mm-1013 === Subject: Rebonato - Interest rate option models 2nd edition - query In Rebonatos book, equation 5.11 appears to me to be incorrect, since Pi(t) on the LHS is a function of t, but the RHS depends on t1, t2, T1 and T2. Is this a typographical error? What should this equation be? === Subject: Re: Rebonato - Interest rate option models 2nd edition - query > In Rebonatos book, equation 5.11 appears to me to be incorrect, since > Pi(t) on the LHS is a function of t, but the RHS depends on t1, t2, T1 > and T2. Is this a typographical error? What should this equation be? You might ask the author, www.rebonato.com gives the email for his Ôteam leader at www.markjoshi.com or otherwise try www.wilmott.com. --- remove the no for mail === Subject: An Inequality Related to the MVT for Integrals The MVT for integrals roughly says that if h has constant sign, and g is continuous then int g(x)h(x) = g(x*) int h(x) for some interior x*. I need a simple (large) class of functions g yielding an integral inequality of the form: (**) int_0^1 g(x)h(x) >= 0 for h single crossing (but not necessarily monotonic) through 0 from + to weakly negative, with int_0^1 h(x) >0 and g increasing (but not necessarily continuous) on [0,1], rising from g(0)>=0 to g(1)=1. I agree that as set up, g and h in some nondescript sense are negatively correlated, but observe that (**) holds if g is constant whenever h<>0. More generally, (**) would also obtain if the MVT (with my modified premises) held with an inequality >= sign (though I do not claim such an extension is valid). I have explored inequalities due to Beesack (1957) and Dallas Banks (1963), but their assumptions are violated here. === Subject: Re: An Inequality Related to the MVT for Integrals >The MVT for integrals roughly says that if h has constant sign, and g >is continuous then int g(x)h(x) = g(x*) int h(x) for some interior >x*. I need a simple (large) class of functions g yielding an integral >inequality of the form: >(**) int_0^1 g(x)h(x) >= 0 for >h single crossing (but not necessarily monotonic) through 0 from + to >weakly negative, with int_0^1 h(x) >0 >and >g increasing (but not necessarily continuous) on [0,1], rising from >g(0)>=0 to g(1)=1. Consider an integrable function g such that int_0^1 g(x) h(x) dx >= 0 whenever h is continuous with only one zero, h(0) > 0 and int_0^1 h(x) dx > 0. By considering a function h that is near 0 except for a narrow positive bump, we find that g >= 0 almost everywhere. By considering a function h that is near 0 except for two narrow bumps, the one on the left positive and the one on the right negative, we find that g(x) >= g(y) a.e. for x < y, i.e. g is non-increasing a.e. On the other hand, its clear that if g is non-increasing and >= 0 almost everywhere, and h is continuous with h(x) >= 0 for x <= p, h(x) <= 0, and int_0^1 h(x) dx >= 0, then int_0^1 h(x) g(x) dx >= g(p) int_0^p h(x) dx + g(p) int_p^1 h(x) dx = g(p) int_0^1 h(x) dx >= 0 If you want to add the additional condition that g is non-decreasing, you certainly dont get a large class of such functions: theyre all constant (except possibly at the endpoints). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Choosing a Math Grad school Hi guys, Id like some info about math PhD programs, if anyone can please help. A concern I have are rankings, in particular: how much do they matter when you make *final* decisions? Ive my hopes fixed on one among: Berkeley, UCLA, Penn State, Chicago, Texas-Austin, Princeton, Illinois-Urbana. My area of interest is operator algebras (thus the first three) or Harmonic analysis and PDE (thus the latter four). Depending on place-offers (and assuming that one of them can be excluded), do any of you guys know reasons for preferring one to the other? (modulo that these are all great places for analysis research). For instance, suppose you get an offer from UCLA and Illinois-Urbana or Penn State, do you obviously go for UCLA because it is higher ranked? My intention is to follow an academic career (professorship), so future placement prospects are crucial to me. have visited the above mentioned universities and may be able to give me some hands-on impressions! Please email comments to: hcanab@yahoo.com Stefan === Subject: Re: Choosing a Math Grad school > Id like some info about math PhD programs, if anyone can please help. A > concern I have are rankings, in particular: how much do they matter when you > make *final* decisions? The rankings matter a whole lot less than the fit with your interests, particularly the faculty. If there is no one at place A that studies something you are (or might be) interested in, its no good for you. Best to go where there are several students working on similar areas -- but not too many. > Ive my hopes fixed on one among: Berkeley, UCLA, Penn State, Chicago, > Texas-Austin, Princeton, Illinois-Urbana. My area of interest is > operator algebras (thus the first three) or Harmonic analysis and PDE > (thus the latter four). Depending on place-offers (and assuming that one > of them can be excluded), do any of you guys know reasons for preferring > one to the other? (modulo that these are all great places for analysis > research). Some of this depends on your preparation. Princeton, in particular, expects you to pick up much of the basic material on your own, but the state schools will have a lot of first-year grad courses to take. > For instance, suppose you get an offer from UCLA and Illinois-Urbana or > Penn State, do you obviously go for UCLA because it is higher ranked? My > intention is to follow an academic career (professorship), so future > placement prospects are crucial to me. Placement does not depend on ranking. It depends on your thesis, mostly, and your advisor. You are talking about good schools here. None of them would be an impediment to your career plans. Find the best fit for you rather than worrying about which is #1 versus #3. Look at the courses, and seminars, that are being offered now (not just what is in the catalogue). Look over the research interests of the faculty. Think about the size of the faculty, and students. Berkeley has a large and excellent faculty, but a lot of grad students to compete for their time. You also have to _live_ during those 4-6 years. If you hate big cities, Chicago would not be right for you. If you would rather die than live in a small town, then maybe Penn State would be a disaster. -- David L. Johnson __o | You will say Christ saith this and the apostles say this; but _`(,_ | what canst thou say? -- George Fox. (_)/ (_) | === Subject: Re: Choosing a Math Grad school >Hi guys, >Id like some info about math PhD programs, if anyone can please help. A >concern I have are rankings, in particular: how much do they matter when you >make *final* decisions? >Ive my hopes fixed on one among: Berkeley, UCLA, Penn State, Chicago, >Texas-Austin, Princeton, Illinois-Urbana. My area of interest is operator >algebras (thus the first three) or Harmonic analysis and PDE (thus the >latter four). Depending on place-offers (and assuming that one of them can >be excluded), do any of you guys know reasons for preferring one to the >other? (modulo that these are all great places for analysis research). >For instance, suppose you get an offer from UCLA and Illinois-Urbana or Penn >State, do you obviously go for UCLA because it is higher ranked? My >intention is to follow an academic career (professorship), so future >placement prospects are crucial to me. >have visited the above mentioned universities and may be able to give me >some hands-on impressions! >Please email comments to: >hcanab@yahoo.com >Stefan budget makes severe cuts to UC funding. Graduate tuition fees are proposed to go up 40%. Less financial aid will be available. There will be fewer freshmen, and hence fewer classes needing Teaching Assistants. -- John Adams served two terms as Vice President and one as President, but lost reelection. Later his son became President despite losing the popular vote. That son lost his reelection attempt badly. Now history is repeating itself. pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: Re: Choosing a Math Grad school > budget makes severe cuts to UC funding. ... Yes, there are some cuts scheduled, but state spending on the Univ. of California nearly doubled between 1995 and 2002. I wouldnt worry about it too much. You can get some more info about the budget here. === Subject: Non-standard analysis Consider the collection of sets I1, I2, I3, ... , In, ... Where I1 = {1}, I2 = {1,2}, I3 = {1,2,3},...,In = {1,2,3,...,n},... for natural numbers n. That is, Ij is the set of natural numbers less than or equal to j. Define a set S to be finite if there is a 1-1 mapping of S onto In for some natural number n. Define a finite set S to be larger than a natural number n, if there is a natural number m and a 1-1 map of S onto Im with m > n. Consider further the following countable collection of sentences: 1. There exists a finite set S1 larger than 1. 2. There exists a finite set S2 larger than 2. .. n. There exists a finite set Sn larger than n. .. Clearly any finite subset of this collection is consistent. [let j be the largest index of a sentence in the finite subset, then, for example, Ij+1 provides the model] By compactness, the whole collection is consistent and therefore there is a model with the property that there exists a finite set S which is larger than n for any n. This is clearly a contradiction. [S finite implies by definition that there exists some natural number n, such that S can be mapped 1-1 onto In. But S larger than n implies, again by definition, that S can be mapped 1-1 onto Im for some natural number m > n] === Subject: Re: Non-standard analysis Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow) >Consider the collection of sets I1, I2, I3, ... , In, ... >Where I1 = {1}, I2 = {1,2}, I3 = {1,2,3},...,In = {1,2,3,...,n},... >for natural numbers n. That is, Ij is the set of natural numbers less >than or equal to j. >Define a set S to be finite if there is a 1-1 mapping of S onto In for >some natural number n. David Ullrich is right that you have to be careful about the language youre working in. Since you appeal to compactness, the most obvious candidates are the first-order language of arithmetic and the first-order language of set theory. S is finite is more straightforwardly expressed in set theory than in arithmetic, so lets think about what happens in that case. You then need to be careful about the concept of the set of all natural numbers. We would like to define this by omega = {1, 2, 3, ...} or something like that, but the trouble is that formulae of infinite length are not allowed in first-order logic. (You can define logics that allow them, but then you dont get a compactness theorem for these logics.) So you need to define omega in some other way, e.g., as having the property that if x is in omega then x U {x} is in omega and that if S is any set with this property then omega is a subset of S. The problem now is that in first-order logic theres no way to force the symbol omega to be interpreted as the true set of natural numbers, i.e., we cant rule out the possibility of nonstandard models of set theory in which omega is interpreted as a set containing nonstandard integers. So now lets look at your definition: S is finite if there is a 1-1 mapping of S onto In for some natural number n. If by some natural number n you mean some n in omega then S could be mapped onto In for some *nonstandard* n, and you dont get any contradiction from your infinite list of conditions. On the other hand, if by some natural number n you mean n = 1 or n = 2 or ... then again you run into difficulties because you dont have infinitely long formulae in first-order logic. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Non-standard analysis > Consider the collection of sets I1, I2, I3, ... , In, ... > Where I1 = {1}, I2 = {1,2}, I3 = {1,2,3},...,In = {1,2,3,...,n},... > for natural numbers n. That is, Ij is the set of natural numbers less > than or equal to j. > Define a set S to be finite if there is a 1-1 mapping of S onto In for > some natural number n. > Define a finite set S to be larger than a natural number n, if there > is a natural number m and a 1-1 map of S onto Im with m > n. > Consider further the following countable collection of sentences: > 1. There exists a finite set S1 larger than 1. > 2. There exists a finite set S2 larger than 2. > .. > n. There exists a finite set Sn larger than n. > .. > Clearly any finite subset of this collection is consistent. [let j be > the largest index of a sentence in the finite subset, then, for > example, Ij+1 provides the model] > By compactness, the whole collection is consistent and therefore there > is a model with the property that there exists a finite set S which is > larger than n for any n. This is clearly a contradiction. > [S finite implies by definition that there exists some natural number > n, such that S can be mapped 1-1 onto In. But S larger than n implies, > again by definition, that S can be mapped 1-1 onto Im for some natural > number m > n] IIRC, the compactness theorem is a theorem of first-order logic, so the domains of the models should always be sets. Either youre trying to put all finite sets in the domain, which would keep this model first-order, except that your domain is now a proper class; or youre trying to quantify over (finite) subsets of the domain, making your sentences second-order, invalidating the applicability of the compactness theorem. -- Jasper === Subject: Re: Non-standard analysis >Consider the collection of sets I1, I2, I3, ... , In, ... >Where I1 = {1}, I2 = {1,2}, I3 = {1,2,3},...,In = {1,2,3,...,n},... >for natural numbers n. That is, Ij is the set of natural numbers less >than or equal to j. >Define a set S to be finite if there is a 1-1 mapping of S onto In for >some natural number n. >Define a finite set S to be larger than a natural number n, if there >is a natural number m and a 1-1 map of S onto Im with m > n. >Consider further the following countable collection of sentences: >1. There exists a finite set S1 larger than 1. >2. There exists a finite set S2 larger than 2. >n. There exists a finite set Sn larger than n. [Detail: for the application of compactness below I think you want to revise these, so they all refer to S, not Sn. And you also dont want the quanitfier There exists; S should be a constant symbol, so the n-th sentence just reads S is finite and larger than n.] >Clearly any finite subset of this collection is consistent. [let j be >the largest index of a sentence in the finite subset, then, for >example, Ij+1 provides the model] >By compactness, the whole collection is consistent and therefore there >is a model with the property that there exists a finite set S which is >larger than n for any n. This is clearly a contradiction. >[S finite implies by definition that there exists some natural number >n, such that S can be mapped 1-1 onto In. But S larger than n implies, >again by definition, that S can be mapped 1-1 onto Im for some natural >number m > n] The contradiction should disappear once you specify exactly what _language_ youre talking about. (In what seems to me to be a reasonable choice of languages, S is finite [by the definition above] is not described by a single formula, so the collection of sentences above is not a collection of sentences and the compactness theorem does not apply. Exactly what language do you have in mind?) ************************ David C. Ullrich === Subject: Regular Transversality Epigone-thread: glabeusnay Hi everybody, What is Regular Transverality? (in the differential geometry context) Where should i look for it? Nir === Subject: Re: Regular Transversality > Hi everybody, > What is Regular Transverality? (in the differential geometry context) > Where should i look for it? A k-plane and an l-plane in R^n intersect transversely if their intersection is a (k+l-n)-plane, which is the generic case, that is, it happens for almost-all such planes. We think of this condition more easily in terms of the codimensions, rather than dimensions. Those planes are of codimension (n-k) and (n-l), respectively, and they are transverse if the intersection is of codimension (n-k)+(n-l). Two submanifolds of, say, R^n intersect transversely if their tangent planes do, at each intersection point. This is the condition that is needed to apply the implicit function theorem, so that the intersection of those two submanifolds is again a submanifold, of the right dimension. So, a plane through the origin in R^3 intersects the unit sphere transversely, and the intersection is a circle (2+2-3=1), but a plane tangent to that sphere intersects it non-transversely. There is an old book by Golubitsky and Guillemin that has a very thorough treatment of this, but any book on differential topology should discuss it some. -- David L. Johnson __o | If all economists were laid end to end, they would not reach a _`(,_ | conclusion. -- George Bernard Shaw (_)/ (_) | === Subject: Regular Transversality Epigone-thread: glabeusnay g:A->B , f:[0,1]->B such that f is transverse to the set of critical values of g. my question is if g^{-1}(im(f)) is a smooth sub-manifold of A. nir === Subject: 7 colors theorem Epigone-thread: trahshampex If we add all edges between two vertices which satisfy the following condition on a planar graph G, then lets denote this graph with G_{m,n}. C. n paths of length m exist between two vertices. I proved the chromatic number of G_{2,3} is 7. [Necessity of the theorem] Consider a planar graph G7 as follows. vertices{0,1,2,3,4,5,6} , edges{02, 03, 04, 05, 06, 23, 34, 45, 56, 62, 12, 13, 14, 15, 16} G7_{2,3} is a perfect graph K_7, so 7 colors are necessary. [Sufficiency of the theorem] It is complicated but no difficulty like Four color problem exists. See my HP. http://boat.zero.ad.jp/~zbi74583/another02.htm It is easy to prove that Kai(G_{2,1})= infinity and Kai(G_{2,2})= infinity. Where Kai(H) means the chromatic number of H. Yasutoshi Kohmoto === Subject: thank you! Epigone-thread: hurrursex solution was the one Gareth McCaughan proposed too; I considered it not pretty since not easy to explain in only two lignes. The perfect difference sets and perfect rulers suggested by Jim Nastos were a little bit too constrained, but allowed me to start the researches on difference sets and rulers, and I found the Golomb rulers which are exactly what I need. Sidon sequences are, as Gerry Myerson noticed, almost what I need but not really the same (because of the condition on 2a_j). This was the first time I put a question on a mathematical forum, but I think not the last one! Irena. === Subject: easy calculation of full conditionals, bayesian statistcs, mcmc the moment I am getting a bit depressed because I am stuck alread at the first ...can be easily found! So I would appreciate it if you can show me, how *easily* the following calculation of full conditionals can be done: Given a random sample of size n from N(mu, sigma^2). We place independent priors on mu and sigma: mu ~ N(xi,1/kappa) sigma^(-2) ~ Gamma(alpha,beta) the resulting posterior is proportional to p(Y|sigma, mu) * p(mu) * p(sigma) so we have (really trivially this time) p(mu,sigma^-2 | Y) proportional to (sigma^-2)^(alpha+n/2-1) * exp{-beta/sigma^2)-(kappa(mu-xi)^2)/2- 1/2sigma^2 * sum(Y_i-mu)^2 } =: POST Ok, now the claim is that it is easy to compute the distribution of the full conditionls mu|sigma,Y and sigma|mu,Y the result is (according to Green): mu|sigma,Y ~ N( (sigma^-2 * sum(y_i) + kappa*xi) / sigma^-2*n+kappa , 1/(sigma^-2*n+kappa)) (and some result for sigma|mu,Y) But how is this computed? I would propose to do it in the following way assuming there is only one Y: p(mu|sigma,Y)= p(mu,sigma | Y) / Int(P(mu,sigma| Y) dp(mu) = p(mu,sigma | Y) / { Int(P(mu,sigma| Y) normal_xi,kappa (mu) } instead of P(mu,sigma| Y) I take POST as the denominators p(mu|sigam,Y) = normal_xi,kappa (mu) * normal_mu,sigma(Y) / int{ normal_xi,kappa(mu) ^2 * normal_mu,sigma (Y) d mu, mu=-infty...infty) Now one would have to solve this. Solving it does not give exactly the claimed solution, but something with a factor of this: exp(-1/2*kappa*(xi-y)^2/((2*kappa*sigma^2+1)*(kappa*sigma^2+1 )))*(-(2*kappa* sigma^2+1)/sigma^2)^(1/2)*2^(1/2)*Pi^(1/2)*(sigma^2/(kappa* sigma^2+1))^(1/2) /((1/kappa)^(1/2)*kappa) away from it. Notice that the factor is independent of mu! So I think by now you say: This cretin! It is so easy, why didnt he see that it can be done much easier (and correctly) in the following way... Well, well, so tell me, I am eager to find out how this really should be done... Marc. -- Institut f.9fr Theoretische Informatik Marc Nunkesser Clausiusstrasse 49 ETH Zentrum CLW B2 CH-8092 Z.9frich === Subject: Irrational numbers Epigone-thread: yodimclim Several participants in this thread have indicated that mathmanÇs second question has the answer no, i.e. you donÇt get anything new if you replace rational coefficients by algebraic ones. This is best understood by considering that a number is algebraic if and only if it is algebraic over Q, the field of rationals. Now if K is a field, and c is an element of an extension of K then c is algebraic over K if and only if it is contained in some field having finite extension degree over K. - Suppose c is a root of a polynomial whose coefficients are algebraic numbers. These coefficients generate a finite extension, say K, of Q. Now, c is algebraic over K, hence contained in some finite extension of K which obviously also has finite degree over Q. === Subject: Re: How to enforce Positive Definiteness ? The easiest way to check for positive semi-definiteness is to attempt to > do Cholesky factorisation and see if it works (no negative numbers in > the square-roots). Hence you can ensure positive semi-definiteness by > adding to the diagonal elements so that you never get negative numbers > in the square-roots when doing the Cholesky factorisation. This is very > easy to implement algorithmically. A minor add-on comment that you probably already know (but maybe other readers dont). A matrix is positive definite, if the diagonal is positive, and the sums of the absolute values of nondiagonal elements in each row and column are less than the corresponding diagonal element in that row or column. So just add a large enough multiple of the identity matrix (for example) and you will make the matrix positive definite with this easy to compute test. Karl Hallowell khallow@hotmail.com === Subject: Re: How to enforce Positive Definiteness ? > A minor add-on comment that you probably already know (but maybe other > readers dont). A matrix is positive definite, if the diagonal is > positive, and the sums of the absolute values of nondiagonal elements > in each row and column are less than the corresponding diagonal > element in that row or column. So just add a large enough multiple of > the identity matrix (for example) and you will make the matrix > positive definite with this easy to compute test. In fact, it suffices if that condition is true for all the rows, *or* for all the columns. -- Gareth McCaughan sig under construc === Subject: Geometric Algebra group to write a scientific review paper concerning mathematics curriculum, with an emphisis on promoting future inclusion of Geometric Algebra. If anyone has the time and is willing to answer a few questions as they arrise; please email me at Funkybside@yahoo.com to discuss. Joe Presson === Subject: Is the non-existence of odd perfect number proved? Im reading some paper written by Simon Davis, A proof of odd perfect number conjecture. I think some people read it already. I didnt done it yet. But I wonder whether the paper is right or wrong. Is there somebody who read it and can comment about it? For that paper, see http://arxiv.org/abs/hep-th/0401052 === Subject: Re: Is the non-existence of odd perfect number proved? Epigone-thread: baislumbleh > Im reading some paper written by Simon Davis, A proof of odd perfect > number conjecture. I think some people read it already. I didnt done > it yet. If you ever do done it, you may be the first. > But I wonder whether the paper is right or wrong. Is there somebody > who read it and can comment about it? Unknown, but Robin Chapman and others tried to read it and got through parts of it. See http://www.mathforum.org/discuss/sci.math/t/569501 . I found Robins analysis useful in understanding what the paper was driving at. I havent heard from anyone who has teased a valid proof out of it. Dan Hoey haoyuep@aol.com === Subject: Re: Is the non-existence of odd perfect number proved? My 2 cents: The paper is poorly written. I cant say its right or wrong since I gave up trying to understand it. IMHO, it should be rewritten and a revised version posted. >Im reading some paper written by Simon Davis, A proof of odd perfect >number conjecture. I think some people read it already. I didnt done >it yet. >But I wonder whether the paper is right or wrong. Is there somebody >who read it and can comment about it? >For that paper, see >http://arxiv.org/abs/hep-th/0401052 === Subject: Re: irrational numbers > Mathematics > at http://members.aol.com/jeff570/mathword.html I copy the following > relevant statements. See the website for a few more details: > ALGEBRAIC NUMBER. According to an Internet web page, this term was used by > Abel. > Algebraic quantity appears in 1673 in Elements of Algebra (1725) by John > Kersey: Two or more Algebraic quantities (OED2). > Algebraic (in the sense of an algebraic number) is found in 1840 in > Mathematical Dissertations (1841) by J. R. Young [James A. Landau]. In Lutzens book Joseph Liouville, 1809-1882: Master of Pure and Applied Mathematics I found this quote from Lagrange in 1794: It is probable that the number pi is not even comprised among the algebraic irrationals, that is, it cannot be the root of an algebraic equation of a finite number of terms whose coefficients are rational, but it seems very difficult to prove this proposition rigorously. Lutzen also refers to Waldschmidt, Cahiers du Seminaire dHistoire des Mathematiques 4 (1983) 93-115, for more on the history of algebraic and transcendental numbers, but I wasnt able to find this in our library. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: irrational numbers >Lutzen also refers to Waldschmidt, Cahiers du Seminaire dHistoire >des Mathematiques 4 (1983) 93-115, for more on the history of >algebraic and transcendental numbers, but I wasnt able to find >this in our library. That paper exists both in French and in Bulgarian, see the following ZentralblattMATH entries: Zbl 0507.01004 Waldschmidt, Michel Les debuts de la th.8eorie des nombres transcendants (a loccasion du centenaire de la transcendance de pi). (French) [J] Cah. Semin. Hist. Math. 4, 93-115 (1983). See printed version [of Zentralblatt f.9fr Mathematik u. ihrer Grenzgebiete] MSC 2000: *01A55 Mathematics in the 19th century 01A50 Mathematics in the 18th century 11J81 Transcendence (general theory) 11-03 Historical (number theory) Keywords: transcendental numbers; algebraic numbers; Ch. Hermite; J. Liouville; J. Cantor; F. v. Lindemann; axiom of choice; constructive proofs Zbl 0658.01007 Waldschmidt, Michel Origins of transcendental number theory. On the centennial of the proof of transcendence of the number pi . (French, Bulgarian) [J] Fiz.-Mat. Spis. 26(59), 167-181 (1984); translation from Cah. S.8emin. Hist. Math. 4, 93-115 (1983). [ISSN 0015-3265] See the review in Zbl 0507.01004. [ see above ] MSC 2000: *01A55 Mathematics in the 19th century 11J81 Transcendence (general theory) 11-03 Historical (number theory) Keywords: transcendental number; algebraic numbers; Ch. Hermite; J. Liouville; J. Cantor; F. v. Lindemann; axiom of choice; constructive proofs Citations: Zbl 0507.01004 The centennial obviously refers to Ferdinand Lindemanns proof in 1882, see the JFM entry http://makeashorterlink.com/?E22A22147 and Lindemanns full paper (in German): http://134.76.163.65/agora_docs/35867TABLE_OF_CONTENTS.html --> p. 213-225 Hermann -- === Subject: Re: irrational numbers |No. Liouville was the first to prove their existence, in his paper, |Sur des classes tr`es-etendues de quantites dont la valeur |nest ni algebrique, ni m^eme reductible `a des irrationnelles |algebriques, C.R. 18 (1844) 883-5. What does the phrase reducible to algebraic irrationalities mean? The way he used it suggests something broader than algebraic. Keith Ramsay === Subject: Re: irrational numbers >|No. Liouville was the first to prove their existence, in his paper, >|Sur des classes tr`es-etendues de quantites dont la valeur >|nest ni algebrique, ni m^eme reductible `a des irrationnelles >|algebriques, C.R. 18 (1844) 883-5. >What does the phrase reducible to algebraic irrationalities mean? >The way he used it suggests something broader than algebraic. Perhaps it means a root of a polynomial with algebraic coefficients. Which of course is the same as algebraic; I suspect Liouville knew this too, but maybe its just stated for the sake of emphasis. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: irrational numbers You get the same set. Suppose your polynomial is sum of a_n*X^n for n=0 to n=N, where a_0,...,a_N are algebraic numbers. For each coefficients a_n, look at all of its Galois conjugates. That is, look at the roots of the minimal polynomial for a_n in Q[X]. Let that set of conjugates be the set A_n = {a_{n,i} for i running from 1 to d_n}. Now look at all possible choices of (N+1)-tuples (b_0,b_1,...,b_N), where b_0 is chosen from A_0, b_1 from A_1, etc. For each such (N+1)-tuple, form the polynomial b_0 + b_1*X + b_2*X^2 + ... + b_N*X^N. Now multiply all those polynomials together and call the result F(X). (1) The roots of the original polynomial are also roots of F(X), since the original polynomial appears as one of the factors of F(X). (2) The coefficients of F(X) are rational numbers, because the coefficients are symmetric polynomials in the conjugates of each of the a_ns (individually). In fancier terms, let K be the field generated by all of the a_{n,i}s. Then the Galois group Gal(K/Q) fixes the coefficients of F(X), which proves that the coefficients are rational numbers. (3) If you want a polynomial with integer coefficients, just multiply F(X) by a large integer to clear the denominators. Joe Silverman PS I dont know anything about the history of the terminology, but you might also want to look at the various different sorts of transcendental numbers, defined by their approximation properties. There are U numbers and Liouville numbers and various other sorts. I know that Kurt Mahler did a lot of work on this topic, but I dont know if he invented the terms. Anyway, your statement that Irrational numbers are divided into two classes, algebraic and transcendental is really just a first approximation!! JS > Irrational numbers are divided into two classes, algebraic (roots of > polynomial equations with integer coefficients) and transcendental. > Im interested in the following questions. > > 2. If we made the coefficients algebraic numbers instead of integers > (in the polynomials), would we get a bigger class? Its clears that > up to fourth degree we wont. === Subject: Re: irrational numbers , > Interesting... I was using MathWorld as a reference, and I will quote > the relevant section: > Georg Cantor was the first to prove the existence of > transcendental numbers. Liouville subsequently showed how to > construct special cases (such as Liouvilles constant) using > Liouvilles approximation theorem. > As usual, we have discrepancies in math history. With all due respect, I think that, as usual, we have a mistake in MathWorld. I trust someone will notify the author. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: when are polynomials equal mod q? Do you mean that the values of the polynomials are equal for all z in Z_q or that they are equal as polynomials (so their values are equal even over infinite extension rings containing Z_q)? >Let f(z) and g(z) be polynomials in z with coefficients in Z_q >and that both factor over Z_q. Im looking for necessary and >sufficient conditions in terms of the multiplicities of >the roots so that f(z) = g(z). For example, mod 4, >(z-1)^2*(z-2)^2*(z-3)^6 = (z-0)^2*(z-1)^8. Are such conditions >known? If not in general what about for prime powers? >-Frank R. === Subject: Re: when are polynomials equal mod q? Hi David, Things like infinite extension rings are currently outside my lexicon. I think of z as an indeterminate. I will not be giving it a value (although a substitution like z <-- 3z is possible). Let me give an example of the sort of think Im looking for: Consider the equation (with arithmetic mod 4) (z-0)^a[1] (z-1)^a[2] (z-2)^a[2] (z-3)^a[3] = (z-0)^b[1] (z-1)^b[2] (z-2)^b[2] (z-3)^b[3], like the example in my original posting. This holds if and only if all of the following three conditions are true: (a) a[i] = b[i] mod 2 (b) a[0]+a[2] = b[0]+b[2] (c) a[1]+a[3] = b[1]+b[3] I.e., expand the products, reduce mod 4, and the coefficients of the two polynomials are then identical. But is it known already or am I covering old ground? What about q=8 instead of q=4? What about q=2^n? What about q=p^n with p prime? about previous results on this or closely related problems. -Frank R. > Do you mean that the values of the polynomials are > equal for all z in Z_q or that they are equal as polynomials > (so their values are equal even over infinite extension rings > containing Z_q)? >>Let f(z) and g(z) be polynomials in z with coefficients in Z_q >>and that both factor over Z_q. Im looking for necessary and >>sufficient conditions in terms of the multiplicities of >>the roots so that f(z) = g(z). For example, mod 4, >>(z-1)^2*(z-2)^2*(z-3)^6 = (z-0)^2*(z-1)^8. Are such conditions >>known? If not in general what about for prime powers? >>-Frank R. === Subject: Re: when are polynomials equal mod q? >Hi David, >Things like infinite extension rings are currently >outside my lexicon. I think of z as an indeterminate. >I will not be giving it a value (although a substitution >like z <-- 3z is possible). Let me give an example of the >sort of think Im looking for: >Consider the equation (with arithmetic mod 4) > (z-0)^a[1] (z-1)^a[2] (z-2)^a[2] (z-3)^a[3] = > (z-0)^b[1] (z-1)^b[2] (z-2)^b[2] (z-3)^b[3], >like the example in my original posting. There are some obvious typos here. >This holds if and only if all of the following three >conditions are true: > (a) a[i] = b[i] mod 2 > (b) a[0]+a[2] = b[0]+b[2] > (c) a[1]+a[3] = b[1]+b[3] >I.e., expand the products, reduce mod 4, and the >coefficients of the two polynomials are then identical. >But is it known already or am I covering old ground? >What about q=8 instead of q=4? >What about q=2^n? What about q=p^n with p prime? Ill call the following conjecture the obvious generalization of your result: Let q = p^n, p prime. Let r = q/p. Let f in R= Z_q[z] be a poly which splits completely over Z_q, f(z) = (z-0)^e0 ... (z-q-1)^e(q-1), and e = (e0,e1,...,e(q-1)). Define the invariant by inv(f) = (e mod r, e0 + e(r), e1 + e(r+1), ..., e(q-r+-1) + e(q-1)). Conj: f(z) = g(z) in R iff inv(f) = inv(g). Ive checked this using maple for some examples. Sorry, I have no references to suggest for this. >about previous results on this or closely related problems. >-Frank R. >>Do you mean that the values of the polynomials are >>equal for all z in Z_q or that they are equal as polynomials >>(so their values are equal even over infinite extension rings >>containing Z_q)? >Let f(z) and g(z) be polynomials in z with coefficients in Z_q >and that both factor over Z_q. Im looking for necessary and >sufficient conditions in terms of the multiplicities of >the roots so that f(z) = g(z). For example, mod 4, >(z-1)^2*(z-2)^2*(z-3)^6 = (z-0)^2*(z-1)^8. Are such conditions >known? If not in general what about for prime powers? >-Frank R. === Subject: Re: when are polynomials equal mod q? >Hi David, >Things like infinite extension rings are currently >outside my lexicon. I think of z as an indeterminate. >I will not be giving it a value (although a substitution >like z <-- 3z is possible). Let me give an example of the >sort of think Im looking for: >Consider the equation (with arithmetic mod 4) > (z-0)^a[1] (z-1)^a[2] (z-2)^a[2] (z-3)^a[3] = > (z-0)^b[1] (z-1)^b[2] (z-2)^b[2] (z-3)^b[3], >like the example in my original posting. > There are some obvious typos here. >This holds if and only if all of the following three >conditions are true: > (a) a[i] = b[i] mod 2 > (b) a[0]+a[2] = b[0]+b[2] > (c) a[1]+a[3] = b[1]+b[3] >I.e., expand the products, reduce mod 4, and the >coefficients of the two polynomials are then identical. >But is it known already or am I covering old ground? >What about q=8 instead of q=4? >What about q=2^n? What about q=p^n with p prime? > Ill call the following conjecture the obvious generalization of your > result: > Let q = p^n, p prime. Let r = q/p. Let f in R= Z_q[z] be a poly which > splits completely > over Z_q, f(z) = (z-0)^e0 ... (z-q-1)^e(q-1), and > e = (e0,e1,...,e(q-1)). Define the invariant by > inv(f) = (e mod r, e0 + e(r), e1 + e(r+1), ..., e(q-r+-1) + e(q-1)). > Conj: f(z) = g(z) in R iff inv(f) = inv(g). What about q = 25, r = 5, f(z) = z (z + 20), g(z) = (z + 10)^2 where e0 + e5 = 1 for f and 0 for g? Or maybe you mean e0+er+...+e((p-1)r), e1 + e(r+1) +...+e((p-1)r+1), ..., e(r-1)+e(2r-1)+...+e(q-1). > Ive checked this using maple for some examples. Sorry, I have no > references to suggest for this. Heres a somewhat different approach. Consider the case n=2 (although I think this can be generalized), and write f(z) = product_{j=1}^m (z + a_j + b_j p) where a_j and b_j are in {0,...,p-1}. Of course the a_j are determined by f (up to a permutation) because of unique factorization in Z_p[z]. Expanding out the product, the coefficient of p is sum_j b_j product_{k <> j} (z + a_k) These uniquely determine the b_j if and only if the matrix A is nonsingular mod p, where A_{ij} is the coefficient of b_i z^j in this expression. Now det(A) is a polynomial, homogeneous of degree m(m-1)/2, in the a_j, which is 0 when any two a_j are equal. So, up to a constant factor, it must be product_{i <> j} (a_i - a_j). I think the constant factor is actually (+/-) 1. If so, then A is nonsingular if and only if at least two of the a_j are equal. For example, if a_1 = a_2, the first two columns of A are equal, and the nullspace of A contains [1,-1,0...0]. This corresponds to the fact that (z + a + b p)(z + a + c p) mod p^2 depends only on b+c mod p. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: when are polynomials equal mod q? >Consider the case n=2 (although I think this can be generalized), and write >f(z) = product_{j=1}^m (z + a_j + b_j p) >where a_j and b_j are in {0,...,p-1}. Of course the a_j are determined by >f (up to a permutation) because of unique factorization in Z_p[z]. >Expanding out the product, the coefficient of p is >sum_j b_j product_{k <> j} (z + a_k) >These uniquely determine the b_j if and only if the matrix A is nonsingular >mod p, where A_{ij} is the coefficient of b_i z^j in this expression. Now >det(A) is a polynomial, homogeneous of degree m(m-1)/2, in the a_j, which is >0 when any two a_j are equal. So, up to a constant factor, it must be >product_{i <> j} (a_i - a_j). I think the constant factor is actually >(+/-) 1. If so, then A is nonsingular if and only if at least two of the >a_j are equal. For example, if a_1 = a_2, the first two columns of A are >equal, and the nullspace of A contains [1,-1,0...0]. This corresponds to >the fact that (z + a + b p)(z + a + c p) mod p^2 depends only on b+c mod p. Things seem to be more complicated for n > 2. In order to have product_{j=1}^m (z + p a_j) = product_{j=1}^m (z + p b_j) mod p^n, you need s_k(a_1,...,a_m) = s_k(b_1,...,b_m) mod p^{n-k} for k = 1 to min(m,n-1) where s_k are the elementary symmetric functions. For example, if r <> 1 is an mth root of 1 mod p, then I think there are a_j, 0 <= j <= m-1, with a_j = b^j mod p, such that product_{j=0}^{m-1} (z - p a_j) = z^m mod p^m. Thus 2 is a 5th root of 1 mod 31, and (z+31+754*31^2)(z+2*31+465*31^2)(z+2^2*31+20*31^2)(z+2^3*31+ 599*31^2) (z+2^4*31+27952*31^2) = z^5 mod 31^5. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: A question about a prime divisor of a number. ********************************************** p, q : prime numbers p > q > 3 Prove that 1 + p + p^2 + ... + p^(q-1) has a prime divisor which is greater than p. ********************************************** Is it possible to prove this question? Or are there papers or theorems or properties whatever that is related or helpful in this question? === Subject: Re: A question about a prime divisor of a number. >********************************************** >p, q : prime numbers >p > q > 3 >Prove that 1 + p + p^2 + ... + p^(q-1) has a prime divisor which is >greater than p. >********************************************** >Is it possible to prove this question? No. Take p=7307 and q=5. Rich === Subject: Re: A question about a prime divisor of a number. >********************************************** >p, q : prime numbers >p > q > 3 >Prove that 1 + p + p^2 + ... + p^(q-1) has a prime divisor which is >greater than p. >********************************************** >Is it possible to prove this question? > No. Take p=7307 and q=5. > Rich 1+7307+7307^2+7307^3+7307^4 = 2851122443841001 = (11)(151)(191)(911)(1481)(6661) === Subject: Re: A question about a prime divisor of a number. >>********************************************** >>p, q : prime numbers >>p > q > 3 >>Prove that 1 + p + p^2 + ... + p^(q-1) has a prime divisor which is >>greater than p. >>********************************************** >>Is it possible to prove this question? >> No. Take p=7307 and q=5. >> Rich >1+7307+7307^2+7307^3+7307^4 = 2851122443841001 > = (11)(151)(191)(911)(1481)(6661) There is also a counterexample with q = 7: |^/| Maple 9 (SUN SPARC SOLARIS) MAPLE / All rights reserved. Maple is a trademark of <____ ____> Waterloo Maple Inc. | Type ? for help. > p := 493397; p := 493397 > ifactor(p); (493397) > ifactor((p^7 - 1)/(p-1)); 2 (29) (127) (1163) (71359) (4229) (138349) (2129) (26041) (50177) For q = 11, well need to try about 10^10 values of (p^11 - 1)/(p - 1) before we find one with all prime factors below p. Add in the condition that p be prime, and we will need p ~= 10^12. -- John Adams served two terms as Vice President and one as President, but lost reelection. Later his son became President despite losing the popular vote. That son lost his reelection attempt badly. Now history is repeating itself. pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: Re: A question about a prime divisor of a number. >********************************************** >p, q : prime numbers >p > q > 3 >Prove that 1 + p + p^2 + ... + p^(q-1) has a prime divisor which is >greater than p. >********************************************** >Is it possible to prove this question? > No. Take p=7307 and q=5. Better yet (smaller number) with q=4 and p=7, the expression factors to (2^4)(5^2), so no prime over 7. J === Subject: Re: A question about a prime divisor of a number. >>********************************************** >>p, q : prime numbers >>p > q > 3 >>Prove that 1 + p + p^2 + ... + p^(q-1) has a prime divisor which is >>greater than p. >>********************************************** > Better yet (smaller number) with q=4 and p=7, the expression factors to >(2^4)(5^2), so no prime over 7. Did you miss the prime numbers? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Rational knots...Transcendent knots? If you circle around a single crossing of a knot and cut that out, you get a (4-)tangle: __/ |t | |__| / Tangle addition: __/--__/ |t1| |t2| |__| |__| / --/ Multiplication: same, only do mirror and turn by 90deg on t2 before adding. (Cant do that properly in ASCII :-) __/--__/ |t1| |N | |__| |&_| / --/ Doing this recursively, you get all rational tangles. Now, assume you invent a new operation root: same as addition, only do turn by 90 deg on t2. Clearly, you now also get nonalternating tangles, so the set of this irrational tangles is bigger. Call it algebraic irrational tangles if you like - but are there transcendent tangles that are not in this set either? (The tangle obtained by cutting out a crossing of the knot 8_18 comes to mind.) -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de als man ankam wollte man werden, die geschichte schreiben, die doofen sollen sterben, der plan als man damals nach hamburg kam (Kettcar) === Subject: manifold buisness Epigone-thread: clurdwongsung Hi everybody, there is the well known theorem: f:N -> P and and Q smooth submanifold of P with f transverse to Q then f^{-1}(Q) is a smooth submanifold of N. is there an analogous of this theorem i.e. something like: for two transverse maps f:N -> P, g:A -> P => f^{-1}(Im(g)) is a smooth submanifold? === Subject: Re: manifold buisness > Hi everybody, > there is the well known theorem: > f:N -> P and and Q smooth submanifold of P with f transverse to Q then > f^{-1}(Q) is a smooth submanifold of N. > is there an analogous of this theorem i.e. something like: for two > transverse maps f:N -> P, g:A -> P => f^{-1}(Im(g)) is a smooth > submanifold? If f is the identity, and g is not an immersion, then, because codim(f_*(T_*(N)))=0, by the usual definitions of transversality f and g are transverse, but f^{-1}(Im(g)) is not a smooth submanifold. -- David L. Johnson __o | Business! cried the Ghost. Mankind was my business. The common _`(,_ | welfare was my business; charity, mercy, forbearance, and (_)/ (_) | benevolence, were, all, my business. The dealings of my trade were but a drop of water in the comprehensive ocean of my 2 can be given through a bivariate polynomial - on other words not every such curve is plane. H === Subject: Complexity of solving N non-linear equations Epigone-thread: frixbeiju I must admit I know a little about it, and I really nead some help. Can anyone tell me what is the complexity of solving a set of N non-linear equations (simultaneous equations - having all variables in each equation). The equations themselves are very complex - contains integrals of a sample maximum functions. I know the question is somehow vague, however an answer of type generally it is exponential is also very helpful for me. David === Subject: Relations between Orthogonal Vectors Epigone-thread: skoxclirclong I have the following problem. Does anyone have a clue on this? Consider two Sets A and B of n-dimensional vectors. A contains ALL the vectors with exactly Ôi 1s (and Ôn-i 0s), and B contains all the vectors with Ôj 1s (and Ôn-j 0s). WLOG, assume i < j <= n. Now consider the folowing game: at each round, you remove one random vector from A (if it is not exhausted yet), and all of its orthogonal vectors (vecotrs whose position wise boolean AND yeild the vecotr 000...000) from B (if there is any left). For example. for i=2, j=3, and n=6, if you remove 110000 from A, you also remove 001110, 000111, 001011, 001101 from B. Note that some of these vectors from B might have already removed by some earlier removal from A. Is it possible to find a closed form of the function F(r), # of vectors remaining in the set B as a function of the number of removals r from A? In case it is not possible, a reasonably tight upper bound would suffice. What makes me confusing is the fact that the required number of removal from B depends on the previous removals from A and sometimes we may not need to remove anything from B (for the vectors need to be emoved has already been removed from B as the result of previous removals from A). I tried to get an upper bound by an straight line from (0, B(n, j)) to (B(n, i), 0), B(n, x) represends the number of n-dimensional vectors with exactly x 1s. This would be sufficient for me, but I could not prove the bound. I tried to attack the problem with the incusion-exclusion principle, but that did not help either. Suman