mm-323 === Subject: : Rebonato - Interest rate option models 2nd edition - queryIn Rebonato's book, equation 5.11 appears to me to be incorrect, sincePi(t) on the LHS is a function of t, but the RHS depends on t1, t2, T1and T2. Is this a typographical error? What should this equation be?=== === Subject: : Re: Rebonato - Interest rate option models 2nd edition - queryIn Rebonato's 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'team leader' at www.markjoshi.com or otherwise try www.wilmott.com.---remove the no for mail=== === Subject: : An Inequality Rela to the MVT for IntegralsThe MVT for integrals roughly says that if h has constant sign, and gis continuous then int g(x)h(x) = g(x*) int h(x) for some interiorx*. I need a simple (large) class of functions g yielding an integralinequality of the form:(**) int_0^1 g(x)h(x) >= 0 forh single crossing (but not necessarily monotonic) through 0 from + toweakly negative, with int_0^1 h(x) >0andg increasing (but not necessarily continuous) on [0,1], rising fromg(0)>=0 to g(1)=1.I agree that as set up, g and h in some nondescript sense arenegatively correla, but observe that (**) holds if g is constantwhenever h<>0. More generally, (**) would also obtain if the MVT (withmy modified premises) held with an inequality >= sign (though I do notclaim such an extension is valid).I have explored inequalities due to Beesack (1957) and Dallas Banks(1963), but their assumptions are viola here.Cheers! - (Prof.) Lones Smith, Economics, Michigan=== === Subject: : Re: An Inequality Rela 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 positivebump, 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, it's 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, thenint_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 >= 0If you want to add the additional condition that g is non-decreasing,you certainly don't get a large class of such functions: they're allconstant (except possibly at the endpoints). Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Choosing a Math Grad schoolHi guys,I'd like some info about math PhD programs, if anyone can please help. Aconcern I have are rankings, in particular: how much do they matter when youmake *final* decisions?I've my hopes fixed on one among: Berkeley, UCLA, Penn State, Chicago,Texas-Austin, Princeton, Illinois-Urbana. My area of interest is operatoralgebras (thus the first three) or Harmonic analysis and PDE (thus thelatter four). Depending on place-offers (and assuming that one of them canbe excluded), do any of you guys know reasons for preferring one to theother? (modulo that these are all great places for analysis research).For instance, suppose you get an offer from UCLA and Illinois-Urbana or PennState, do you obviously go for UCLA because it is higher ranked? Myintention is to follow an academic career (professorship), so futureplacement prospects are crucial to me.Thanks for any comments, particularly from mathematicians that have been orhave visi the above mentioned universities and may be able to give mesome hands-on impressions!Please email comments to:hcanab@yahoo.comStefan=== === Subject: : Re: Choosing a Math Grad school> I'd 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 studiessomething you are (or might be) interes in, it's no good for you. Bestto go where there are several students working on similar areas -- butnot too many.> I've 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 thestate 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 themwould be an impediment to your career plans. Find the best fit for yourather than worrying about which is #1 versus #3. Look at the courses, and seminars, that are being offered now (not justwhat is in the catalogue). Look over the research interests of thefaculty. Think about the size of the faculty, and students. Berkeley hasa large and excellent faculty, but a lot of grad students to compete fortheir 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 ina 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,>I'd 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?>I've 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.>Thanks for any comments, particularly from mathematicians that have been or>have visi the above mentioned universities and may be able to give me>some hands-on impressions!>Please email comments to:>hcanab@yahoo.com>>Stefanbudget makes severe cuts to UC funding. Graduate tuition feesare 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 lostreelection. 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: : Non-standard analysisConsider 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 lessthan or equal to j.Define a set S to be finite if there is a 1-1 mapping of S onto In forsome natural number n.Define a finite set S to be larger than a natural number n, if thereis 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 bethe largest index of a sentence in the finite subset, then, forexample, Ij+1 provides the model] By compactness, the whole collection is consistent and therefore thereis a model with the property that there exists a finite set S which islarger than n for any n. This is clearly a contradiction.[S finite implies by definition that there exists some natural numbern, 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 naturalnumber m > n]=== === Subject: : Re: Non-standard analysisOriginator: 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 languageyou're working in. Since you appeal to compactness, the most obviouscandidates are the first-order language of arithmetic and the first-orderlanguage of set theory. S is finite is more straightforwardly expressedin set theory than in arithmetic, so let's think about what happens inthat case. You then need to be careful about the concept of the setof all natural numbers. We would like to define this by omega ={1, 2, 3, ...} or something like that, but the trouble is that formulaeof infinite length are not allowed in first-order logic. (You can definelogics that allow them, but then you don't get a compactness theoremfor 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 omegaand 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 there's no way to force thesymbol omega to be interpre as the true set of natural numbers,i.e., we can't rule out the possibility of nonstandard models of settheory in which omega is interpre as a set containing nonstandardintegers. So now let's look at your definition: S is finite if thereis a 1-1 mapping of S onto In for some natural number n. If by somenatural number n you mean some n in omega then S could be mappedonto In for some *nonstandard* n, and you don't get any contradictionfrom your infinite list of conditions. On the other hand, if by somenatural number n you mean n = 1 or n = 2 or ... then again you runinto difficulties because you don't have infinitely long formulae infirst-order logic.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe 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 you're 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 you're 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 thinkyou want to revise these, so they all refer to S, not Sn.And you also don't want the quanitfier There exists;S should be a constant symbol, so the n-th sentencejust 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_ you're talking about. (In what seems to me to be areasonable choice of languages, S is finite [by the definitionabove] is not described by a single formula, so the collection ofsentences above is not a collection of sentences and thecompactness theorem does not apply. Exactly what languagedo you have in mind?)************************=== === Subject: : Regular TransversalityEpigone-thread: glabeusnayHi everybody,What is Regular Transverality? (in the differential geometry context)Where should i look for it?Thanks Nir=== === Subject: : Re: Regular TransversalityHi 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 theirintersection is a (k+l-n)-plane, which is the generic case, that is, ithappens for almost-all such planes. We think of this condition moreeasily in terms of the codimensions, rather than dimensions. Thoseplanes are of codimension (n-k) and (n-l), respectively, and they aretransverse if the intersection is of codimension (n-k)+(n-l). Twosubmanifolds of, say, R^n intersect transversely if their tangent planesdo, at each intersection point. This is the condition that is needed toapply the implicit function theorem, so that the intersection of those twosubmanifolds is again a submanifold, of the right dimension. So, a planethrough the origin in R^3 intersects the unit sphere transversely, and theintersection is a circle (2+2-3=1), but a plane tangent to that sphereintersects it non-transversely. There is an old book by Golubitsky and Guillemin that has a very thoroughtreatment of this, but any book on differential topology should discuss itsome.-- David L. Johnson __o | If all economists were laid end to end, they would not reach a _`(,_ | conclusion. -- George Bernard Shaw (_)/ (_) | === === Subject: : 7 colors theoremEpigone-thread: trahshampex If we add all edges between two vertices which satisfy thefollowing condition on a planar graph G, then let's denote this graphwith 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 complica but no difficulty like Four color problemexists. 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: hurrursexThank you to all of you, who answered me. My not-pretty-at-allsolutionwas the one Gareth McCaughan proposed too; I considered it not prettysince not easy to explain in only two lignes. The perfect differencesets and perfect rulers sugges by Jim Nastos were a little bit tooconstrained, but allowed me to start the researches on difference setsand rulers, and I found the Golomb rulers which are exactly what Ineed.Sidon sequences are, as Gerry Myerson noticed, almost what I needbut not really the same (because of the condition on 2a_j).This was the first time I put a question on a mathematical forum, butI think not the last one! Thank you again and have a lot of fun on the forum!Irena.=== === Subject: : easy calculation of full conditionals, bayesian statistcs, mcmcthe moment I am getting a bit depressed because I am stuck alread at thefirst ...can be easily found!So I would appreciate it if you can show me, how *easily* the followingcalculation 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 } =: POSTOk, now the claim is that it is easy to compute the distribution of the fullconditionlsmu|sigma,Y and sigma|mu,Ythe 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 compu? I would propose to do it in the following wayassuming 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 cancel out.Now also the prior on sigma cancels out and I am left with: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 theclaimed solution, but something with a factor ofthis: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 didn't he seethat 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 bedone...Marc.-- Institut f.9fr Theoretische Informatik Marc NunkesserClausiusstrasse 49 ETH Zentrum CLW B2CH-8092 Z.9frich=== === Subject: : Irrational numbersEpigone-thread: yodimclimSeveral participants in this thread have indica that mathmanssecond question has the answer no, i.e. you dont get anything newif you replace rational coefficients by algebraic ones. This is bestunderstood by considering that a number is algebraic if and only if itis algebraic over Q, the field of rationals. Now if K is a field, andc is an element of an extension of K then c is algebraic over K if andonly if it is contained in some field having finite extension degreeover K. - Suppose c is a root of a polynomial whose coefficients arealgebraic numbers. These coefficients generate a finite extension, sayK, of Q. Now, c is algebraic over K, hence contained in some finiteextension of K which obviously also has finite degree over Q. Peter Flor, Graz (Austria).=== === Subject: : Re: How to enforce Positive Definiteness ?> An idea which may or may not be useful: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 otherreaders don't). A matrix is positive definite, if the diagonal ispositive, and the sums of the absolute values of nondiagonal elementsin each row and column are less than the corresponding diagonalelement in that row or column. So just add a large enough multiple ofthe identity matrix (for example) and you will make the matrixpositive definite with this easy to compute test.Karl Hallowellkhallow@hotmail.com=== === Subject: : Re: How to enforce Positive Definiteness ?> A minor add-on comment that you probably already know (but maybe other> readers don't). 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 therows, *or* for all the columns.-- Gareth McCaughansig under construc=== === Subject: : Geometric Algebragroup to write a scientific review paper concerning mathematicscurriculum, with an emphisis on promoting future inclusion ofGeometric Algebra. If anyone has the time and is willing to answer afew questions as they arrise; please email me at Funkybside@yahoo.comto discuss.Thanks!Joe Presson=== === Subject: : Is the non-existence of odd perfect number proved?I'm reading some paper written by Simon Davis, A proof of odd perfectnumber conjecture. I think some people read it already. I didn't doneit yet.But I wonder whether the paper is right or wrong. Is there somebodywho read it and can comment about it?For that paper, seehttp://arxiv.org/abs/hep-th/0401052=== === Subject: : Re: Is the non-existence of odd perfect number proved?Epigone-thread: baislumbleh> I'm reading some paper written by Simon Davis, A proof of oddperfect> number conjecture. I think some people read it already. I didn'tdone> 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 throughparts of it. See http://www.mathforum.org/discuss/sci.math/t/569501 .I found Robin's analysis useful in understanding what the paper wasdriving at. I haven't heard from anyone who has teased a valid proofout of it.Dan Hoeyhaoyuep@aol.com=== === Subject: : Re: Is the non-existence of odd perfect number proved?My 2 cents: The paper is poorly written. I can't say it'sright or wrong since I gave up trying to understand it.IMHO, it should be rewritten and a revised version pos.>I'm reading some paper written by Simon Davis, A proof of odd perfect>number conjecture. I think some people read it already. I didn't 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 Lutzen's 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 d'Histoire des Mathematiques 4 (1983) 93-115, for more on the history of algebraic and transcendental numbers, but I wasn't able to find this in our library.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: irrational numbers|No. Liouville was the first to prove their existence, in his paper, |Sur des classes tr`es-'etendues de quantit'es dont la valeur |n'est ni alg'ebrique, ni m^eme reductible `a des irrationnelles |alg'ebriques, 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.=== === Subject: : Re: irrational numbers>|No. Liouville was the first to prove their existence, in his paper, >|Sur des classes tr`es-'etendues de quantit'es dont la valeur >|n'est ni alg'ebrique, ni m^eme reductible `a des irrationnelles >|alg'ebriques, 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; Isuspect Liouville knew this too, but maybe it's just sta for the sake of emphasis.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2=== === Subject: : Re: irrational numbersYou get the same set. Suppose your polynomial is sum of a_n*X^n for n=0 to n=N, wherea_0,...,a_N are algebraic numbers. For each coefficients a_n, look atall of its Galois conjugates. That is, look at the roots of theminimal polynomial for a_n in Q[X]. Let that set of conjugates be theset 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), sincethe original polynomial appears as one of the factors of F(X).(2) The coefficients of F(X) are rational numbers, because thecoefficients are symmetric polynomials in the conjugates of each ofthe a_n's (individually). In fancier terms, let K be the fieldgenera 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 arerational numbers.(3) If you want a polynomial with integer coefficients, just multiplyF(X) by a large integer to clear the denominators.Joe SilvermanPS I don't know anything about the history of the terminology, but youmight also want to look at the various different sorts oftranscendental numbers, defined by their approximation properties.There are U numbers and Liouville numbers and various other sorts. Iknow that Kurt Mahler did a lot of work on this topic, but I don'tknow if he inven the terms. Anyway, your statement that Irrationalnumbers are divided into two classes, algebraic and transcendental isreally just a first approximation!! JS> Irrational numbers are divided into two classes, algebraic (roots of> polynomial equations with integer coefficients) and transcendental. > I'm interes in the following questions.> 2. If we made the coefficients algebraic numbers instead of integers> (in the polynomials), would we get a bigger class? It's clears that> up to fourth degree we won't.=== === 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 Liouville's constant) using> Liouville's 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.-- === === Subject: : Re: when are polynomials equal mod q?Do you mean that the values of the polynomials areequal for all z in Z_q or that they are equal as polynomials(so their values are equal even over infinite extension ringscontaining Z_q)?>Let f(z) and g(z) be polynomials in z with coefficients in Z_q>and that both factor over Z_q. I'm 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 currentlyoutside my lexicon. I think of z as an indeterminate.I will not be giving it a value (although a substitutionlike z <-- 3z is possible). Let me give an example of thesort of think I'm 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 threeconditions 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 thecoefficients 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?Thanks for any insight you or other readers can giveabout previous results on this or closely rela 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. I'm 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: : A question about a prime divisor of a number.**********************************************p, q : prime numbersp > q > 3Prove that 1 + p + p^2 + ... + p^(q-1) has a prime divisor which isgreater than p.**********************************************Is it possible to prove this question? Or are there papers or theoremsor properties whatever that is rela 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. 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.caDepartment 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 t2before adding. (Can't do that properly in ASCII__/--__/|t1| |N ||__| |&_|/ --/ Doing this recursively, you get all rational tangles.Now, assume you invent a new operation root: same asaddition, only do turn by 90 deg on t2. Clearly, you nowalso get nonalternating tangles, so the set of thisirrational tangles is bigger. Call it algebraicirrational tangles if you like - but are there transcendenttangles that are not in this set either? (The tangle obtainedby cutting out a crossing of the knot 8_18 comes to mind.)-- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.deals man ankam wollte man werden, die geschichte schreiben,die doofen sollen sterben, der plan als man damals nach hamburg kam(Kettcar)=== === Subject: : manifold buisnessEpigone-thread: clurdwongsungHi everybody,there is the well known theorem:f:N -> P and and Q smooth submanifold of P with f transverse to Q thenf^{-1}(Q) is a smooth submanifold of N.is there an analogous of this theorem i.e. something like: for twotransverse maps f:N -> P, g:A -> P => f^{-1}(Im(g)) is a smoothsubmanifold? thanks=== === 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, becausecodim(f_*(T_*(N)))=0, by the usual definitions of transversality f and gare 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