mm-1509 === Subject: Re: on the existence of injective compact bounded linear operators >However,now, I feel uncorfortable with the compact operator definition >(while before this thread, I was OK): can the image of a hilbert space >through a compact operator be an infinite dimensional space? > Certainly. For example T(x_n) = (x_n/x) on H = l^2. (I agree that we have infinite dimensional space for compact operators. I just wanted to understand where my argumentation was wrong. However, with the help of your reply, I know now where I am wrong: I used the property of a continuous linear operator in the wrong sense: i.e. |T(x)| >= M|x| that is wrong (we have |T(x)| <= M|x|), thus, we have K such that B(K.T(H)) included in Closure[T(B(H))] and not the inverse. However, now, I have an other impossibility: Suppose that T is a compact injective operator with a *dense* image on a hilbert space (i.e. closureT(H)=T(H))(Now, we know that H must be separable to have such possibility). Can we say that such operator does not exist? If the answer is yes, can we lower the restrictions such that the existence remains impossible? Seratend. === Subject: Re: on the existence of injective compact bounded linear operators >[...] >However, now, I have an other impossibility: >Suppose that T is a compact injective operator with a *dense* image on >a hilbert space (i.e. closureT(H)=T(H))(Now, we know that H must be >separable to have such possibility). Can we say that such operator does >not exist? I don't understand the question. Partly because you say with a *dense* image on a hilbert space (i.e. closureT(H)=T(H)), but closureT(H)=T(H) does not say T(H) is dense, it says T(H) is closed. >If the answer is yes, can we lower the restrictions such that the >existence remains impossible? >Seratend. ************************ David C. Ullrich === Subject: Re: on the existence of injective compact bounded linear operators <75e7u051070co8mn3o88ob473rkal2hr4f@4ax.com> well, It is again a typo error : ). A closed image must be sufficient for the impossibility (i.e. T(H) is a closed space). So my question is: Can we say that there isn't any injective compact operator with a *closed* image on an infinite dimensional Hilbert space? (the closed image brings the impossbility to get such an operator on an infinite dimensional hilber space) (I think the impossibility should also work this time for the banach space case) Seratend. P.S. For me closure in H of T(H)= T(H) means T(H) is dense in T(H). However, it is very confusing in the way I have written it: the hilbert space may be H or T(H) and it is not very important for the impossibility question. === Subject: Re: on the existence of injective compact bounded linear operators >well, It is again a typo error : ). A closed image must be sufficient >for the impossibility (i.e. T(H) is a closed space). >So my question is: Can we say that there isn't any injective compact >operator with a *closed* image on an infinite dimensional Hilbert >space? Yes, this follows by the argument a few people gave very early in this thread. >(the closed image brings the impossbility to get such an operator on an >infinite dimensional hilber space) >(I think the impossibility should also work this time for the banach >space case) >Seratend. >P.S. For me closure in H of T(H)= T(H) means T(H) is dense in T(H). Well this is ridiculous. For _me_ black means white - I don't see why nobody understands what I mean... >However, it is very confusing in the way I have written it: the hilbert >space may be H or T(H) and it is not very important for the >impossibility question. ************************ David C. Ullrich === Subject: Re: on the existence of injective compact bounded linear operators I have forgotten: on infinite dimensional hilbert spaces. === Subject: Re: General cubic solution > y^3 + py + q = 0 > it is very complicated. Have a look at http://mathworld.wolfram.com/CubicFormula.html -Michael. === Subject: spherical valued functions I am looking for pointers on some theory for spherical-valued functions (analysis, approximation, compression, algorithms ...). Gabriel === Subject: Well-ordered series of ordinals Given a family (ai) of ordinals indexed by a well-ordered set I. Define the sum sum_i ai as the unique ordinal isomorphic to the well-ordered set: X = /_{i in I} {i} x ai where the good order on X is given by: [(i,x) <= (j,y)] <-> [ (i Given a family (ai) of ordinals indexed by a well-ordered set I. > Define the sum sum_i ai as the unique ordinal isomorphic to the > well-ordered set: > X = /_{i in I} {i} x ai > where the good order on X is given by: > [(i,x) <= (j,y)] <-> [ (i I would like someone to confirm that provided I is isomorphic > to a limit ordinal, we have the equality: > sum_i ai = sup_{J where J === Subject: Re: Well-ordered series of ordinals > This is a statement of set equality. You can prove two sets are equal by > showing that each is a subset of the other. One direction is trivial, > and the other is an easy consequence of the definition of a limit > ordinal. -- Noel. === Subject: Re: a polynomial with positive real roots >> >> Dealing with an algorithm, I came across the following problem: >> >> I have a polynomial P of degree>1 and real coefficients. I know >that, >> if z is a root of P, then Re(z)>=0. It follows that, if r is a real >> root of P', then r>=0. >> >> I can affirm this because all roots of P lie in the closed half >plane >> to the right of the imaginary axis. According to Lucas' theorem, it >> follows all the roots of P' lie in this same half plane, which >> implies >> that, if r is a real root of P', than r>=0. >> This conclusion is correct, isn't it? >> >> The conclusion is correct, though I'm not sure what Lucas' Theorem >> is. The roots of P' lie in the convex hull of the roots of P. >Lucas' theorem says that, if all roots of P are in a same open half >plane of the complex plane, then the roots of P' are in this same half >plane. I think Lucas' theorem implies the roots of P' lie in the convex >hull of the roots of P. > Yes, that's equivalent, because a convex set is an intersection > of half-planes. Lucas's theorem deals only with open half planes, right? That is, it's proof assumes the roots of P are in an open half plane, isn't it? Amanda === Subject: Re: a polynomial with positive real roots ... > Lucas's theorem deals only with open half planes, right? That is, it's > proof assumes the roots of P are in an open half plane, isn't it? But a closed half plane is the intersection of infinitely many open half planes, so it follows that the theorem also holds for closed half planes. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: ANSWER THE FCKING QUESTION GHOST by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0BCmU031453; >> Neither follows from my answer, which involves all finite prefixes only. >> While all digits do appear (given any digit, it has a finite index), >> the number does not. >> It's the same problem as your favorite sequence >> <.3> >> <.33> >> <.333> >> <.3333> >> <.33333> >> ... >> which has all digits of 1/3. Does it contain 1/3? No. >> Your logic does not work. > YOU ARE STUPID. >you say ALL THE DIGITS OF PI APPEAR IN THE LIST. >THEN YOU SAY >only a finite number of them do. >SO JUST IGNORE ALL MY POSTS >SEE IF SOMEONE ELSE CAN ACTUALLY REASON. >Anyone else comment on those 2 statements. >Herc I can't say I have any insight to add to the problem, but you really need to get laid, Herc. Honestly, chill the hell out. Your blood pressure must be through the roof just from reading these forums. === Subject: Re: ANSWER THE FCKING QUESTION GHOST >> >> Neither follows from my answer, which involves all finite prefixes only. >> While all digits do appear (given any digit, it has a finite index), >> the number does not. >> >> It's the same problem as your favorite sequence >> >> <.3> >> <.33> >> <.333> >> <.3333> >> <.33333> >> ... >> >> which has all digits of 1/3. Does it contain 1/3? No. >> >> Your logic does not work. >> > YOU ARE STUPID. >you say ALL THE DIGITS OF PI APPEAR IN THE LIST. >THEN YOU SAY >only a finite number of them do. >SO JUST IGNORE ALL MY POSTS >SEE IF SOMEONE ELSE CAN ACTUALLY REASON. >Anyone else comment on those 2 statements. Sure. It seems you have two problems : 1) You refuse to use enough rigour to ask the questions in any comprehensible way, and thus the answers cannot be simple. 2) The only way to fully answer your question is to say that For any positive integer n, the first n digits of pi is a computable sequence. does NOT imply that The sequence consisting of all the digits of pi is a computable sequence. Duncan === Subject: Re: ANSWER THE FCKING QUESTION GHOST > I am The Truman of Jim Carrey fame. Please help stop me being tortured since April 2002 > with constant microwave laser from the Truman satelite splitting my head and tormenting me > and people around me. Not a prank, I am not crazy, The Truman Show made you think that > ---------------------------------------------s-o-s-------------------------- - -------------- >DONT PICK ON MY SPELLING AS A COPOUT YOU LYING FOOL. >ANSWER THE QUESTION OR SHUT UP ABOUT CARDINALITY >How many digits of pi appear in order in ><3> ><31> ><314> ><3141> ><31415> >.. >Remember pi is here <314159265.......................................> >not in the list above! How many digits OF PI appear IN ORDER? It would appear that any finite number of digits apears on the list. It looks like a trick question because if I say 5 digits you can clearly show that the next term presents 6 digits. On the other hand, if I say infinitely many digits, you will point out that each element of the list has only finitely many digits. Because there is an infinite list of finite strings, it is possible for the *list* to contain all the digits of pi even though no element on the list does. Having said all that, the above also seems to be an example of the type of answer you don't like because it uses extra words to point out the subtleties of the different perspectives that can be taken when answering the question. The simple answers you often ask for have, in the past, resulted in discussions that don't always respect those subtleties. >Funny thing that NONE OF YOU CAN ANSWER ANY SIMPLE QUESTIONS It can be answered, but a precise answer is not necessarily a simple answer, and vice versa. >>Perhaps they have, and you simply do not like/agree with the answers. > Over half of them are wrong. all of them aren't correct answers since they > come with clauses and explanations, and half of them give 2 different answers > to the same question, one in graphical format one in text, and EVERYONE > who does answer wont reply when I correct them. Funny, I get the impression you are trying to force people into false-dichotomies, or otherwise prevent them from giving a complete/correct answer. Some of your questions are loaded in such a way that very statement of the question reveals that you don't understand what the words you are using mean (the bit about how many digits of pi are computable based on a list comes to mind). > Like a kindergarten of delinquents, I'd get better reponses from 10 years olds easily. > We'll see how many weeks on usenet with all your caniving tricks it takes to give > a brief demonstration. Unfortunately, 10 year olds will not necessarily give responses that have anything to do with how computability works, even if they are more in line with how you think it should work. You seem to be trying to force people to abandon one set theory in favor of another, when both are logically consistent. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: ANSWER THE FCKING QUESTION GHOST ---------------------------------------------s-o-s-------------------------- --------------- > >> >> Neither follows from my answer, which involves all finite prefixes > only. >> While all digits do appear (given any digit, it has a finite index), >> the number does not. >> >> It's the same problem as your favorite sequence >> >> <.3> >> <.33> >> <.333> >> <.3333> >> <.33333> >> ... >> >> which has all digits of 1/3. Does it contain 1/3? No. >> >> Your logic does not work. >> > > > YOU ARE STUPID. > > > >you say ALL THE DIGITS OF PI APPEAR IN THE LIST. > >THEN YOU SAY > >only a finite number of them do. > > > > > >SO JUST IGNORE ALL MY POSTS > >SEE IF SOMEONE ELSE CAN ACTUALLY REASON. > > > >Anyone else comment on those 2 statements. > Sure. It seems you have two problems : > 1) You refuse to use enough rigour to ask the questions in any > comprehensible way, and thus the answers cannot be simple. > 2) The only way to fully answer your question is to say that > For any positive integer n, the first n digits of pi is a computable > sequence. > does NOT imply that > The sequence consisting of all the digits of pi is a computable sequence. > Duncan how can you miss the appears in order is for 1 to n when there is a diagram? Pi = <314159..................................................................... . ...............> |<------how many digits---------->| Herc === Subject: Re: ANSWER THE FCKING QUESTION GHOST > ---------------------------------------------s-o-s-------------------------- - -------------- >Neither follows from my answer, which involves all finite prefixes >>only. >While all digits do appear (given any digit, it has a finite index), >the number does not. > >It's the same problem as your favorite sequence > ><.3> ><.33> ><.333> ><.3333> ><.33333> >... > >which has all digits of 1/3. Does it contain 1/3? No. > >Your logic does not work. > >> >> >> YOU ARE STUPID. >> >> >> >>you say ALL THE DIGITS OF PI APPEAR IN THE LIST. >> >>THEN YOU SAY >> >>only a finite number of them do. >> >> >> >> >> >>SO JUST IGNORE ALL MY POSTS >> >>SEE IF SOMEONE ELSE CAN ACTUALLY REASON. >> >> >> >>Anyone else comment on those 2 statements. >>Sure. It seems you have two problems : >>1) You refuse to use enough rigour to ask the questions in any >>comprehensible way, and thus the answers cannot be simple. >>2) The only way to fully answer your question is to say that >>For any positive integer n, the first n digits of pi is a computable >>sequence. >>does NOT imply that >>The sequence consisting of all the digits of pi is a computable sequence. >>Duncan > how can you miss the appears in order is for 1 to n when there is a diagram? > Pi = <314159..................................................................... . ...............> > |<------how many digits---------->| Because the diagram and the words appears in order convey almost no meaning when put in with all the other stuff. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: ANSWER THE FCKING QUESTION GHOST > I can't say I have any insight to add to the problem, but you really need to get laid, Herc. Honestly, chill the hell out. Your blood pressure must be through the roof just from reading these forums. yeah I know, test dummy of psychotronic weaponry from space going on 3 years. The Pretender === Subject: Re: f ^ [f(x)] (x) = (22x^2-75x+64) / (10x^2 -34x +29) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0BCmVL31463; >Is this kind of equation known? >^[f(x)] means iteration f(x) times. >How do we 'deal' with such a thing? >>Starting with f(x)=(5x-8)/(2x-3) ou 1/(f(x)-2)=1/(x-2)+2 ,a form >>very simple to iterate f^[r](x)=> 1/(f^[r](x)-2)=1/(x-2)+2*r , >>with r=f(x) 1/(f^[f(x)](x)-2)=1/(x-2)+2*f(x) , >> all computing done -> >> f ^ [f(x)] (x) = (22x^2-75x+64)/(10x^2 -34x +29) >> This case was built on purpose ;I just wanted to show >> the given equation had a plain solution: >> f(x)=(5x-8)/(2x-3) >More generally, if f(x) = (ax+b)/(cx+d) (a fractional-linear >transformation), let A be the 2 x 2 matrix [[a,b],[c,d]]. >Then f^[n] is the fractional-linear transformation corresponding >to the matrix A^n. >Suppose (as in your case of [[5,-8],[2,-3]]) the matrix A >has Jordan canonical form J = [[1,1],[0,1]]. Thus there is an >invertible matrix S such that A = SJS^(-1), and >A^n = S J^n S^(-1). If S = [[r,s],[t,u]], then >S J^n S^(-1) = (ts-ru)^(-1) [[nrs + ts-ru, -nr^2], [ns^2, ts-ru-nrs]] >If we formally take n=f(x) = ((rs+ts-ru)x - r^2)/(s^2 x + ts-ru-rs) >we find that >f^[f(x)](x) = >((-r^2*u*s+r^2*s^2+t*s^2*r+t*s^3-r*u*s^2)*x^2 > +(-2*r*u*t*s+r^2*u*s+r^2*u^2-r^2*t*s+t^2*s^2-t*s^2*r-2*r^3*s+r^3*u)*x+r^4)/ >((-r*u*s^2+s^3*r+t*s^3)*x^2+(-t*s^2*r-2*r^2*s^2+t*s^3-r*u*s^2+r^2*u*s)*x > +r^2*u^2+t^2*s^2-t*s^2*r-2*r*u*t*s+r^2*u*s+r^3*s) >Setting this equal to (22x^2-75x+64) / (10x^2 -34x +29) and solving as >an identity in x (i.e. put the difference over a common denominator and >make the coefficients of all powers of x in the numerator 0), the only >nontrivial solution according to Maple is r = 2 s and u = t/2 - s/4, >yielding your solution f(x)=(5x-8)/(2x-3). >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia Vancouver, BC, Canada I do appreciate your reply. Sometimes 'confronting' wiews is a good thing; I've dealt with iteration of fractional-linear functions with closed formulas built from Abel(s) (two cases one or two fixed points), Alain, Cou.8fron France. === Subject: Re: On Hardy's essay: by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0BCmW831506; >> If he does not get money by studying mathematics, his apology is obviously okay. >> Everyone can do whatever he likes to do as long as it's legal and not against >> moral. >> The fact was that he did get money from the government. >Hardy got a salary from a university, as I recall, and I suppose he also >gained royalty income from his books. Most of what he did day by day to >fulfill university duties (or to revise his book manuscripts) was only >partially related to what Hardy really liked to do, which was to find >beautiful mathematical discoveries, right? >> He did not know that number theory would be applied in cryptology, >> but this is beside his point. >He gloried in the uselessness (as he supposed) of the mathematics he >investigated most thoroughly. Now his mistaken supposition about the >uselessness of number theory is taking as a paradigmatic case by a new >generation of mathematicians who also can't see immediate applications >for their work--but who do not suppose that their work is therefore >incapable of being applied to future practical problems. Good remark in my opinion. Why not stating things as they are? Mathematical research is only partially done because the results can be useful now or in the future. Like other science mathematics is also done because we humans are curious; we want to know. There is nothing bad about that. And nothing bad about the fact that a part of the tax payer's money is spend into such activities. Mathematicians don't need to hide themselves behind state- ments explaining how useful number theory is in cryptography. Why not telling the whole story? >A Mathematician's Apology is a very interesting read. I think I was >nudged to finally get it from a library and read it after reading >Michael Spivak's list of suggested readings in his calculus textbook. >-- >Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 >Learn in Freedom (TM) http://learninfreedom.org/ >remove .de to email H === Subject: Re: Sum of a series ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0BCmUr31426; >|> My aim is to find /if possible/ the sum of the series with many exact >digits. >According to Maple: >> e_k:=(1+1/k)^k: E_k:=(1+1/k)^(k+1): > B_k:= e_k *( 1- (ln(e_k))*ln(E_k) )/k: > evalf(Sum(B_k,k=1..infinity), 300); >.10594277245004458544757692246184080874897656051926608855429929408610734655 6817024465565948518829322121595126061979425690043137355208629911191246061020 7 6729827692331396663215621777369536344117264876486716014532155677042810860781 1 675427994094144601114575796801802526570248756984817132697276878476036642 >Enough digits for you? >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia Vancouver, BC, Canada Note also the asymptotic expansion for B_k obtained with asympt(B_k,k,10): exp(1)*(1/12/k^3-1/8/k^4+73/480/k^5-11/64/k^6+18125/96768/k^7-5525/27648/k^8 +5233001/24883200/k^9) + O(1/k^10) V. Anisiu === Subject: cooperation in the area of English proofreading and language editing Central European Science Journals www.cesj.com Publisher of scientific e-journals from Central and Eastern Europe is interested in setting up a cooperation in the area of ENGLISH PROOFREADING AND LANGUAGE EDITING with persons who are: - mathematicians - English native speakers - interested in new scientific publications We offer: - Opportunity to participate in a developing e-publishing project - Opportunity to access new scientific papers during the peer-review process - Chance to cooperate with the scientific community from Central and Eastern Europe We do not offer the remuneration. If you are interested in our proposal, please do not hesitate to send us your CV and motivation letter at rekrutacja@cesj.com. === Subject: Re: Analytic functions, n-th derivative zero everywhere, Laurent series In <200501100449.j0A4nEl14709@proapp.mathforum.org>, on 01/10/2005 > 2)I am confused with the fact that we can rewrite > any one function as a Laurent series for different > regions, and just by rewriting it, the region of > convergence changes: Convergence is a property of a series, not of a function. If you pick two points such that the function is analytic in a neighborhood of each, and express the function as a series around each point, you get two different series, and there is no a priori reason to expect them to have the same radius of convergence. > Given f(z)=1/z(z+1), > If we want the Laurent series to converge in |z|<1 > then we write : f(z)=-1/z(1-z)=-1/z(1+z+z^2+z^3+...) That's a series around 0, but you could have picked any other z for which |z|<1 and obtained a different series, with a different radius of convergence. The closer the point you pick is to -1, the smaller the radius of convergence. > But it is the same function after all, how can the region of > convergence change by writing it differently? It doesn't; there is no region of convergence for a function. Each series is different, and has its own radius of convergence. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Arkhimedes theorem and cardianlity In <200501101208.j0AC86s31221@proapp.mathforum.org>, on 01/10/2005 at 01:04 PM, not.adomain@foo.com (joccis) said: >The corollary, a-e < x < a < y < a+e : a belongs to R and e>0 and x,y >belong to Q, is a direct consequence of the Arkimede's theorem. Are you trying to say that for any a in R and any e>0 there exist x and Y in Q such that a-e < x < a < y < a+e? If so, that is indeed a consequence of the Axiom of Archimedes. However, x and y are not unique. >Clearly we have a relation R->Q. No, a relation R:QxQ, given as {(x,(a,b)), X in R, a in Q, b in Q| aAt first glance this looks very much lika a bijection, No, it doesn't even look like a function, and it is not 1- in either direction. >could someone explain why the relation actually is not a bijection >allowing us to list the real numbers. See above; not only is it not a bijection, but for given rational a and b the inverse image of {(a,b)} is uncountable. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Definition of In <200501100649.j0A6n2n19509@proapp.mathforum.org>, on 01/10/2005 >As far as I know the notion of a >meromorphic function< >exists since a long time. Certainly more than the dozens of years that Rand Yates mentions. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: what is SU(2) algebra In , on 01/10/2005 at 03:12 PM, idyllic.bbs@bbs.math.nctu.edu.tw (.) said: >I'm reading a paper of quantum mechanics. >It mentioned SU(2) coherent state recently, but I don't know what it >is. You've actually got three very different questions there. Please note that there is a significant distinction between su(2) and SU(2), normally indicated by a very different type face. SU(2), one of the classical groups, is the special unitary group of 2x2 unitary matrices with unit determinant. Lower case gothic su(2) is the Lie algebra of SU(2). The third question, the relevance of SU(2) and su(2) to Physics, is Model of Quantum Theory involves the group U(1)xSU(2)xSU(3), and if you ignore the Strong Interaction then you get the U(1)xSU(2) of the Electro-weak Interaction. >I wanna to learn that gradually, is there any book recommended about >SU(2) algebra(has something to do with Lie algebra?)?? I doubt that any book specifically about su(2) or SU(2) would be elementary. For a Mathematics major I'd probably recommend getting a Dover reprint of The Classical Groups, but that might be to advanced for you. Look for something on Lie groups and Lie algebras with, e.g., elementary, introduction, in the title. Have you studied manifolds yet? If not, you may find it heavy going. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Probably an easy group theory question keep me in line. Each of your points is helpful... in the last step gH=Hg -> gHg^(-1)=H Joe === Subject: Re: How many digits is pi computable to? > I am The Truman of Jim Carrey fame. Please help stop me being tortured since April 2002 > with constant microwave laser from the Truman satelite splitting my head and tormenting me > and people around me. Not a prank, I am not crazy, The Truman Show made you think that > ---------------------------------------------s-o-s-------------------------- - -------------- >each x is a some digit from {0,2,5,6,7,8,9} >List 1 >0.3xxxxxx.. >0.31xxxxxx.. >0.314xxxxxx.. >0.31xxxxxx.. >0.3xxxxxx.. >0.31xxxxxx.. >0.314xxxxxx.. >0.31xxxxxx.. >0.3xxxxxx.. >0.31xxxxxx.. >0.314xxxxxx.. >0.31xxxxxx.. >... >pi is computable here to _____ digits. >>pi is almost computed here to 3 digits, but that doesn't address how >>many digits it is computable to. Note: the decimal point is in the >>wrong place. >List 2 >0.3xxxxxx.. >0.31xxxxxx.. >0.314xxxxxx.. >0.3141xxxxxx.. >0.31415xxxxxx.. >0.314159xxxxxx.. >.. >pi is computable here to _____ digits. >>Pi is computed to 6 digits here, at least from what I can see, but again >>fails to address how many digits it is computable to. >>Pick a (finite) number of digits, and you can compute it to that many >>digits. > You are a deliberate moron. What does the .. stand for? > 6 is the stupidist answer, only 5 year olds would say that for real, > any 6 year old would see it cointinues. Sometimes I am. However, you are the one who doesn't appear to understand that the number of digits of pi that are computable does not depend on a particular presentation of a sequence of values. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Difference between AN UNKNOWN and UNKNOWN > I am The Truman of Jim Carrey fame. Please help stop me being tortured since April 2002 > with constant microwave laser from the Truman satelite splitting my head and tormenting me > and people around me. Not a prank, I am not crazy, The Truman Show made you think that > ---------------------------------------------s-o-s-------------------------- - -------------- Herc, claiming to be the Truman does not help your credibility. >Nothing if you subscribe to sci.math >Here's omega. Its impossible to always determine if a program halts. >But.... it either does or it doesn't. that means we can assign its halt value >to a variable. >You can't tell the difference between AN UNKNOWN and UNKNOWN. >>Sure I can, the Unknown is the variable. The value of the variable is >>unknown. One is a noun, the other is an adjective. > and what do you assign to unknowns? *Sometimes* a value is assigned to an unknown. Not always. The power of a variable is the ability to work with it without ever assigning a value to it. > whether machine n halts is UNKNOWN, right? It depends on n, but in general, yes. > Here's an apple, here's another apple.. > NOW, that makes 2 apples, right? > Here's is an unknown value, halt(1000). > Here is another unknown value, halt(2000). > halt(1000) and halt(2000) don't make anything! > 2 things that are unknown is just something that is unknown. > They aren't things. Its UNKNOWN. Having an unknown output doesn't make it not a thing. > Here's a set of unknowns { (*& (*(*& ^&(* )(( ^& (()* &**& } encoded so you can't see > Here is another set of unknowns { *&&*(**& (*(**( *&*& ((*)} > They are UNKNOWN. > You can't union them, you can't make them members, you can't do anything with them. > the result ISNT KNOWN. YOU DONT KNOW. And that doesn't bother me. > Here's a maths puzzle. > 2 x + 3 = 7 x is an_unknown > 2x = 4 > x = 2 NOW YOU KNOW X No, now I know that x=2 *if and only if* the original equation is true. > This is not a maths puzzle. > x = halt(1000) x is NOT KNOWN > Sure it *MAY* halt, hence your confusion with it being AN_UNKNOWN. > but its not AN_UNKNOWN. > for some values of H(n), you WILL NEVER KNOW. However, x is AN_UNKNOWN, it's value is UNKNOWN. > the program may halt IN THE FUTURE. > it has no HALT *value*. Time is not a factor in the value of halt(1000). Why do you think it is? > Things that are unknown are not objects. Why do you think that? Perhaps I just have a different notion the possible state of things than you. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: A Limit Involving Permutations > Is it true that: > The fraction of all permutations of {1,2,3,...,n^2}, > which are such that no multiples of n are adjacent in the permutations, > approaches 1/e as n approaches infinity. > In other words, > if P(n) = the number of permutations of {1,2,3,....,n^2} > which are such that no multiples of n are adjacent, > then limit{n->oo} P(n)/(n^2)! = 1/e. Yes. There's nothing special about the multiples of n, except that there are n of them in {1,2,3,...,n^2}. More generally, we can consider a random sample of n elements (without replacement) from {1,2,3,...,N}, and ask for the probability P(n,N) that the sample contains no adjacent elements. Since there are (N choose n) ways of taking the sample, we have P(n,N) = F(n,N)/(N choose n) where F(n,N) is the number of ways to choose the sample so that it contains no adjacent elements. Now F(n,N) satisfies the recurrence F(n,N) = F(n,N-1) + F(n-1,N-2) for n >= 1, N >= 2 with F(0,N) = 1, F(1,1) = 1, and F(n,N) = 0 for n > N, from which we can show that F(n,N) = (N-n+1 choose n) for N >= n. Thus P(n,N) = (N-n+1 choose n)/(N choose n) for N >= n. In particular, P(n,n^2) = Gamma(n^2-n+2) Gamma(n^2-n+1)/(Gamma(n^2-2n+2) Gamma(n^2+1)) Using the usual asymptotic expansion of Gamma I get (with Maple's help) P(n,n^2) = exp(-1) (1 + 5/(6 n^2) + 5/(6 n^3) + O(1/n^4)) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: approximate a pdf function with exponential distribution by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0BK0CN09662; >I have got the probability density function(pdf) of a continuous >non-negative random variable. Now, I am hoping to approximate this pdf >function with an exponential distributed random variable. >1. in such problem, which criteria is the quantity we should minimize? >2. Can you please suggest a way to determine the mean of the exponential >pdf? Any comments, keywords, references are highly appreciated. >-- >Yan ZHANG >http://www.nict.com.sg/zhang/ Find the one with the same entropy as the original distribution. This way you do not introduce any unwarranted information into the exponential distribution. === Subject: Re: Integration in R^2 >Let D = [0,+oo[ x [0,1[ and f be a real-valued function (i) absolutely >integrable on bounded measurable subsets of D. Assume that: (ii) lim int(f, >A_k x B_k) = 0 for every pair of ascending sequences: (A_k) and (B_k) of >measurable sets such that: U A_k = [0,+oo[ and U B_k = [0,1[. Must f be >absolutely integrable over D ? > Of course not. Think about int_1^infinity sin(x)/x. I'm thinking that isn't a counterexample (even if we assume the improper integral is 0, is it?). You can take unions of intervals under the graph of sin(x)/x that grow at different rates under the positive and negative parts ... === Subject: Re: Integration in R^2 > Of course not. Think about int_1^infinity sin(x)/x. > I'm thinking that isn't a counterexample (even if we assume the > improper integral is 0, is it?). You can take unions of intervals > under the graph of sin(x)/x that grow at different rates under > the positive and negative parts ... That's why I mentioned the answer is yes assuming we do not restrict ourselves to sets of the form A_k x E_k (and in particular in the real case; absolute integrability of f on bounded measurable subsets of D is decisive in order to repeat your argument mentioned above), but I really suspect the answer is no and there's some tricky counterexample -- Julien Santini === Subject: Re: Ayn Rand's greatest idea >> > No, you are apparently confusing Rand's ideas with those of the > no-government feudalists like David Friedman. Rand was clearly >not > one of them. >> >> Someone else asked. >> > Is feudalism the best description of the system such folks >> advocate, or is > a little too specific? >> >> The usual term is anarcho-capitalism. Unlike feudalism, it >doesn't >> assume any close link between geographical location, rights >enforcement >> and dispute resolution, nor does it assume the sort of hierarchical >> structure we associate with feudalism. >> >> For a more deatiled description, see: >> >http://www.daviddfriedman.com/Libertarian/Machinery_of_Freedom/MofF_Chapter _29.html >> And other material on my site. >> We've been through this before. >If that's so, then you should know better. He does know better. Than you. >> It's feudalistic/aristocratic/however you want to call it, because >you >> believe in a privileged class---landowners---who have the right to >extract >> Ricardian rent from others >There are landowners in the present-day United States, who extract >rent. But we do not call the United States feudalistic. Because their titles to the land are at the public sufferance, and not defended by themselves. >> (i.e., extract economic rents for something they >> didn't create), presumably backed up by some form of violence which >you >> consider legitimate, >In the US, it is considered legitimate for landowners to use force to >defend their property. But that is not the _basis_ of their title. >> simply because they (or their ancestors) were there >> first. >You are calling feudalistic and aristocratic something which >anarcho-capitalism has in common with present-day US (and present-day >Great Britain and present-day Japan). No. A-C does not recognize the public's ultimate ownership of the land. >Your usage is therefore >idiosyncratic and you are not really writing in English, you are >writing in an idiosyncratic language. Nonsense. He is identifying a real difference between A-C land ownership and land ownership as implemented in civilized countries. -- Roy L === Subject: Re: Problem in Complex Analysis, meromorphic functions >How can I exhibit a merormorphic function on C having a simple pole at z = n >with residue sqrt(n) for n = 1,2,3,... and no other poles? >Brett > This is a direct application of the Mittag-Leffler's theorem. > You can impose even the principal parts for each pole. Obviously, but if the OP knew Mittag-Leffler, why would he be asking the question? === Subject: Re: Focussing Lens without Spherical Aberration > A single element lens can be completely corrected for third order coma > and all orders of spherical aberration with a single aspheric surface. The other being...spherical, planar? By corrected for all orders of SA do you actually mean stigmatic on axis for a source at infinity? > In a fast lens there will be higher order coma, which can be > effectively corrected with a bi-aspheric design. This is the basis of > the objective lenses used in all CD/DVD players. Concerning the CD/DVD lenses how do they take care of the large SA caused by the plastic layer, especially at high NA? By introducing a certain amount of SA of the opposite sign? === Subject: Re: Focussing Lens without Spherical Aberration <10tp70s79ad9m05@corp.supernews.com> <10u8dodpb2jvhc8@corp.supernews.com> > A single element lens can be completely corrected for third order coma > and all orders of spherical aberration with a single aspheric surface. The other being...spherical, planar? In general spherical, although if you are free to vary the refractive index you could make the rear surface plano. By corrected for all orders of SA do you actually mean stigmatic on axis for a source at infinity? Yes, stigmatic on axis for a source at infinity. > In a fast lens there will be higher order coma, which can be > effectively corrected with a bi-aspheric design. This is the basis of > the objective lenses used in all CD/DVD players. Concerning the CD/DVD lenses how do they take care of the large SA caused by the plastic layer, especially at high NA? By introducing a certain amount of SA of the opposite sign? The cover layer just becomes a part of the design, and poses little difficulty. The only time it really becomes an issue is if you want to reduce the lens size as much as possible, and you have to start worrying about the BFL. Brian www.caldwellphotographic.com === Subject: Re: HOW MANY DIGITS OF PI HAVE PROPERTY X ? > Pi = <314159..................................................................... . ...............> > |<------how many digits---------->| > Let property X = the digit, and every preceding digit up to that digit occur in order in the right spot on > (a member of) list Y > Let Y = { > <333333333..> > <300000000..> > <399999999..> > <314314314..> > .. > Y has infinite members. > The above is just a sample. > Y is computed by UTM(row, col) mod 10 > Y includes all computable numbers for some numeric representation > The answer should be a quantity > |<------how many digits---------->| > that is not related, dependant, or refers to Y. The answer should be a quantity only if you allow infinity as a quantity. > Example : > Pi = <314159..................................................................... . ...............> > |<------how many digits---------->| > Let Y = { > <31400000..> > <31411111..> > <31322222..> > <31433333..> > Answer : 3 -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: How many numbers in N have property is_positive? > oo > that's a clue, 5000 so called mathematicians singing about superinfinities all day > don't know how many naturals there are. Or, to be more precise, Aleph_0 of them, since oo might refer to Aleph_1, Aleph_2, etc. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Who originated this proof of uncountability of reals? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0BL4rf16069; Michael Orion http://mathforum.org/discuss/sci.math/m/670461/670461 > Does anyone know who originated the proof below for the > uncountability of the reals? I found it in Kuratowski, > Introduction to Set Theory and Topology, 1962 Pergamon. > It is different than Cantor's diagonalization proof. The second post below has a lot of information about this. The other posts contain things you might also be interested in. The essence of Cantor's 1874 proof. http://mathforum.org/discuss/sci.math/m/388412/388425 Comments and references about Cantor's 1874 nested interval proof and Cantor's 1891 diagonalization proof. http://mathforum.org/discuss/sci.math/m/470447/470655 A discussion of Hankel's 1870 Baire category-like proof that a Riemann integrable function has a dense set of continuity points, and whether this might have inspired Cantor's 1874 proof that the reals are uncountable. http://mathforum.org/discuss/sci.math/m/484926/485136 Dave L. Renfro === Subject: Re: Analytic functions, n-th derivative zero everywhere, Laurent series by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0BLBuL16759; >> I would appreciate your help with the following. Please have> patience with me, as I am an Economics student and not >> a mathematician. >> 1)If f is analytic in a region R, and the n-th derivative >> of f is zero for every point in R, can we conclude that >> f is 0? >Hint: think about polynomials. Yes, sorry, I meant to ask if it follows from the above condition that f is a polynomial of degree n-1. >> 2)I am confused with the fact that we can rewrite >> any one function as a Laurent series for different >> regions, and just by rewriting it, the region of >> convergence changes: ( I apologize if this question >> is dumb, but I seem to be missing something, and I don't >> know what that is) >> Given f(z)=1/z(z+1), >> If we want the Laurent series to converge in |z|<1 >> then we write : f(z)=-1/z(1-z)=-1/z(1+z+z^2+z^3+...) >> and if we want the series to converge in |z|>1, then >> we write it as: >> f(z)=-1/[z^2(1-1/z)]=1/z^2[1+1/z+1/z^2+...+1/z^n+...) >> Aren't we saying that the convergence properties of >> f(z)=1/z(z+1) are changing, just by writing it differently? >It's not the function that converges, it's the series. There are two >different series, and they converge in different regions. It just >happens that there is one function that is the sum of series #1 in >region #1 and is also the sum of series #2 in region #2. by manipulating the same expression, i.e, 1/z(z+1) in different ways?. Doesn't the fact that we start from an identical starting point and use valid mathematical manipulations mean that the two expressions should be identical?. How can we start with something of the form A=A (A=1/z(z+1)) , manipulate both sides separately and end up with different expressions in each side? I don't know if the analogy is valid, but if we write 4=2*2 or 4=3+1, which are different ways of writing 4, we would still have 2*2=3+1. >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia Vancouver, BC, Canada === Subject: Re: Problem in Complex Analysis, meromorphic functions by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0BLFYP16936; >>How can I exhibit a merormorphic function on C having a simple pole at z = n >>with residue sqrt(n) for n = 1,2,3,... and no other poles? >> >> >>Brett >> >> This is a direct application of the Mittag-Leffler's theorem. >> You can impose even the principal parts for each pole. >Obviously, but if the OP knew Mittag-Leffler, why would he be >asking the question? In have also posted the reference to Rudin's book. V. Anisiu === Subject: Commutator subgroup of Sn. Hello. I've read on the internet that : for all n>=2, the commutator subgroup of the symmetric group S_n, is the alternating group A_n ; where the commutator subgroup of a group G is : G' = < { [x,y] = x.y.x^(-1).y^(-1), x and y in G } >. But I haven't found any proof. Could you help me ? === Subject: Re: Commutator subgroup of Sn. > I've read on the internet that : > for all n>=2, the commutator subgroup of the symmetric group S_n, is the > alternating group A_n ; > where the commutator subgroup of a group G is : > G' = < { [x,y] = x.y.x^(-1).y^(-1), x and y in G } >. Clearly the commutator of S_n is included in A_n. Also, taking x = (k,i) and y = (k,j), [x,y] = (i,j)(k,j), whence setting: z = (l,j) and t = (l,k) gives: [z,t] = (j,k)(l,k); eventually, [x,y][z,t] = (i,j)(l,k), which was to be proved. -- Julien Santini === Subject: Re: Commutator subgroup of Sn. days. My association with the Department is that of an alumnus. >I've read on the internet that : >for all n>=2, the commutator subgroup of the symmetric group S_n, is the >alternating group A_n ; >where the commutator subgroup of a group G is : >G' = < { [x,y] = x.y.x^(-1).y^(-1), x and y in G } >. >But I haven't found any proof. >Could you help me ? There are a few ways to go, depending on how much you assume of S_n. For example, if you are willing to assume that the only normal subgroups of S_n for n=2,3 and n>4 are {1}, A_n, and S_n, then since G/[G,G] is the largest abelian quotient of G, it is easy to see that [S_n,S_n] must equal A_n. In the case of n=4 you would need to check that the subgroup {1, (12)(34), (13)(24), (14)(23)} does not yield an abelian quotient. To show it directly for n>=3 (you can do n=2 by hand), start by showing that [S_n,S_n] is certainly contained in A_n, since every commutator is an even permutation ([x,y]=xyx^{-1}y^{-1}, so regardless of the parity of x and y, this product is expressible as a product of an even number of transpositions). Next, note that [x,y] is the unique element of G such that xy = [x,y]yx. Using this, show that you can obtain a 3-cycle as a commutator, by a suitable choice of x and y. Then, since [S_n,S_n] is normal and contains one 3-cycle, it contains all 3-cycles; and the 3-cycles generate A_n, which would show that A_n is contained in the commutator subgroup, giving equality. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Inverting quotients G/N. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0BLTHn17764; Is there any way of somehow inverting quotients, i.e, given, say: R/Z~[0,1), or T=|z|=1 is the standard quotient where R is the reals, Z is the integers, T is the unit circle, could we somehow solve for R, by inverting the quotient?. i.e, if we only knew that for some unknown $, we had, as above $/Z~[0,1), is there any way of finding out what $ is? could we do it for general quotients? === Subject: Re: Inverting quotients G/N. > Is there any way of somehow inverting quotients, i.e, > given, say: > R/Z~[0,1), or T=|z|=1 > is the standard quotient where R > is the reals, Z is the integers, T is the unit circle, could > we somehow solve for R, by inverting the quotient?. i.e, > if we only knew that for some unknown $, we had, as above > $/Z~[0,1), is there any way of finding out what $ is? > could we do it for general quotients? I'm a little confused, since you keep referring as R/Z ~[0,1) which isn't true, R/Z ~ S^1, the unit circle. Also, here you seem to be talking about the action of Z on R, i.e. x~y <-> x-y is in Z. For general quotient spaces, trying to invert is pretty much hopeless. Remember that in general, the equivalence relation to define a quotient space X/~ doesn't have to be 'nice'. If you mean, given a space X and X=Y/G where G is some group acting on Y then you might have a chance with some additional structure ,e.g. X is a C-W complex and G is assumed to act cellularly on Y. === Subject: detail of proof in algebra Hi I need some help to understand the following equality : Let H be a subgroup of a group G and let x,g in G. x{ Intersection_[g in G] gHg^(-1) }x^(-1)=Intersection_[g in G] xgHg^(-1)x^(-1) Why is this true? === Subject: Re: detail of proof in algebra days. My association with the Department is that of an alumnus. >Hi >I need some help to understand the following equality : >Let H be a subgroup of a group G and let x,g in G. >x{ Intersection_[g in G] gHg^(-1) }x^(-1)=Intersection_[g in G] xgHg^(-1)x^(-1) >Why is this true? Because the sets are equal. This is an equality of sets, so you can always try proving it by double inclusion. Let y be in the left hand side. That means that y = xux^{-1}, where u is an element of the intersetion of gHg^{-1} for each g in G. That is, for each g in G, there exists h_g in H such that u = g*h_g*g^{-1}. To show that y lies on the right hand side, you need to show that for every g in G, there exists m_g in H such that y = x*g*m_g*g^{-1}*x^{-1}. Given what we already deduced about y, that should be trivial. Conversely, if z lies in the right hand side, then for every g in G there exists m_g in H such that z = x*g*m_g*g^{-1}x^{-1}. You want to show that x^{-1}*z*x lies in the intersection of all gHg^{-1} (g ranging over all of G) in order to show that z is contained in the left hand side (there is a small step to be done to see why this is what you want, by the way). If you can show both, then it follows that both sets are equal. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: just 5 quick answers then I can summarise and GO > Yup, pi is a real number. Note that this entails that pi can be > represented in a decimal expansion: a series of the form > x = Sum (x_i / 10^i) where each x_i is in the set {0,1, ... 9}. > Intuitively, a real number x is computable if there is a Turing > machine > T such that if i is a natural number, the output of T with input i is > x_i. (OK, this is slightly stronger than computability, but it's > much > more intuitive than the general notion, with all its limits) > There are reals that aren't computable, though; are these the so-called > transcendental numbers? > and yes, that seems stronger than computability. > Ken In this (simplified) context, a number a is transcendental if there no polynomial q(x) with rational coefficients such that q(a) = 0. (In the general case, we say that a number b is transcendental over a field F if there is no polynomial f with coefficients in F so that f(b) = 0) pi, for instance, is transcedental *and* computable. I think you're looking for uncomputable numbers. ;-) 'cid 'ooh === Subject: Re: just 5 quick answers then I can summarise and GO > A) > SEQUENCE = <314159265.................................................................. . .......>> > <--- HOW MANY DIGITS???---> > (B) > COMPUTABLES > 1 <398498498.................> > 2 <484849848.................> > 3 <383873838.................> > .. > How many digits of (A) appear in correct sequence in (B), guaranteed? > 1 _______ Assuming you mean that B is a list of all computable numbers, since A is computable, all of its digits appear in B. > (A) > RANDOM SEQUENCE = <654445676764545............................................................ . .> > <--- HOW MANY DIGITS???---> > (B) > COMPUTABLES UTM(row, col) mod 10 > 1 <398498498.................> > 2 <484849848.................> > 3 <653873838.................> > .. > How many digits of (A) appear in correct sequence in (B), guaranteed? > (Randomness could be introduced from outside the computer). > 2 __________ Arbitrarily many, but you will not necessarily find all of them in a particular location on the list B. Note: they may appear on the diagonal. > So given a coin sequence > How many flips (at least) of cs appear on the computable number list in order? > CNL = UTM(row, col) mod 2 > <0010101010100101..> > <1010110110101001..> > <0000000000000000..> > <1111111111000000..> > <1010101010101010..> > .. > 3 __________ I'm going to guess you want 1=H, 0=T, or something similar. Same answer as for 2. >>all coin sequences are computable to infinite length ? >>false > 4 = all coin sequences to infinite length appear (all flips in order) in any UTM computable list. > 4 <-> _________ Huh? All finite sequences will appear. Some infinite sequences will not appear. >What about this one, any Cantorians want to assign it T or F? >There is a maximum to the number of coins in any given oo coin sequence, that can be computed >>false too > What is the opposite as a true proposition? > 5 ___________________________________________ For any given oo coin sequence, there is no maximum length computable subsequence. > -- > I am The Truman of Jim Carrey fame. Please help stop me being tortured since April 2002 > with constant microwave laser from the Truman satelite splitting my head and tormenting me > and people around me. Not a prank, I am not crazy, The Truman Show made you think that > ---------------------------------------------s-o-s-------------------------- - -------------- No, you aren't Truman. No, you haven't been tortured by satelite. Yes, you are crazy. And to save you some effort, no I don't care what links you offer to substantiate your claims that such things exist. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: There are uncountably many irrationals was Re: abundance of irrationals <41e31163$13$fuzhry+tra$mr2ice@news.patriot.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> 01/10/2005 >>If there is an irrational number left, then there is a rational one >>left between the irrational and the rational next to it. >> There is no the rational next to it. > By construction, the set Q_n contains finitely many rationals. Finitely many? How many would those be? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: abundance of irrationals > So you claim that irrationals do not exist, but that the rationals > and the irrationals have the same cardinality? Isn't that > a bit contradictory? > Not the same cardinality! I proved that Card(IQ) is not less than > Card(IX). This result remaines correct, after we find out that there is > not one single irrational existing. If no irrationals exist, as you appeear to be claiming, then you must be also claiming that x^2 = 2 has no real solutions. Or are you claiming that sqrt(2) is rational? === Subject: Re: abundance of irrationals > You must be able to identify and distinguish a number from any other > number - not its name from any other name. That is the criterion of > reality! What about all those numbers which are inaccessible? === Subject: Re: abundance of irrationals > No, not for any positive epsilon. That is the big mistake. x^2 - 2 > = 0 > cannot be solved to a precision of 1/10^10^100 in x. > X = sqrt(2) and x = -sqrt(2) are solutions with better precision than > that. They are just not expressed as decimal expansions. > That are names for desired solutions, not numbers. If you compare > Cantor's work you will find that this was his position too. Decimal expansions are also just names, not numbers, too. There is always a distinction between the number and any of its representations. That some representations are more useful thatn others for certain purposes does not mean that they are anything more than representations. Depending on one's model, a real number is either a certain type of partitioning of the rationals (Dedekind) or a collection of infinite sequences of rationals (Cauchy sequences whose differences are null sequences). It is not, in any truly mathematical model, merely a decimal expansion. > Could you decide whether sqlmn < sqrlp, unless you have rational > approximations for those names? That would be the minimum requirement > for a number. For sqrt(2) you know a lot of rational approximations > which enable you to decide the <, =, or > -question for a lot of other > rational numbers. But you can't leave the rational domain. Given either the partition form or the family of sequences form, as referred to above, of both numbers, such questions are trivial. === Subject: Re: abundance of irrationals > You stated the square root of two does not exist. And if that's > independent of your thinking, what do you believe? To what do you > assign the diagonal of the unit square and the solution to the > equation > x^2-2 = 0? > Excuse me, it's not my fault and not my responsibility that we have not > enough storage space in the universe to represent these numbers with > arbitrary precision. Hence, we cannot distinguish them from > infinitely many rational numbers. > The diagonal and above solution can be found in a fairly good > approximation, but not with arbitrary precision. I responded previously to this and have yet to receive a reply. Here's my response:You don't believe the square root of two exists and so by extension no algebraic irrationals exist. Why, then, at the begining of this thread did you try to prove that the rationals and irrationals have the same cardinality? I can construct with straight edge and compass alone a line that is the square root of two units long. Draw a line segment with your straightedge and use your compass to bisect the line segment with another line segment perpendicular to the original line segment. Draw a circle with your compass using the point of intersection of the two line segments as the center of the circle. Call the line segment from the center of the circle to the circle itself along any of the line segments one unit. The diagonal line connecting any two of these points is exactly the square root of two units long. Because you don't believe this number exists, then the number line has a gap at that point. A line segment can also be trisected with straightedge and compass. If you say that no irrationals exist because their decimal expansion is infinite, then no rational numbers with infinite decimal expansion exist. This logic leads to more gaps in the number line. Further, you say that numbers are limited by storage capacity in the universe. That leads to the non-existence of very large positive and negative integers. These integers are the building blocks of rational numbers, therefore more gaps in the number line. If this is your agenda, please be more direct and say so. === Subject: Re: There are uncountably many irrationals was Re: abundance of irrationals <87zmzia3o5.fsf@phiwumbda.org> <87y8f1i3vf.fsf@phiwumbda.org> Discussion, linux) > > > Er, for well-ordering uncountable sets? >> >> For countable sets no well-ordering theorem is required! >> Yes, right. That's why I said that the theorem is good for >> well-ordering uncountable sets. > But what good does it do here? Pass through all the countable > ordinals and you're sure to exhaust the rationals and you are also > sure you have not exhausted the irrationals. >> >> Continue! There is always a rational between two reals. >> So? If you've exhausted the rationals then you've exhausted them. >> You cannot continue. > The conclusion is correct. But you believe that the rationals can be > exhausted. That's wrong. You have proved no such thing. My proofs are easy and standard proofs that (1) Q is countable and (2) R isn't. Thus, > Disagree with the second claim? Then show which step in any > (well, > properly, in each) of the proofs |N| < |R| is invalid. >> >> Cantors first proof (1873) uses the fact that limits do not belong > to >> sequences. Nevertheless his proof fails, if applied to the set of >> irrational numbers alone. The reason is, that any infinite > sequence >> need not converge to an irrational limit. Already the absence of a >> single number, zero for instance, cannot be tolerated, because it > is >> the limit of several sequences. >> Lordy, Lordy, Lord. >> You agree that R = Q u (R Q) and that R Q is the set of >> irrationals, yes? >> Theorem: |N| = |Q|. >> Theorem: |Q| < |R|. >> Theorem: A union of two countable sets is countable. >> These three facts allow one to conclude that R Q is uncountable. >> Cantor's proof doesn't have to apply directly to R Q. > But don't you see some reason to be sceptic? What would be the > conclusion if the countability of Q had not yet been established in > advance? No conclusion at all, unless I happened to be clever enough to prove Q is countable by myself. But so what? No one ever said that the fact that Q is countable and R is uncountable is intuitively obvious (or at least *I* didn't say so). > Your second theorem is wrong. Further, are you sure that classical > logic holds in the infinite? Classical arithmetics doesn't. My second theorem is easy to prove, as you know. You haven't demonstrated any flaw at all. Am I sure that classical logic holds in the infinite? What on earth would that mean? >> Try it again, why don't you. Pick one of the above theorems and pick >> a standard proof of the theorem. Show which step is invalid. > No, look at these proofs yourself and try to find the flaws. Okay, wait just a sec... Nope. Can't find any flaws. Now what? -- When you go to class today, if your professor talks about algebraic number theory, or misuses Galois Theory[,] I want you to carefully notice how you feel. Hold on to that feeling so that you never forget it. --James S. Harris, on channeling rage via Galois theory. === Subject: Re: THIS STATEMENT HAS NO PROOF IN ANY SYSTEM = true or false? >>I *did* say it was an assumption. > You said that A is defined to include.... So what I'm questioning > is not an assumption - something that may or may not be true - but a > proposed definition. The question is whether you are in fact referring > to anything. You knew *perfectly* well what I meant. Are you questioning the assumption that something meeting the definition of A exists, or are you not? > I cannot sensibly say Fnorx and add that I'm assuming > that what I'm saying makes sense. Are you claiming that the effective procedure used by mathematicians to determine mathematical truth is so nonsensical that it cannot *possibly* exist? > To hold that mathematical truth, or historical truth, > or astronomical truth, is objective is not to hold that it is in the > least well-defined what today constitutes knowledge of mathematical, > historical, or astronomical truth, let alone that we can speak > sensibly of what people can know in principle in these fields. Is it your position that the question are there true facts that to which we can never know the answer? is itself so ill formed, or poorly defined that it can not have a meaningful answer? If so, why do you waste time on it? > People may hold all sorts of things to be true or evident or valid, > and opinions about what is true or evident or valid may vary due to > a number of subjective factors, even though the facts are > perfectly objective. Here you verge into non-sequetir. I don't think I said anything about what people may hold. Given a set of 100 facts that I cannot determine the truth of, I can *hold* the correct evaluation of about 50 of them by guessing at random. That does not mean I *know* the truth of any of them. Ralph Hartley === Subject: Re: THIS STATEMENT HAS NO PROOF IN ANY SYSTEM = true or false? >>There are stronger versions of the CT thesis that imply it. For instance if >>the laws of physics are computable. Then any physical object, including >>mathematicians, are effectively describable. > By the laws of physics are computable, I'll take you to mean something > roughly like the following: The universe is finite and discrete, with > one time dimension and some number of spatial dimensions; it has an > initial state, and for all t, the state at time t is determined by a > deterministic and recursive function of the states at times less than t. No, I don't mean roughly that. Most, or perhaps all, of the assumptions you list are not necessary. > The real point is that it does not follow from this model of the universe > that a given subset of the universe is going to be recursive, or even > recursively enumerable. By your characterization of the thesis the universe only has finitely many states. You may have a definition of subset of the universe that allows it not to be finite, but that would be odd. All finite sets are recursive. Not accepting that characterization, I don't think it matters, but it certianly shows that I have no idea what you could possibly mean. > What does this truth correspond to in our model of the universe? This is highly > unclear. The question is not what truth corresponds to. Perhaps you meant to say what does knowledge correspond to? There are enough ways to parse definitions that I don't think that is an argument I want to have. In any case it is mostly irrelevant to the main point: If the laws of physics do not permit any physical process *whatsoever* to decide some question, then it is hard to see *any* sense in which inhabitants of that universe can be said to know the answer. Of course the universe may have laws that don't have that property, or no laws at all, but the property is much less restrictive than being recursive. >>There *could* be objective methods that are not effective, but I don't know >>of any. > Even theoremhood in ZFC is not effectively decidable, but it's surely > objective. I said objective *methods*. A proof in ZFC is an effective method. Being a non-theorem in ZFC may be an objective fact, but it is hardly a method. It is one of those facts we might not be able to know. Ralph Hartley === Subject: Re: THIS STATEMENT HAS NO PROOF IN ANY SYSTEM = true or false? Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow) >I'm sorry. I misspoke. What I meant was that I think that there exist >arguments for why the axiom of choice should be true. So the result >of polishing is something like a formal proof, together with informal >arguments as to why the axioms themselves are acceptable. But your original argument was that the ability to recognize solid mathematics is as good as the ability to create it: You just enumerate all possible character strings, and for each one, you check to see if it is a solid mathematical argument. What it seems that you've conceded is that to recognize solid mathematics, you must also check to see if the informal arguments for the axioms are correct. We have no reason to suppose that this can be done in any systematic manner even by professional mathematicians, let alone students or AI programs. -- 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: too much information! >Be warned. RAND's data is complete crap. It's trivially >compressible by a non-negligible fraction of a percent due >to incompetant unbiassing. See a comp.compression archive >for estimates how non-random it is. > Interesting. I haven't found those estimates yet, and would like > a more specific reference, but this reminds me of Stanislaw Lem's > SF comedy His Master's Voice, where alien radio transmissions > go unnoticed at first and the punch cards holding the data get > thrown out... then used to compile a table of random numbers... > but then, by analyzing the data in the table, they realize it's > a fascinating communication! I don't want to give the rest of the > story away, but it's a wacky reflection on the concepts of randomness > and information. > ................................................................... > Generation of random numbers is too important to be left > to chance... > Donald Knuth http://www.quotationspage.com/quote/461.html What Knuth *did* say is Random numbers should not be generated with a method chosen at random. ÖDonald E. Knuth Vic === Subject: Re: too much information! <41e24162$0$256$edfadb0f@dread12.news.tele.dk> I think so; a nicely compressed file looks random; you can't compress it much more. But it get's worse. An efficiently compressed file is an example of efficient communication, since the redundancy is removed. (Thus jpeg and zip files.) Redundancy is needed of you have a communication channel with loss. (Shannon.) It's not great of you have an efficient channel. A new efficient communication form is Ultra Wide Band radio. It's power spectra looks like white noise on top of the background noise. === Subject: Re: too much information! Originator: baez@math-cl-n03.math.ucr.edu (John Baez) > In the 1950's, RAND produced a table of a million > random digits, and another one with > 100,000 Normal Deviates. ( link is below) >>Be warned. RAND's data is complete crap. It's trivially >>compressible by a non-negligible fraction of a percent due >>to incompetant unbiassing. >Hmm... I remember reading some experiments in parapsychology, and >correlations were found here and there. Hi! You mean they used psychics to unearth correlations in the RAND table of random digits? :-) >in-boxes, but they fear a proposed national 'Do Not Spam' registry will >make it impossible to use e-mail as a marketing tool. year, which in information terms equals a pickup truck full of books. === Subject: Re: too much information! Originator: baez@math-cl-n03.math.ucr.edu (John Baez) >In the 1950's, RAND produced a table of a million >random digits, and another one with >100,000 Normal Deviates. ( link is below) I keep on reading this as 100,000 normal deviants, which seems like a description of usenet news. Apparently the librarians thought similarly, since this book was first filed under psychology. >Having downloaded the 1 million digits (line numbers included), >12,500 lines at 80 digits per line. >This got compressed to a 484,760-byte *.zip file. >>Someone else claimed that these random digits were >>complete crap and could be significantly compressed; let's see >>if your zip file reflects that. >>1 million decimal digits should be about 3.3 million bits; >>divide by 8 and get about 410,000 bytes. >>Wait a minute - that's LESS than your supposedly compressed >>file! What's going on? It can't be those line numbers. >>Am I making some dumb mistake? >Are you still assuming there's 5 bits/char? No, since the data was decimal digits, I was assuming ln(10)/ln(2) or about 3.3 bits per character. But maybe you've solved the puzzle: maybe whatever data compression Bernier used to generate his .zip file was not smart enough to take full advantage of the fact that the data was just numbers! Seems dumb, but... At 5 bits/char we'd get 5 million bits or 625,000 bytes; then some not-terribly-smart program might compress that down to 484,760 bytes and feel proud of itself. >Also the data was on cards. There's a different encoding with >cards that has to do with two fields and one hole punch in each. >Geography of the holes was how characters were defined. Oh-oh - now this is getting too complicated for me to understand. >Also 20,000 cards ain't very many cards. It's only 10 boxes. That's irrelevant, your honor! What's at stake here is the number of bits, and whether the defendent compressed the data in an ill-advised manner. === Subject: Re: too much information! Originator: baez@math-cl-n03.math.ucr.edu (John Baez) > 7) Pixels on a computer screen. 16,000,000 colors for each pixel. > How many different pictures are possible? >>For a 640x480 screen, it's 10^2219433, slightly larger than a googol. I'd say you're engaging in understatement, but that would be an understatement. Slightly larger than 10^100?? This number makes a googol look like one. That is, 1. === Subject: Re: phi Lets call think of any rectangle(not square) and call its larger arm the length(l) and the other side its breadth(b). Note the ratio l/b. Now cut from this rectangle a square of length b. You are left with a new rectangle of breadth (l-b) and length b. Again note the length/breadth ratio (now it is b/(l-b)). Keep on doing the same operation. It can be found by solving for l/b such that we always arrive at similar rectangles(length/breadth is same) that l/b must be 1.618 aprrox. This ratio is referred to as the Golden Ratio. When I was a kid I heard somewhere(but not in a school) that some Roman pyres used to have such dimensions.