check facts as what I said IS correct.Actually, no. You said three things, and you implied a fourth. Of thethree things you said, one is false. Therefore, what you said is NOTcorrect.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 trying to hide a correctargument with meaningless operations.Statements (b) and (c) are correct. Statement (a) is false. Yourdenition of b_3 MEANS that youve added and subtracted 3 to go from5a_3(x)+7 to 5b_3(x) + 22. Just because you did not say so explicitlydoes thing.You said three things. One of them is false. Therefore, you claim thatwhat you said IS correct is false. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidinmagidin@math.berkeley.eduthe *only* time that I ever made a mistake,was joining the JSH Show. maybe the matforprot.blog is really a sign,that hes been getting paid for this experiment, all-along,per character of his would-be help-meets,including all the silly contextual quoting,which is supposed to be avialable in the thread.alas, Usenet was too small for the proof . >Note that all you have done is add and subtract 3 to dene 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 mathematically. Do you still make>>that claim? Just curious.>>That is a lie from Arturo Magidin as in fact he just subtracted and--ils duces dEnron!http://tarpley.net/bush8.htmhttp://www.wlym.com/ PDF-SpReps/SPRP13.pdf How do I integrate this?f(x) = sin(x)/x> Indenite integral? That is not elementary. It is known as the sineintegral function, written Si(x).-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ =Tal P says...How do I integrate this?f(x) = sin(x)/xThere is a special case that can be found by fourier transforms:Integral from x = -innity to +innity sin(x)/x dxFirst, note that sin(x)/x = integral from k= -1 to +1 of 1/2 cos(kx) dkNow, if we dene F(k) according to F(k)=1/2 (for k between -1 and +1) =0 (otherwise)then we have sin(x)/x = integral from k= -innity to +innity of F(k) cos(kx) dkThats just a Fourier transform of the function F(k). We cantake the inverse transform tofind an integral representationof F(k): F(k) = 1/(2 pi) integral from x= -innity to +innity of sin(x)/x cos(kx) dxIn the particular case k=0, we have (the cos(kx) drops out) F(0) = 1/(2 pi) integral from x= -innity to +innity of sin(x)/x dxSince we dened F so that F(0) = 1/2, we have integral from x= -innity to +innity of sin(x)/x dx = pi--Daryl McCulloughIthaca, NY =Here are two geometry questions from a curious layperson. If theyre unclear,please tell me. Ditto if theyre addressed to the wrong forum.1) Are there surfaces that can be bent to each other (the way a at piece ofpaper is bent to a cylinder) but not to anything at?2) Besides planes and spheres, is there any other surface S such that a piece ofS can be moved around adlibitum while each of its points remains in contact withS? Ive wondered about things like this, off and on, for a long time; but Ihavent done much studying about it, because the going has always seemed toodense. If anybody knows a good place to make an entrance into this body ofknowledge, Id be grateful to know about it. Here are two geometry questions from a curious layperson.I think these are wonderful questions; such a curious layperson iswelcome 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 always clear, but let meassume that S is what would technically be called a smooth2-dimensional submanifold of R^3. For piece of S I will take anopen neighborhood around a point P.If you zoom in very near to P, you will see that this piece of Slooks at, that is, there is a well-dened tangent plane to S atthe point P, which approximates the shape of S very well near P. (The existence of such a plane is the denition of differentiability or smoothness.) For convenience, spin S around in space so that this plane is horizontal.Now zoom out a little bit -- not too far! -- and you will see that theregion around P isnt really at. It will have hills and valleysnear P, looking very much like the graph of a function f denedon the horizontal plane. (Thats the Implicit Function Theorem at work.)If you see several hills and complicated valleys, zoom in a bit more.In a small enough region around P, you can approximate the shapewith the graph of a _quadratic polynomial_ Q instead of f itself.(Thats Taylors theorem -- the right Q can be computed if youknow the partial derivatives of f at P.) With a bit of high schoolgeometry you can rotate coordinates on the horizontal plane so thatQ 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 horizontal plane,so if this plane -- which I have never yet specied! -- is just thetangent plane mentioned earlier, then c = 0.)The constants a and b are pivotal. They measure how markedly thegraph of Q (and so of f, and so of S itself) twists up or downas you travel along the axes. Moreover, these two directions are theextremes: along 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 theshape of S at P. Although they can be computed from variouscoordinates or other formulas, they are clearly geometric in natureand uniquely specify what is known as the _curvature_ of S at P.(Be careful of that word; its also used for other quantities whichsummarize a and b , e.g. reporting only their sum or product.)So your request comes down to this: you want to know what are thesurfaces which have the same curvature (tensor) at every point.You have provided two important examples: on planes, we have a=b=0 ateach point; on spheres we also have a=b (but nonzero, and constant) at each point. It turns out there are no other surfaces of thisparticular type: if at the point P we have a=b, then P is saidto be an _umbilic point_ of S . If every point is an umbilic point(with, a priori, possibly different values of a=a(P) at each point)then it turns out that a is indeed constant, and the surface ispart of a plane or sphere.Youre forgetting another obvious example: on a cylinder we havea=0 at each point P, and a nonzero constant value for b.(Unlike the previous case, here it IS now possible for every pointto be parabolic in this sense but to have b vary with P,giving generalized cylinders. But they dont have the propertyyou asked for.)Any other examples of surfaces with the property you seek would haveat each point a pair of distinct, nonzero values of a and bwhich remain constant on the surface. In particular, their productwould remain constant, which means your surface would be a surface of constant Gaussian curvature. Their sum is also constant,making the surface a constant mean curvature space. These areinteresting categories of surfaces, each containing some beautifulexamples. (Soap bubbles stretching across a wire frame form CMC surfaces.)But it turns out that you cant nd 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 at > piece of paper is bent to a cylinder) but not to anything at?Youre asking for a one-parameter family of local isometries.I dont know if you intended at in the technical sense but youvegot it right: at means Gaussian curvature is zero, i.e. ateach point P we have a=0 in the discussion above.(The Moebius strip is at in this sense too, but maybe isnt whatyou had in mind as being at. You can certainly bend a Moebiusstrip into many congurations.)Heres a nontrivial example, using what are called _minimal surfaces_.(These are surfaces of constant mean curvature equal to zero, that is,at each point we have a = -b.) One such surface is the helicoid, formed by taking a double helixwrapping around a vertical line and then joining points in one helixwith a line segment stretching horizontally to the other helix. Theysell such things as yard ornaments, packaged with all the horizontalline segments lying in a plane, but the helicoid is not really at:the true surface would have some fabric stretching from one horizontalline to the next, with more fabric near the helices, so that whenyou try to lay the thing at there will be rufes of fabric whichare very noticeable far from the central vertical line.The other surface is the catenoid, formed by rotating a catenary,e.g. the graph of y=(1/2)( exp(x)+exp(-x) ), around the horizontal axis.This is really a different surface from the helicoid. But you canmake one from exactly one full turn of the helicoid by forming eachof the helices into a circle; the central vertical line of the helicoidthen also forms a circle (smaller than the other two circles). Thehorizontal parallels have been bent into copies of the catenary, andthe fabric in the rufes smoothes out to give the surface of thecatenoid.You can perform this transformation in a such a way that at everyintermediate moment the fabric is fully stretched out, that is,this is a continuous bending.Cool, huh?dave Here are two geometry questions from a curious layperson. If theyre unclear,> please tell me. Ditto if theyre addressed to the wrong forum. 1) Are there surfaces that can be bent to each other (the way a at piece of> paper is bent to a cylinder) but not to anything at?Im not sure I understand very well... it depends on whatkind of bending would be allowed. For instance, woulda cube and a sphere be bendable into each other?Spheres and eggs?If you think about it, I guess you will have a hard timeexplaining (and even dening) what you mean exactlywith 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 along the axisor perpendicular to it. You cannot rotate it like you*can* on a plane or sphere.> Ive wondered about things like this, off and on, for a long time; but I> havent done much studying about it, because the going has always seemed too> dense. If anybody knows a good place to make an entrance into this body of> knowledge, Id be grateful to know about it.Both questions can be handled by a real tough partof mathematics, called differential geometry.Keyword: radius of curvature.The second question might be partly treated in themore accessible part of analytical geometry of3D-space, but only partly.But Im afraid that both parts cover an area thatis much and much larger than what you have inmind :-(So... good questions, but very difcult to answerappropriately.Dirk Vdm Here are two geometry questions from a curious layperson. If theyre unclear,>> please tell me. Ditto if theyre addressed to the wrong forum.>> 1) Are there surfaces that can be bent to each other (the way a at piece of>> paper is bent to a cylinder) but not to anything at?Im not sure I understand very well... it depends on what>kind of bending would be allowed. 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 dening) what you mean exactly>with bending.I assumed she meant bending without stretching:For example theres a natural map from part of the planeto part of a cylinder that preserves distances exactly(if distances are measured along 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 along the axis>or perpendicular to it. You cannot rotate it like you>*can* on a plane or sphere.> Ive wondered about things like this, off and on, for a long time; but I>> havent done much studying about it, because the going has always seemed too>> dense. If anybody knows a good place to make an entrance into this body of>> knowledge, Id be grateful to know about it.Both questions can be handled by a real tough part>of mathematics, called differential geometry.>Keyword: radius of curvature.>The second question might be partly treated in the>more accessible part of analytical geometry of>3D-space, but only partly.>But Im afraid that both parts cover an area that>is much and much larger than what you have in>mind :-(So... good questions, but very difcult to answer>appropriately.Dirk Vdm>************************David C. Ullrich =How can I prove, that the following inequation is always 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 AustriaMartin Ranzmaier How can I prove, that the following inequation is always 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 xing an arbitrary b > 0? Then you want to show the above holds for all 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 dene f(a) = (b+1)log(b+a) - [log(a) + (1+b)log(1+b)].You want to show f(a) >= 0 for all a > 0. So youve got a calculus problem, namely, showing the minimum value (if it exists) of f over (0,oo) is 0. =How can I prove, that the following inequation is always true.(b+1/a)^(b+1) >= (1/a)*(1+b)^(b+1)You want to be a bit careful; Im not sure if you mean(b + (1/a) )^(b+1)or((b+1)/a)^(b+1)though Im 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 youcan 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 itshould be clear. Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.eduHow can I prove, that the following inequation is always true.>(b+1/a)^(b+1) >= (1/a)*(1+b)^(b+1)You want to be a bit careful; Im not sure if you mean(b + (1/a) )^(b+1)or((b+1)/a)^(b+1)though Im pretty sure you mean the latter, given what happens.The latter is false: The factor (1/a), which fails for a > 1. =Sorry 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 always true.>(b+1/a)^(b+1) >= (1/a)*(1+b)^(b+1) You want to be a bit careful; Im not sure if you mean (b + (1/a) )^(b+1) or ((b+1)/a)^(b+1) though Im 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. == = Its not denial. Im just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> == == Arturo Magidin> magidin@math.berkeley.edu> Sorry for the double meaning in the inequation.I ment (b + (1/a) )^(b+1)I still suggest taking logarithms and playing with those. == ==Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu 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 real roots of a>polynomial. But from this we cannot (always) determine if a polynomial>of degree n has n real roots.>Now we can apply Sturms Theorem to determine the exact number of>distinct real roots of a polynomial on an interval(a,b). However we>must limit ourselves to an interval here.You can let the end points of the interval approach plus orminus innity. The signs are constant far out and easilydetermined. Most textbooks give examples of this.Also, it is easy to get an upper bound on the magnitude ofthe roots of an equation. The sum of the magnitudes of theother coefcients, assuming the leading coefcient is 1,is more than enough.>So is there any specic condition whereby given a polynomial of>degree n we can always state that, given this specic condition>holds, all its roots are real?>Maybe what i am asking is rather silly/uneducated and i am missing>some basic points here, if so please tell :)>>hi,>>could anyone tell me what conditions are neccessary in order for a>>polynomial to have real roots only?>> Look for Sturm sequences in an old Theory of equations>> book, or elsewhere. They tell you exactly how many>> real roots a real polynomial has.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University What are the conditions that guarantee, given a polynomial of degree>n, that the polynomial has n real zeros?For example given polynomial of degree 2, > ax^2 + bx + c, >The condition that, > b^2 - 4ac > 0>guarantees that this polynomial shall have 2 real zeros.Does there exist such a condition that holds for polynomials of>arbitary degree?There is a similar set of conditions which guarantee that the cubic ax^3 + bx^2 + cx + d has three real roots, expressed asconditions on signs of certain polynomials in a,b,c,d. There is a similar set of conditions which guarantee that a polynomialof degree 100 has 100 real roots. It is expressed in terms of the101 coefcients. And so on.Now, what exactly are you expecting for a condition that holds forpolynomials of arbitrary degree? Do you expect to see a formula withinnitely many variables in it, which can be used equally well forquadratics (which have d=e=f=... = 0) and cubics (which have e=f=...=0)and all other degrees? How exactly would we write down such a thing?Once you x the degree n, the family of all polynomials forms avector space of dimension n+1. The polynomials with n distinctreal roots form an open subset of this space. The boundary of thisregion is contained in the variety which picks out the polynomialswith multiple roots; this variety is dened by the vanishing of thediscriminant, which is a polynomial in the n+1 coefcients. For polynomials of high degree, though, the open subset dened bythe condition discriminant > 0 includes several components, andnot just the region where all the roots are real. So you need multiple conditions on the coefcients to describe the region youare interested in.Apart from all this, I should point out that you keep requestingconditions which guarantee the existence of n real roots, andnot conditions which exactly describe the existence of n realroots. Is that what you really want? If so, I can give you conditionslike this: The cubic ax^3 + bx^2 + cx + d has three real rootsif the point (a,b,c,d) is within [some distance] of (1,6,11,6).Obviously other cubics also have three real roots too!dave What are the conditions that guarantee, given a polynomial of degree>n, that the polynomial has n real zeros?>For example given polynomial of degree 2, > ax^2 + bx + c, >The condition that, > b^2 - 4ac > 0>guarantees that this polynomial shall have 2 real zeros.>Does there exist such a condition that holds for polynomials of>arbitary degree?There is a similar set of conditions which guarantee that the > cubic ax^3 + bx^2 + cx + d has three real roots, expressed as> conditions on signs of certain polynomials in a,b,c,d. There is a similar set of conditions which guarantee that a polynomial> of degree 100 has 100 real roots. It is expressed in terms of the> 101 coefcients. And so on.Now, what exactly are you expecting for a condition that holds for> polynomials of arbitrary degree? Do you expect to see a formula with> innitely many variables in it, which can be used equally well for> quadratics (which have d=e=f=... = 0) and cubics (which have e=f=...=0)> and all other degrees? How exactly would we write down such a thing?Once you x the degree n, the family of all polynomials forms a> vector space of dimension n+1. The polynomials with n distinct> real roots form an open subset of this space. The boundary of this> region is contained in the variety which picks out the polynomials> with multiple roots; this variety is dened by the vanishing of the> discriminant, which is a polynomial in the n+1 coefcients. > For polynomials of high degree, though, the open subset dened by> the condition discriminant > 0 includes several components, and> not just the region where all the roots are real. So you need > multiple conditions on the coefcients to describe the region you> are interested in.Apart from all this, I should point out that you keep requesting> conditions which guarantee the existence of n real roots, and> not conditions which exactly describe the existence of n real> roots. Is that what you really want? If so, I can give you conditions> like this: The cubic ax^3 + bx^2 + cx + d has three real roots> if the point (a,b,c,d) is within [some distance] of (1,6,11,6).> Obviously other cubics also have three real roots too!daveDamn Dave you sound brainy!I hear what your saying about being able express for every polynomialsome condition in terms of its coefcients that will guarantee youhave real 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 iwas just curious as to whether there was some general condition thatwould hold, analagous to Descartes Sign method, i.e. something simplethat even i could understand :)...But i suppose using the Sturmtechnique to show n real zeros exist or the Routh-Hurwitz Criterion isjust 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 itbecomes clear that guarantee is not what i was looking for andexactly describe sounds more like what i am. But i think i did say iwas looking for a general condition that must hold for polynomials ofarbitary degree. So your example just holds for this one specicexample of a polynomial of one specic degree(yes i know you couldjustfind a similar point for another polynomial and specify [somedistance]), or maybe i took you up wrong and your example was to showwhere i was way.pat >>Now we can apply Sturms Theorem to determine the exact number of>>distinct real roots of a polynomial on an interval(a,b). However we>>must limit ourselves to an interval here.> Yes, but one can effctivelyfind a number M such that all zeroes>> of the polynomial satisfy |z| < M. Apply Sturm to (-M,M).Yes obviously we can set our range to (-innity, +innity).> But that is just a method forfinding out the number of real zeros.What are the conditions that guarantee, given a polynomial of degree> n, that the polynomial has n real zeros?For example given polynomial of degree 2, > ax^2 + bx + c, > The condition that, > b^2 - 4ac > 0> guarantees that this polynomial shall have 2 real zeros.Does there exist such a condition that holds for polynomials of> arbitary degree?For a given polynomial you can build a particular matrix from itscoefcients, such that you can read off the information about thenumbers of (real) zeroes from the rank and signature.I have forgotten the details, it should be in any book on real algebra.Marc Now we can apply Sturms Theorem to determine the exact number of>distinct real roots of a polynomial on an interval(a,b). However we>must limit ourselves to an interval here.>> Yes, but one can effctively nd a number M such that all zeroes> of the polynomial satisfy |z| < M. Apply Sturm to (-M,M).>>Yes obviously we can set our range to (-innity, +innity).>But that is just a method forfinding out the number of real zeros.>>What are the conditions that guarantee, given a polynomial of degree>n, that the polynomial has n real zeros?>>For example given polynomial of degree 2, > ax^2 + bx + c, >The condition that, > b^2 - 4ac > 0>guarantees that this polynomial shall have 2 real zeros.>>Does there exist such a condition that holds for polynomials of>arbitary degree?For a given polynomial you can build a particular matrix from its> coefcients, such that you can read off the information about the> numbers of (real) zeroes from the rank and signature.I have forgotten the details, it should be in any book on real algebra.MarcIs this not the Routh-Hurwitz Criterion? =I would like to point out that if you have any odd-degree polynomial,you can be gurenteed that there will be at least one real zero. Thisis because the structure of x^3, x^5, x^7... will naturally cause itto cross the X-axis. I know this only answers that you will be certainof one real zero some of the time, but I hope this helps!Anthony> hi,> could anyone tell me what conditions are neccessary in order for a> polynomial to have =I would like to point out that if you have any odd-degree polynomial,> you can be gurenteed that there will be at least one real zero. This> is because the structure of x^3, x^5, x^7... will naturally cause it> to cross the X-axis. I know this only answers that you will be certain> of one real zero some of the time, but I hope this helps!Anthony>hi,>could anyone tell me what conditions are neccessary in order for a>polynomial to have nice piece of info, cheers! =While there is no simple rule to determine whether a polynomial of arbitrarydegree has only real zeros, from a stability perspective, it is very unlikely.In addition, the coefcients must be big.In particular, I believe that the following is true for polynomials withpositive coefcients :if the polynomial has the formp(z) = sum_{k=0}^n a_k z^k, where all a_k ge 0 and a_n=a_0=1, then it cannothave all real zeros if a_k < {n choose k} for any k--irascible since 1957 While there is no simple rule to determine whether a polynomial of arbitrary>degree has only real zeros, from a stability perspective, it is very unlikely.What does this mean, unlikely? There is the family of all polynomials(of a given degree, perhaps?) and within it the family of all polynomialsall of whose roots are real. Do you have in mind a nite measure on therst space in which the measure of the subspace is zero? I am notfamiliar with such a result.Looking at the family of quadratics ax^2 + bx + c, for example, I seea 3-dimensional vector space with an unbounded open subset of quadratics having two real roots. We could, for example, scale allpolynomials down to the unit sphere in R^3 (i.e., assume a^2+b^2+c^2=1),in which case the ones with two real roots ( b^2-4ac > 0) lie inthe portion of the sphere lying between the two sheets of the hyperbolic surface a^2 + c^2 + 4ac = 1. This certainly includes halfthe sphere (the whole region where a and c have opposite signs)and a little more. In this way I compute that the quadratics with two real roots constitute about 51.234% of all quadratics, using thismeasure.Thats not unlikely, so I guess the claim is that this fractiondecreases to zero as the degree rises. Is that true? Proof? Reference?davehe reminded me of a result of Polya which notes that if f has all roots real, so does t0 f + t1 f for every real t0 and t1.(That is, our all-roots-real region includes whole planes passingthrough each of its points.) So this region isnt all that small.(Proof of Polyas result: apply Rolles theorem to exp(t x) f(x).) >For example given polynomial of degree 2, > ax^2 + bx + c, >The condition that, > b^2 - 4ac > 0>guarantees that this polynomial shall have 2 real zeros.>>Does there exist such a condition that holds for polynomials of>arbitary degree?noSo basically for some low order polynomials we have conditions thatallow us state that a polynomial satisfying this condition has onlyreal roots.But in general, there is no such condition that guarantees polynomialsof arbitary size have only real roots, and the only way offinding outif a polynomial of degree n has n real roots (without actually solvingit) is to apply some technique like a Sturm function to it, and if ittells us the polynomial has n real roots, then, well we know it has nreal roots :)Right? =bcc new advanced physics listPaul, I suspect the key CLASSICAL argument against a LOCAL vacuum gravity stress-energy tensor different from Ruv itself in Einsteins vacuum eld equationRuv = 0goes something like this1. If ordinary matter, radiation and near EM elds (all in Tuv) are locally present then Einsteins CLASSICAL local eld equationGuv = Ruv - (1/2)Rguv = - (8piG/c^4)TuvCan be written astuv(vac) + Tuv = 0Wheretuv(vac) = (c^4/8piG)Guv = (String Tension)(Marble Curvature)i.e. in Einsteins colorful termstuv(Marble) + Tuv(Wood) = 0End of story, simple solution.2. Another aspect of the problem is to add new nonlinear terms liketuv(vac) =Lp^2 (c^4/8piG)RuwlsRv^w^l^sto the eld equation. I did not check BTW to see if that particular expression is a symmetric tensor. I am just trying to show the general idea.3. Quantum aspect of the problem is that random zero point vacuum uctuations of all physical quantum elds contribute to the cosmological term /zpfguvthat must be added to Einsteins eld equations to getGuv + /zpfguv = - (String Tension)^-1TuvNote that SakharovsMetric Elasticity = (String Tension)^-1 = Space-Time StiffnessThe tighter the string tension the less elastic is the geometry./zpf = exotic vacuum unied Dark Energy/Matter LOCAL eld, which in large scale FRW regime is Einsteins cosmological constant but on smaller scale is a local inhomogeneous dynamic spin 0 eld that stabilizes the spatially-extended hidden variable lepto-quarks against their explosive self-electric charge. The lepto-quarks obey a micro-geon quasi Kerr-Newmann metric with additional topological handles for the SU(2)xSU(3) charge generators in the extra space dimensions (i.e. Calabi-Yau spaces of M theory in engineers language via black hole - membrane dualities). The lepto-quarks shrink from 10^-13 cm at low energy to 10^-18 cm for high energy large magnication 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 centrifugal & Coriolis inertial forces in equilibrium. All this is IT in the Bohm Pilot BIT Wave theory.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 subsidiary classical marching orders from geometry.The geometry is from the macro-quantum coherence, i.e.,guv = (Minkowski)uv + (Metric Elasticity)(Modulation of Goldstone Vacuum Superuid Phase)uv/zpf = 0 is classical 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 galactic spheres) oat in the Hubble ow like w ~ 0 for us distant observers.Lp*^2 = G*h/c^3Universal Regge slope of hadronic resonances on micro-scale isG*/hc^5 inJ = (G*/hc^5)E^2 + Jo = n/2Blackett Effect seen in many rotating astrophysical objects with anomalous magnetic dipole is reduced toG^1/2m = e(papers by Saul-Paul Sirag)at large scale where G* ---> G.Now Matt Vissers review paper has impractical perturbative quantum eld theory formulae for both large-scale G(Newton) and cosmological constant /.However my model is better because it is intrinsically non-perturbative like the BCS theory is.I have NON-PERTURBATIVEVisser does not have that non-perturbative More is different (PW Anderson) Vacuum Coherence whose generalized phase rigidity is precisely String Tension (in Blackhole-Membrane Duality) = Sakharovs (Metric Elasticity)^-1.I also explain how LOCAL MACRO PHYSICS emerges from quantum eld theory whose entangled Fock space states are nonlocal. I also explain why post-inationary early universe has low entropy from collapse of vacuum phase space volume hence how the cosmic irreversibility Arrow of Time is pointed. I also give a micro-QED dynamical instability for the large scale FRW chaotic ination eld of Linde et-al, which previously was only a useful fudge factor with no fundamental dynamics (according to PaulThe MTW argument against Yilmaz is simply that one cannot make a localclassical vacuum pure gravity stress-energy density tensor from thetorsion-free non tensor connection without violating the equivalenceprinciple.But we have to be very careful about what we mean by the equivalence principle.There is the so-called EEP; there is Einsteins original equivalence principle; whichcan be viewed as an interpretation of the EEP; and there is the alternative interpretationof the EEP in terms of *inertial compensation* which rejects Einsteins fundamentalidentication of gravitational and inertial elds.What you need to do is spell out carefully all of these different versions in formal terms and in clearoperational terms to see if there is really any scientic difference of any importance. I do not understandwhat you mean by inertial compensation above. I think the gravity force, i.e. the ctitious or inertial g-force i.e. the metrictorsion-free connection for parallel transport in the original 1915 theory is LOCALLY equivalent to agravity force. Whether or not that gravity force is real depends on whether or not there is a local tidal curvature differentialin the g-force as described by the geodesic deviation equation between two neighboring timelike geodesic LIF free oat observers.This curvature tensor tidal differential force cannot be eliminated locally.All I mean effectively by EEP is the set of following formal statements.1. There exist local tetrads ea^u(P) at point P such thatnab = ea^u(P)eb^v(P)guv(P)2. The local symmetric torsion-free connection from rst derivatives of the metric guv(P) can be set to zero locally at P.This denes the distinction between LIFs on timelike geodesics and LNIFs on timelike non-geodesics both intersecting at same P.Note timelike non-geodesics could not exist without electrical 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 assuggested by Sakharovs metric elasticity.Part of the confusion is that curvature = eld strength in the modern ber bundle picture of the spin 1 gauge forces.The connection for parallel transport in the internal dimensions beyond space-time (or extra space dimensions of Kaluza-Klein et-al) is theYang-Mills potential like the EM 4-vector potential Au(P) in the U(1) EM simplest case.On the other hand it is the metric, specically goo(P) ~ (1 + 2U(Newton)/c^2) in weak eld slow motion static spherically symmetric vacuum case outside Mfrom source mass Mleading to notion of gravity force as connection from Newtons 2nd law F = -m GradU(Newton)F is locally eliminitable on the LIF timelike geodesics in curved space-time.Tidal inhomogeneous differentials F(P+dP) - F(P) are not.EEP, as I understand it, merely requires that the net gravitational-inertial forcestidal effects canalways be locally neglected.This last statement needs clarication. It is strictly not true. It is only a weak eld approximationthat the tidal effects are of order (L/r)^2 where r is the local scale of radius of curvature. MTW makethis approximate nature of the remark very clear when I read them. You have misinterpreted themas 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 dont know, it is of no real consequenceexcept in the history of the evolution of the idea.This can be consistently interpreted in at least two different ways: (1) as the consequenceof a (local) *fundamental physical identity* of gravitational and inertial elds (Einstein, Pauli);or (2) as merely the result of inertial compensation of the separately existing gravitationaleld (Eddington et al.) in conjunction with the natural attenuation of tidal effects withinarbitrarily small neighborhoods of spacetime.Either way , the theory remains generally covariant, and a weak correspondencerelationship with SR holds -- locally measurable effects that locally discriminate betweenthe predictions of SR (at spacetime) from those of GR (curved spacetime) can beregarded as practically negligible in a sufciently small neighborhood.Yes, in the weak eld limit, but not at a black hole, which Puthoff wrongly claims does not exist.I would argue that a weak correspondence relation is all that one can reasonablydemand of the re-interpreted theory. There is in reality no at spacetime *anywhere*.Fine, thats how I read MTW anyway. Also in my theory, globally at space-time is intrinsicallydynamically unstable from quantum electrodynamics.Also, it is important to understand that none of this depends on the empirical correctness ofYilmazs particular theory vis a vis GR.Ifind Yilmazs idea completely obscure intuitively. For example, the alternative approach is taken in anumber of bimetric theories that agree empirically with GR. These employ a backgroundmetric that accounts for inertial effects in conjunction with a separate (though commensurable)gravitational metric that accounts for physical gravitation.Very ugly. Less with more.In the alternative interpretation, I believe the way is open to consistently dening a truegravitational eld stress-energy density, which is fully localizable, and an inertially-compensated*net effective* stress-energy density which corresponds to the force eld that is actuallymeasured in an accelerated frame of reference.Show me the math otherwise the words have no meaning to me.This means that Machs principle is effectively abandoned -- along with his rather loony brandof strict empiricism (remember, Mach atly rejected the atomic theory).Machs idea reenters in the Holographic World whereLp* = Lp^2/3(c/H)^1/3 = 1 fermi now.But this may have a fatal aw since H is cosmic time dependent in the FRW regime.The connection for parallel transport locally vanishes inLIFs on timelike geodesics.Of course it does. But we can (at least locally) distinguish between the inertial andthe gravitational *components* of the canonical connection eld (distinct contributionsto the net metric gradients). That comes right out of the differential geometry (also, see e.g.Weinbergs Gravitation and Cosmology).I think I said that above.The changes observed in net gravitational-inertial forces under acceleration of ones frame ofreference are then viewed not *merely* as an abstract mathematical property of the laws of GRunder certain class of spacetime coordinate transformations, but also as a *real physical effect*arising from the interaction of ponderable matter with the physical vacuum (AKA the luminiferousaether).The Urstoff!Non-material, yet physical...As I understand it, this is precisely the way the electromagnetic eld is treated in GR whenobserved in accelerated frames -- as inertially compensated. There is in canonical GR noquestion of deriving EM forces from a geometrodynamic connection eld, and so we are left(in standard GR) with inertial forces derived from an *inertial* connection eld that can physicallycompensate the EM forces in the EM analog of free-fall.Of course, we then no longer have general relativity in Einsteins sense, since wedo not have strict gravitational-inertial equivalence (gravitational and inertial eldsconsidered fundamentally indistinguishable).Einsteins point may then be taken as this: the non-material aether is so inextricably bound upwith empty space, that the physical effects that result from interaction of ponderable matterwith the vacuum might as well be viewed as arising from a certain kind of geometry. We thenPOV, we can alwaysdecide as an alternative approach to investigate the properties of empty space as propertiesof a physical entity -- the vacuum -- which is to be distinguished from a void. The voidof Einstein with its Riemannian geometry of spacetime manifolds then appears merely as theshadow cast by a deeper reality -- the physics of the *physical vacuum* (ghost of the departedaether).However, we do still have general covariance of the laws. The strength of the gravitational eld-- and, I would argue, its associated gravitational eld stress-energy density -- is then no longerdependent on ones frame of reference, and the true gravitational stress-energy should itself thenbe represented by a tensor quantity (Diff(4) world-tensor).What *is* dependent on ones frame of reference in this alternative interpretation is the *inertialcontribution* to the net forces experienced by massive bodies. Thus the inertial stress-energywill then obviously NOT be represented by a tensor.So we are talking about a (local) decomposition of the Christoffel connection into its inertial(acceleration-dependent) and gravitational (acceleration-independent) parts.In this view, for a single gravitating mass, the true gravitational eld is the force eld that isobserved in a frame at rest with respect to the source.In the Einstein approach, in contrast, *all* frames of reference are on an equal physical footing(general relativity) and there is no true gravitational force eld -- any more than there is in SR atrue inertial frame in which clocks and rods assume their actual properties, or a true electricor magnetic eld considered separately.If the connection is not zero at a pointthen one is in a LNIF at that point feeling weight or g-force on atimelike non-geodesic world line.Of course. None of this is disturbed by the shift of interpretation, nor by the (local)decomposition of the connection eld into its inertial and gravitational parts.We simply make a fundamental physical distinction between inertial and gravitational forces,while fully recognizing the close mathematical analogy between them that was pointed out byEinstein (both are derivable from metric-tensor elds) . In the alternative paradigm, we look forthe ground of this analogy at a deeper physical level, rather than basing it on some mysticGalilean-Platonistic faith in the reality of the forms presented to us at rst glance by tensoranalysis and differential geometry.In the alternative (neo-Lorentzian) paradigm, gravitation is a physical effect that dependson the existence of gravitating matter and its time-dependent distribution in space. Inertialforces, on the other hand, are separate and distinct physical effects that do not depend onthe existence of material gravitational sources but only on the accelerative interaction ofponderable matter with the physical vacuum (and not, *contra* Mach, with other materialobjects)-- as I would assume any workable eld theory of quantum gravity would predict.While the curvature tensor tidalforce between two neighboring time-like LIF geodesic observers issmall in the weak eld limit it is not locally zero in general andcannot be eliminated exactly.But there is the awkward question of the multipole effects experienced Riemannian gravitational elds -- which I understand do not scale down with spacetime volume asdo tidal effects.And of course, there is also the closely related absolute character of spacetime curvature, bywhich I mean Riemannian curvature. Its either there or it isnt. Einstein was only able to savehis general relativity thesis by pretending that Riemann curvature has no direct empiricalmeaning -- at least locally -- thus xing all attention on the connection eld as physically real.That now strikes me as a classic *ad hoc* stratagem in what Imre Lakatos might have called adegenerating problemshift.These kinds of questions were thoroughly thrashed out between Einstein and von Laue in the early1920s. I believe later Einstein quietly came to accept the fact that his concept of general relativity hadfailed.Which raises fundamental questions about the true character and status of special relativity, sinceEinsteins theory of relativity program required both special *and* general relativity to succeed inorder to be fully vindicated.One could try to make a local classicalvacuum stress-energy density tensor from contracted quadratic forms ofthe 4th rank curvature tensor, but then one has a eld equation likeRuv - (1/2)Rguv + /zpfguv + Quadratic Functional of Ruvwl =8pi(G/c^4)Tuv(matter)The problem then is to conserve the local currents.Wouldnt it be easier to look rst at the denitions of gravitational and inertial stress-energy densitiesfor at elds?What is the stress-energy density of the pure inertial eld (no sources)? What are its mathematicalproperties?What is the acceleration-independent gravitational eld stress-energy density in the far eld (asymptoticallyat spacetime)? What are its mathematical properties?Z.I will get back to you on last half later. said:>>Just as the >>answer to a mathematical question is either right or right, and a>>proof is either valid or invalid, >Not where I went to school. A perfectly valid 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 isdone; rst get the proof, then MAYBE look for elegance. Also,it is quite common for elegant proofs to hide the concepts;using Liouvilles Theorem to prove the Fundamental Theorem ofAlgebra is such.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University = Herman went to school. A perfectly valid prrof would be markeddown if it was not elegant; quite properly, IMHO. I see no reason for this. This is not the way mathematics is> done; rst get the proof, then MAYBE look for elegance.Taking away the elegance from ANY proof, transforms the proof into atedious, boring and mechanical procedure, which should interest no truemathematician.> Also,> it is quite common for elegant proofs to hide the concepts;Id rather have an elegant proof which hides some of the concepts yet givesthe overall idea, than a mechanistic and non-interesting proofwhich bogglesthe mind with inane details which are of no interest to the reader.> using Liouvilles Theorem to prove the Fundamental Theorem of> Algebra 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--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------ ------------------------Eventually, _everything_ is understandable > Well, thats already been proved wrong. > Since if youfind a chemist that knows anything > about science, youll be a rst.>Are you saying that chemists are not scientists, therefore brain >chemistry does not inuence 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 didnt have have anything interesting to say about science. Its just the same old Gaussian test for low IQ. And well ignore the reference the term classic since a mathetoid wouldnt know the difference between classic and modern if Cantor rose from the dead and bit him in the ass. And thirdly given that the Turing test wasnt invented by Turing, but rather second rate Van Neumann-wannabee mathemaphilosophers, its irrelevent. - Randy =[added sci.edu, comp.edu]>> I dont get it.Im a perfect 0 at math.> Some people have no problems at all wrong question -- youre 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 youre much better>newsgroup and just dont know it, or maybe youre a>high-expectation kind of person, and then anything>below Newton, or Gauss, or Fouriers brains means>too dumb for math in your mind? :-))Ive come across various students who viewed that they were missing themathematics gene (or programming gene, or whatever the particularsubject happened to be). In those cases it was uniformly the case thattheir difculty was emotional/attitudinal, rather than cognitive. As youmention, one unhelpful attitude is perfectionism, especially in hard-edgedsubjects where some answers are clearly objectively *wrong* and thus thestudent has no wiggle room to avoid the conclusion that they made anerror.>Anyway, this, plus many of the things that have>been already said (mainly about math being genuinely>hard -- the more sophisticated level of math, the>harder, of course)Several of the missing gene students had the unhelpful attitude thatthey expected maths to be easy, since they had found their schooling easyso far.ISTM that cognitive issues do kick in when dealing with high levels ofabstraction where there are no readily-accessible concrete models. Forexample, my brain hit the wall trying to visualise non-Hausdorff spaces,and my painful memory of the rest of that topology course is of generallymindless memorizing and proof cranking.>HTH,Carlos-- ---------------------------| 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 dont know how to quantify it, but it seems that a largesubset of the people who get stuck in mathematics at somepoint actually missed some key point earlier in their studyof mathematics without realizing it. The way some of themget unstuck is by going back to the point that they didntreally understand.One seemingly common version of this is the good studentwho is able for awhile to compensate for the weakness oftheir understanding of some group of concepts in math bymemorizing more and learning more techniques. Ironicallya good ability to deal with mathematics on the levelof meaningless manipulations can betray these students later.They may be trying their best to succeed, as they imaginethey should be, but it gets harder and harder. Unfortunatelythe way we teach students too often conduces to this strategyon their part.Again, backing up to what they didnt quite understand seemsto be one of the best ways to deal with it. Of course, thiscan be hard when youre on a xed schedule and seem not tohave the time to go back and learn things properly, but evenso, tracking down a key concept and nally getting it cansave a lot of wasted effort in trying to cope without quiteunderstanding.This raises the issue of pacing. I once read a claim which Ithought was interesting and somewhat plausible, that the bestmath students in school have a well-developed ability toadjust their pace in accord with how fast they need to go.This means rst being aware when you actually are understandingand when you arent, and second being able to slow way downwhen you need to. Smart students are often accustomed to beingable to read quickly and compose essays pages at a time, but inmathematics actually digesting a few brief statements sometimestakes hours and hours. Of course there are parts that good mathstudents can then breeze through rapidly too, especially oncethe concepts are clear.Best wishes to struggling students. Its important to exercisepatience, especially not panicking when it seems not to begoing well. Take a few slow deep breaths. Get other studentsand have a group study after working on the problems yourselves.Ask your teachers questions in and out of class.Keith Ramsay =I am writing a computer algorithm that calculates the APR forxed-rate and adjustable-rate mortgages (ARM).The xed-rate APR algorithm is working, and it is uses the actuarialmethod that is dened in Title 12 Truth in Lending (Regulation ZArticle J, section 226.36, pp. 43-46):http://www.federalreserve.gov/regulations/title12/ sec226/12cfr226_01.htmDoes anyone know or know where tofind the procedure/formula forcalculating the APR of an ARM?While I cannotfind a detailed description for the ARM APR procedure,I have pieced some of it together, but Im not sure how accurate itis...An example 3/1 ARM:Loan Amt: $129,500Initial Rate: 3.25%Years: 30Unit Periods: 12 (i.e., monthly payments)Fees: $3,753.01Index: 2.75%Margin: 1.13%Six-Step ARM APR Procedure (I dont think it is correct):1. Calculate the monthly payment for the rst three years based onthe following inputs:Loan Amt: $129,500Rate: 3.25 (initial rate)Years: 30Unit Periods: 122. Calculate principal paid out over rst three years based on thefollowing inputs:Loan Amt: $129,500Payment: (previously calculated payment)Rate: 3.25 (initial rate)Years: 3Unit Periods: 123. Calculate the remaining balance:New balance: $129,500 less principal paid out over rst threeyears4. Calculate fully-indexed rate:Rate: 3.88 (index + margin)5. Calculate the new monthly payment based on the new input values:Loan Amt: (new balance)Rate: 3.88 (index + margin)Years: 27 (30 - 3)Unit Periods: 126. Calculate the APR:Loan Amt: (new balance)Payment: (previously calculated payment)Years: 27Unit Periods: 12Fees: $3,753.01Estimated APR: 3.88 (input fully-indexed rate here as estimated APR -- the acturatial method is a convergence algorithm) Is it possible to determine what the uncountable product of all real>numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1,>but is this concept studied in general anywhere (e.g. the concept of>uncountably innite products or sums of real numbers)?Lookup the term summable family. This is a concept which allows denition> of convergence for any numbers a_i with i in I where I is an arbitrary> index set. By applying exp-log-formalism (transform your product into a> sum) you get the analogue theory for products.> A necessary condition for such a family to converge is that its support (the> largest set J subset I such that a_j <> 0 forall j in J) is countable -> this means that your product cannot converge.> Because equally well than your reasoning I could argue that it converges to> any other real number <> 1. Cf. Riemanns rearrangement theorem.This is something that I have wondered about. Occasionally inprevious threads the idea of summing an uncountable set of numbers hascome up. It is usually commented this is not possible unless all buta countable number are zero. However my reaction has always been: isthere any denition for innite sums of any sort other than thecommon series.I wondered about summing the values of a function f from a set to R orC. I wondered what sort of structure the set would have to have andwhat class of functions could be handled. Even when the set iscountable, the answer is not obvious since the sequence can affect theresult. I mentioned in a recent thread, the distinction betweenabsolutely and conditionally convergent series.This is the rst reference that I have noticed to a generalisation onnite sums. Unfortunately I could notfind much on Summablefamilies on the net. A search of the groups revealed little but thisthread. A search of the web found more but many seemed to mention itonly in passing or were in formats that I could not read. Mathworlddid not seem to have anything.Can you point me somewhere or give a bit more detail?For non-countable sets, could measure theory be considered ageneralisation 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 Im sure is still wrong). What if|you consider the innite product of the numbers in the interval|A=[2/3,3/2].The product of the numbers, just as a product of numbers, isundened as has been explained by others. I think youvemoved to considering something which bears the samerelationship to products as taking an integral 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 all of them is 1.||Or is there some problem with this?For this sense of continuous product, it fails for the samereason that the following doesnt work:Whats the average of x^2 from 0 to 1? Well, for each valueof x^2 > 1/2 theres a corresponding value 1-x^2 at thepoint sqrt(1-x^2). So the values of x^2 match up in pairsadding to 1, so the average value is 1/2.This fails because the horizontal spacing also counts.Transforming from x to sqrt(1-x^2) or 1/x stretches orcompresses intervals on the x axis, which affects theresult.Keith Ramsay The product of the numbers, just as a product of numbers, is> undened as has been explained by others. I think youve> moved to considering something which bears the same> relationship to products as taking an integral does to sums.This is known as a product integral. in the case thatthe values are positive real numbers, it may be doneby taking log, integrating, then exponentiating.It gets more interesting when the values are in some space withnon-commutative multiplication. Matrices, say.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ = [.snip.]>Natural numbers pose a special problem since they may be viewed in two>different ways. Actually, they can be dened in many, in many, many different ways,and viewed in many, many, many different ways.>We may view them either as something dened by>induction, or as the smallest algebraic structure which is closed>under the operations 0, 1 , +, and * (theese being subject to the>usual laws).As far as I can tell, {0,1} satises the hypothesis you give, throughsuitable denitions. Presumably, you mean the smallest subset ofR/Q/Z which contains 0, 1 and is closed under + and *...But there are plenty of other ways to characterise the naturalnumbers; they are the smallest non-nite ordinal; they are the rstsingular cardinal; they are a set together with a function thatsatises the recursion theorem, and many others. =Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu .....................>> 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 natural numbers I gave, unlike this one, there is no risk of>misunderstandings.>Two questions whose answers I am seeking are:>1) What exactly might we mean by representation?>2) How do we dene a mathematical object if we do not want to>identify it with some representation of it?This is done quite often. The Peano Postulates CHARACTERIZEthe positive (or non-negative) integers. There are many otherways to characterize them. >Natural numbers pose a special problem since they may be viewed in two>different ways. We may view them either as something dened by>induction, or as the smallest algebraic structure which is closed>under the operations 0, 1 , +, and * (theese being subject to the>usual laws).Only two? There is also the cardinal version, which is stilldifferent. They are also the smallest subset of the realnumbers (suitably dened) containing 1, which is dened asthe real numbers form a eld, and closed under addition.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University >>A natural number is a sequence of symbols in which only the symbol>>| occurs.> That is ONE representation of a natural number.>Indeed. But identifying mathematical objects with what are really>>representations of them is very common, and I was just doing what>>everyone else does, though maybe I should not have.Its crass formalism, a major impediment to undertstandingmathematics. And not everyone else does it. >> This is an extremely poor way of attempting to understand>> mathematical objects.I agree. The rst thing I do is always to abstract away from> inessentials, which includes representations (or are there cases where> it is necessary 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 nite string of symbols>> from a collection of 15 symbols, of which one of the>> symbols cannot be the rst element of the string.> Now what led me to produce this statement? You might>> consider that it came from the base-15 representation of>> integers, but 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 ve>> 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 natural numbers I gave, unlike this one, there is no risk of> misunderstandings.You identication of natural numbers with a sequence ofsymbols is already a msiunderstanding.Natural numbers pose a special problem since they may be viewed in two> different ways. We may view them either as something dened by> induction, or as the smallest algebraic structure which is closed> under the operations 0, 1 , +, and * (theese being subject to the> usual laws).Isnt the eld of two elements a smaller algebraic structure which is closed under the operations 0, 1 , +, and * (theese (sic) being subject tothe usual laws) than the natural numbers.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) = >A natural number is a sequence of symbols in which only the symbol>>| occurs.>> That is ONE representation of a natural number.>Indeed. But identifying mathematical objects with what are really>representations of them is very common, and I was just doing what>everyone else does, though maybe I should not have.Its crass formalism, a major impediment to undertstanding> mathematics. And not everyone else does it.> This is an extremely poor way of attempting to understand> mathematical objects.>>I agree. The rst thing I do is always to abstract away from>inessentials, which includes representations (or are there cases where>it is necessary 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 nite string of symbols> from a collection of 15 symbols, of which one of the> symbols cannot be the rst element of the string.>> Now what led me to produce this statement? You might> consider that it came from the base-15 representation of> integers, but 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 ve> 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 natural numbers I gave, unlike this one, there is no risk of>misunderstandings.You identication of natural numbers with a sequence of> symbols is already a msiunderstanding.>Natural numbers pose a special problem since they may be viewed in two>different ways. We may view them either as something dened by>induction, or as the smallest algebraic structure which is closed>under the operations 0, 1 , +, and * (theese being subject to the>usual laws).Isnt the eld of two elements a smaller algebraic structure which is > closed under the operations 0, 1 , +, and * (theese (sic) being subject to> the usual laws) than the natural numbers.You are right! Of course, Z_2 is the smallest such structure. Oneneeds to add some axiom that forces there to be innitely manyelements.Mattias =In Set Theory, is an element of N also a proper subset of N?For example, I want to prove that a cardinal k belonging to N is lessthan the cardinal of N itself, aleph null. If an element of thenatural numbers also a proper subset of N, then I can show k < alephnull by showing that a set A with cardinal k is a proper subset of thenatural numbers. Since a natural number n cant be mapped onto aproper subset of itself, I think I can say that aleph null cant bemapped onto a proper subset of itself. In Set Theory, is an element of N also a proper subset of N?Depends on the denition, but under the usual one (e.g., the one inHalmoss _Naive Set Theory_), then yes. The natural numbers are anexample of a transitive set, which means a set that satises thatany element is also a subset. It is then not hard to show that in thecase of N, they are in fact proper subsets.>For example, I want to prove that a cardinal k belonging to N is less>than the cardinal of N itself, aleph null. Thats true for any limit cardinal; in particular for N.>If an element of the>natural numbers also a proper subset of N, then I can show k < aleph>null by showing that a set A with cardinal k is a proper subset of the>natural numbers.Thats not enough; the natural numbers contain proper subsets whichhave the same cardinality as the full subset. In fact, thats one wayto dene innite (at least, in the presence of the Axiom ofChoice). > Since a natural number n cant be mapped onto a>proper subset of itself, I think I can say that aleph null cant be>mapped onto a proper subset of itself.That would be false, though. The natural numbers can be mapped ontothe set of all even natural numbers, a proper subset of itself. =Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) =Arturo Magidinmagidin@math.berkeley.edu >In Set Theory, is an element of N also a proper subset of N?Depends on the denition, but under the usual one (e.g., the one in>Halmoss _Naive Set Theory_), then yes.Im using Jechs Inroduction to Set Theory, 3rd Ed.>Thats true for any limit cardinal; in particular for N.Whats a *limit* cardinal???>Thats not enough; the natural numbers contain proper subsets which>have the same cardinality as the full subset. In fact, thats one way>to dene innite (at least, in the presence of the Axiom of>Choice). Youre right, Arturo . . . I forgot about that . . .And my professor wont cover the Axiom of Choice until late November Ithink.>> Since a natural number n cant be mapped onto a>>proper subset of itself, I think I can say that aleph null cant be>>mapped onto a proper subset of itself.That would be false, though. The natural numbers can be mapped onto>the set of all even natural numbers, a proper subset of itself.Now I see that a jump from a natural number to aleph null would be toomuch. But how can I *show* k < aleph null? Its like proving that 1 In Set Theory, is an element of N also a proper subset of N?>>Depends on the denition, but under the usual one (e.g., the one in>>Halmoss _Naive Set Theory_), then yes.Im using Jechs Inroduction to Set Theory, 3rd Ed.>Thats true for any limit cardinal; in particular for N.Whats a *limit* cardinal???Oops. Looks like I screwed up, at least with respect to Jechsbook. He only denes limit ordinals, and of course, every innitecardinal is a limit ordinal when considered an ordinal.I was trying to refer to cardinals which were not successor cardinals(like 0, aleph_0, aleph_omega, etc). But that is rather silly. TheAxiom of regularity guarantees that if an element of a set A is also asubset, then it must be a PROPER subset. Otherwise, {A} would have noepsilon-minimal element.>>Thats not enough; the natural numbers contain proper subsets which>>have the same cardinality as the full subset. In fact, thats one way>>to dene innite (at least, in the presence of the Axiom of>>Choice). Youre right, Arturo . . . I forgot about that . . .And my professor wont cover the Axiom of Choice until late November I>think.I only mentioned the axiom of choice because there are many differentways to dene innite and nite, and Im not positive which oneis usually used when you are in ZF. I know that whichever it is, if weare in ZFC then the denition of innite is equivalent to there isa one-to-one bijective correspondence between A and a proper subset ofA.> Since a natural number n cant be mapped onto a>proper subset of itself, I think I can say that aleph null cant be>mapped onto a proper subset of itself.>>That would be false, though. The natural numbers can be mapped onto>>the set of all even natural numbers, a proper subset of itself.Now I see that a jump from a natural number to aleph null would be too>much. But how can I *show* k < aleph null? Its like proving that 1 <>innity. Its obviously true, but how do you prove it?Looking at my copy of Jechs Set Theory, 2nd edition, a set isdened to be nite if it has the same cardinality as some n inN. Aleph_0 is dened as the least innite cardinal; and the orderof cardinals/ordinals is by inclusion. So if a is in N, then a is asubset of N, and thus a<=N. If you had a=N, then N would be an elementof itself, which contradicts regularity, so a distribution. Determine the joint density of X = maj(U,V) and Y= U+V.I get this:P(Xnoticed 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.Thats your impression. In _fact_ you have never trouncedanyone here.>It occurs to me that with all those posting in all those threads that>I create, theres lots of you who probably dont keep up, and dont>realize that I usually win.You usually win. Right. Except for the times you dont, whichhappen to be always.It must be nice, living in a world where everything is the wayyou want it to be...>So Im thinking of ways to nail these posters down, force them to have>to stand by their positions in clear sight, and when theyre trashed,>make certain that they cant just wait for a while and come back later>with the same ol crap.Now like the sneaky bastards they are, they try to act like Im the>one whos keeping things confusing by being a moving target, but Im>only moving towards simplication.That is, when Ifind that the argument is too complicated for most>readers, and that other posters have it easy confusing people, I work>to simplify it, which is how Ive gone fromP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)toP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3and *still* Ifind posters trying tofind ways to confuse, and as>usual, piling up posts in my threads!!!So, Im looking for ways to end their games.Those who want a stable version of my argument, can just go to a blog>Ive created, where you can read it, without distractions from dumb>posters!!!http://mathforprot.blogspot.com/Go, you know you want to go read it. Go ahead, no one will know.Just go and read it already. Why are you still here reading? Go!!!>James Harris>http://mathforprot.blogspot.com/********************* ***David C. Ullrich There are quite a few posters who reply in my threads. Now Ive> 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 all those posting in all those threads that> I create, theres lots of you who probably dont keep up, and dont> realize that I usually win. So Im thinking of ways to nail these posters down, force them to have> to stand by their positions in clear sight, and when theyre trashed,> make certain that they cant just wait for a while and come back later> with the same ol crap. Now like the sneaky bastards they are, they try to act like Im the> one whos keeping things confusing by being a moving target, but Im> only moving towards simplication. That is, when Ifind that the argument is too complicated for most> readers, and that other posters have it easy confusing people, I work> to simplify it, which is how Ive 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* Ifind posters trying tofind ways to confuse, and as> usual, piling up posts in my threads!!! So, Im looking for ways to end their games. Those who want a stable version of my argument, can just go to a blog> Ive created, where you can read it, without distractions from dumb> posters!!! http://mathforprot.blogspot.com/ Go, you know you want to go read it. Go ahead, no one will know. Just go and read it already. Why are you still here reading? Go!!! James Harris> http://mathforprot.blogspot.com/Does this mean you will now stop posting the same crap in sci.math? Theonly thing useful to come out of you would be of more interest to organicfarmers than mathematicians.--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com There are quite a few posters who reply in my threads. Now Ive> 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 algorithms since you got your ass kicked there... There are quite a few posters who reply in my threads. Now Ive>noticed that I trounce one poster and then another pops up and starts>yapping. Ive noticed that you never trounce anyone, except in your head. Youusually misunderstand what is said, engage in red herrings, strawmen,and ad hominems, and then declare unilateral victory.> 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 beto repeat your incorrect argument over and over, start new threads,claim victory, and lie about what has happened before.Here you go again.>It occurs to me that with all those posting in all those threads that>I create, theres lots of you who probably dont keep up, and dont>realize that I usually win.You do not usually win, except maybe in your imagination. Youroutinely fail to even comprehend what is written, let alone answerit; in fact, you often fail to even read it. [.snip.] == ==Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidinmagidin@math.berkeley.edu> There are quite a few posters who reply in my threads. Now Ive> 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.Dirk Vdm =Which number should replace the question mark so that each group of fournumbers has the same logical mathematical property:2240 3534 4451 8765 994?It is a quiz question and I just cant seem to work it out. I assume that a)it is a single digit that replaces the question mark and b) the numbers arenot four digit numbers, but are in fact groups of single digit numbers thatare somehow related.Someone must have an idea about this! Lots of us have been trying to work itout but just cant get anything that makes sense.Dom F =from your address with no luck. Ive been ltered! But I guess mymessage is suitable for general viewing, so here goes...)I was thinking about these (and CFs in other number elds)and came across your posts on sci.crypt; it seems like thegeneral consensus was use the Euclidean algorithm foralpha/1 and nearest neighbor in place of greatest integer. Thenearest neighbor guarantees the magnitude of the remainer is less thanone for Z[i] but not for general number elds.Conjugates of alpha have the same norm; is there always a conjugate ofalpha with purely positive coefcients? If so, then one could takethe algebraic integer nearest zero in the cell alpha lies in. Thisguarantees that the coefcients remain positive, but has thedisadvantage that the difference can be greater than 1. There is anupper bound on the magnitude of the remainder r, though; call it b. Then we can take b/r and be guaranteed that it will be at least 1.Something similar happens with cfs over the reals; if you subtractmultiples of some integer n, then r < n; and rather than simplyinverting, take n^2/r > n. Then the convergents p/q satisfyp^2 - alpha^2 q^2 < (2 n alpha)I havent worked out yet whether there are similar bounds for theGaussian integers given the integer nearest zero greatest integerfunction above.Did youfind anything more after receiving those replies to yourposts?Mike Stay .... Im curious as to how Euler determined Eulers constant .... If youd really like to dig into the history, heres a reference. published in Commentarii academiae Petropolitanae ad annum 1736, Tom.VIII, pp.14-16. Unfortunately Cajori doesnt give the title of the Ken Pledger. actually it was... ever heard of the Abacus?>Williamous Gatius most likly bought them out. >2) 6000 plus indeed! Are you always off by a factor of one? Squash and>> potatoes were domesticated 10-11 KYA*, avocados 9-10 KYA, peppers 9>> KYA, chiles 8-9 KYA, cotton 7-8 KYA, corn 7 KYA, beans 5-7 KYA, and>> fertilizers rst used 5-7 KYA. And thats just in the Americas. Seed>> selectcion began at 9-10 KYA in the Americas.>*KYA = thousand years ago.> even if you double it. it still a big gap... Sorry, I rst thought you were a creationist.nope I am not >>The Theory of Evolution does not depend on> your understanding, or on my ability to explain it. It depends on> scientic observations, and the ability to make predictions aboutthose> observations.> and yet evolution as the origin of man, does not pass thescientic>>method..>And special creation does? *poof* God did it! How can it be tested. We>> can test any evolutionary theory, but creationism (or its soft-line>> cousin, intelligent design) is inherently untestable. And its not>> very creative; its like the Sims.> you are confusing me with a creationist... I am not.>It seem to me you are equating evolution as a science theory that does>not need the support of the scientic method. that is very odd. Then what are you? (The 6000 years claim made me think you were a> creationist.) And last I checked, the scientic method had been> around for less than 500 years. Generally speaking, using only the> scientic method as the end-all-and-be-all is considered a Bad Thing> among scientists.Scientist refusing the scientic method... woe to the world... woe to theworld... need to kill someone else (i.e., reproduce the experiment) in order to> solve a homicide. Instead, what does a detective do? Gather evidence.> And theres a vast amount of evidence favoring evolution.As with any detective work, wrong conclucions can be drawn from theevidence. ....> According to http://members.aol.com/jeff570/mathword.html :The terms HYPERBOLIC GEOMETRY, ELLIPTIC GEOMETRY, and PARABOLIC> GEOMETRY were introduced by Felix Klein (1849-1925) in 1871 in .86ber> die sogenannte Nicht-Euklidische Geometrie (On so-called> non-Euclidean geometry), reprinted in his Gesammelte mathematische> Abhandlungen I (1921) p. 246 (Ken Pledger and Smart, p. 301). I have not read that reference, but.... The reference has been translated into English, in John Stillwell,Sources of Hyperbolic Geometry, 1996, p.72 (3rd paragraph andfootnote). Id rather not type it all now, but anyway Stillwells littlebook is well worth looking at. :-) Ken Pledger. =Actually, there is no mistake there as I say that the as from the>given factorization are roots of the given expression.>>And where the can applies is to the factorization as there are an>*innity* of factorizations for P(x), but Im pointing out that one>of them is the one that follows.>>So, theres a direct statement telling *exactly* what the as 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 really a problem, or at least it is a problem that is> easily xed. Theres no problem there. Now when a poster makes mistakes Ive gotten to where I just stopthere, but since this one is pressing the issue, lets see what hasgotten Nora Baron so excited. > But as I mentioned at the top there are two problems. The > second one is the real one. You will not be able to x it. But you for some reason deleted off all 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 all objections. Deleted part of my post:This problem is easy to x. Just replace 5 by y or some other symbol.> You have then restored P(x) to being a polynomial 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 as > *must be* roots of [**].Thus this is one place where you have a mistake, but it is easily xed.> I will pretend that you have made the right x and proceed.Did that make sense to anyone?>Notice it *appears* that the constant terms for the three factors are>all 7, which cant be right, as the constant term of P(x) is 1078, so>setting x=0, reveals>P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)>as the cubic dening the as with x=0 is>a^3 - 3a^2, which has roots, 0, 0 and 3, and Ive picked a_1 and a_2>for 0, so that leaves a_3 with a value 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.> All this is basically OK, conditional on the x described above.This poster Nora Baron is as usual, annoying.But P(x) has 49 as a factor as every 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. Also 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 thats clearly wrong as it follows from the previous step.Now the poster Nora Baron has done this before as I challenged theposter to point to an error in the argument, and the poster pointed atthe 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 thatthe constant terms of the factors of P(x)/49 dont have 7 as a factorbecause 22 doesnt have 7 as a factor.Notice how short and direct that is.Ive connected the conclusion to what came before, and its obviousenough, unless you wish to challenge the assertion that 7 is not afactor of 22.James Harris >Actually, there is no mistake there as I say that the as from the>>given factorization are roots of the given expression.>>And where the can applies is to the factorization as there are an>>*innity* of factorizations for P(x), but Im pointing out that one>>of them is the one that follows.>>So, theres a direct statement telling *exactly* what the as 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 really a problem, or at least it is a problem that is>> easily xed. Theres no problem there. Now when a poster makes mistakes Ive gotten to where I just stop>there, but since this one is pressing the issue, lets see what has>gotten Nora Baron so excited.>> But as I mentioned at the top there are two problems. The >> second one is the real one. You will not be able to x it.> But you for some reason deleted off all 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 all objections.> == => Deleted part of my post:> This problem is easy to x. Just replace 5 by y or some other symbol.>> You have then restored P(x) to being a polynomial 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 as >> *must be* roots of [**].> Thus this is one place where you have a mistake, but it is easily xed.>> I will pretend that you have made the right x 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>>all 7, which cant be right, as the constant term of P(x) is 1078, so>setting x=0, reveals>P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)>as the cubic dening the as with x=0 is>a^3 - 3a^2, which has roots, 0, 0 and 3, and Ive picked a_1 and a_2>for 0, so that leaves a_3 with a value 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.>> All this is basically OK, conditional on the x described above.This poster Nora Baron is as usual, annoying.And, as usual, correct.>>But P(x) has 49 as a factor as every 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. Also 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 thats clearly wrong as it follows from the previous step.No, it does not follow from the previous steps. Thats PRECISELY whatwe are objecting to. >Now the poster Nora Baron has done this before as I challenged 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 followsfrom the previous ones. It does NOT follow from the previous ones. Itis NOT true that it is the only way that the constant terms keepmatching. I already have given you many other ways, but here is theKEY points again:(1) Given ANY function f(x) from the algebraic integer to the complex numbers, and given any algebraic a and any algebraic integer b, there exists a function g(x) from the algebraic integers to the complex numbers, and 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 algebraic 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 every algebraic 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 equal to 5a_1(x)/7 if and only if w_1(x)=7 forall x; the function H_2(x) is equal to 5a_2(x)/7 if and only ifw_2(x)=7 for all x; and the function H_3(x) is equal to 5b_3(x)+22 ifand only if w_3(x)=1 for all x. This is clearly not the only waythings can happen.In addition, we may always choose w_1(x), w_2(x), w_3(x) to havevalues which are ALGEBRAIC INTEGERS, and such that they divide (in thering of all algebraic integers) 5a_1(x)+7, 5a_2(x)+7, and 5b_3(x)+22,for each x. These values need not be constant (and in fact, cannot beconstant). In any case, it should be clear that your claim that it isthe only way is false, since there are innitely many other ways equality, andwhich 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 theprevious statements. It is false.>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 dont have 7 as a factor>because 22 doesnt 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.>Ive connected the conclusion to what came before,By a leap of logic.> and its obvious>enough,that you have no idea what a logical argument is.> unless you wish to challenge the assertion that 7 is not a>factor of 22.No, your assertion is NOT based merely on the fact that 7 is not afactor of 22 in the ring of all algebraic integers. It is based on theincorrect and unspoken assumption that there is only one possiblechoice for functions w_1(x), w_2(x), and w_3(x) that will allow you towrite the functions with constant terms 1, 1, and 22; you thenassume that since you found one way, and there is only one way, yourway is the only and therefore the correct way. The unspoken assumptionthat there is only one way is incorrect. It has nothing to do with 7not dividing 22. =Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of gures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indenite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidinmagidin@math.berkeley.edu =Methinks I nally disproved CH byfinding something with cardinality of C,all think by looking in places noone else looks.amateur -----Original Message-----> Conversation: How to prove that sqrt(2) exist?>The fact that you dontfind that convincing is a>good thing, from a purely mathematical point of view.:-)>There are various ways to show that a real number>exists - the simplest is rst to give a rigorous>denition 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 between>a and b, and then apply this theorem with>f(x) = x^2 - 2, a = 0, b = 2.I actually 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 very notion of continuity> relies on the existence of numbers... So, the> function has a chance at being continuous because> the values exist -- how could I use that to prove> that a value exists such that f(x) = 0?What Ifind curious is that this continuity issue> seems to rely on the fact that between any two> distinct real numbers, there is at least one other> real number -- but this fact is not a sufcient> condition for the existence of such number for> which f(x) = 0: it is also true for rational> numbers that between any two distinct rational> numbers, there is at least one other rational> number.> The thing is that the reals are complete (no gaps). Its the connectedness that makes it work. The rationals satisfy the other properties of the reals that dont relyon completeness.> (at this point, Im guessing my doubt becomes> sort of philosophical? 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?)Actually none of it exists, but some notions are a little easier to swallow than others. = I.e., how to prove that there exists a real number x> such that x^2 = 2? Im getting started with the rigourous side of maths> (as a hobby -- being an engineer, with several years> of experience, you understand that I havent really> needed it), and really enjoyed reading a proof that> sqrt(2) can not be a rational number. However, the justication 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 dontfind that convincing at> all (I mean, sure, intuitively yes -- I wouldnt> even need that evidence to convince me -- I have> always accepted that sqrt(2) exists, just because,> well, it has to be there, somewhere between 1.4> and 1.5 ... You know... engineers!! :-)) But anyway, they dont really prove that sqrt(2) is> a real number: they prove that there is no rational> number p/q such that (p/q)^2 is 2. So, Im curious: how can one prove that sqrt(2)> exists and it is a real number?> Carlos> -->I think this whole thing could be resolved if Carlos just bought, orborrowed, a book on Introductory Real analsis.Lurch I think this whole thing could be resolved if Carlos just bought, or> borrowed, a book on Introductory Real analsis.And what do you think prompted my question?!! :-)I do have (just bough a month or so ago) Russell GordonsReal Analysis - A rst course, and Im enjoying readingit.Its precisely there where I saw the proof that sqrt(2)is not a rational number. But as I said, they justargue (in the intuitive sense, I think), that sqrt(2)exists, using the triangle as argument.Now that I think about it (after reading Mike Kentsmessage) that maybe my skepticism was due to the factthat I dont have a clue on the rigorous side ofgeometry. It is true that the triangle is an actualconstruct that shows you that that length (whichexists, because the triangle exists) is such thatits square must be 2.But I guess my doubt came from the fact that I feelthat whatever results in geometry (including Pythagoras)relies on existence and other properties of real numbers;so, using that as an argument to claim existence of aparticular real number sounded odd to me...After reading all the posts that you guys have sokindly written, I see that the issue is indeed complex(well, I have more or less understood the arguments,but the whole analysis side of maths is new to me,so I guess itll take me some time to fully digestthe ideas).Anyway, Im very grateful for all the replies and theCarlos-- =Analysis, and higher math, can have very subtle arguements. It sometimestakes a few readings to even get the idea of what is going on. They saythere are stages to understanding proofs. Firstly, one understands thelogical progression. Secondly, one understands the proof enough toreproduce 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 Carlos just bought, or>borrowed, a book on Introductory Real analsis. And what do you think prompted my question?!! :-) I do have (just bough a month or so ago) Russell Gordons> Real Analysis - A rst course, and Im enjoying reading> it. Its precisely there where I saw the proof that sqrt(2)> is not a rational 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 Mike Kents> message) that maybe my skepticism was due to the fact> that I dont have a clue on the rigorous side of> geometry. It is true that the triangle is an actual> 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 geometry (including Pythagoras)> relies on existence and other properties of real numbers;> so, using that as an argument to claim existence of a> particular real number sounded odd to me... After reading all 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 analysis side of maths is new to me,> so I guess itll take me some time to fully digest> the ideas). Anyway, Im very grateful for all the replies and the> Carlos> --> =In sci.math, Carlos Moreno:>> Or one can go the slightly silly route: sqrt(2) exists because>> (sqrt(2)^2) = 2You can always dene it that way (for instance, the way> you dene the imaginary unit). But that does not mean that> that number you just dened is in the set of real numbers.In particular, I could arbitrarily come up with such denition> for the rational numbers: I could simply say: sqrt(2) is the> rational number X such that X^2 is 2. I could claim that the> number exists because I dened it, and that means that it> exists. You can, however, prove that if such number exists,> it is not a rational number (and thus, not an integer or natural> number either).I can prove a contradiction from your denition, so yourclaim is bogus. However, X *is* real, for a properlyrigorous denition thereof -- just not rational.But lets prove X isnt rational, 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. Thismeans p is even. Write p = 2 * r, and we getp^2 / q^2 = 4 * r^2 / q^2 = 2. We ip this andnote that q^2 / r^2 = 2, or q^2 = 2 * r^2, so that q iseven as well.Oops...this means gcd(p,q) cannot be 1 and thereforethe assumption that p and q exist at all leads us intoa contradiction.Therefore X, if it exists at all, is not in Q.Obviously X doesnt necessarily exist until someonesuch as Dedekind or Cauchy step up and prove thatit makes sense to treat such numbers with the usualarithmetic operations.The question remains -- when trying to be rigourous, at least> the way Im understanding it, as a beginner in the rigourous> side of mathematics, you ask yourself the question: how do I> really 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 very clear on what you might mean by the term number.Peano, in particular, went the rather interesting route ofstarting with 1 and deriving all the other natural numberstherefrom, using 4 axioms and 1 axiom which I prefer to calla meta-axiom, his weak induction axiom.We therefore get N, the set of natural numbers. If thats notrigorous 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 thenprove that S_0 != S_1 != S_2 != ... by set theory, perhaps.Since S_{n+1} = S_n union {S_n} Peanos successor operationis well-dened here as well, and we get 0 as a bonus.Once one has the whole numbers one can construct the integersby assuming -a is such that (a + (-a)) = 0. Fractions areanother extension: p/q, q != 0, gives us Q.Now it gets tricky. Can we extend Q in a rigorous fashion,dening real numbers with a consistent denition of +,-, *, and /? I would think that we can but would haveto research the issue somewhat. However, Ive no problemwith pi existing, as one can easily approximate it byvarious sequences, all of which are Cauchy. (Most of themare derived from innite series.) e is even easier, asthe sequencee = 1/0! + 1/1! + 1/2! + 1/3! + ... + 1/n! + ...is absolutely convergent and generates a Cauchy sequence, namelye_n = 1/0! + 1/1! + 1/2! + 1/3! + ... + 1/n!One can even dene sqrt(2) in such a manner. Note thatthe binomial 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 indenitely extended for non-integral powers.Setting x= -1/2, one gets sqrt(1/2), which is sqrt(2)/2 --if Ive done this right.The crowning achievement, after one more extension (i=sqrt(-1)),is that our numbers are now complete -- we dont get anymore by trying to solve cubics, quartics, quintics, etc., althoughwe may have trouble with a precise formulation of the roots,resorting to numerical approximations instead for degree 5 and up.>> Or one goes the engineering route, which youve stated you>> dont like: sqrt(2) = 1.4142; pi = 3.1416; e=2.71828.Well, yes. Theres the practical side to it, which is what> I have all my life lived with -- its not like now I want to> become an actual mathematician; but Im enjoying that side> of maths, and am trying to adapt my mind to work in the> skeptical mode.It is a good question, and one that has bedeviled theoristsfor years, if not centuries (it feels a bit like ZenosParadox, in a way). Im not entirely certain weve xedCantors rather famous diagonal proof yet, althoughat least one webpage Ive seen purports such. The mainproblem appears to be that Cantors diagonal is not purelyconstructive; one can x that by force-selecting 0 or1 in base 10, avoiding the .999... problem. Since therequirement is to show that an enumeration of all realsisnt possible by constructing a new one, I dont see thisas a big problem, but then, Im not all that rigorous,either. :-)Innities, however, are like that: tricky.Carlos> --> -- #191, ewill3@earthlink.netIts still legal to go .sigless. The fact that you dontfind that convincing is a>good thing, from a purely mathematical point of view.> :-)>There are various ways to show that a real number>exists - the simplest is rst to give a rigorous>denition 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 between>a and b, and then apply this theorem with>f(x) = x^2 - 2, a = 0, b = 2.I actually 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 very notion of continuity> relies on the existence of numbers... So, the> function has a chance at being continuous because> the values exist -- how could I use that to prove> that a value exists such that f(x) = 0?What Ifind curious is that this continuity issue> seems to rely on the fact that between any two> distinct real numbers, there is at least one other> real number -- but this fact is not a sufcient> condition for the existence of such number for> which f(x) = 0: it is also true for rational> numbers that between any two distinct rational> numbers, there is at least one other rational> number.(at this point, Im guessing my doubt becomes> sort of philosophical? 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?)> Carlos> --Many posters have touched on the real core of the issue, which is howto dene a real number. As Ullrich says, you are right to distrustthe approach that simply constructs a triangle one of whose sides hasa length which squares to 2.There are a variety of ways to dene real numbers, none of which goesinto terribly murky philosophical issues. We assume the existence ofrational numbers together with their operations of addition andmultiplication and the usual rules relating to these operations. Itis also important to include the order properties so that we are clearon the meaning of statments like a > b when a and b are real numbers.Beyond that we have an intuition for what the fundamental propertiesof real numbers are, and from this intuition we can construct a setthat does indeed have these properties. Having done that we soonreach the conclusion that every nonempty set of real numbers (whatevertrivial to deduce a square root of 2 by applying to the set of realnumbers whose squares are less than 2.All of the above evades the question of how to actually construct thereal numbers. The one that works best for me is the approach of usingCauchy sequences which was developed by Cantor and not by Cauchy. Asequence {a_0, a_1, ...) is called Cauchy if for every epsilon > 0,there is an N such that if m and n are both greater than N, then a_mand a_n are within epsilon of each other. Backtracking a llttle, asequence is said to converge to a limit L if for every epsilon > 0there is an N such that if n > N, then a_N is within epsilon of L. Itis easy to show that a sequence can converge to at most one limit, andthat if it converges to a limit, then the sequence is a Cauchysequence. At least it is easy after some experience. I recall havingto wrestle pretty hard with these concepts when they were new to me. It is less clear, but one would hope, that every Cauchy sequenceconverges to some limit. This is not true for the rational numbers. We intend to turn the collection of Cauchy sequences into the set ofreal numbers. It is obvious how to add, subtract, multiply, anddivide Cauchy sequences, namely term by term. There is a minor, butsolvable problem, with dividing Cauchy sequences when the dividend haszeros. We think intuitively of the Cauchy sequence as representingits limit. The rational number q is turned into the Cauchy sequence{q, q, q, q, q, ...}. The only remaining problem to resolve is thatmany Cauchy sequences may converge the same limit, if they areconvergent, but if that is the case, the difference in these sequencesmust converge to 0. that is the key to the nal step. Real numbersare dened as equivalence classes of Cauchy sequences, two sequencesbeing equivalentif and only if their difference converges to 0. It isthen straightforward to extend the ordering to these real numbers, toprove that all Cauchy sequences of real numbers (how about that,Cauchy sequences of Cauchy sequences) which can be dened because theorder properties carried across, and then the least upper boundproperty. It is a fair amount of effort, the end result of which isthat you drop all the baggage, think of the real numbers just the wayyou always did, but are now condent that the real numbers have avariety of extremely useful properties such as least upper bounds,nested intervals, convergence of all Cauchy sequences, and others. Anal benet is that you are in a very good postion to grasp thenature of p-adic numbers without a great deal of trouble.Dedekind chose another path. I never worked it out in great detail,but both the Dedekind cut approach and the Cauchy sequence approachcan be found in a great many books with names like Real Analysis, orIntroduction to Real Analysis, etc.Achava = > The fact that you dontfind that convincing is a>good thing, from a purely mathematical point of view. :-) > There are various ways to show that a real number>exists - the simplest is rst to give a rigorous>denition 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 between>a and b, and then apply this theorem with>f(x) = x^2 - 2, a = 0, b = 2. I actually 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 very notion of continuity> relies on the existence of numbers... So, the> function has a chance at being continuous because> the values exist -- how could I use that to prove> that a value exists such that f(x) = 0? What Ifind curious is that this continuity issue> seems to rely on the fact that between any two> distinct real numbers, there is at least one other> real number -- but this fact is not a sufcient> condition for the existence of such number for> which f(x) = 0: it is also true for rational> numbers that between any two distinct rational> numbers, there is at least one other rational> number. (at this point, Im guessing my doubt becomes> sort of philosophical? 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 dont think so. Of course, you cant prove everything. Some things youjust have to assume. We call such things axioms.Maybe you want to go back to the beginning and dene what the naturalnumbers, the integers, the rational numbers, and then the real numbers are.Until you know what they are, you cant really prove anything about them.Now, basically, all these different things are is stuff that satisescertain axioms. Different people have different axioms, and they are allequivalent. (Except the ones that arent, which are called non-standard,or intuitionist [I can buy non-standard, but intuitionism leaves me cold].There may be others that Im not aware of.) What equivalent means is thatthe axioms of one system are either also axioms in another system, or aretheorems in that system. (That means provable.)So, you need to go to your axioms for the reals (whatever they are), andprove something called the least upper bound theorem.Lets say you dene a real number to be a pair of sets of rationals, (L,R)such thata. Every rational number is in exactly one of L and R andb. Every member in L is less than every member of R.This is called a Dedekind cut. Now, if you can (in the language of youraxioms) say x^2<2, then the set {x rational | x^2<2 or x < 0}will be your L,and {x rational | x^2 > 2 and x > 0} will be your R. Then the set (L,R)will be the square root of 2.Gee, this doesnt look like a real number that Im familiar with, does it?Well, thats the price of formalism. What you construct from simpler partsdoesnt look like what youre familiar with. But what does sqrt(2) looklike anyway?If you think of reals as being innite decimals, then you probably want touse a Cauchy sequence construction (real numbers are equivalence classes ofCauchy sequences of rationals). But what do equvalence classes of Cauchysequences look like? Or you could just cut to the chase and dene realnumbers to be innite decimals (assuming you can dene what an innitedecimal is -- most people use the denition that an innite decimal isreally an innite sum, which is just a particular type of [you guessed it]Cauchy sequence).Now, since youve got axioms, you might well ask Is there in fact anythingthat really satises these axioms? Mathematically, the answer to that isbasically We dont know. It turns out that the best we can hope for is toprove consistency or inconsistency. But even that isnt always possible --this is the content of Goedels Incompleteness Theorem. However, we canprove that our axiomatization of the reals is consistent with ouraxiomatization of the rationals (this is called relative consistency), whichis in turn consistent with our axiomatization of the natural (counting)numbers. These seem intuitively to exist (well, we do count, dont we?), sothe real concern (if there is any concern at all) would be that we haventaccurately axiomatized them. (Or that logic doesnt really work at all.)But hey, youre an engineer, so youre comfortable with not being absolutelysure that all your assumptions are strictly true.I hope I havent snowed you here. Ive tried to keep all your options open,so you can use the approach you want (Im strongly pro-choice [even up toaxiomatization]), but also want to give you a vague idea of what thedifculties you might encounter are. If Ive only confused you, ask again.Either I or someone else can really help -- but a lot depends on where youare and what you want.Jon Miller >I.e., how to prove that there exists a real number x>>such that x^2 = 2?>>Im getting started with the rigourous side of maths>>(as a hobby -- being an engineer, with several years>>of experience, you understand that I havent really>>needed it), and really enjoyed reading a proof that>>sqrt(2) can not be a rational number.>>However, the justication 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 dontfind that convincing at>>all ...> If we allow that real numbers are in one-to-one> correspondence with points on the x-axis, then> we can mark off sqrt(2) on this real number line,> using compass and straightedge. Voila. There it is.Thats not a solid convincing argument -- its likesaying it must exist because I can picture it in mymind.That may be a good way to understand something, butdenitely not to prove it. You could have thoughtthe same about it with rational numbers. In fact,I can most denitely picture in my mind a fractionwhose square is two (not picture as in telling youthe numerator and denominator, but picturing as invisualizing that there is one) -- if it werentbecause I know, and I can prove, that there can notbe a fraction whose square is two.After all, you could say well, any point in thereis a fraction of the length of a unit interval; sothere, I put a mark in there, and that must be afraction... And yet, one woudl be wrong.There are plenty of arguments that would otherwiseconvince me that there is a rational number whosesquare is 2, such as between any two distinct rationalnumbers, there is at least one other rational number;with this, you could reason that whatever additionaldecimals you put, theres always a fraction thatrepresents that, and thus, no matter how far wego, there will always be a fraction, and thus, thereis a fraction that is equal to sqrt(2)... Voila, itsright there, one might say. And no, intuition failsin that particular case.I know, it feels strange trying to argue that yourattempt at convincing me of something that not onlyIm convinced, but that I know for a fact is true,is not convincing enough... (when I say I know fora fact, I mean that any mathematician or mathematicsbook would tell you so, and would tell you that itcan be proven).Still, if we play the game of being skeptical andproving everything, then lets be truly skeptical.:-)Carlos-- I.e., how to prove that there exists a real number x>such that x^2 = 2?>Im getting started with the rigourous side of maths>(as a hobby -- being an engineer, with several years>of experience, you understand that I havent really>needed it), and really enjoyed reading a proof that>sqrt(2) can not be a rational number.>>However, the justication 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 dontfind that convincing at>all ...Why not? The segment has a length, and the squarewith sides of that length has twice the area of asquare with sides of length 1. What would thatlength be, if not the square root of 2?There is a good (and inexpensive, in paperback) book by L. Blumenthal, A Modern View of Geometry, that has a quite rigorous development of co-ordinate systemsand algebraic structures from purely geometric ideas. I.e., how to prove that there exists a real number x>>such that x^2 = 2?>>However, the justication 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 dontfind that convincing at>>all ...>Why not? The segment has a length, and the square>with sides of that length has twice 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 real? I might know that there exists anumber z such that z = Sqrt(-1), but is it a real number? And if youdene Sqrt(2) like that, what happens in non-euclidean geometries?Does the value of Sqrt(2) change? saucisse et au marteau:> I think the issue was, is it real? I might know that there exists a> number z such that z = Sqrt(-1), but is it a real number? And if youNo, there is _not_ any number z such that z = sqrt(-1). There is z suchthat 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 = -1Every number, positive or negative, has two square roots. For positivereals, we dene THE square root by the positive one. On negative reals,as there is no order on C, we cant pick one to be THE square root.Therefore, we decide that there are two square roots, i and -i.-- Nicolas .88 la saucisse et au marteau:> I think the issue was, is it real? I might know that there exists a>> number z such that z = Sqrt(-1), but is it a real number? And if youNo, 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 its still a valid denition.http://mathworld.wolfram.com/ImaginaryUnit.html = > I think the issue was, is it real? I might know that there exists a> number z such that z = Sqrt(-1), but is it a real 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 its still a valid denition.Not so! The square root of -1 does not exist! A square root does, in fact, if any square rootof -1 exists then two of them do, with no way of distinguishing between them. What we sometimes label i is picked arbitrarily to be one of them and -i the other. saucisse et au marteau:> No, but its still a valid denition.> http://mathworld.wolfram.com/ImaginaryUnit.htmlIve already seen mistakes on MathWrold and I would not take it as areference. But if you have more serious references, I may change mymind.-- Nicolas grava .88 la saucisse et au marteau:> No, but its still a valid denition.> http://mathworld.wolfram.com/ImaginaryUnit.htmlIve already 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.Im not equipped with a lot of serious complex analysis books, but atleast some physics texts (Brehm-Mullin: Introduction to the Structureof Matter, for one) dene i like that. Most mathematical approachestake i as the ordered pair of reals (0, 1) and work from there. If we allow that real numbers are in one-to-one>correspondence with points on the x-axis, then>we can mark off sqrt(2) on this real number line,>using compass and straightedge. Voila. There it is.> Thats not a solid convincing argument -- its 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 argumentthat theoretically can be made rigorous.> That may be a good way to understand something, but> denitely not to prove it.... If we allow that real numbers are in one-to-one>>correspondence with points on the x-axis, then>>we can mark off sqrt(2) on this real number line,>>using compass and straightedge. Voila. There it is.> Thats not a solid convincing argument -- its 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 theoretically can be made rigorous.But as far as I can see making it rigorous involvesa lot more work than with the other approaches thathave been suggested (the _simplest_ way to makeit rigorous seems to be to _include_ one of thoseother approaches...)>> That may be a good way to understand something, but>> denitely not to prove it....>************************David C. Ullrich If we allow that real numbers are in one-to-one>correspondence with points on the x-axis, then>we can mark off sqrt(2) on this real number line,>using compass and straightedge. Voila. There it is.>Or do you think this line has holes or gaps?> Thats not a solid convincing argument -- its like> saying it must exist because I can picture it in my> mind.> That may be a good way to understand something, but> denitely not to prove it....>Maybe we disagree on what proof is.>I see a proof as being a convincing argument>that theoretically 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...)Carlos asked: How to prove that sqrt(2) exists? Right?I assumed he had an idea what real numbers are; i.e.,maybe a completion of the rationals (thats not p-adic)or maybe a continuous extension, formally constructedby taking limits of Cauchy sequences of equivalenceclasses of rationals 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)falls within his understanding of the reals.However, he has revealed that he knows little or no analysis,even at the level of baby Rudin, etc. And his question isbeing answered by posters dening the reals. All-in-all, IMHO,his question was not well-posed. =Hey allWhen we say a function f(t) is smooth, does this mean thatf has innite differentials with respect to t?Or any other formal denition on this?Fred Hey allWhen we say a function f(t) is smooth, does this mean that> f has innite differentials with respect to t?Or any other formal denition on this?The number of continuous derivatives that a function possesses is a measure of its smoothness.Bob Kolker > Hey all> When we say a function f(t) is smooth, does this mean that>> f has innite differentials with respect to t?> Or any other formal denition 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) issmooth and dont specify anything further, we often meanthat f is innitely differentiable. This is particularlycommon among differential topologists. I have a feeling thatreal analysts, for instance, will tend to be more precise here;but maybe not.Lee Rudolph I am trying to calculate the probability that a gambler with capital C> and who uses a Martingale 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 calculate 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 satisfactory for the purpose could be arrived at in a simpler> way. All I need is the probability that there would be AT LEAST one> run of length AT LEAST n.If in n independent trials the probability of a loss at a single trial> is q, the probability of all losses in n trials would be q^n. A string> of n losses could begin at any trial from the rst 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) isthe probability of at least 1 run of at least length n in m trialsthenP(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 material in this thread, I assume Ive made a simplemistake or that a similar formula is given in Feller but notconsidered a practical method of calculation in 1968.Admittedly, you wouldnt want to use this denition for handcalculations, but its pretty easy to program up. A spreadsheet isne if the values for m and n are not too big (as was suggested bythe mention of a number of gambling games).Ian SmithP.S. I have tried with sanity checks to make sure the formula isntludicrous so Id appreciate an indication of where the following logicis faulty or an example which shows up the error rather than dont bestupid or see Feller (I havent got access to a copy).The logic is simple. Either the rst n trials are losses or the rsttrial is a success and then we have n losses or we have no runs of nor more in i trials followed by a success and then n losses (for i =1.. m-n-1). I think this enumerates all the possibile combinationswithout duplication. There have been a number of threads on this topic in sci.math, e.g.>Consecutive runs in Bernoulli trials in June 1997.> |> In Bernoulli trials, an event A occurs with probability p, and the|> probability|> of A occurring exactly M times in N trials 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 trials (with p being the probability of A|> occurring|> at any given trial)? I am assuming that the trials are independent.Let f(n) be the probability that the rst run of M consecutivesuccesses occurs at trial n (i.e. that trials n-M+1, n-M+2, ..., n aresuccesses, and there is no previous run of M consecutive successes).The generating function of this isF(s) = sum_{n=r}^innity f(n) s^n = p^M s^M (1-ps)/(1-s+q p^Ms^(M+1)) = p^M s^M/(1 - qs - qps^2 - ... - q p^(M-1) s^M)This can be expanded in partial fractions:F(s) = -p/q + sum_r a_r/(s-r)where the sum runs over the roots of the denominator (this must bemodied in the case of multiple roots). The probability of no run of M consecutive successes in N trials issum_{n=N+1}^innity 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 small, the smallest root in absolute value, which isthe unique positive root, gives the main contribution, and theresulting approximation is quite good.See Feller, An Introduction to Probability Theory and ItsApplications,Vol. 1, sec. XIII.7.Robert Israel israel@math.ubc.caDepartment of Mathematics (604) 822-3629University of British Columbia fax 822-6074Vancouver, BC, Canada V6T 1Y4Comment: This is one of several standard approaches and I am shingfor an original one or one that culminates in a set of formulas thatcan be put into a reference book and be useful to non-mathematicians.Supposedly this problem was rst propounded and solved by AbrahamDeMoivre and could well be the deepest problem in all of Probability.Sam Allen [Shadrack] =Since making the rst post I made a post to alt.sci.math.probabilityand 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 rst studied by Abraham DeMoivre, and acollection of his works might be a good place to start for anyone whois interested. References may be found athttp://mathworld.wolfram.com/Run.htmlI also have looked atBurnside, WilliamThe Theory of Probability (1928)Dover Publications, 1959Chapter IIIAn approximate solution is arrived at by means of a differenceequation.Uspensky, J. V.Introduction to Mathematical ProbabilityChapter V, Section 3A difference equation and a generating function are discussed and anapproximation using a Lagrange series is derived.Burnside gives a lower bound and his derivation is much simpler thanUspenskys.A collection of probabilities of runs is a distribution of runs andsometimes these are used in tests for randomness.