mm-195 === Subject: What is the Origin of Space and Time?What is the origin of space and time?Space and time is the 4D fabric within which all lifehas evolved. It is such a common part of everydayexperience that it is taken for granted. But is it hereforever?Gravity curves space and slows time. Relativisticspeeds contract space and dilate time.Is space and time constantly being generated at theoutermost parts of the universe? If so, what is beyondthe outermost parts of the universe? Do timelessnessand spacelessness enclose the universe?This is related to mathematics, since Cartesiancoordinates describe space, and functions of timemap events into that space. === Subject: Re: What is the Origin of Space and Time?> What is the origin of space and time?It's beginning, of course... > Space and time is the 4D fabric within which all life> has evolved. I think you mean space-time.> It is such a common part of everyday> experience that it is taken for granted. But is it here> forever?I'll get back to you in about in'nity.> Gravity curves spaceRight.> and slows time. Wrong.> Relativistic> speeds contract space and dilate time.> > Is space and time constantly being generated at the> outermost parts of the universe? If space is curved, then how can there be outermost parts? You assume thatreality has edges in order for there to be outermost parts. On what do youbase such an assumption?> If so, what is beyond> the outermost parts of the universe? Why does there have to be something outside? What *is* the de'nition ofoutside of the universe? Can you check to see if it also describes what'snorth of the north and also what was before time?> Do timelessness> and spacelessness enclose the universe?What is timelessness and spacelessness?-- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people think, Mr. Feynman? === Subject: Re: What is the Origin of Space and Time?> What is the origin of space and time?> > Space and time is the 4D fabric within which all life> has evolved. Wrong. It is in fact the 8th dimentional time-space fabric.See OThe adventurures of Buckaroo Banzai across the 8th dimension'.Do your homework, kid! === Subject: Re: What is the Origin of Space and Time?Porky Pig Jr space and time?>> Space and time is the 4D fabric within which all life> has evolved.>> Wrong. It is in fact the 8th dimentional time-space fabric.>> See OThe adventurures of Buckaroo Banzai across the 8th dimension'.>> Do your homework, kid!Uh, I think it is you who needs to do the homeowork. Everyone knows thereare exactly FIVE dimensions. Every hear about a little something called'Twilight Zone' (aka the 'fth dimension) !?!?! I bet you feel pretty dumbright about now.l8r, Mike N. Christoff === Subject: Re: What is the Origin of Space and Time?>> Wrong. It is in fact the 8th dimentional time-space fabric.>> See OThe adventurures of Buckaroo Banzai across the 8th dimension'.>> Do your homework, kid! Uh, I think it is you who needs to do the homeowork. Everyone knows there> are exactly FIVE dimensions. Every hear about a little something called> OTwilight Zone' (aka the 'fth dimension) !?!?! I bet you feel pretty> dumb right about now.And here I was, pondering the collapse of a 10- or 26-dimensional universein a false vacuum state into 4-dimensional, more stable universe with theremaining dimensions curled upon themselves... Silly me. ;)-- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people think, Mr. Feynman? === Subject: Re: The Clearest by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 13:30:21 -0500I had some dif'culty understanding this paper.In places that I found unclear, other readers might also have dif'culty, so it might be worth your time to try to clear them up.De'nition 2.4, is the union of the J' supposed to equal theinterval N_a^b_{2k=1}?Page 3 line 4, given any nth point of A ... there exists the corresponding nth point of B ... Is this an equivalence of sets?(That is, union J'_{r_i,p_i}^{a,b} = union J'_{r'_i,p_i}^{c,d} - (c-a)where (c-a) is subtracted from each element of the second set?)Or is the individual covering J'_{r_i,p_i}^{a,b} on the left supposed to be related somehow to the corresponding coveringJ'_{r'_i,p_i}^{c,d} on the right?Same place, given any nth point of A ... counted from the extreme left... nth point of B ... counted from the extreme left.Are they both to be counted from the left simultaneously, or is one to be counted from the left and the other from the right?Lemma 2.6. Here's where I really get lost.F = union_{i=1}^x J'_{r_i,p_i}^{3,q}.So in particular F contains J'_{r_1,3}^{3,q} for some value r_1,so it contains all the odd numbers between 3 and q congruent tor_1 modulo 3. Then there is no way that equation (1) could be satis'ed, since P'^{3,q} does not contain such a large arithmetic progression. I must be missing something.I see P' as a set of primes, and F as a union of arithmetic progressions, and P' cannot contain an arithmetic progression with pitch 2*3=6 because it is too sparse (as soon as q is large enough).Other suggestions on notation:Use N_{odd} instead of N_{2k+1}.Reason: other quantities like J_{k,p} that have k in their notationactually depend on the value of some integer k, but this one does not.In the de'nition of N_{2k+1} (with presubscript a and presuperscript b), include x odd in the formula:{ x : a leq x leq b , x odd }Similarly in the de'nition of P (presubscript 3, presuperscript p_l)say { p : 3 leq p leq p_l , p prime }.In both cases it has been stated in words beforehand, but it's handy to be part of the formula in case the reader checks back for thede'nition.The de'nition of a covering, should the restriction two or more points be part of the de'nition?Remark 2.2, what is a linear graph?>The idea is explained in the clearest way. This would be last trial.>>Please review the paper on Goldbach Conjecture at>>http:// www.geocities.com/erdosfan/solution.htmlHisanobu Shinya === Subject: Re: e is transcendental (was: classes of transcendental numbers ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id guys have a copy of the proof of e is transcendental? ..id really>> appreciate if somebody could help me. ive been searching and found nothing.>>Spivack, CALCULUS, 2nd edition, Chapter 21.>Are you sure that the proof there is for e being transcendental and not>>just irrational? I know that the proof of the irrationality of e is>>simple enough for a Calculus book, but that the transcendence is quite a>>bit more complicated.>>In my version of Spivak, the transcendence of e is proven in Chapter 20,>and the irrationality of pi (or actually, the stronger result, the>irrationality of pi^2) is proven in Chapter 16. Both chapters are marked>with an asterisk, presumably denoting that the chapters are intended for>the more advanced student. I note that Spivak describes these chapters as>optional. As others have noted, another good book is Irrational Numbers>by Ivan Niven, in which a proof of the transcendence of e, pi, and many>others, is provided in the 'rst chapter. Speci'cally, Niven gives the>proof of the result that if a_1, ..., a_n are distinct algebraic numbers,>then exp(a_1), ..., exp(a_n) are linearly independent over the algebraic>numbers. The transcendence of e is then proven by taking a_1 = 0, a_2 = 1, >and the transcendence of pi is proven by taking a_1 = 0, a_2 = i pi.The Greek book that I, hold by M.A. MPRIKAS (The famous InsolubleProblems of Antiquity -1970 ),refers to Lambert (1766)stating that his proofs (for the insoluble part of circle's quadrature simply)were not complete,and that Legendre (1794)completed the proofs for proving that Pi and Pi^2 are not rational.But ,still,his proofs did not show the imposibility of circle's quadrature simply.Liouville confronted ,again the same problem in 1840 , and his proposition was simpli'ed in 1873 byG. Cantor.However ,since the transcendental numbers formulated by Liouville and G. Cantor,were somehow TECHNICAL,Charles Hermite proved that number e ,the basis of the natural logarithms ,is not algebraic That was important since it was being proved that it wastranscendental ,and related to pi via Euler's relationshipe^[iPi]+1 =0[ Eyler gave the general formula : e^[ix]=cosx+isinx in 1748 , gave e and its value 2.718 ,and new since 1728 the relationship: e^[iPi]+1 =0 ]The proof of the transcendence of e by Ch.Hermite was given in 1873(published 1874 Comptes Rendus )Hermites proof was very complicated and lengthy ,covering 31 pages in hi APANTA (all) VOL III,page 150-181.A 'rst simpli'cation was given by Weierstrass(Berliner Berichte ,1885) and then followed others, in 1893 by Hilbert,Hurwitz and Gordan(Mathematical Annalen,1893).Lindemann gave in 1882 his proof that number Pi is transcendental [Math.Annalen,20 (1882),p 213] a generalization of Hermites proof , which is alsolengthy ang perplexed.Simpli'ed proof was given by Hilbert [Math.Annalen(1893)P.216-219].In his proof makes use of the relationship e^[ipi]+1=0 , ore^[ipi]=-1 My comments---------------Since e^[iPi]=cosPi+isinPior , e^[iPi]=-1+i[0]then there are two solutions here, to the given equatio:A) e^[ipi]=-1 the real part solution and B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio.My question :What is the implication of this second value of Stefanideshttp://www.stefanides.gr>David McAnally>> Despite anything you may have heard to the contrary,> the rain in Spain stays almost invariably in the hills. === Subject: Re: e is transcendental (was: classes of transcendental numbers ?panamars@otenet.gr (Eur Ing Panagiotis Stefanides) e^[ipi]+1=0 , or>e^[ipi]=-1 >My comments>--------------->Since e^[iPi]=cosPi+isinPi>or , e^[iPi]=-1+i[0]>then there are two solutions here, to the given equatio:>A) e^[ipi]=-1 the real part solution and >B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio.The reasoning behind A and B is not evident to me. Please explainwhere B comes from. e^[i pi] = -1 is right of course, but the last step that you mentioned in your derivation looks completely unjusti'ed.>My question :>What is the implication of this second value of e^[ipi]=0 ?No implication. It is not true.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. === Subject: Re: e is transcendental (was: classes of transcendental numbers ?D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) e^[ipi]+1=0 , or>>e^[ipi]=-1 >>My comments>>--------------->>Since e^[iPi]=cosPi+isinPi>>or , e^[iPi]=-1+i[0]>>then there are two solutions here, to the given equatio:>>A) e^[ipi]=-1 the real part solution and >>B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio.>The reasoning behind A and B is not evident to me. Please explain>where B comes from. e^[i pi] = -1 is right of course, but the last >step that you mentioned in your derivation looks completely >unjusti'ed.>>My question :>>What is the implication of this second value of e^[ipi]=0 ?>No implication. It is not true.Or rather, e^[i pi] = 0 implies anything that you wish to name. A false statement implies anything.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. === Subject: Re: construct a circle of unit area? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id Panagiotis Stefanides a .8ecrit:>> e^[iPi]=-1>> However the complete form for THETA=Pi is:>> e^[iPi]=-1+i[0] , so>> Only the real pert was considered ,and>> the imaginary part i.e. : >> e^[iPi]=i[0] , or>> >> e^[iPi]= 0 >>well, let'solve equation x = 3 + 4>there are two solutions>either x=3 or x=4>no comment...philippe 92 : x = 3 + 4 , gives X=7 You see there are no imaginary parts here philippe 92 ,is it not Stefanideshttp://www.stefanides.gr/theo_circle.htmhttp:// www.stefanides.gr/gmr.htm === Subject: Re: construct a circle of unit area?Eur Ing Panagiotis Stefanides a .8ecrit:> e^[iPi]=-1>However the complete form for THETA=Pi is:> e^[iPi]=-1+i[0] , so>Only the real pert was considered ,and>the imaginary part i.e. : > e^[iPi]=i[0] , or> > e^[iPi]= 0 >>well, let'solve equation x = 3 + 4>>there are two solutions>>either x=3 or x=4>>no comment...> > philippe 92 : x = 3 + 4 , gives X=7 > You see there are no imaginary parts here philippe 92 ,> is it not true?> It's the same for you :e^[iPi]=-1+i[0] does not meane^[iPi]=-1ore^[iPi]=0x=-1+0*i is not x=-1 *or* x=0*i but just x = -1+0 = -1Just in case you can understand (don't feed trolls and cranks)full stop (no further posts)-- === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational>>Hello>>How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect>>this has something to with the golden ratio, and maybe with Fibonacci>>numbers or continued This boils down to me ,to prove whether arcos[IRRATIONAL-G.R.] is IRRATIONAL and if so the ratio:>> >> IRRATIONAL devided by Stefanides>> http://www.stefanides.gr>I haven't worked out the details as yet, but a possible answer might>lie in the fact>that sin(pi/10) = (sqrt(5)-1)/4 = cos(2*pi)/5)>>But>> IRRATIONAL divided by IRRATIONAL(Pi) = IRRATIONAL>is simoly not true! Try pi/pi .> > > Ray > I,thank You.> What I stated includes the words WEATHER ......AND.> Possibly ,AND ALSO should have been be more appropriate.> > > number divided by a transcendental number always>irrational?>>Note that we also have>sin(pi/6)= (sqrt(4) -0)/4>and>sin(pi/12)= (sqrt(6) -sqrt(2) )/4>>Challenge from an old trig book:>>Show that if d > 12 and sin(pi/d)= (sqrt(n) - sqrt(n-4) )/4 then d is>trranscendental.>Hint: Gelfond's of the original question.. So let's clear the board.It depends on the following lemma of Lindemann and Hermite. A proof can be found inany good book on transcendence theory.If a is an algebraic number then e^a is transcendental.I will use this to show that if cos x = (1 + sqrt(5))/2 then x is transcendental. (Hence, it is irrational.)Let's let (1 + sqrt(5))/2)= b/2Using DeMoivre's theorem, e^(ix) = cos x + i sin x, and the hypothesis,we getcos x = (e^ ix + e^(-ix) )/2 = b/2,which leads toe^(2ix) -be^(ix) +1 = 0.This is a quadratic in e^(ix), so we can solve it by the usual formulaand gete^(ix) = c or d, say, both of which are algebraic.But this implies that x is transcendental, by the lemma, and the result follows.Actually, b can be any algebraic number. The same proof will === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational> I 'nally found a proof of the original question.. So let's clear > the board. It depends on the following lemma of Lindemann and > Hermite. A proof can be found in any good book on transcendence > theory.> > If a is an algebraic number then e^a is transcendental.> > I will use this to show that > if cos x = (1 + sqrt(5))/2 then x is transcendental. (Hence, it is > irrational.)But that wasn't the original question. The original question was whether x / pi is irrational. A whole nother kettle of 'sh.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational this to show that > if cos x = (1 + sqrt(5))/2 then x is transcendental. (Hence, it is > irrational.)> > But that wasn't the original question. The original question > was whether x / pi is irrational. A whole nother kettle of 'sh. To make this question actually look interesting, note that if cos(x)=( (1+sqrt(5)) / 4 ), then x/pi is equal to 1/5. (Also, the original question has sqrt(5)-1 instead of sqrt(5)+1 . Again, cos(x) = ( (sqrt(5)-1)/4 ) implies x/pi = 2/5 . The question is to show that this 2/5 value turns transcendental when that O4' on the denominator changes to a O2.')J === Subject: Re: Question for the math >> Brian Doyle terms, say n is 100 digits, m is 50 digits, and I want to>> produce a c that is no more than 10 digits long. Is it possible to>> generate a function (and importantly, if so, how) that translates m to m'>> such that m'^e mod n produces a c that is no more than 10 digits long?>> >> That's not possible on it's face. 10^50 is much larger than 10^10 so the>> function could no longer be a bijection [or at least have an inverse].>> At the very least you'd have to make ||c|| >= ||m||.>>[OT:] As seen with my server, the reference count of>your post is 1 for all three groups concerned. Had>possibly my server discarded OP's post or where did>the original post that you brian@balancesoftwarecom>Rob Johnson take out the trash before replying === Subject: Re: need help!!try http://www.integrals.com/it can integrate almost everything> hi> > i can not integrate from arcsin[2/(3+cosx)]> if u can ,please say me by hupo19@yahoo.com> thank you> hupo> === Subject: Re: need help!!> try http://www.integrals.com/> it can integrate almost everything > > hi> > i can not integrate from arcsin[2/(3+cosx)]> > if u can ,please say me by hupo19@yahoo.com> > thank you> > hupo>i know answer by i dont know how it solve thank you hupo === Subject: Re: need help!! not integrate from arcsin[2/(3+cosx)]> > if u can ,please say me by hupo19@yahoo.com> > i know answer by i dont know how it solve> thank you > hupoWhat answer do you have?J === Subject: Re: need help!!> > > > i can not integrate from arcsin[2/(3+cosx)]> > > if u can ,please say me by hupo19@yahoo.com> > i know answer by i dont know how it solve> thank you > hupo> > What answer do you have?> Jdear jimmy answer is =x*arcsin[2/(3+cosx)] === Subject: Re: need help!!hupo integrate from arcsin[2/(3+cosx)]> > > if u can ,please say me by hupo19@yahoo.com> >> > i know answer by i dont know how it solve> > thank you> > hupo>> What answer do you have?> my answer is =x*arcsin[2/(3+cosx)]Well, what's the derivative of that?Doug === Subject: Re: need help!! > hupo message> >> > >> > > > i can not integrate from arcsin[2/(3+cosx)]> > > > if u can ,please say me by hupo19@yahoo.com> >> > i know answer by i dont know how it solve> > thank you> > hupo> >> > What answer do you have?> > my answer is =x*arcsin[2/(3+cosx)]> > Well, what's the derivative of that? It is de'nitely not arcsin[2/(3cosx)] . I am guessing that Ohupo' has plugged in his expression into integrals.com (or whatever that site it) and it interpreted (cosx) as just a variable name, since it will not parse it into cos(x). If hupo changes his cosx and plugs it back in with cos(x) he will 'nd that even the programs can not come up with an inde'nite integral (probably because there is no closed form for one.)J 10:57:10 -0700, Jim Nastos > It is de'nitely not arcsin[2/(3cosx)] .> I am guessing that Ohupo' has plugged in his expression into >integrals.com (or whatever that site it) and it interpreted (cosx) as just >a variable name, since it will not parse it into cos(x).>I looked brie¤y at that site and was puzzled because it said theintegral of (1/x)*sin(x) was sin(x). Upon reading a little more abouthow to enter a function I discovered that built in functions must becapitalized and use square brackets e.g. (1/x)*Sin[x]. A bit unusualand it makes you wonder how many students get and believe ridiculousanswers.--Lynn === Subject: Re: need help!!> > > > It is de'nitely not arcsin[2/(3cosx)] .> I am guessing that Ohupo' has plugged in his expression into >integrals.com (or whatever that site it) and it interpreted (cosx) as just >a variable name, since it will not parse it into cos(x).>> > I looked brie¤y at that site and was puzzled because it said the> integral of (1/x)*sin(x) was sin(x). Upon reading a little more about> how to enter a function I discovered that built in functions must be> capitalized and use square brackets e.g. (1/x)*Sin[x]. A bit unusual> and it makes you wonder how many students get and believe ridiculous> answers.> > --Lynn> > I beleive that those notational requirements are standard for Mathematica. === Subject: Re: need help!! > On > > > It is de'nitely not arcsin[2/(3cosx)] .> I am guessing that Ohupo' has plugged in his expression into >integrals.com (or whatever that site it) and it interpreted (cosx) as just >a variable name, since it will not parse it into cos(x).>> > I looked brie¤y at that site and was puzzled because it said the> integral of (1/x)*sin(x) was sin(x). Upon reading a little more about> how to enter a function I discovered that built in functions must be> capitalized and use square brackets e.g. (1/x)*Sin[x]. A bit unusual> and it makes you wonder how many students get and believe ridiculous> answers. Ha, I didn't look that far... so I just tried plugging in the integral of: ArcSin[ 2/(3*Cos[x]) ] dx and it returns: Integral ( ArcSin[2*Sec[x]/3] ) dx... at least we know it is correct interpreting the cosine function.J V1+rfqZQQe5bq9ifSJ4sImR5J2u0wcOp-Bqopk04yfLJ-qMfrzoLo+ Lynn that site and was puzzled because it said the>integral of (1/x)*sin(x) was sin(x). Upon reading a little more about>how to enter a function I discovered that built in functions must be>capitalized and use square brackets e.g. (1/x)*Sin[x]. A bit unusual>and it makes you wonder how many students get and believe ridiculous>answers.This is hilarious. But the integrator is right on the point, ofcourse. Reminds me of the time I taught calculus...ThomasP.S. But it's kind of strange, that he allows sin as a multiplicativeconstant or is sin = s*i*n ? BTW: Do christian fundamentalists studytrigonometry [SCNR] === Subject: Re: R-Integration> f[x],g[x]:[0,1]->R de'ned by> > f[x]=1/n if x=1/n> =0 otherwise> g[x]=n if x=1/n> =0 otherwise> then which is function R-integrableWell, what do you think? Did you think at all about this? Show us some work. === Subject: Re: How to determine whether a point is contained inside an object/polygon?> I am trying to determine whether a given point is contained inside a> tetrahedron, and I believe I am following the correct method in order to> do so but need it double-checked just in case!A bit of trivia about this kind of problem: if you happened to want a bookthat would cover this, you might think you'd have to 'nd a bookstore with apretty good math selection, but actually, most big mainstream bookstoreswill actually have what you need. Not in the math section, but over in thecomputer section. There are a whole bunch of books on game programming,and 'nding when a point is inside something is an important problem in gameprogramming.In particular, there are two series of books that often have a lot o'nteresting mathematics in them, which will be found over in the gamesprogramming section: Game Programming Gems (currently 4 volumes, I believe) Graphics Gems (5 volumes, I believe)These books are collections of individual chapters by different authors,gathered together in related sections, covering a wide range of game orgraphics topics.-- --Tim Smith === Subject: Re: Dense Linear Ordering> >> I'm curious, I know how to show dense linear orderings without>> endpoints are isomorphic. However, how do you show that dense linear>> orderings without any advice.> > Are you aware that the rationals and the reals have dense linear> orderings without endpoints but are not isomorphic?Countable dense orderings w/o endpoints are all isomorphic.Are all such orderings of a given cardinality isomorphic?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Dense Linear OrderingX-DMCA-Noti'cations: 19:00:02 +0000, Robin Chapman> > I'm curious, I know how to show dense linear orderings without> endpoints are isomorphic. However, how do you show that dense linear> orderings without any advice.>> >> Are you aware that the rationals and the reals have dense linear>> orderings without endpoints but are not isomorphic?>>Countable dense orderings w/o endpoints are all isomorphic.>>Are all such orderings of a given cardinality isomorphic?Surely not...For example R and R with an interval removed and the rationalsin that interval stuck back in.************************David C. Ullrich === Subject: Re: Dense Linear OrderingI'm pretty con'dent that the rational and Robin Chapman> > > I'm curious, I know how to show dense linear orderings without> endpoints are isomorphic. However, how do you show that dense linear> orderings without endpoints advice.>> Are you aware that the rationals and the reals have dense linear>> orderings without endpoints but are not isomorphic?>>Countable dense orderings w/o endpoints are all isomorphic.>>Are all such orderings of a given cardinality isomorphic?> > Surely not...> > For example R and R with an interval removed and the rationals> in that interval stuck back in.> > > > ************************> > David C. Ullrich === Subject: Re: Newbie: Integration? <16bWb.20296$l04.17724@newssvr16.news.prodigy.com> <%0cWb.15477$Lj5.8599@news01.roc.ny> is actually just it. When> Math Tutor mentioned trapezoids, that is really just 'tting a> linear polynomial between a pair of neighbouring points. What you> get is a (ver slim) trapezoid. Calculating the area under that is> pretty simple. The hard part is that you have to add up all these> slim trapezoids. This is best done using a computer program, if only> because it gets very tedious.Perhaps the OP means that by adding up the areas of the trapezoidswould not mean to integrate to calculate the area under the curve.Maybe he has a semantic question about integrals. Not sure.There are methods of integration; one of them is to slice a shape inin'nitely many slices and take the limit of the sum (of the slices).A nice tutorial is http://www.karlscalculus.org/calc10 0.htmlHTH,Daniel === Subject: Re: Dogs, ¤eas, and hairy balls> No, really, I haven't ...I'm not a topologist, so someone might have to correct me. There is a 'xed point theorem that says something like: given a continuous mapping from the surface of a sphere to the same sphere there must be at least one 'xed point, i.e. a point that is mapped into itself. The hairy ball idea follows - take a hairy ball, say a tennis ball, and try to comb the hair in such a way that it is smoothly combed everywhere. You can't do it, there will be two points (source and sink) where there hair is not locally all combed in the same direction.Gib === Subject: Re: Dogs, ¤eas, and hairy balls> >> No, really, I haven't ...> > I'm not a topologist, so someone might have to correct me. There is a> 'xed point theorem that says something like: given a continuous mapping> from the surface of a sphere to the same sphere there must be at least> one 'xed point, i.e. a point that is mapped into itself.The antipodal map is continuous on the surface of the sphere :-)> The hairy> ball idea follows - take a hairy ball, say a tennis ball, and try to> comb the hair in such a way that it is smoothly combed everywhere. You> can't do it, there will be two points (source and sink) where there hair> is not locally all combed in the same direction.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Dogs, ¤eas, and hairy balls> > > [snipped a few lines]>I've noticed that there is a spot on her belly where there are always>>several ¤eas congregated. This is a convergence point for the lay of>>her fur, in other words all the hairs around this point are oriented>>towards it.>>I think ¤eas know about the hairy ball theorem, i.e. they know that if>>they follow the direction of the fur they will eventually get to a 'xed>>point. My theory is that this innate behaviour enables them to come>>together to mate - the 'xed point is a like a singles bar for ¤eas.>>BTW, the other convergence point (reached by going upstream) is at the head.>>Gib> > > An interesting observation (and a possible explanation). Could the phenomenon> also be explained by the possibility that it is easier for the ¤eas to travel> along the ¤ow of the fur. When they reach a sink, they are stuck (statistically> anyway:)I don't think they are stuck in any sense. When surprised they ¤ee in all directions ;), and in a few seconds they are all gone. I get the strong impression that there is something social going on. They are not feeding. I think it may be easier for them to go with the ¤ow, but of course they go where they like. They seem to prefer the back legs, presumably because there is plenty of blood close to the surface and it isn't so easy for the dog to catch them there.I think this is in the domain of bio-mathematics :)Gib === Subject: in message> I'm intrigued by the questions raised by your recent posts, as for> years you were this guy who came up with rather creative ways to> insult me, and now I 'nd it hard not to 'gure you're just doing so> again.>> Now I'll embarrass you a bit as from what I've read on the web you> have one of the highest IQ's out there, so it seems to me there's> probably some reason to what you're doing, and possibly I'm wrong> about what it is.>> Therefore, I'm going to give you the opportunity that I feel *I* don't> get, which is the bene't of the doubt.>> Tell me succinctly and in a way that will minimize potential> embarrassment for both of us, what it is that your up to, and no, none> of this wild stuff about how great I supposedly am, or how I've proven> FLT or any of that, as I just want you to say something that 'ts into> a worldview that makes sense.>> What's your intent?But the fact is, Jim Ferry really is everything he claims to be, so yourpoint is moot. How do I know that Jim is what he is? It's as simple asthis: he claims to be, so I believe him.The other day he said:> xf[u_,m_] := Sqrt[1-m*Sin[u]^2]/m;> yf[u_,m_] := ((m/2-1) EllipticF[u,m] + EllipticE[u,m])/m;> u0 = 1.140659153420324; m0 = Csc[u0]^2;> fac = 1 / Re[EllipticK[m0]];> ParametricPlot[fac {xf[u,m0], yf[u,m0]}, {u, -u0, u0},> AspectRatio -> Automatic];I can't follow his Mathematica, but that's just because it's too complex forme. But this doesn't matter. I know that Jim is right.I realize how such glori'cation of Jim must seem to you James. It soundsabsurd. It sounds, in fact, like I'm mocking Jim by praising him overmuch.But I sincerely believe that Jim's future reputation will be so great -- fargreater than any human being currently enjoys -- that to speak of him in anylesser tones would simply be inaccurate. What I'm trying to do here, James,is give you some idea of just how important a 'gure Jim Q. Ferry will be inthe future, and of how vital it is to your own reputation that you stopassailing him so.-- Clive === Toothhttp://www.clivetooth.dkSubject: Re: Looking for primes of a particular form. > > Hugo Pfoertner an old version of OpenPFGW (which> currently seems to be unavailable).> > It's not unavailable at all, it's very much alive. It's just that> the original author, the one who has the DNS record, is no longer> involved in teh project, and noone else can host that old site.> > The project's still in active develpment, and recent builds> (only weeks old) can be picked up from the 'les area of the> openpfgw (for beta versions, that's our developers' forum)> and primeform (stable versions, that's the users' forum)> yahoogroups at groups.yahoo.com.> > PhilThere are many people who hesitate to sign up in something likeyahoogroups, which says about itsselfYahoo! Groups is an advertising supported service, just to get a copyof a program, that was conveniently accessible via a webpage before.I myself had used OpenPFGW to create some sequences in Neil Sloane'sOEIS giving a link to http://www.primeform.net/openpfgw/ , which is nowdead. Unless the PFGW project will be able to provide a reasonablelink for reference (and not just a pointer to a login mask), I wouldhave to ask Neil Sloane to remove the dead links from the ~15 sequencesreferencing OpenPFGW.Hugo === Subject: Re: Looking for primes of a particular form.Hugo Pfoertner with an old version of OpenPFGW (which> > currently seems to be unavailable).>> It's not unavailable at all, it's very much alive. It's just that> the original author, the one who has the DNS record, is no longer> involved in teh project, and noone else can host that old site.>> The project's still in active develpment, and recent builds> (only weeks old) can be picked up from the 'les area of the> openpfgw (for beta versions, that's our developers' forum)> and primeform (stable versions, that's the users' forum)> yahoogroups at groups.yahoo.com.>> Phil>> There are many people who hesitate to sign up in something like> yahoogroups, which says about itsself> Yahoo! Groups is an advertising supported service, just to get a copy> of a program, that was conveniently accessible via a webpage before.>> I myself had used OpenPFGW to create some sequences in Neil Sloane's> OEIS giving a link to http://www.primeform.net/openpfgw/ , which is now> dead. Unless the PFGW project will be able to provide a reasonable> link for reference (and not just a pointer to a login mask), I would> have to ask Neil Sloane to remove the dead links from the ~15 sequences> referencing OpenPFGW.Primeform seems to be available at:http://pages.prodigy.net/chris_nash/primeform.html Brian Gladman === Subject: Re: Topology> let R with usual topology;U={x in R/-1<=x<=1;x!=0}> then U is T-2 space? and U is First countable?WWW gave you da best hint ! === Subject: Re: Accumulation and Limit Point> I am trying to clarify the difference between accumulation point and > limit point (or maybe there is not a substantial difference). In > analysis the terms seem to be used almost interchangeably, but now I'm > told a limit point should be considered the point of convergence of a > limit while an accumulation point of a set S exists if given any eps>0 > there is a point x in S with x =0 and |x - x_0| of a d-limit point in topology without dealing with limits of sequences > or functions.> > I'm confused. Can someone help me?> > Tom AdamsI'm not sure if this is standard terminology, but I have seen thefollowing de'nitions for limit and accumulation points of a set S:x is a limit point of S if every neighborhood of x contains at leastone element of S distinct from x. (I think this is standard.)x is an accumulation point of S if every neighborhood of x containsin'nitely many elements of S. (Not sure if this is standard)In metric spaces the 2 de'nitions turn out to mean the same thing.But in general topological spaces they are not the same concept.Artur === Subject: Re: Accumulation and Limit Point> > I'm not sure if this is standard terminology, but I have seen the> following de'nitions for limit and accumulation points of a set S:>> x is a limit point of S if every neighborhood of x contains at least> one element of S distinct from x. (I think this is standard.)> Yes standard> x is an accumulation point of S if every neighborhood of x contains> in'nitely many elements of S. (Not sure if this is standard)> seems true according to mathworld> In metric spaces the 2 de'nitions turn out to mean the same thing.> But in general topological spaces they are not the same concept.> In any T1-space, it's also true.Lets see a counterexample :E={a,b} with the trivial topology (The only open sets are the empty set and E)b is an limit point of {a} but not an accumulation point.If you increase the topology adding {b} as new open set,you 'nd a Kolmogorov (T0) space with the same counterexample. === Subject: Please help prove! about modules. by support1.mathforum.org (8.11.6/8.11.6/The Math cannot 'gure out two proofs about modules.1. S-module M is projective <=> each short exact sequence 0->A->B->M->0 splits.2. Q, the abelian group of rational numbers, is an injective Z-module.If anyone could help me, it would be === Subject: Re: Please help prove! about modules.> I am now studying algebra by myself.> > I cannot 'gure out two proofs about modules.> > 1. S-module M is projective <=> each short exact sequence 0->A->B->M->0> splits.It rather depends on the de'nition of projective you are using.(Some would regard this property as the de'nition of projectivity.)> 2. Q, the abelian group of rational numbers, is an injective Z-module.The best way of doing this involves proving Baer's criterionfor injectivity, and then showing that Q satis'es it.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Please help studying algebra by myself.> > I cannot 'gure out two proofs about modules.> > 1. S-module M is projective <=> each short exact sequence 0->A->B->M->0> splits.> > 2. Q, the abelian group of rational numbers, is an injective Z-module.> > If in advance.Not the answer you were looking for but these two results are prettystandard and can be found in most Algebra texts - I recommendHungerford. (They are exercises in Lang.) If you can't locate a suitable text, ask again.-- Paul SperryColumbia, SC (USA) === Subject: Re: construct a circle of unit area? by support1.mathforum.org (8.11.6/8.11.6/The Math .8ecrit:> e^[iPi]=-1>>However the complete form for THETA=Pi is:>> e^[iPi]=-1+i[0] , so>>Only the real pert was considered ,and>>the imaginary part i.e. : >> e^[iPi]=i[0] , or>> >> e^[iPi]= 0 >>well, let'solve equation x = 3 + 4>there are two solutions>either x=3 or x=4>no comment...>> >> philippe 92 : x = 3 + 4 , gives X=7 >> You see there are no imaginary parts here philippe 92 ,>> is it not true?>> >>It's the same for you :>e^[iPi]=-1+i[0] does not mean>e^[iPi]=-1>or>e^[iPi]=0>>x=-1+0*i is not x=-1 *or* x=0*i but just x = -1+0 = -1>Just in case you can understand (don't feed trolls and cranks)>full stop (no further posts)>-- Well philippe 92 You may see the following and think what I am implying:The Greek book that I, hold by M.A. MPRIKAS (The famous InsolubleProblems of Antiquity -1970 ),refers to Lambert (1766)stating that his proofs (for the insoluble part of circle's quadrature simply)were not complete,and that Legendre (1794)completed the proofs for proving that Pi and Pi^2 are not rational.But ,still,his proofs did not show the imposibility of circle's quadrature simply.Liouville confronted ,again the same problem in 1840 , and his proposition was simpli'ed in 1873 byG. Cantor.However ,since the transcendental numbers formulated by Liouville and G. Cantor,were somehow TECHNICAL,Charles Hermite proved that number e ,the basis of the natural logarithms ,is not algebraic That was important since it was being proved that it wastranscendental ,and related to pi via Euler's relationshipe^[iPi]+1 =0[ Eyler gave the general formula : e^[ix]=cosx+isinx in 1748 , gave e and its value 2.718 ,and new since 1728 the relationship: e^[iPi]+1 =0 ]The proof of the transcendence of e by Ch.Hermite was given in 1873(published 1874 Comptes Rendus )Hermites proof was very complicated and lengthy ,covering 31 pages in hi APANTA (all) VOL III,page 150-181.A 'rst simpli'cation was given by Weierstrass(Berliner Berichte ,1885) and then followed others, in 1893 by Hilbert,Hurwitz and Gordan(Mathematical Annalen,1893).Lindemann gave in 1882 his proof that number Pi is transcendental [Math.Annalen,20 (1882),p 213] a generalization of Hermites proof , which is alsolengthy ang perplexed.Simpli'ed proof was given by Hilbert [Math.Annalen(1893)P.216-219].In his proof makes use of the relationship e^[ipi]+1=0 , ore^[ipi]=-1 My comments---------------Since e^[iPi]=cosPi+isinPior , e^[iPi]=-1+i[0]then there are two solutions here, to the given equatio:A) e^[ipi]=-1 the real part solution and B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solution.My question :What is the implication of this second value of e^[ipi]=0 ?C'est la difference.N'est Stefanideshttp://www.stefanides.gr === Subject: Re: InequalityHI There,>Consider a and b :>a.> -1 <= 2/x <= 1>b.> -1 <= 2/x > and > 2/x <=1>Is there any difrence between two inequalities ? >I mean when we write them togheter and when write them alone.> > No, there's no difference, (b) is exactly what (a) means. > (If you said -1 <= 2/x or 2/x <=1 than _that_ would be different...)> >Solve them for me please.> > Too hard, UllrichSo how do you write-1 <= 2/x or 2/x <=1Mathematically === Subject: Re: InequalityX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>> >>HI There,>>Consider a and b :>>a.>> -1 <= 2/x <= 1>>b.>> -1 <= 2/x >> and >> 2/x <=1>>Is there any difrence between two inequalities ? >>I mean when we write them togheter and when write them alone.>> >> No, there's no difference, (b) is exactly what (a) means. >> (If you said -1 <= 2/x or 2/x <=1 than _that_ would be different...)>> >>Solve them for me please.>> >> Too hard, David C. Ullrich>>So how do you write>-1 <= 2/x or 2/x <=1>Mathematically LIke so: -1 <= 2/x or 2/x <=1.>(i.e without or ) ?Huh? Who told you that the word or wasnot mathematical? you>Omid Hezaveh************************David C. Ullrich === Subject: Re: Riemann IntegrationDavid C. Ullrich be continious. then set of all such f's such that>> inegration with limits 0 to 1 of t^n*f9t) dt=0 is>> Please re-read your posts before posting them; it should be t^n*f(t).>> a) single element>> b)contably in'nite>> C) in'nite>> The 'rst possibility is the correct one.>>If your f satis'es this restriction, what about cf? It seems thatthere's>an uncountable number if there's a single nonzero f that satis'es the>criterion.>> That's correct. It's also true that (a) is the right answer, if we> assume that the equation is supposed to hold for all> non-negative integers n.>Thinking about this, I began to think I could show that because thiscontinuous f is orthogonal to all {x^n}, whose linear combinations generatethe space of polynomials, the closure of which includes the continuousfunctions, that f was necessarily either not continuous or equal to 0. Doesthat kind of reasoning hold up?>How did you interpret the n? As 'xed or Carlos Santos>> ************************>> David C. Ullrich === Subject: Re: Riemann IntegrationX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>>David C. Ullrich continious. then set of all such f's such that> inegration with limits 0 to 1 of t^n*f9t) dt=0 is>> Please re-read your posts before posting them; it should be t^n*f(t).>> a) single element> b)contably in'nite> C) in'nite>> The 'rst possibility is the correct one.>If your f satis'es this restriction, what about cf? It seems that>there's>>an uncountable number if there's a single nonzero f that satis'es the>>criterion.>> That's correct. It's also true that (a) is the right answer, if we>> assume that the equation is supposed to hold for all>> non-negative integers n.>>Thinking about this, I began to think I could show that because this>continuous f is orthogonal to all {x^n}, whose linear combinations generate>the space of polynomials, the closure of which includes the continuous>functions, that f was necessarily either not continuous or equal to 0. Does>that kind of reasoning hold up?Yes. An easy way to see this would be like so: Suppose that f iscontinuous but not identically 0. Then int_0^1 f^2 > 0.Choose a sequence of polynomials P_n with P_n -> f uniformly.Then f*P_n -> f^2 uniformly, but int_0^1 f*P_n = 0 for all n,and you've proved 0 > 0.>>How did you interpret the n? As 'xed or varying over all>>positive integers?> Best ************************>> David C. Ullrich>************************David C. Ullrich === Subject: Re: :: more focus on the constructive discussion ::: What, exactly, are you trying establish here?: Why have you cross-posted this to alt.fan.noam-chomsky?: Or have I just been suckered in by a troll who has WAY too much time: on their hands, and needs to get a life?Besides Chomsky's distinctive political theories, he also contributed to thelogical school of linguistic analysis, which my post deals with.I have made enough time in my busy schedule to put some initial focus on theconcepts I discuss over these forums, but I have a life with many otherjoys, demands, and other signs of life.As to being a troll, I have seen many more unrelated posts in all thenewsgroups I posted to. Instead, I have kept within the topicality of allnewsgroups I posted too.-- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea: prankster, fablist, magician, liar === Subject: Re: :: more 13:31:44 -0800, galathaea >: What, exactly, are you trying establish here?>: Why have you cross-posted this to alt.fan.noam-chomsky?>: Or have I just been suckered in by a troll who has WAY too much time>: on their hands, and needs to get a life?>>Besides Chomsky's distinctive political theories, he also contributed to the>logical school of linguistic analysis, which my post deals with.>>I have made enough time in my busy schedule to put some initial focus on the>concepts I discuss over these forums, but I have a life with many other>joys, demands, and other signs of life.>>As to being a troll, I have seen many more unrelated posts in all the>newsgroups I posted to. Instead, I have kept for responding.I found your original post confusing as to it's intent. === Subject: Re: :: more focus on the constructive I thought it might be useful to post more references and reorganise into>more focused questions. I appologise for the more playful 'rst post. It's>sometimes hard for me to put aside creativity for more formality.>>There are Heyting algebras in a lot of 'elds. They are one of the more>useful logical forms for our models in these 'elds.>>In the study of cognitive research we have models of neural interaction, and>many experiments match well to the models. There is a school of thought>that says concepts are attractors in the neural models dynamics. This>allows us to pose dynamical propositions in the neural models. The>propositional structure would then likely be Heyting in such a model.>> :>> What, exactly, are you trying establish here?> Why have you cross-posted this to alt.fan.noam-chomsky?Curiously, when I open Introduction to Formal Languages by Gyorgy Revesz,Chapter 1 Section 2 is entitled The Chomsky Heirarchy of Languages.The back of the Dover title advertises the book with the following introductoryparagraph,Although formal languages have long played afundamental role in mathematics, logic, and othersciences, the computer revolution has resultedin an unprecedented proliferation of arti'ciallanguages. Formal language theory attempts tobring a semblance of order to this modern Towerof Babel. This is possible in one respect becausethere is a close relationship between formal grammarsand other abstract notions used in computer science,such as automata and algorithms.Galathaea is restating intent made known in a previous thread. This issue isactually educational criteria and questionable curriculum priorities (That is whyI included the quote referring to a Tower of Babel.).Nearly all my friends went through logic classesas prerequisites in their undergraduate studies, andthese never explained anything outside of quanti'edclassical logic (often without much algebra). And itseems to me that with just a little restructuring, sucha course could include constructive logic as it buildsthe axiom set.There is a question here as to whether curriculum priorities should be changed.The existing standard curriculum is grounded in historical prejudices.Galathaea's threads are intended to be substantive with respect to an actiondecision. Any newsgroup participant who may be in a position to in¤uencecurriculum committees is being asked to consider this de'ciency. For everyoneelse, there is essentailly an offer of edi'cation on how Heyting algebras relatein their 'elds.There are many different kinds of trolls.:-)mitch === Subject: Re: :: more focus on the -0800, galathaea I thought it might be useful to post more references and reorganise into>>more focused questions. I appologise for the more playful 'rst post. It's>>sometimes hard for me to put aside creativity for more formality.>>There are Heyting algebras in a lot of 'elds. They are one of the more>>useful logical forms for our models in these 'elds.>>In the study of cognitive research we have models of neural interaction, and>>many experiments match well to the models. There is a school of thought>>that says concepts are attractors in the neural models dynamics. This>>allows us to pose dynamical propositions in the neural models. The>>propositional structure would then likely be Heyting in such a model.>> :>> What, exactly, are you trying establish here?>> Why have you cross-posted this to alt.fan.noam-chomsky?>>Curiously, when I open Introduction to Formal Languages by Gyorgy Revesz,>Chapter 1 Section 2 is entitled The Chomsky Heirarchy of Languages.>>The back of the Dover title advertises the book with the following introductory>paragraph,>>Although formal languages have long played a>fundamental role in mathematics, logic, and other>sciences, the computer revolution has resulted>in an unprecedented proliferation of arti'cial>languages. Formal language theory attempts to>bring a semblance of order to this modern Tower>of Babel. This is possible in one respect because>there is a close relationship between formal grammars>and other abstract notions used in computer science,>such as automata and algorithms.>Galathaea is restating intent made known in a previous thread. This issue is>actually educational criteria and questionable curriculum priorities (That is why>I included the quote referring to a Tower of Babel.).>>Nearly all my friends went through logic classes>as prerequisites in their undergraduate studies, and>these never explained anything outside of quanti'ed>classical logic (often without much algebra). And it>seems to me that with just a little restructuring, such>a course could include constructive logic as it builds>the axiom set.>There is a question here as to whether curriculum priorities should be changed.>The existing standard curriculum is grounded in historical prejudices.>>Galathaea's threads are intended to be substantive with respect to an action>decision. Any newsgroup participant who may be in a position to in¤uence>curriculum committees is being asked to consider this de'ciency. For everyone>else, there is essentailly an offer of edi'cation on how Heyting algebras relate>in their 'elds.>>There are many different kinds of Galathaea's post was an attempt to clarify an issue, or issues,then it failed abysmally for me. === Subject: The role of in'nity in mathWhat is the role of in'nity in math:How is it de'ned?Why is it needed?At what precition does math work?These are questions which seem to have to accepted answer, what do === youthink?Subject: Re: The role of in'nity in > What is the role of in'nity in math:For part of your answer, I suggest How is it de'ned?It depends on the context. In calculus, it is usually a concept, not a value. In complex analysis (as modeled by the Riemann sphere), it is a number. In set theory, there are multiple distinct in'nities corresponding to various cardinalities.> > Why is it needed?It's easier to write x -> oo than as x grows larger than any arbitrary value epsilon.> > At what precition does math work?That depends on what the acceptable error is in your statistical analysis/approximation.On a more serious note, I'm not sure what you're trying to ask. My inclination is to so something like in'nite precision or absolute precision.> These are questions which seem to have to accepted answer, what do you> think?Huh? This question doesn't make sense.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: Re: The role of in'nity in mathWill Twentyman of in'nity in math:>> For part of your answer, I suggest How is it de'ned?>> It depends on the context. In calculus, it is usually a concept, not a> value. In complex analysis (as modeled by the Riemann sphere), it is a> number. In set theory, there are multiple distinct in'nities> corresponding to various cardinalities.> Why is it needed?>> It's easier to write x -> oo than as x grows larger than any> arbitrary value epsilon.> At what precition does math work?>> That depends on what the acceptable error is in your statistical> analysis/approximation.>> On a more serious note, I'm not sure what you're trying to ask. My> inclination is to so something like in'nite precision or absolute> precision.Yes this was in'niteprecision.> These are questions which seem to have to accepted answer, what do you> think?> Huh? This question doesn't make sense.You are right, it doesn't. I'll try again:These are questions which seem not to have an accepted answer, what do youthink?>> -- > Will Twentyman> email: wtwentyman at copper dot net> === Subject: Re: The role of in'nity in math> What is the role of in'nity in math:> > How is it de'ned?> > Why is it needed?> > At what precition does math work?> > These are questions which seem to have to accepted answer, what do you> think?That is a very controversial question. On the one hand, there is mydisturbed brother-in-law Ben (who posted a few threads and believesthat there are only a 'nite number of number). He is of theultra-'nitist philosophy and he is not alone, believe it or not.Although, most people of that philosophy would not go so far as to doa computer search for the largest number. Seehttp://en.wikipedia.org/wiki/UltraintuitionismThen there is the intuitionist philosophy which is a little moremainstream but is still not the majority opinion. Their objection isnot based on whether in'nity exists or not, but on whether we caneven say anything about in'nity if it did exist. What separates themfrom the classic mathematicians is that they reject Aristotle's Law ofthe Excluded Middle that the proposition (A or not A) is a tautology,since one can never know for sure if for each x, (A(x) or not A(x)) isa tautology when x ranges over an in'nite domain. While this may seemcrazy and counter-intuitive, intuitionists might respond that thetheorems that result from this assumption are also counter-intuitive -i.e., Cantor's aleph null and aleph one. Why should there be manytypes of in'nities? In'nity is in'nity. And also, look at all ofthe paradoxes of set theory - Is there a set of all sets, etc. Whenone eliminates the Law of the Excluded Middle, one avoids a lot of thecrazy paradoxes associated with in'nity.The classic philosophy (Platonism) says that all of this stuff(in'nity) is in fact real. And just because we cannot perceive itdoes not mean that it does not exist. This is the majority opinion,not necessarily because the majority believe it, but more because itis the most practical way of thinking of mathematics, since theconclusions have applications in natural sciences. If we were to go towhat my brother-in-law proposes, replacing differential equations withdifference equations, etc., we would get the same theorems, but theywould be much more dif'cult to write and describe. The derivativeof x^2 would not be 2x but would be 2x+1/M, where M is the largestnumber. In practice, this number 1/M would be so small that itwouldn't even matter, so why write it down on paper? Also, thePythagorean theorem would not hold, and it would be much more messy todescribe this relationship. But this way of thinking still avoidsparadoxes.Craig === Subject: Re: The role of in'nity in math> What is the role of in'nity in math:It's only a supporting cast member. But,it won an Oscar for Best SupportingTrans'nite Number In A Scienti'c Discipline.> These are questions which seem to have to accepted answer, what do you> think?I think.....very dirty thoughts about Rachael Ray.-- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people think, Mr. Feynman? === Subject: Re: The role of in'nity in math> What is the role of in'nity in math:None at all. Mathematicians (and teachers) would probably do everyone a greatservice if they never made reference to in'nity; leave themysticism to someone else (Buzz Lightyear, perhaps?)Certainly we have great need of some sets which are not 'nite;the set of counting numbers comes to mind. Since the sets are not'nite, we call them in-'nite sets, but notice that this is a negative word: we know what the sets are _not_ (they're not 'nite)but we don't waste energy thinking about what they _are_ (that is,we don't try to grapple with in'nity per se).When you have a set that's not 'nite, and people ask you howmany things you've got, you can say in'nitely many, but they'restill just ordinary things (points, numbers, whatever); there'sjust a lot of them -- not a 'nite number of them at all, so, well,an in-'nite number of them.Mathematicians WILL sometimes use the phrase in'nity as a sort ofslang to refer to something that takes more words to describe properly,but that's just between consenting adults. There is really nothingin'nite about most of these things. For instance, if you puncturea balloon, you have some rubber which you can stretch out to give a model of the (complex) plane; running this backwards, you can roll upthe plane into a sphere, but you have to add back the point of thepuncture. That extra point isn't part of the rubber plane, but it ispart of the sphere. For some reason a more pedestrian name is notused; instead, it's call the point at in'nity (or just in'nity).But it's still just a === point.daveSubject: Re: The role of in'nity in > > >>What is the role of in'nity in math:> > > None at all. How many integers are there?> > Mathematicians (and teachers) would probably do everyone a great> service if they never made reference to in'nity; leave the> mysticism to someone else (Buzz Lightyear, perhaps?)Modern set theory deals a great deal with in'nity as a core concept.> > Certainly we have great need of some sets which are not 'nite;> the set of counting numbers comes to mind. Since the sets are not> 'nite, we call them in-'nite sets, but notice that this is a > negative word: we know what the sets are _not_ (they're not 'nite)> but we don't waste energy thinking about what they _are_ (that is,> we don't try to grapple with in'nity per se).> > When you have a set that's not 'nite, and people ask you how> many things you've got, you can say in'nitely many, but they're> still just ordinary things (points, numbers, whatever); there's> just a lot of them -- not a 'nite number of them at all, so, well,> an in-'nite number of them.You are focusing on the individual elements, but when people talk about how many elements there are in a set, the desire is to get a meaningful value. In fact, simply saying that it is in'nite is not always enough detail. Thus the cardinals to measure levels of in'nity.> > Mathematicians WILL sometimes use the phrase in'nity as a sort of> slang to refer to something that takes more words to describe properly,> but that's just between consenting adults. There is really nothing> in'nite about most of these things. For instance, if you puncture> a balloon, you have some rubber which you can stretch out to give a > model of the (complex) plane; running this backwards, you can roll up> the plane into a sphere, but you have to add back the point of the> puncture. That extra point isn't part of the rubber plane, but it is> part of the sphere. For some reason a more pedestrian name is not> used; instead, it's call the point at in'nity (or just in'nity).> But it's still just a point.The Riemann sphere has a point on it that *is* in'nity. It is a value, not a concept, in that model of the complex numbers.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: Re: The role of in'nity in math>> >What is the role of in'nity in math:>>None at all. >>Mathematicians (and teachers) would probably do everyone a great>service if they never made reference to in'nity; leave the>mysticism to someone else (Buzz Lightyear, perhaps?)>[...]Hersh,_The_Mathematical_Experience_:The constructivists regard as genuine mathematics only what can be obtained by a 'nite construction. The set of real numbers, or any other in'nite set, cannot so be obtained.Loc. cit.,Constructivists are a rare breed, whose status in the mathematical word sometimes seems to be that of tolerated heretics surrounded by orthodox members of an established church.Personally, I am *not* of this rare breed.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: The role of in'nity in math> >> >What is the role of in'nity in math:>> >>None at all. >>Mathematicians (and teachers) would probably do everyone a great>service if they never made reference to in'nity; leave the>mysticism to someone else (Buzz Lightyear, perhaps?)>> [...]> > Hersh,_The_Mathematical_Experience_:Not necessarily.For example, I'm not a constructivist. I use ordinals and cardinals allthe time. I like classical math and set theory, and after readingD. Rusin's posts I think I am in complete agreement with him. Leonard Blackburn> > The constructivists regard as genuine mathematics only what can be > obtained by a 'nite construction. The set of real numbers, or any > other in'nite set, cannot so be obtained.> > Loc. cit.,> > Constructivists are a rare breed, whose status in the mathematical word > sometimes seems to be that of tolerated heretics surrounded by orthodox > members of an established church.> > Personally, I am *not* of this rare breed. === Subject: Re: The role of in'nity in math>What is the role of in'nity in math:I responded:>None at all. >Hersh,_The_Mathematical_Experience_:>>The constructivists regard as genuine mathematics only what can be >obtained by a 'nite construction. The set of real numbers, or any >other in'nite set, cannot so be obtained.I don't think I'm a constructivist (though I don't claim to have anauthoritative delineation of the schools of mathematical thought).I happily use existence proofs in which the thing whose existence isclaimed is never, in fact, constructed. I merrily work with in'nitesets and other betes noires of the constructivists.The example given in D&H's quote is probably perfect: I 'nd thereal numbers to be very useful. All of them, individually, and thewhole set of them. It's an in'nite set. But inside that set I'nd only numbers. I don't 'nd a thing called in'nity. Well,OK, so in'nity is not a real number. What is it, then, and whyshould I care about it? (You can argue that it's a complex number,as I did in a prior post. But then, it just sits in my bag of complexnumbers next to 0 and i and sqrt(2). It doesn't glow and sparkleand boggle my mind. It's just a point with a funny name, one whosearithmetic rules are a little different.) Take a look at the theoremsof mathematics: are any of them about in'nity? (Not the pointsat in'nity on a curve; they are ordinary points in projective space.Not limits in calulcus which equal in'nity; those are limits whichfail to exist but have an easily-described behaviour. Not in'nitesets; they're studied as sets, or ordered sets, or something, andthey just happen not to be 'nite. I just don't think I've ever seenOne can argue that sets exist in a platonic realm, and the wholedevelopment of set theory is an attempt to capture, in concreteaxioms, the features that we know are there. Sometimes real-analysiscourses build on students' prior knowledge of numbers and simply present the de'ning axioms for R as a means of capturing the essenceof that prior experience, so that there is a foundation on which tobuild proofs. Broadly speaking, this is how most productive mathematicsfunctions: we have an object in mind, and we set some axioms to mirrorits behaviour, then investigate that model. Well, great. Now whatsort of previously-conceived object is in'nity, that we shouldstudy it?I'm all in favor of enlivening the minds of school kids who aresympathetic to mathematics. Physicists can talk about black holes,and mathematicians can talk about vastly huge sets. Fine. But weget a lot of gobbledygook in sci.math and elsewhere which wouldprobably be avoided if we stopped getting the students to thinkabout in'nity as a (mathematical) object. That's how I interpretedthe OP's 'rst question; that's why I responded as I did.dave === Subject: Re: The role of in'nity in mathStephen J. Herschkorn only what can be > obtained by a 'nite construction. The set of real numbers, or any > other in'nite set, cannot so be obtained. This is denounced as pure mysticism by those who want to actuallybe able to obtain things in mathematics by a 'nite construction. === Subject: Re: The role of in'nity in F1vU=Jm9LchOj|&)y/y'n33$K{,'|+j:PBvlXn]+u&LG@{ zY!w}Yp)i57ga,S] B3`Xj1LZGa)d5?ma~E_^>yu_xS(E:wqD.(nw&)SmG)LQK:H;^S;&*0!.uuW+ 59UwyslR=xGd RtodT9/H-B.I_>&f9CXI;dty3> > What is the role of in'nity in math:> > None at all. > > Mathematicians (and teachers) would probably do everyone a great> service if they never made reference to in'nity; leave the> mysticism to someone else (Buzz Lightyear, perhaps?)> > Certainly we have great need of some sets which are not 'nite;> the set of counting numbers comes to mind. Since the sets are not> 'nite, we call them in-'nite sets, but notice that this is a > negative word: we know what the sets are _not_ (they're not 'nite)> but we don't waste energy thinking about what they _are_ (that is,> we don't try to grapple with in'nity per se).> > When you have a set that's not 'nite, and people ask you how> many things you've got, you can say in'nitely many, but they're> still just ordinary things (points, numbers, whatever); there's> just a lot of them -- not a 'nite number of them at all, so, well,> an in-'nite number of them.> [rest snipped]Dave, I understand part of what you're saying here: that in'nite oftenmeans not 'nite-- but I'm quite puzzled by your overall point.What if someone asks me how many hyperbolic 3-manifolds there are, andI reply Omega to the Omega power. I'm not only saying there are anot-'nite amount of them, I'm saying the collection can be ordered asa particular order type. And note I'm not doing this to be weird, butthat is actually the way you would count them. First count all thehyperbolic 3-manifolds with 0 cusps, then those with 1 cusp, then thosewith 2 cusps, etc. It would seem I am in fact grappling with what theyare, not just what they are not -- to use your wording.So I guess I'm not understanding how ordinals 't into your explanationof why in'nity plays no role in math. It seems to me that I amclassifying different types of in'nite numbers. === Subject: Re: The role of in'nity in mathIn response to my stand on a have a set that's not 'nite, and people ask you how>> many things you've got, you can say in'nitely many, but they're>> still just ordinary things (points, numbers, whatever); there's>> just a lot of them -- not a 'nite number of them at all, so, well,>> an in-'nite number of them.>> [rest snipped]>What if someone asks me how many hyperbolic 3-manifolds there are, and>I reply Omega to the Omega power. I'm not only saying there are a>not-'nite amount of them, I'm saying the collection can be ordered as>a particular order type. >So I guess I'm not understanding how ordinals 't into your explanation>of why in'nity plays no role in math. It seems to me that I am>classifying different types of in'nite numbers.That's odd; I wouldn't say you're dealing with numbers at all. You'vegot a collection of things which you've classi'ed in big book, andthe table of contents looks like this:* * * * * * ***... * * * * * ***... * * * *...[etc]. That is, you've got a pattern in mind which you can think of asan ordered set, or a kind of tree or something. But it's the manifoldsyou're studying, right? Or ordinals [order types] more generally.But not in'nity.I think in fact your example shows we're on the same wavelength here.You know the set of manifolds is not 'nite, so then you ask, what is it?The answer in'nity is correct as a synonym for not 'nite, butyou know that there are many non-'nite things, so you want morespeci'city. By contrast the naive questioner who wants to know whatin'nity is seems to think that it's a single, 'xed thing that hasa nature sui generis. There are 'nite sets and non-'nite sets; many species of well-ordered sets, only the most boring of which are 'nite;sets which can be parameterized by the counting numbers, or by thereal numbers, or by some other set. We know something when we knowthat these sets are not 'nite. That's one thing that they all _are not_;what is the thing that all of these _are_? What is this thing calledin'nity? I don't think that adds anything useful (besides sayingnot 'nite) and I think that you know that --- that's why you countthe items in your set differently instead of just saying we've reached in'nity.Maybe I mis-interpreted the OP's question. There are plenty ofnon-'nite things of interest in mathematics: ordinals and cardinals,sequences and series, non-terminating algorithms, etc., and ifthat's what the question was about I apologize. I thought the personwas talking like a movie producer. (Oh, those mathematicians are so different from us; they can study Oin'nity'!)dave === Subject: Re: The role of in'nity in What is the role of in'nity in math:>As Arturo mentioned, it depends on the branch of math your are in. But forexample, take the set of natural number, or sometimes refered to as thecounting numbers 1,2,3,... etc... How many of these are there? If you tellme some number M is the largest (hence there would be exactly M countingnumbers), I could always say but what about M+1?. So there are in fact anin'nite number of counting numbers since no Olargest' counting numberexists.> How is it de'ned?>Well there are actually levels of in'nity. The set of all counting numbersis what is called a Ocountably in'nite' set. The link below may be ofhelp:http://en.wikipedia.org/wiki/Countable> Why is it about the Osize' ofthe set of all counting numbers, we need to get into in'nities since no'nite set of counting numbers will ever be able to contain _all_ of thecounting numbers.As mentioned above, another application is found in computer science. Onecould talk about the set of all computer programs which is also Ocountably'in'nite set. This is because there is a 1 to 1 correspondence between thecounting numbers and each possible computer program.> At what precition does math work?>Despite what your encyclopedia or dictionary may tell you, math is not allabout making measurements. Take theoretical computer science. There aremany branches of math that do not even use numbers at all as theirfundamental objects!> These are questions which seem to have to accepted answer, what do you> think?>Each branch of math will have its own possibly unique Oaccepted answer'.l8r, Mike N. Christoff === Subject: Re: The role of in'nity in math Adjunct Assistant Professor at the University of Montana.>What is the role of in'nity in math:>>How is it de'ned?Depends on the 'eld of math you are working on. It means differentthings in calculus than it does in set theory, for example.>Why is it needed?Because it is useful. In Calculus, for example, it is a very usefulshorthand for a number of related phenomena, that are important andthat we want to be able to talk about. In set theory, it is thefoundation of a lot of interesting mathematics that could not be donewithout the concept of in'nity as it is used there. In other areas(such as functional analysis) it is downright essential.>At what precition does math work?That does not seem to be a very clear question. And it will againdepend on what you mean by precision, and what 'eld of math you aretalking about.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: The role of in'nity in math>What is the role of in'nity in math:>>How is it de'ned?> > Depends on the 'eld of math you are working on. It means different> things in calculus than it does in set theory, for example.> >Why is it needed?> > Because it is useful. In Calculus, for example, it is a very useful> shorthand for a number of related phenomena, that are important and> that we want to be able to talk about. In set theory, it is the> foundation of a lot of interesting mathematics that could not be done> without the concept of in'nity as it is used there. In other areas> (such as notion of in'nity really not be dispensed with inmathematics? If not, can it be restricted to a very small area ofstudy?In calculus (and functional analysis?), the formal de'nitions o¤imits, etc. do not make explicit use of the notion of in'nity. Asfar as I know, it is only used as a convenient shorthand notationthere. In number theory, we say only that each number has a suitablyde'ned successor. What does in'nity really get you outside of,say, the theory of cardinal numbers?DanVisit DC Proof Online at http://www.dcproof.com === Subject: Re: The role of in'nity in math Adjunct Assistant Professor at the University of Montana.>What is the role of in'nity in math:>>How is it de'ned?>> >> Depends on the 'eld of math you are working on. It means different>> things in calculus than it does in set theory, for example.>> >>Why is it needed?>> >> Because it is useful. In Calculus, for example, it is a very useful>> shorthand for a number of related phenomena, that are important and>> that we want to be able to talk about. In set theory, it is the>> foundation of a lot of interesting mathematics that could not be done>> without the concept of in'nity as it is used there. In other areas>> (such as the notion of in'nity really not be dispensed with in>mathematics? Depends on what you mean by notion of in'nity. You can certainlyexcise the word in'nity everywhere that it appears in mathematics;but in most places, you would still need the concept codi'ed by thatword.>In calculus (and functional analysis?), the formal de'nitions of>limits, etc. do not make explicit use of the notion of in'nity.What is the notion of in'nity? Functional analysis deals mainlywith in'nite dimensional vector spaces, so I do not see how you couldget away from it.>In number theory, we say only that each number has a suitably>de'ned successor. There is more to number theory than the natural numbers. There'sanalytic number theory, that requires complex analysis.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: The role of in'nity in mathArturo Magidin it de'ned?>> Depends on the 'eld of math you are working on. It means different> things in calculus than it does in set theory, for example.Ok, math related to measurement, calculus, algebra, number theory, etc.>Why is it needed?>> Because it is useful. In Calculus, for example, it is a very useful> shorthand for a number of related phenomena, that are important and> that we want to be able to talk about. In set theory, it is the> foundation of a lot of interesting mathematics that could not be done> without the concept of in'nity as it is used there. In other areas> (such as functional analysis) it is downright essential.It is practical, yes. But why use it instead of a high number, it would notproduce as nice results, but what is it about the concept of in'nity thatmakes it usefull? (When it comes to math dealing with measurement)>At what precition does math work?>> That does not seem to be a very clear question. And it will again> depend on what you mean by precision, and what 'eld of math you are> talking about.Again all measurement work with === presition, but what about math in itself?Subject: Re: The role wtwentyman@copper.net > Arturo Magidin role of in'nity in math:>>How is it de'ned?>>Depends on the 'eld of math you are working on. It means different>>things in calculus than it does in set theory, for example.> > > Ok, math related to measurement, calculus, algebra, number theory, etc.real analysis, combinatorics, statistics, differential equations, logic, ...> > >Why is it needed?>>Because it is useful. In Calculus, for example, it is a very useful>>shorthand for a number of related phenomena, that are important and>>that we want to be able to talk about. In set theory, it is the>>foundation of a lot of interesting mathematics that could not be done>>without the concept of in'nity as it is used there. In other areas>>(such as functional analysis) it is downright essential.> > > It is practical, yes. But why use it instead of a high number, it would not> produce as nice results, but what is it about the concept of in'nity that> makes it usefull? (When it comes to math dealing with measurement)Because no matter how large a number you pick, it is possible to create a function with a behavior as x -> oo that does not exhibit that behavior prior to your large number.Also, in'nity is the only useful way to start talking about how many integers there are.> > >At what precition does math work?>>That does not seem to be a very clear question. And it will again>>depend on what you mean by precision, and what 'eld of math you are>>talking about.> > Again all measurement work with presition, but what about math in itself?It depends on the branch and how much error is acceptable. For many branches, error=0. For others, precision doesn't make sense. What do you mean by precision in logic?Ob pet-peeve: please spell precision correctly. Arturo corrected you and you responded without the correction.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: Re: The role of in'nity in mathX-DMCA-Noti'cations: 23:02:29 +0100, Martin Johansen>>[...]>At what precition does math work?>> That does not seem to be a very clear question. And it will again>> depend on what you mean by precision, and what 'eld of math you are>> talking about.>>Again all measurement work with presition, but what about math in itself?Math in itself works to a prisishun of 3 signi'cant digits.(That doesn't make much sense, does it? No. Well, likeArturo's been saying, you need to clarify what you mean bythe question.)************************David C. Ullrich === Subject: Re: The role of in'nity in math Adjunct Assistant Professor at the University of Montana.>>Arturo the role of in'nity in math:>>How is it de'ned?>> Depends on the 'eld of math you are working on. It means different>> things in calculus than it does in set theory, for example.>>Ok, math related to measurement, calculus, algebra, number theory, etc.It still means different things; in algebra, the usual meaning will becloser to that of set theory: a set is in'nite if and only if itcan be put into one-to-one bijective correspondence with a properset of itself; and 'nite otherwise. An in'nite dimensional vectorspace is a vector space where any 'nite set is not a basis, or where(assuming the Axiom of Choice) every base forms an in'nite set. Innumber theory, it will depend: if you are doing analytic numbertheory, in'nity will probably be related to the notion of in'nityfrom complex analysis (the one point compacti'cation of the complexplane). In algebraic number theory, in'nity has to do with thearchimedean norms that can be put on a number 'eld (they are calledthe Oplaces at in'nity').In standard calculus, in'nity is used as shorthand for certainthings. Saying a function goes to in'nity as x approaches y meansthat for every value N, there exists a positive integer e such that if0<|x-y|0 there is an N>0 such that if x>N, then |f(x)-v|>Why is it needed?>> Because it is useful. In Calculus, for example, it is a very useful>> shorthand for a number of related phenomena, that are important and>> that we want to be able to talk about. In set theory, it is the>> foundation of a lot of interesting mathematics that could not be done>> without the concept of in'nity as it is used there. In other areas>> (such as functional analysis) it is downright essential.>>It is practical, yes. But why use it instead of a high number, it would not>produce as nice results, but what is it about the concept of in'nity that>makes it usefull? (When it comes to math dealing with measurement)When you are measuring things in the real world, you don't usein'nity. As for very high numbers, it is not the same thing to saythat something gets larger than anything we might possibly name (goesto in'nity) than to say that it gets larger than a speci'c largenumber we mention. They do not mean the same thing.>>At what precition does math work?>> That does not seem to be a very clear question. And it will again>> depend on what you mean by precision, and what 'eld of math you are>> talking about.>>Again all measurement work with presition, but what about math in itself?Again, your means something that only makes sense when you talkabout measurements. It is something related to the margin of errorsand so on. But what does it mean to ask whether math is precise?Axiomatic mathematics is precise in the sense that all non-primitivenotions must be explicitly and very clearly de'ned if they are to beused, but is not precise in any sense realted to margin of errors.Your question, as stated, does not make sense. You need to be veryclear about just what you mean by precision before it can beanswered, and you need to be clear about what kind of math are youtalking about.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: Aspiring mathematicians, send me your proofs!> Won't run on Mac, eh?> Not without Windows (95 or later, and IE4 or later). I understand thatsome Macs can actually run Windows.Dan === Subject: Any properties for such matrixHi here we de'ne M as a mxm complex matrixand M^n = Iis there any lot! === Subject: Re: block matrix, determinant zero> Please forgive my mistake.> Coming back to my original question,> how is it proven that det([A,B;C,D])= det(A)det(D) when B or D is zero matrix ?> I think that you've _still_ got this misstated. What's true is the following: Suppose the nxn matrix M is partitioned as M = [A,B;C,D} with A, D square submatrices, Then if either B or C is a zero matrix, det(M) = det(A) det(D). This is most easily proved using the _de'nition_ of the determinant: if M = [ m(i,j) ], then det(M) = Sum (-1)^sign(p) m(1,p(1))m(2,p(2))...m(n,p(n)) p in S_n where S_n is the group of permutations of {1, 2, ..., n}. The point here is this: Lemma. Let r_A = {1, 2, ..., k}, r_D = {k+1, k+2, ..., n} and let p be a permutation in S_n. Unless p(r_A) = r_A and p(r_D) = r_D, there are i in r_A, j in r_D such that p(i) in r_D, p(j) in r_A. To apply the lemma, let r_A be the indices of the rows in M occupied by A and r_D the corresponding set for D. Note that, if B (say) happens to be a zero matrix and p is any permutation that sends some i in r_A into r_D, m(i,p(i)) lies in B and so is 0. It follows that all of the terms in the sum de'ning det(M) vanish _except_ those that correspond to permutations p satisfying p(r_A) = r_A *and* p(r_D) = r_D. Any such permutation arises from a unique pair of permutations p_A in S_k, p_D in S_{n-k}, in such a way that sign(p) = sign(p_A)+sign(p_D). And, given such a pair of permutations, they determine a unique permutation in S_n. It's then immediate that (in this case) det(M) = det(A) det(D). The > Not so. The determinant of a triangular matrix is the product of the> diagonal entries, which can be directly seen from the cofactor formula. === Subject: [Help] Borel A_2, ... be events in probability space. De'ne X_n=A_1+A_2+...+A_nand s_n = E(X_n). Suppose s_n ->inf and ||X_n/s_n||_2 ->1.Show that {X_n=0} <= (k - X_n)(k+1 - X_n)/k(k+1) for each positive integer k.Your help is === Subject: Re: [Help] Borel CantelliX-DMCA-Noti'cations: showing next problem;>>Let A_1, A_2, ... be events in probability space. De'ne X_n=A_1+A_2+...+A_n>and s_n = E(X_n). Suppose s_n ->inf and ||X_n/s_n||_2 ->1.??? You say X_n is the sum of some _events_, ie _sets_, and thenX_n appears to be a random variable, ie a _function_. Thismakes no sense. Was X_n actually supposed to be the sum ofthe indicator functions of those sets?>Show that> {X_n=0} <= (k - X_n)(k+1 - X_n)/k(k+1) >for each positive integer k.Again, you ask us to show that a set is <= a function; Idon't know what this means (do you really want thatthe indicator function of that set is <= the right sideor what?)Also I wonder if the above is _really_ what you wantto prove. The reason I wonder is that it has nothingwhatever to do with the hypotheses s_n ->inf and ||X_n/s_n||_2 ->1...>Your help is Ullrich idea of showing next problem;>>Let A_1, A_2, ... be events in probability space. De'ne X_n=A_1+A_2+...+A_n>and s_n = E(X_n). Suppose s_n ->inf and ||X_n/s_n||_2 ->1.> > ??? You say X_n is the sum of some _events_, ie _sets_, and then> X_n appears to be a random variable, ie a _function_. This> makes no sense. Was X_n actually supposed to be the sum of> the indicator functions of those sets?Yes, X_n is sum of indicator function of those sets.> >Show that> {X_n=0} <= (k - X_n)(k+1 - X_n)/k(k+1) >for each positive integer k.> > Again, you ask us to show that a set is <= a function; I> don't know what this means (do you really want that> the indicator function of that set is <= the right side> or what?)This is the same, it is the indicator function. > > Also I wonder if the above is _really_ what you want> to prove. The reason I wonder is that it has nothing> whatever to do with the hypotheses s_n ->inf > and ||X_n/s_n||_2 ->1...> I would also like to show following;2)By appropriate choice of k, deduce that Sum_0^inf A_i >= 1 a.s.(again it is indicator function)3)Prove that Sum_m^inf A_i >=1 a.s. for 'xed m. (again indicatorfunction)4)Deduce that P{omega in A_i i.o.} = 1To summarize I used the same symbol for a set and its indicatorfunction. === Subject: Re: Fourier to help me with a question about> the spectum of a radarsignal.> A radar sends out a pulsed sinosoidial waveform.> The carrier cos(wt-kz) which is a cosine that propagates in the> z-direction> is modulated by a pulsetrain which is a typical square-wave wich is a> sum of cosine with its odd harmonics cos(wt)+(1/3)*cos(wt)+ ...> > If I do the amplitude modulation (multiplication) of the pulse train> and the carrier. I get a discrete spectrum...>> But somebody told me that the following rule exists:> A continous signal has a discrete spectrum and a discrete signal has a> continous spectrum.>> The radarsignal is a discrete signal. The emitter is on and off, on> and off,...> it should according to the rule of thumb have a continous spectrum.>> What have I missed Fourier transform swaps large-scale and small-scale structure.A periodic input produces a sampled output.A sampled input produces a periodic output.Both the input and output of an FFT are both sampled and periodic.A non-periodic input produces a continuous-spectrum output.A continuous input produces a non-periodic output.Multiplication becomes convolution and vice versa.On the small scale, the sharp edges of a stairstep signal transformto become, on the large scale, a spectrum that tends to decreaseaccording to the reciprocal of the frequency.The smallest scale feature of the radar signal is the carrier.It is multiplied by pulses. These transform to have a reciprocaldecay which is offset so that it appears surrounding the carrierfrequency instead of the origin. The large-scale periodic natureof the pulses means that the reciprocal decay will be chopped upinto 'ne lines representing the harmonics of the period. === Subject: Re: Fourier analysis of a radarsignalStephen M. Fortescue and small-scale structure.>> A periodic input produces a sampled output.> A sampled input produces a periodic === Subject: Re: Fourier analysis of a radarsignal> Stephen M. Fortescue Fourier transform swaps large-scale and small-scale structure.>>A periodic input produces a sampled output.>>A sampled input produces a periodic output.> > > Undiluted R.Tschaggelar - http://www.ibrtses.com& commercial newsgroups - http://www.talkto.net === Subject: Re: Fourier analysis of a rule of thumb saying that a continous signal shouldhave a discrete spectrum, and a discrete signal a continous spectrum?I found a review of radarsignals in a book: Airborn radar by Stimsonpage 206.There one can see that a single pulse of a carrier has a continousspectrum which looks exactly the same as a non-coherent pulsetrain ofthe carrier (random starting phase)A coherent(same starting phase) pulsetrain of in'nite lengthhas a perfect discrete spectrum, and as you guys pointed out to me, atrain of coherent pulses with limited length has a similar lookingdiscrete spectrum but with widened spectal peaks...Isn't there any way I can rede'ne the word discrete ? Could a coherent pulsetrain be said to be continous, or is a signaldiscrete as soon as there are longer periods in the signal whichhave zero amplitude ?Kindest requires an in'ntely> long signal. As soon as the signal is shorter, basically the> before in'nitely thin frequency lines widen.> A frequency can only be represented as accurate as many cycles> you have witnessed. A 1000 cycles of a sine make the linewidth> in the order of 1/1000 th or so.> > Rene === Subject: Re: existence of bounded linear functionalHow about this:Since the problem starts out by letting {a_k} be any sequence of scalars,declare a_1 = M ||x_1||.Then|f(x_1)| = |a_1| = M ||x_1||and we have to have equality in the norm: ||f||=M. Does that do it?J. Woodward you've got f(x)=b_1*a_1 + ... for x = b_1*x_1 + ... in the> subspace generated by all of the x_i.>> Then |f(x)| <= M ||x|| for all x in the subspace.>> Before we even bother extending this functional to the rest of X, how> do we know ||f|| = M? The OP must have misstated his problem... (maybe> he means ||x|| <= M continuous. I'm of'cially an idiot.> blah this norm stuff). Suppose> >> > | b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||> >> > for some M for each n and for scalars b_1,...,b_n. We want toconstruct a> > functional x' in X' such that x'(x_j)=a_j for ALL j = 1,2,....> >> > Consider the subspace spanned by ALL the x_k, and call it S. Then anyx in> S> > can be written> >> > x = b_1x_1 + b_2x_2 + ....> >> > De'ne f(x) = b_1a_1 + b_2a_2 + ....> >> > Then f(x_j)=a_j for all j = 1,2...> >> > But is f bounded? The given inequality only works for 'nite n, and> >> > |f(x)| = |b_1a_1 + ... | = |lim b_1a_1 + ... + b_na_n| = lim |b_1a_1+> ...> > + b_na_n| < = lim M ||b_1x_1 + ... b_nx_n||, but> >> > lim ||b_1x_1 + ... + b_nx_n|| = ||lim b_1x_1 + ... b_nx_n||, so howdo> you> > show f is bounded???> >> >> > Robert Israel still have one lingering question, though. Given{x_k}> a> sequence in X and {a_k} a sequence of scalars, de'ne a boundedlinear> > >functional on the subspace generated by x_1,...,x_n by f(b_1x_1 +... +> > >b_nx_n) = b_1f(x_1) + ... + b_nf(x_n) , f(x_k)=a_k.>> > >How does the requirement that>> | b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||> >guarantee that ||f|| = M? It's obvious that ||f|| <= M, but how doyou> > >assert equality?> >> > Oh. It doesn't, of course, and I don't. If x_1,...,x_n already> > span X, you're clearly out of luck, and your statement is false.> > Hahn-Banach gives you x' with ||x'|| <= M. If x_1,...,x_n don't> > span X (e.g. if X is in'nite-dimensional), it also gives you> > y' <> 0 such that y'(x_j) = 0 for j=1..n. For suitable t,> > x' + t y' will be what you want.> >> > Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel> > University of British Columbia> > Vancouver, BC, Canada V6T it's dual. a_1,...,a_n somescalars,> > >>x_1,...,x_n some elements of X. How do you show that>> > >>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n||>> > >>for some M for each n and for scalars b_1,...,b_n implies thatthere> exists> >x' in X' satisfying x'(x_k)=a_k and ||x'|| = M.>> > >>i don't know how to show this without explicitly constructingthe> bounded> > >>linear functional on X, and of course i'm stuck on constructingit> (hence> > >>the problem in the 'rst place)..>> > >>if X is 'nite dimensional, then each x in X can be written x =b_1> x_1 +> > >>... + b_n x_n on some basis {x_k}..then de'ne x'(x)=b_1 x'(x_1)+> ...> +> b_n> > >>x'(x_n) and de'ning x'(x_k)=a_k. This is bounded by the given> constraint.> > > > >> Careful: the x_n's were given, and might not be a basis.> > > > >>But what to do if X is in'nite-dimensional???> > > > >> Use the Hahn-Banach Theorem.> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> >> > === Subject: Limit Of # Of Coprime IntegersLet H(m;k,j) = the number of positive integers <= m which are coprimeto *both* k and j.(So, for instance, H(m;k,j) = H(m;kj,1).) (And H(kj;j,k) = phi(jk), the Euler phi function.) I believe thatlimit{m-> oo} m m --- --- 1 --- > > H(m;k,j)m^2 / / --- --- k=1 j=1 12 1= ---- - ------- , pi^2 zeta(3)which is, in linear-mode,limit{m-> oo}(1/m^2) *sum{k=1 to m} sum{j=1 to m} H(m;k,j)= 12/pi^2 - 1/zeta(3).(zeta(r) is, generally, sum{j=1 to oo} 1/k^r.)And, more generally,limit{m-> oo} m m m --- --- --- 1 ------- > > ... > H(m;k_1,k_2,..,k_n)m^(n+1) / / / --- --- --- k_1=1 k_2=1 k_n=1 n+1 --- 6*n 1= ---- - / ------- , pi^2 --- zeta(j) j=3which is, in linear-mode:limit{m-> oo} (1/m^(1+n))* sum{k_1=1 to m} sum{k_2=1 to m}...sum{k_n=1 to m} H(m;k_1,k_2,..,k_n)= 6*n/pi^2 - sum{j=3 to n+1} 1/zeta(j) .And H(m;k_1,k_2,..,k_n) =the number of integers, j, coprime with (k_1)*(k_2)*(k_3)*...*(k_n)and such that 1 <= j <= m.Am I Quet === Subject: Re: Limit Of # Of Coprime IntegersLeroy Quet number of positive integers <= m which are coprime> to *both* k and j.>> (So, for instance, H(m;k,j) = H(m;kj,1).)>> (And H(kj;j,k) = phi(jk), the Euler phi function.)> I believe that>> limit{m-> oo}>> m m> --- ---> 1 > --- > > H(m;k,j)> m^2 / /> --- ---> k=1 j=1> 12 1> = ---- - ------- ,> pi^2 zeta(3)> which is, in linear-mode,> limit{m-> oo}>> (1/m^2) *sum{k=1 to m} sum{j=1 to m} H(m;k,j)> = 12/pi^2 - 1/zeta(3).>> (zeta(r) is, generally, sum{j=1 to oo} 1/k^r.)>> And, more generally,> limit{m-> oo}>> m m m> --- --- ---> 1 > ------- > > ... > H(m;k_1,k_2,..,k_n)> m^(n+1) / / /> --- --- ---> k_1=1 k_2=1 k_n=1>> n+1> ---> 6*n 1> = ---- - / ------- ,> pi^2 --- zeta(j)> j=3> which is, in linear-mode:>> limit{m-> oo}>> (1/m^(1+n))*> sum{k_1=1 to m} sum{k_2=1 to m}...sum{k_n=1 to m} H(m;k_1,k_2,..,k_n)> = 6*n/pi^2 - sum{j=3 to n+1} 1/zeta(j) .> And H(m;k_1,k_2,..,k_n) =>> the number of integers, j, coprime with (k_1)*(k_2)*(k_3)*...*(k_n)> and such that 1 <= j <= m.> Am I right? At least, is my n=2 case right?> book/compendium of some sort containing all ofyour various games, proofs and conjectures? There are a lot of on-demandpublishers out there and you could even get it listed at amazon.com.l8r, Mike N. Christoff === Subject: Re: Request for F1vU=Jm9LchOj|&)y/y'n33$K{,'|+j:PBvlXn]+u&LG@{ zY!w}Yp)i57ga,S] B3`Xj1LZGa)d5?ma~E_^>yu_xS(E:wqD.(nw&)SmG)LQK:H;^S;&*0!.uuW+ 59UwyslR=xGd RtodT9/H-B.I_>&f9CXI;dty3> Chan-Ho Suh book expects far far more knowledge of general topology than Lefschetz.> > For Hatcher, a typical quarter or semester course in general topology> will suf'ce. I didn't realize you were lacking in this since usually> one doesn't study algebraic topology until one has studied some basic> topology -- the basics are also covered in undergrad analysis courses> that cover some generalities like metric spaces.>> I can believe Lefschetz requires less topology background; the point of> all that star/link business is to avoid some topological messiness. On> the other hand, I have to question how much you will get out of it if> you haven't studied the basic general topology. I mean, can you> appreciate the idea of constructing families of groups as topological> invariants of certain topological spaces when you haven't learned about> simple invariants of topological spaces? I suppose it's possible, but> there's a good reason people teach the basic stuff 'rst.>> > I have a basic understanding of general topology. Things that I don't know> include quotient spaces, notation like Of | A' with regard to functions,> e-neighbourhood (is this a neighbourhood about a subset e?), and> orientability. Is this all? Then it seems like you know enough to look at a book likeHatcher and bene't quite a bit. The things you listed are reallyquite easy to pick up. Quotient spaces may seem kind of bizarre at'rst, but if you pick up a book like say, Munkres undergrad topologybook and read the section on it, should only take you a day or two toget the idea; of course your understanding will deepen in time, but mypoint is that it's not gonna be this marathon effort to learn it. Speaking of quotient spaces, that's one reason a book like Lefschetzcan be kind of deceptive. It may appear more friendly since heexplains quotient spaces in the 'rst chapter (on general topology),but he only takes half a page to explain it! So you may retort, seehe requires less background, Hatcher doesn't even explain what aquotient space is! On the other hand, hopefully you (will) realizethat if you don't know about quotient spaces, having a half a page ofexplanation is only minimally helpful. You see, back in Lefschetz'sday, books were often written to be self-contained. That means thata very industrious and clever reader would not have to read some otherbook to understand something in the 'rst book. But of course, havinga preliminary chapter where you shove everything into a paragraph orless, makes a book self-contained, but there are real disadvantageswith that approach.Of course, if you like how Lefschetz has those introductory chapters ongeneral topology and group theory, then by all means use them. There'sno rule saying you can only read one book at a time. But here I wouldalso caution you. As another poster has remarked, there are bettertreatments of this material elsewhere.As for orientability, part of understanding that will come from readingHatcher. Orientability is sometimes de'ned in terms of (co)homologyand you'll learn about that. Other things meant by orientability couldbe referring to orientable surfaces versus nonorientable surfaces. Forthings like that, there's a very concrete way to view orientability. You can regard a surface as orientable if it contains no Moebius band. But again, that's not hard to pick up (probably learning about surfaceswill be more effort than learning about quotient spaces). In any case,you ought to know about surfaces before you tackle something likeLefschetz's book. They are the basic examples of combinatorialmanifolds, i.e. the simplicial theory of manifolds.What I'm trying to get at with this post and the previous, is that youwill bene't more if you have the background. What's puzzling,however, is that from what you write, it seems like you have more thanenough background, but you seem to be panicking, overwhelmed for somereason. It's normal to be missing a few things that are prerequisites,like not knowing about quotient spaces, for example, but that'sactually really good, since most people never come close to satisfyingall the prerequisites. The main problem I see with Lefschetz's book is that he develops a verygeneral theory 'rst, before specializing to simplicial complexes. Ithink this would be very confusing for people not even familiar withsimplicial complexes. Pedagogically, there are several books that onlyteach simplical homology, or at least *start* with that, beforegeneralizing. I would recommend you start with one of those if youinsist on taking that route.Of course, the funny thing is that Hatcher teaches simplicial homology,or at least a variant of it. [Technical note: what he actually de'nesare a kind of pseudo-simplicial complex, what he calls delta-complexes;this gives him more ¤exibility than with simplicial complexes whilebasically having a very similar developement] If you just start withChapter 2, you will be basically starting at the same point asLefschetz does. The nice thing about Hatcher's book, though, is thathe has a Chapter 0, that kind of sets the stage for the rest of thebook; it's a chapter that's not at the same level of rigor as the restof the book and is intended to motivate.> What I do know includes things like the euclidean topology,> metric spaces, continuous mappings, homeomorphisms, non-homeomorphic spaces,> subspaces, connectedness, compactness, completeness, contraction mappings,> Baire spaces, Heine-Borel spaces, product spaces, Tychonoff's theorem, the> fundamental theorem of algebra, etc... but doesn't cover seemingly basic> things like the genus of a space. I suppose a half-semester course would> cover material beyond these things.I'm not sure why you put the fundamental theorem of algebra in there,unless you are referring to a topological proof of it that you saw inclass.By genus of a space are you referring to a surface's genus?As I said, I 'nd your trepidation quite puzzling. I would think youknow enough to start studying a book like Hatcher. In the end, you'reeither going to have to believe the advice you get and accept thatthere is something wrong with your mindset, or do what you would havedone anyway. But if you do the latter and regret it, don't blame me! === Subject: Re: Request for comments on antiquated algebraic topology online-bookChan-Ho Hatcher's book expects far far more knowledge of general topologythan> > Lefschetz.> >> >> > For Hatcher, a typical quarter or semester course in general topology> > will suf'ce. I didn't realize you were lacking in this since usually> > one doesn't study algebraic topology until one has studied some basic> > topology -- the basics are also covered in undergrad analysis courses> > that cover some generalities like metric spaces.> >> > I can believe Lefschetz requires less topology background; the pointof> > all that star/link business is to avoid some topological messiness.On> > the other hand, I have to question how much you will get out of it if> > you haven't studied the basic general topology. I mean, can you> > appreciate the idea of constructing families of groups as topological> > invariants of certain topological spaces when you haven't learnedabout> > simple invariants of topological spaces? I suppose it's possible,but> > there's a good reason people teach the basic stuff 'rst.> > I have a basic understanding of general topology. Things that I don'tknow> include quotient spaces, notation like Of | A' with regard to functions,> e-neighbourhood (is this a neighbourhood about a subset e?), and> orientability.>> Is this all? Then it seems like you know enough to look at a book like> Hatcher and bene't quite a bit. The things you listed are really> quite easy to pick up. Quotient spaces may seem kind of bizarre at> 'rst, but if you pick up a book like say, Munkres undergrad topology> book and read the section on it, should only take you a day or two to> get the idea; of course your understanding will deepen in time, but my> point is that it's not gonna be this marathon effort to learn it.>> Speaking of quotient spaces, that's one reason a book like Lefschetz> can be kind of deceptive. It may appear more friendly since he> explains quotient spaces in the 'rst chapter (on general topology),> but he only takes half a page to explain it! So you may retort, see> he requires less background, Hatcher doesn't even explain what a> quotient space is! On the other hand, hopefully you (will) realize> that if you don't know about quotient spaces, having a half a page of> explanation is only minimally helpful. You see, back in Lefschetz's> day, books were often written to be self-contained. That means that> a very industrious and clever reader would not have to read some other> book to understand something in the 'rst book. But of course, having> a preliminary chapter where you shove everything into a paragraph or> less, makes a book self-contained, but there are real disadvantages> with that approach.>> Of course, if you like how Lefschetz has those introductory chapters on> general topology and group theory, then by all means use them. There's> no rule saying you can only read one book at a time. But here I would> also caution you. As another poster has remarked, there are better> treatments of this material elsewhere.>> As for orientability, part of understanding that will come from reading> Hatcher. Orientability is sometimes de'ned in terms of (co)homology> and you'll learn about that. Other things meant by orientability could> be referring to orientable surfaces versus nonorientable surfaces. For> things like that, there's a very concrete way to view orientability.> You can regard a surface as orientable if it contains no Moebius band.> But again, that's not hard to pick up (probably learning about surfaces> will be more effort than learning about quotient spaces). In any case,> you ought to know about surfaces before you tackle something like> Lefschetz's book. They are the basic examples of combinatorial> manifolds, i.e. the simplicial theory of manifolds.>> What I'm trying to get at with this post and the previous, is that you> will bene't more if you have the background. What's puzzling,> however, is that from what you write, it seems like you have more than> enough background, but you seem to be panicking, overwhelmed for some> reason. It's normal to be missing a few things that are prerequisites,> like not knowing about quotient spaces, for example, but that's> actually really good, since most people never come close to satisfying> all the prerequisites.>> The main problem I see with Lefschetz's book is that he develops a very> general theory 'rst, before specializing to simplicial complexes. I> think this would be very confusing for people not even familiar with> simplicial complexes. Pedagogically, there are several books that only> teach simplical homology, or at least *start* with that, before> generalizing. I would recommend you start with one of those if you> insist on taking that route.>> Of course, the funny thing is that Hatcher teaches simplicial homology,> or at least a variant of it. [Technical note: what he actually de'nes> are a kind of pseudo-simplicial complex, what he calls delta-complexes;> this gives him more ¤exibility than with simplicial complexes while> basically having a very similar developement] If you just start with> Chapter 2, you will be basically starting at the same point as> Lefschetz does. The nice thing about Hatcher's book, though, is that> he has a Chapter 0, that kind of sets the stage for the rest of the> book; it's a chapter that's not at the same level of rigor as the rest> of the book and is intended to motivate.>> What I do know includes things like the euclidean topology,> metric spaces, continuous mappings, homeomorphisms, non-homeomorphicspaces,> subspaces, connectedness, compactness, completeness, contractionmappings,> Baire spaces, Heine-Borel spaces, product spaces, Tychonoff's theorem,the> fundamental theorem of algebra, etc... but doesn't cover seemingly basic> things like the genus of a space. I suppose a half-semester coursewould> cover material beyond these things.>> I'm not sure why you put the fundamental theorem of algebra in there,> unless you are referring to a topological proof of it that you saw in> class.>Yes.>> By genus of a space are you referring to a surface's genus?>Yes again.> As I said, I 'nd your trepidation quite puzzling. I would think you> know enough to start studying a book like Hatcher. In the end, you're> either going to have to believe the advice you get and accept that> there is something wrong with your mindset, or do what you would have> done anyway. But if you do the latter and regret it, don't blame me!Well I've pretty much decided not to invest too much time with Lefschetz.However, initially, I only need to learn a small subset of algebraictopology (AT). Speci'cally, the same topics as found in the 'rst chapterof Munkres' OElements of AT'. The application is to distributed computingwhere the evolution of 'nite asynchronous distributed protocols are modeledas high dimensional simplicial complexes (if thats the correct terminology).combinatorial. The seminal explanatory paper on the approach can be foundat:http://www.cs.brown.edu/people/mph/HerlihyR96/sv.pdfI have taken another look at Hatcher, and you're right - it doesn't seemnearly as intimidating as it did before I bulked up on general topology abit. Hopefully I won't need to shell comments.l8r, Mike N. Christoff === Subject: Re: Request for comments on antiquated algebraic topology online-bookOriginator: israel@math.ubc.ca (Robert Israel)> example, if you look at the de'nition of complex at the beginning> of Chapter 3, it is (almost) equivalent to the modern de'nition of a> chain complex (over the integers), but looks completely different. > > You've completely misinterpreted what is going on here. It's not> supposed to be the de'nition of a chain complex or a precursor of that> concept. It's supposed to be (and is) a de'nition that generalizes> the notion of a simplicial complex. It is general enough to encompass> singular homology, not only simplicial homology.> > I> didn't see singular homology theory in there (but I didn't have time> to download and scan the whole book) while this is an essential part> of most modern treatments. Cf my comment above, singular homology is indeed in there.Sorry, I guess I should have looked at it more carefully. Anyway mypoint is that even where it is equivalent to modern material, thelanguage and notation are (at least based on a super'cial glance)somewhat different.> I don't agree. For someone who is a rank beginner, learning simplicial> homology the old-fashioned way can be quite instructive. Seifert and> Threlfall is extremely well-written (I can only speak for the English> translation, but the German is just as well-written I hear).I'll have to check that book of Quantum Mechanicsis one of my all-time favorites.> A lot of the basics can be learned without exact sequences.Sure, but since they clarify and simplify so many things, why wouldone want to?> lot depends on what the OP is aiming for. It's possible to get into a> lot of the abstract theory with little grounding in the old-fashioned> stuff, but if the OP wants to work in low-dimensional topology, s/he> will need to have a 'rm understanding of things like simplicial> complex and concrete interpretations of duality.I completely that one should know something about simplicialcomplexes, simplicial homology, and Poincare's original intuition forduality in terms of dual complexes. But many modern books, forexample Hatcher, still cover these notions, perhaps not in as muchdepth as older books, but enough to give you the basic idea.My only complaint about Hatcher's book in this regard is that it doesnot (AFAIK) explain (aside from a brief hint in the introduction tothe chapter on cohomology) the duality between cup product andintersection of submanifolds. I guess I like to emphasize smoothmanifolds more than many algebraic topology books do, which is onereason why I like Bredon's book so much. > Hatcher's book is not only great, it's free, available from his webpage.> > As you say, it's sophisticated, but that doesn't mean it's not> accessible.Sure. I guess my point here is that Hatcher's book has some rathersophisticated digressions into topics that are not (AFAIK) ordinarilycovered in an introductory course, so the guidance of an instructormight be needed to select out the core material. === Subject: Re: Average geometric value of a real function> > The formula for the average arithmetic value of a function is well known.> What about the average geometric value of a function over an interval.> Consider the followin de'nition:> Given a function f over an interval [a,b] such that f(x)>0 for every x in> this > interval.One can de'ne the average geometric value of f over [a,b] as> > e^(1/(b-a)*(Integral of log(f) between the limits a to b)),> provided the integral> exists which is always the case if f is continuous. > Is this formula known in the mathematical circles?> > Yes. I would call it a de'nition for the geometric mean of f> (with respect to Lebesgue measure on [a,b]).And I bet that the occasional Putnam problem has used the above toshow, after distilling things down, that:For b > a >= 0, f(x) >= 0 for all real x between a and b,1/(b-a) integral{a to b} f(x) dx is always >=exp(1/(b-a) integral{a to b} ln(f(x)) === Subject: Re: :: towards a constructive education :: (news server friendly): In your story, Heyting algebras are just propositional,: right? No quanti'ers, at least not necessarily?Well, each 'eld has its own level of sophistication of what types ofsentences can be written in their models. Quite a number of them do haveformulations with quanitifers, but I am looking mostly at the commonpropositional algebra as the universal.-- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea: prankster, fablist, magician, liar === Subject: Linear System across software that solves systems of the form [A](x)= (b), where the elements of the [A] Matrix, (x) vector, and (b)vector are complex numbers.I am familiar with such systems for REAL elements, but I have nopersonal experience with a system comprised of COMPLEX elements.I am curious how such a system would arise. Could somebody in thisgroup please point out a few situations in which such a system wouldbe created (or please also recommend documents, web sites, etc. whichmight present some examples).When such a system does arise, does it have typical sizes (forexample, would such a system be comprised of, say, a 3 x 3 [A] Matrix80% of the time Hill === Subject: Re: Linear System [A](x) = (b) with COMPLEX === Elements>Subject: Linear System [A](x) = (b) with COMPLEX Elements>Message-id: all.>>I recently came across software that solves systems of the form [A](x)>= (b), where the elements of the [A] Matrix, (x) vector, and (b)>vector are complex numbers.>>I am familiar with such systems for REAL elements, but I have no>personal experience with a system comprised of COMPLEX elements.>>I am curious how such a system would arise. Could somebody in this>group please point out a few situations in which such a system would>be created (or please also recommend documents, web sites, etc. which>might present some examples).>When such a system does arise, does it have typical sizes (for>example, would such a system be comprised of, say, a 3 x 3 [A] your help.>Michael Hill>>Michael,This type of problem would be found in random signal process where A is theinverse of correlation matrix of noise plus interference, x is the signal o'nterest (the signal, and b is weight vector that maximizes the signal to noiseplus interference ratio. The value would be complex in radar signal processing problems where the received signal is compared against a known signal (i.e.quadature downconversion)Scott === Subject: Mathematics, in general, equals a MazeIn selm=78kbmti88zn8%40forum.mathforum.com>I know mathematics is a maze.>But it is a maze without end.>It is a maze where goals shift>And directions change.>It is almost impossible,>If it is not impossible,>To climb upon its walls>And view everything whole.I would add that mathematical proof isA maze of in'nite complexity,of in'nite exits and entrances(each valid),and of in'nite dead-ends(each invalid),a maze of opaqueness and transparency,of obviousness and mystery,of unappreciated beautyonly seen as it 'nally becomes knownto esoteric eyes...But seriously,...When I sometimes try to prove a math result, I will often getsidetracked, following other paths, paths not originally intended, tosome other related results instead. (I might as well prove*something*!...)Does anyone else have an excellent example of this?(I know the maze-metaphor applies to Wiles' {et al} proof Fermats Lasttheorem, since it turned out that a once-abandoned path was actuallythe way to Wiles' {et al} ultimate goal.)And of course, in mathematics, if mathematicians fail to solvesomething as originally planned, still their work need not becompletely for nothing, since other related results can still comefrom their partial successes.And more ambitiously, as far as the maze metaphor is concerned, canviewing mathematics and proofs as a maze or as a graph (one resultleads to another leads to another), lead to any new ways to solvemath-problems in general?(I bet the Oright-hand-rule' algorithm for solving mazes has noobvious analog that can be applied to proving math results, as anexample, but you never abuse@meganetnews.com === Subject: (Reference) Book for learning interest in learningmathematics, but since I have completed high-school andwon't be going into college for another few years, I thinkthe best way to learn would be to do it by myself. I don'tlike pressure either, and I'm not hoping to major inmathematics in the future. I started thinking about gettingmyself a book, but since I've completed the most basichigh-school maths, it would be unnecessary to buy a wholeseries of books starting with explaining additioncompletely. A reference book or some sort of dictionarywith examples and complete de'nitions would be the ideal.Is mathematics a far too great 'eld to summarize into onereference book? It would be great to have a book that I canlook things up in when I get to an advanced level. Ofcourse, the best way to learn math is to solve problems andexercises, but that might be out of scope for a referencebook. I might add that I wish to learn mathematics up to,let's say, a university level.So my question is: given my need, could any of you pleaserecommend some good reading for me? My wallet-size isn'tin'nite, so one or two major works is === Subject: Re: (Reference) Book for learning mathematics from the ground recently gained an interest in learning> mathematics, but since I have completed high-school and> won't be going into college for another few years, I think> the best way to learn would be to do it by myself. I don't> like pressure either, and I'm not hoping to major in> mathematics in the future. I started thinking about getting> myself a book, but since I've completed the most basic> high-school maths, it would be unnecessary to buy a whole> series of books starting with explaining addition> completely. A reference book or some sort of dictionary> with examples and complete de'nitions would be the ideal.> Is mathematics a far too great 'eld to summarize into one> reference book? It would be great to have a book that I can> look things up in when I get to an advanced level. Of> course, the best way to learn math is to solve problems and> exercises, but that might be out of scope for a reference> book. I might add that I wish to learn mathematics up to,> let's say, a university level.> > So my question is: given my need, could any of you please> recommend some good reading for me? My wallet-size isn't> in'nite, so one or two advance.Proof, Logic and Conjecture The Mathamatician's ToolboxRobert S WolfIsbn 0-7167-3-5--2Carl === Subject: Re: (Reference) Book for learning mathematics from the ground upI second Ken's suggestion of Courant and Robbin's (and Stewart's) What IsMathematics? Ian Stewart also has a Dover book, Concepts of ModernMathematics. And Stan Ulam and somebody else have a good survey, again putout by Dover, which I can't lay my hands on immediately.John-- John T Lowry, PhDFlight Physics5217 Old Spicewood Springs Rd, #312Austin, Texas 78731(512) 231-9391jlowry100@earthlink.netTim Cambrant gained an interest in learning> mathematics, but since I have completed high-school and> won't be going into college for another few years, I think> the best way to learn would be to do it by myself. I don't> like pressure either, and I'm not hoping to major in> mathematics in the future. I started thinking about getting> myself a book, but since I've completed the most basic> === Subject: Re: (Reference) Book for yes mathematics, but since I have completed high-school and> won't be going into college for another few years, I think> the best way to learn would be to do it by myself. I don't> like pressure either, and I'm not hoping to major in> mathematics in the future. I started thinking about getting> myself a book, but since I've completed the most basic> high-school maths, it would be unnecessary to buy a whole> series of books starting with explaining addition> completely. A reference book or some sort of dictionary> with examples and complete de'nitions would be the ideal.> Is mathematics a far too great 'eld to summarize into one> reference book? It would be great to have a book that I can> look things up in when I get to an advanced level. Of> course, the best way to learn math is to solve problems and> exercises, but that might be out of scope for a reference> book. I might add that I wish to learn mathematics up to,> let's say, a university level.> > So my question is: given my need, could any of you please> recommend some good reading for me? My wallet-size isn't> in'nite, so one advance. Richard Courant and Herbert Robbins, What is Mathematics? A classic, brilliantly written to open your eyes to wide mathematical vistas, yet covering a range of signi'cant basic ideas, with exercises too. Ken Pledger. === Subject: Re: (Reference) Book for learning recently gained an interest in learning> mathematics, but since I have completed high-school and> won't be going into college for another few years, I think> the best way to learn would be to do it by myself.I know it's doable, because I've done it, through undergraduate level and alittle beyond. But using reference books for the purpose is tough. And suchbooks aren't cheap to produce, and they have a rather small market. A lotcan be said for used textbooks, and paperback reissues of good older books.Dover is helpful for the latter.http://store.doverpublications.com/> It would be great to have a book that I can> look things up in when I get to an advanced level.Dieudonn.8e's 9-volume _Treatise on Analysis_ is expensive but very good (foranalysis). Lang's _Algebra_ (now published by Springer) is prettycomprehensive and readable. Such books assume that the reader is accustomedto the canon of de'nitions-theorems-proofs. And of course no two books tellthe whole story.Applied mathematical writings (on phsyics, engineering, andprobability/statistics) have been helpful to me many times.LH alt.sci.physicsX-NewsOnePostHost: bridge designer, I found that the majority of those whoapply the principles of physics to engineering, don't really understand allof it, but have a pretty good working knowledge of basic mechanics and thesimple algebraic equations which apply to it.Like most scientists they were in awe of physics and physicists, andsubscribed to the belief that physics was abstractly complicated beyond theken of ordinary people like themselves: Believing that only geniuses couldcomprehend it all.Nor did many of them think very much of my propounding of (anonymous) Usenet News via the Web ----- http://newsone.net/ -- Free reading and anonymous posting to 60,000+ groups other postsmade through NewsOne.Net violate posting guidelines, email abuse@newsone.net === Subject: Re: Applied physics> > While working as a bridge designer,Bridge *ornamentation* designer, Dumb Donny Head. Your entirecareer was drawing frou-frou.> I found that the majority of those who> apply the principles of physics to engineering, don't really understand all> of it, but have a pretty good working knowledge of basic mechanics and the> simple algebraic equations which apply to it.You are looking in a mirror, Dumb them think very much of my propounding of ideas about> physics(;^~(You do not have ideas, Dumb Donny Head. You sufferhallucinations. If empirical reality says you are an ass, Dumb DonnyHead, then you are an emprical way,http://www.mazepath.com/uncleal/effete6.jpg The entire remainder of the planet sees you this way,http://www.mazepath.com/uncleal/effete3.png--Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: Applied physics> While working as a bridge designer, I found that the majority of those who> apply the principles of physics to engineering, don't really understand all> of it, but have a pretty good working knowledge of basic mechanics and the> simple algebraic equations which apply to it.> > Like most scientists they were in awe of physics and physicists, and> subscribed to the belief that physics was abstractly complicated beyond the> ken of ordinary people like themselves: Believing that only geniuses could> comprehend it all.> > Nor did many of them think very much of my propounding of ideas about> physics(;^~(Sort of like around here, huh?Double-A === Subject: Re: Applied physics Nor did many of them think very much of my propounding of ideas about> physics(;^~(Some of us don't understand why you are propounding them in sci.math.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Applied physicsRobin Chapman>> Nor did many of them think very much of my propounding of ideas about> physics(;^~(>> Some of us don't understand why you are propounding them in sci.math.Because of sci.math's growing reputation, no doubt :) === Subject: Three Body Problem IdeasIn the Three Body Problem enigma, it is desired tocalculate the trajectories of three masses as theyattract each other through time. They can havedifferent initial velocities and different masses. Theonly changes are the coordinates of the masses asthey converge to zero potential, and time.The region around and between the masses could be divided into a nearlyin'nite number of cubes. Thereare enough of them so that only one trajectory intersectsonly one cube at a time. It may be due to the motion ofthe 'rst mass, maybe the second or third. The state ofthe system is adjusted due to the resulting anomaly, andthe masses are set free until one of them intersectsanother cube, and the process is repeated. Error compounds exponentially,though.Suppose each cube consists of a space and timelasting 1 msec of 1 x 1 x 1 mm. When the trajectoryintersects the cube, the cube gives the trajectory anangle representing the space-time nature of thetrajectory. All the other cubes are updated with thisangle and the new 'eld is mapped that governs the newmotions of the masses. As they move, one more cubeis intersected and the process repeated.Suppose the trajectory passes through a cube in2 msec over the span of (1,1/2,0) mm within thecube. Its velocity is (1/2,1/4,0) m/sec. The cube'sconstant velocity is (1,1,1) m/sec. The angle couldbe the angle between these two velocities.However, there are errors in this. The trajectory onlygoes part way into each cube, and the time needed totraverse the full cube is lost. The angle needsyet another angle added or it only represents a coneof velocities to the other cubes. Velocity tells whereone mass is headed to the other cubes, but all that isneeded for the resultant force of two masses on thethird mass, are their coordinates, not their velocities.Error is exponentially compounded again, and thismethod doesn't seem much better than the 'rst.Jon Giffen === Subject: Re: branch of log z> What exactly is to choose a branch of log z for complex z? I'd> appreciate if someone help me picture this.If you're familiar with the complex exponential function, just notethat it is not one-to-one, it is periodic, in fact. So if you want totake the inverse of the exponential function, you really can't. Youcan take a local inverse (my language is loose, but hopefullyintuitive) which works for strips of the complex plane with width 2pi. So choosing a branch point corresponds to which strip of width 2piyou want the Log function to output to. (You're basically restrictingthe range of the Logarithm relation so that it becomes a bijectivefunction on that range)This stuff can get messy later. Ocid Oooh === Subject: Silly question on limits, tensor productsIn a nutshell, I see this line in my text : Let u(t) be a smooth curve inthe vector space U and v(t) be a smooth curve in the vector space V.lim [ u(t+h) tensor [v(t+h) - v(t)]/h ]h--> 0=u(t) tensor dv/dtFine and dandy. But I am stepping back a bit and asking myself thefollowing: There was one step that was skipped. That is :lim [ u(t+h) tensor [v(t+h) - v(t)]/h ] =h--> 0lim u(t+h) tensor lim ( [v(t+h) - v(t)]/h )h ----> 0 h --> 0Why is that true? (I know, it is a stupid stupid question) The way Ijustify this is by basically hand-waving and saying it makes sense, but Iam not convinced that you can do this type of thing for every occasion inwhich a limit breaks up into two limits. When can you do this sort ofthing with limits where one breaks up into two and when can't you? I knowI'm being a bit vague, but if someone has a quick counterexample or canunderstand what I'm trying to say then I'd really appreciate the insight.THanks,Moshe === Subject: Re: The Clearest by support1.mathforum.org (8.11.6/8.11.6/The Math http://www.geocities.com/erdosfan/solution.html, the new Letme ask the questions here. >I had some dif'culty understanding this paper.>In places that I found unclear, other readers might also have >dif'culty, so it might be worth your time to try to clear them up.>>De'nition 2.4, is the union of the J' supposed to equal the>interval N_a^b_{2k=1}?No. N_a^b_{odd} was introduced so that J' is de'ned precisely.In a set-theoretic explanation, J' is a subset of N_a^b{odd}.This notion of subsets was newly de'ned in the new version.>>Page 3 line 4, given any nth point of A ... there exists thecorresponding nth point of B ... Is this an equivalence of sets?>(That is, >union J'_{r_i,p_i}^{a,b} = union J'_{r'_i,p_i}^{c,d} - (c-a)>where (c-a) is subtracted from each element of the second set?)>Or is the individual covering J'_{r_i,p_i}^{a,b} on the left >supposed to be related somehow to the corresponding covering>J'_{r'_i,p_i}^{c,d} on the right?>>Same place, given any nth point of A ... counted from the extremeleft... nth point of B ... counted from the extreme left.>Are they both to be counted from the left simultaneously, or is oneto be counted from the left and the other from the right?The answer to the question ... counted from the left simultaneouslywould be enough. Yes. The answer is yes; they are both countedsimultaneously from the same direction only.>Lemma 2.6. Here's where I really get lost.>F = union_{i=1}^x J'_{r_i,p_i}^{3,q}.>So in particular F contains J'_{r_1,3}^{3,q} for some value r_1,>so it contains all the odd numbers between 3 and q congruent to>r_1 modulo 3. Then there is no way that equation (1) could besatis'ed, since P'^{3,q} does not contain such a large arithmeticprogression. I must be missing something.>I see P' as a set of primes, and F as a union of arithmeticprogressions, and P' cannot contain an arithmetic progression withpitch 2*3=6 because it is too sparse (as soon as q is large enough).This is totally my careless mistake. Changing notation fromsubset to [equivalent] seemed quite harmless. But not at all,actually. Please see the new version. >Other suggestions on notation:>>Use N_{odd} instead of N_{2k+1}.>Reason: other quantities like J_{k,p} that have k in their notation>actually depend on the value of some integer k, but this one does not.>>In the de'nition of N_{2k+1} (with presubscript a and presuperscript b), include x odd in the formula:>{ x : a leq x leq b , x odd }>>Similarly in the de'nition of P (presubscript 3, presuperscript p_l)>say { p : 3 leq p leq p_l , p prime }.>>In both cases it has been stated in words beforehand, but it's handyto be part of the formula in case the reader checks back for the>de'nitionThese were also taken care in the new version.>The de'nition of a covering, should the restriction two or more points be part of the de'nition?Yes. Rewriting the paper again, I de'nitely was convinced thatthe restriction is very important; without it, all logic in the paperseem to collaspe.>Remark 2.2, what is a linear graph?I just felt that this was unnecessary. The new version reallyis really good in explaining what I got.I really of whether the paper is accepted by the Annals or not,your advices really helped me understand the idea by myself.This is what's called Way to go, right?Hisanobu Shinya === Subject: Please help prove! ~~~~~~~~~~>_<~~~~~~~~~ by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 22:45:03 -0500I am learning algebra by myself.I cannot 'gure out some proofs.Any one could help??Proof1. Q, the abelian group of rational numbers, is an injective Z-module.2. A system of linear equations over a 'eld F has 5 solutions => F === Subject: Re: Please help prove! ~~~~~~~~~~>_<~~~~~~~~~> > 2. A system of linear equations over a 'eld F has 5 solutions => F is> isomorphic to Z_5.> Fill in the blank:The set of solutions of a system of linear equations is a ...... .The number of elements of a ..... is a power of the characteristic.5 is prime. === Subject: Re: Postal Lottery: Turn $6 into $60,000 in 90 days, MONEY!!!MAKE MONEY!!!> MAKE THOUSANDS!!!>> I found this on a bulletin board and decided to try it: I don't care> about the useless pre-fabricated crap this message usually says. All I> say is, it works. Continue pre-fab crap.>You are an ass hole and a thief. === Subject: Re: Postal Lottery: Turn $6 into $60,000 in 90 days, GUARANTEEDTurn $6 into $60,000 in 90 days, **GUARANTEED**Ok. Send me a legally binding document stating that if I don't make $60,000in the 90 days you will personally send me $60,000 out of your own pocket.You should have no problem with this since your lottery always works.l8r, Mike N. Christoff === Subject: Is it known whether pi + e is irrational?I saw this problem on an unsolved problem archive. Why isn't thisproblem solved? I know it must be deeper than I think, so couldsomeone explain why this simple proof doesn't work:Assume pi + e is irrationalThen let pi + e = q, where q is a rational number. Let i = sqrt(-1).Then i*(pi + e) = i*qthene^(i*(pi + e)) = e^(i*q)thene^(i*pi)*e^(i*e) = e^(i*q)then-e^(i*e) = e^(i*q)then squaring both sides:e^(2*(i*e)) = e^(2*(i*q))and since e is a strictly increasing fuction we have(2*i)*e = (2*i)*qmeaninge = q.But this would mean e is rational, which is not true. Therefore pi + eis not irrational.Why hasn't this proof been proposed, and if it has, why doesn't itsuf'ce? === Subject: Re: Is it known whether pi + e is irrational?> I saw this problem on an unsolved problem archive. Why isn't this> problem solved? I know it must be deeper than I think, so could> someone explain why this simple proof doesn't work:> > Assume pi + e is irrational> Then let pi + e = q, where q is a rational number. Let i = sqrt(-1).> Then > i*(pi + e) = i*q> then> e^(i*(pi + e)) = e^(i*q)> then> e^(i*pi)*e^(i*e) = e^(i*q)> then> -e^(i*e) = e^(i*q)> then squaring both sides:> e^(2*(i*e)) = e^(2*(i*q))> and since e is a strictly increasing fuction we have> (2*i)*e = (2*i)*qThe exponential function is only strictly increasing for strictly real arguments.And e^(i*x) = e^(i*y), for real x and y, does not allow one to conclude that x = y. The best one can conclude from e^(i*x) = e^(i*y) is that x-y = 2*pi*n for some integer n.> meaning> e = q.> But this would mean e is rational, which is not true. Therefore pi + e> is not irrational.> > Why hasn't this proof been proposed, and if it has, why doesn't it> suf'ce? === Subject: Re: Is it known problem on an unsolved problem archive. Why isn't this>>problem solved? I know it must be deeper than I think, so could>>someone explain why this simple proof doesn't work:>>Assume pi + e is irrational>>Then let pi + e = q, where q is a rational number. Let i = sqrt(-1).>>Then >>i*(pi + e) = i*q>>then>>e^(i*(pi + e)) = e^(i*q)>>then>>e^(i*pi)*e^(i*e) = e^(i*q)>>then>>-e^(i*e) = e^(i*q)>>then squaring both sides:>>e^(2*(i*e)) = e^(2*(i*q))>>and since e is a strictly increasing fuction we have>>(2*i)*e = (2*i)*q> > > > >>meaning>>e = q.If this result would be right, it actually proved something more surprising: pi+e = q (from the assumption) e = q (from the argumentation)coming to... pi = 0 (as a new ISO-norm for the exhausted student.... );-)Gottfried Helms === Subject: Re: Is it +0100, Gottfried Helms>coming to...>> pi = 0 (as a new ISO-norm for the exhausted student.... )I heard an anecdote of an expert in Fourier analysis who started hispresentation by stating: I'm sure the audience won't mind if wedeclare 2Pi = 1 for the remainder of this presentation.-- I'm not interested in mathematics that might have anythingto do with reality. -- Russell Easterly, in sci.math === Subject: Re: Is it known whether pi + e is irrational?>>I saw this problem on an unsolved problem archive. Why isn't this>>problem solved? I know it must be deeper than I think, so could>>someone explain why this simple proof doesn't work:>>Assume pi + e is irrational>>Then let pi + e = q, where q is a rational number. Let i = sqrt(-1).>>Then>>i*(pi + e) = i*q>>then>>e^(i*(pi + e)) = e^(i*q)>>then>>e^(i*pi)*e^(i*e) = e^(i*q)>>then>>-e^(i*e) = e^(i*q)>>then squaring both sides:>>e^(2*(i*e)) = e^(2*(i*q))>>and since e is a strictly increasing fuction we have>>(2*i)*e = (2*i)*q>>meaning>>e = q.>> If this result would be right, it actually proved something more> surprising:>> pi+e = q (from the assumption)> e = q (from the argumentation)>> coming to...>> pi = 0 (as a new ISO-norm for the exhausted student.... )from which we can easily deduce that _all_ numbers are zero.Therefore I suggest that, instead of calling it a new ISO-norm,we should call it the new is0-norm. ;-)David === Subject: Re: Is it known whether pi + e is irrational?> I saw this problem on an unsolved problem archive. Why isn't this> problem solved? I know it must be deeper than I think, so could> someone explain why this simple proof doesn't work:> > Assume pi + e is irrational You mean rational? > Then let pi + e = q, where q is a rational number. Let i = sqrt(-1).> Then > i*(pi + e) = i*q> then> e^(i*(pi + e)) = e^(i*q)> then> e^(i*pi)*e^(i*e) = e^(i*q)> then> -e^(i*e) = e^(i*q)> then squaring both sides:> e^(2*(i*e)) = e^(2*(i*q))> and since e is a strictly increasing fuction we have **********************************Now that is absolute nonsense, because you are talking about exp (x) with complex arguments. > (2*i)*e = (2*i)*q> meaning> e = q.> But this would mean e is rational, which is not true. Therefore pi + e> is not irrational.You mean therefore pi + e is not rational? Why hasn't this proof been proposed, and if it has, why doesn't it> suf'ce?You didn't use any properties of e except that it is irrational. So the same proof can be used to show pi + x is irrational if x is any irrational number. Which is clearly wrong if x = 5 - pi, for example. === Subject: Re: Is it known whether pi + e is irrational?...> e^(2*(i*e)) = e^(2*(i*q))> and since e is a strictly increasing fuction we haveStrictly increasing *on the real line. In the complex plane, e^(2*i*pi)=1 so exponential is notinjective: for all complex x, integer ne^(x+n*2*pi*i) = e^x === Subject: pi + e is irrational?Ok, complete disregard my last two posts. Here is what I meant to say:Are you sure that it is unknown if pi + e is unknown?assume pi + e is rationalThen let pi + e = q, where q is a rational number. Let i = sqrt(-1).Then i*(pi + e) = i*qthene^(i*(pi + e)) = e^(i*q)thene^(i*pi)*e^(i*e) = e^(i*q)then-e^(i*e) = e^(i*q)then squaring both sides:e^(2*(i*e)) = e^(2*(i*q))and since e is a strictly increasing fuction we have(2*i)*e = (2*i)*qmeaninge = q.But this would mean e is rational, which is not true. Therefore pi + eis not rational.Why hasn't this proof been proposed, and if it has, why doesn't itsuf'ce? === Subject: Re: pi + e is irrational?> Ok, complete disregard my last two posts. Here is what I meant to say:> Are you sure that it is unknown if pi + e is unknown?> > assume pi + e is rational> Then let pi + e = q, where q is a rational number. Let i = sqrt(-1).Look here.> (2*i)*e = (2*i)*q> meaning> e = q.> But this would mean e is rational, which is not true. Therefore pi + e> is not rational.Even assuming all of your steps to this point are accurate, you started offwith an unsupported assumption that q (the sum of pi and e) was rational.Your conclusion is non sequitor because your premise is unsupported.> Why hasn't this proof been proposed, and if it has, why doesn't it> suf'ce?Because it's not supported.-- Darryl L. Pierce Visit the Infobahn Offramp - What do you care what other people think, Mr. Feynman? === Subject: Re: pi + e is irrational?Darryl complete disregard my last two posts. Here is what I meant to say:> Are you sure that it is unknown if pi + e is unknown?>> assume pi + e is rational> Then let pi + e = q, where q is a rational number. Let i = sqrt(-1).>> Look here.>> (2*i)*e = (2*i)*q> meaning> e = q.> But this would mean e is rational, which is not true. Therefore pi + e> is not rational.>> Even assuming all of your steps to this point are accurate, you startedoff> with an unsupported assumption that q (the sum of pi and e) wasrational.> Your conclusion is non sequitor because your premise is unsupported.That's not the problem. There are problems, but that's certainly not it.He attempted a proof by contradiction, assuming pi+q was rational andderiving a contradiction.The fact that his steps are invalid doesn't mean that proofs bycontradiction do not work.Doug === Subject: Re: pi + e is irrational?> e^(2*(i*e)) = e^(2*(i*q))> and since e is a strictly increasing fuction we haveexp is strictly increasing on the reals. not on the complexes...As others pointed out, exp is not one-to-one on the complexes. === Subject: Re: pi + e is irrational?> e^(2*(i*e)) = e^(2*(i*q))> and since e is a strictly increasing fuction we have> (2*i)*e = (2*i)*q> meaning> e = q.> But this would mean e is rational, which is not true. Therefore pi + e> is not rational. e^(2*(i*e)) = e^(2*(i*q)) gives e = q + k*pi where k is an integer:k = ...-3,-2,-1,0, 1, 2, 3...Remember that the complex function f(z)=e^(i*z) is periodic of period2*pi, and that in the set of complex numbers there is no ordering ofthe number, and hence you cannot speak of increasing or decreasingfunctions. Perhaps you can de'ne some order, but that won't help yourproof. === Subject: Re: pi + e is irrational?> Ok, complete disregard my last two posts. Here is what I meant to say:> Are you sure that it is unknown if pi + e is unknown?> > assume pi + e is rational> Then let pi + e = q, where q is a rational number. Let i = sqrt(-1).> Then > i*(pi + e) = i*q> then> e^(i*(pi + e)) = e^(i*q)> then> e^(i*pi)*e^(i*e) = e^(i*q)> then> -e^(i*e) = e^(i*q)> then squaring both sides:> e^(2*(i*e)) = e^(2*(i*q))> and since e is a strictly increasing fuction we have> (2*i)*e = (2*i)*q> meaning> e = q.> But this would mean e is rational, which is not true. Therefore pi + e> is not rational.> > Why hasn't this proof been proposed, and if it has, why doesn't it> suf'ce?Because e^(i*x) = e^(i*(x+2*pi*n)) for all integers n so your conclusion that 2*i*e = 2*i*q is unfounded. === Subject: Re: pi + e is irrational?> assume pi + e is rational> Then let pi + e = q, where q is a rational number. Let i = sqrt(-1)....> meaning> e = q.> But this would mean e is rational, which is not true. Therefore pi + e> is not rational. Note that you did not use the fact that Oq' is rational anywhere. This should have told you that your algebra went wrong somewhere, since if you were claiming that you just want to 'nd the (irrational) value of pi+e, you set pi+e=q (for any q), then you conclude that e=q, then you should be able to conclude that pi = 0 (no matter what q was, rational or irrational.) Does this prove that pi = 0?J === Subject: Re: message> Ok, complete disregard my last two posts. Here is what I meant to say:> Are you sure that it is unknown if pi + e is unknown?>> assume pi + e is rational> Then let pi + e = q, where q is a rational number. Let i = sqrt(-1).> Then> i*(pi + e) = i*q> then> e^(i*(pi + e)) = e^(i*q)> then> e^(i*pi)*e^(i*e) = e^(i*q)> then> -e^(i*e) = e^(i*q)> then squaring both sides:> e^(2*(i*e)) = e^(2*(i*q))> and since e is a strictly increasing fuction we have> (2*i)*e = (2*i)*qNo this means2*i*e= 2*i*q+2*n*pi*ifor some integer nand that integer in this case is just -1so e=q-pi which is what you started with> meaning> e = q.> But this would mean e is rational, which is not true. Therefore pi + e> is not rational.>> Why hasn't this proof been proposed, and if it has, why doesn't it> suf'ce?Notice that in constructing this proof, you did not use the fact that q isrational anywhere before the last line which is a contradiction by itselfwhatever this q is. === Subject: Re: 3-D analogue of pythagorean theorem> ... The theorem seems analogous to me to the> pythagorean theorem but somewhat unknown as the three dimensional> extension of the pythagorean theorem. Can the theorem be generalized> to n-dimensional Euclidean space? ...> Equivalently, if the three right triangles are in coordinate planes,> A_i = n_i * A,> where A_i is the area of the right triangle with normal in the ith> direction, n_i is the ith component of the unit normal to the base (the> fourth face), and A is the area of the base. This property is used in> elasticity theory to derive the formula relating the stress vector to> Cartesian components of stress.Yes, direction cosines (l,m,n) give the 3 D extension of Pythogorustheorem normalized with respect to areas as l^2+m^2+n^2=1, with 2 Dspecial case n = cos[Pi/2]=0, l = cos(th) , m = cos(Pi/2- th). === Subject: Re: 3-D analogue of pythagorean theoremI had found it in Altshiller-Court's book,which was cited in the Wolframite Index,Altshiller-Court, Nathan, Modern Pure Solid Geometry.New York: Chelsea, pp. 92 and 300, 1979. I think that there's a more-recent reprintfrom Chelsea, which I missed after I'd talked about iton a synergetics-l, because it had already donewhat Bucky said was so important. that is why I also call it,_The Tetrahedron and How to Use It_. there's also the related generalization,which is easier to prove than for this trirectangular tetrah.,for a quadrirectangular tetrah., sinceit just deals with the XYZ edges (XX+YY+ZZ=WW). > de Gua's theorem...> > http://mathworld.wolfram.com/deGuasTheorem.html as for productivity, forget it. we have a *huge* de'citwith China e.g., and the funding for the mortgagesof the housing bubble are investments from abroad (as wellas a lot of the bigger houses being those of wealthy foreigners). of course, a lot of this is moot, if we allow Cheenyto grind us into Tony's McCrusade (Usama's MacJihad). see my sig.--Give the World a Trickier Dick Cheeny -- out of of'ce after GIGA years.http://www.benfranklinbooks.com/http://larouchepub.com/ www.rand.org/publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanachttp://www.wlym.com/PDF-68 soros.html === Subject: Re: A good source calculus help.> Stacy,> > There, I knew somebody here would be able to help you. There are ppl here> who can integrate stuff with one hand tied behind their back!Can someone recommend a site or book that could be used in tandem with my textbook? I'm an engineering student in calc 2, and while I don't have any speci'c questions at the moment, I get lost in the steps of problem solving often.> A few general pointers...> > # top posting [writing your reply before what you are replying to] is> frowned upon. Please use bottom posting, as I am now.> > # Whenever you post asking for help, always explain what efforts you have> made so far and the results of those efforts.> > # You can, if you want to, just ask for a hint, rather than the full> solution.> > # Some ppl here are very picky. They love to 'nd small errors in what> others post. Annoy them! Check and double-check your posts just to thwart> them.Chris Leonardhttp://www.ae.utexas.edu/~leonarct === Subject: Re: A good source calculus help.X-DMCA-Noti'cations: 17:23:26 -0000, The Last Danish Pastry>[...]>>Stacy,>>There, I knew somebody here would be able to help you. There are ppl here>who can integrate stuff with one hand tied behind their back!>>A few general pointers...>># top posting [writing your reply before what you are replying to] is>frowned upon. Please use bottom posting, as I am now.>># Whenever you post asking for help, always explain what efforts you have>made so far and the results of those efforts.>># You can, if you want to, just ask for a hint, rather than the full>solution.>># Some ppl here are very picky. They love to 'nd small errors in what>others post. Annoy them! Check and double-check your posts just to thwart>them.Dammit, don't tell her to do that! I just _hate_ it when people makeposts that don't have any errors.************************David C. Ullrich === Subject: Re: A good source calculus help.The Last Danish Pastry 17:23:26 -0000, The Last Danish Pastry>> >There, I knew somebody here would be able to help you. There are pplherewho can integrate stuff with one hand tied behind their back!>> You need both hands to integrate?>> What a stupid thing for me to say!!> Of course, you can integrate with just one hand.But what if your other hand was tied behind your back so tightly that itwas painful? Or that it restricted blood circulation? I bet it'd beharder to integrate, especially by parts.Doug === Subject: Re: A good source calculus help.Doug Norris 17:23:26 -0000, The Last Danish Pastry> >> >There, I knew somebody here would be able to help you. There are ppl> herewho can integrate stuff with one hand tied behind their back!> >> > You need both hands to integrate?>> What a stupid thing for me to say!!> Of course, you can integrate with just one hand.>> But what if your other hand was tied behind your back so tightly that it> was painful? Or that it restricted blood circulation? I bet it'd be> harder to integrate, especially by parts.Oh boy. You're right of course. Also, suppose you are right handed and theb*st*rds tie your RIGHT hand behind your back!-- Clive Toothhttp://www.clivetooth.dk === Subject: Four Roots to any Polynomial0Sum a[n]x^n = 0 nth degree polynomialn=0N=(a[1],a[2],a[3],..,a[n]) Normal to Plane -a[0]Q= --------- N Perpendicular from origin to plane |N|^2X[n] = unit vectors along the n axes n=1,2,3,...Z[1]=N.. n-1 X[n-1]*Z[p]Z[n]= X[n-1] - Sum(--------------- Z[p] ) n=2,3,4,... p=1 |Z[p]|^2Z[n] are the axes of a basis orthogonal with N.Z is not a unit vector but all of its componentsZ[1],Z[2],Z[3],Z[4],.. are orthogonal.Z[2]=(z[21],z[22],....,z[2n] )Z[3]=(z[31],z[32],....,z[3n] )using the formula, a[1]^2z[21]= 1 - -------- |N|^2 a[1]a[2]z[22]= - ---------- |N|^2..------------ a[1]a[2] 1 a[1]^2z[31]= ---------(---------- - --------------- - 1) |N|^2 |Z[2]|^2 |N|^2|Z[2]|^2 a[2]^2 a[1]^2 a[2]^2z[32]= 1 - -------- - --------------- |N|^2 |N|^4 |Z[2]|^2 ==Z[2]*T a[0]a[1]------- = t + ---------- call this f2(t)|Z[2]| |N|^2Z[3]*T a[1]a[2] a[0]a[1]^2 a[2] a[0]a[2]------- = t^2 + --------------- t + ---------------- + ---------|Z[3]| |N|^2 |Z[2]|^2 |N|^4 |Z[2]|^2 |N|^2call this f3(t)Dimensional Analysis is off due to taking the dot product ofunit vectors in the numerator. t=x in the polynomial, andT=(t,t^2,t^3,...,t^n)If Q=(q[1],q[2],...,q[n]) Z[2] Z[3]T-Q = f2(t)-------- + f3(t) --------- |Z[2]|^2 |Z[3]|^2decoding, z[21] z[31]t - q[1] = f2(t) --------- + f3(t) ----------- |Z[2]|^2 |Z[3]|^2 z[22] z[32]t^2 - q[1] = f2(t) --------- + f3(t) ----------- |Z[2]|^2 |Z[3]|^2These two equations precipitate 4 roots to any degreepolynomial. Although it is the solution for only one sliceof all possibilites, it is one of the possibilities that it mustat least satisfy.Jon Giffen === Subject: Re: Four Roots to any Polynomial> 0> Sum a[n]x^n = 0 nth degree polynomial> n=0> > N=(a[1],a[2],a[3],..,a[n]) Normal to Plane> > -a[0]> Q= --------- N Perpendicular from origin to plane> |N|^2> > X[n] = unit vectors along the n axes n=1,2,3,...> > Z[1]=N> .> . n-1 X[n-1]*Z[p]> Z[n]= X[n-1] - Sum(--------------- Z[p] ) n=2,3,4,...> p=1 |Z[p]|^2> > Z[n] are the axes of a basis orthogonal with N.> Z is not a unit vector but all of its components> Z[1],Z[2],Z[3],Z[4],.. are orthogonal.> > Z[2]=(z[21],z[22],....,z[2n] )> Z[3]=(z[31],z[32],....,z[3n] )> > using the formula,> > a[1]^2> z[21]= 1 - --------> |N|^2 a[1]a[2]> z[22]= - ----------> |N|^2> .> .> ------------> > a[1]a[2] 1 a[1]^2> z[31]= ---------(---------- - --------------- - 1)> |N|^2 |Z[2]|^2 |N|^2|Z[2]|^2> > a[2]^2 a[1]^2 a[2]^2> z[32]= 1 - -------- - ---------------> |N|^2 |N|^4 |Z[2]|^2> > ==> > Z[2]*T a[0]a[1]> ------- = t + ---------- call this f2(t)> |Z[2]| |N|^2> > Z[3]*T a[1]a[2] a[0]a[1]^2 a[2] a[0]a[2]> ------- = t^2 + --------------- t + ---------------- + ---------> |Z[3]| |N|^2 |Z[2]|^2 |N|^4 |Z[2]|^2 |N|^2> > call this f3(t)> > Dimensional Analysis is off due to taking the dot product of> unit vectors in the numerator. t=x in the polynomial, and> T=(t,t^2,t^3,...,t^n) If Q=(q[1],q[2],...,q[n])> > Z[2] Z[3]> T-Q = f2(t)-------- + f3(t) ---------> |Z[2]|^2 |Z[3]|^2> > decoding,> z[21] z[31]> t - q[1] = f2(t) --------- + f3(t) -----------> |Z[2]|^2 |Z[3]|^2> > z[22] z[32]> t^2 - q[1] = f2(t) --------- + f3(t) -----------> |Z[2]|^2 |Z[3]|^2> > These two equations precipitate 4 roots to any degree> polynomial. Although it is the solution for only one slice> of all possibilites, it is one of the possibilities that it must> at least satisfy.> > Jon Giffen> > What are the 4 roots of the polynomial f(x) = x ? === Subject: Correct Address by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id meanthttp://www.geocities.com/erdosfan/solution.htmlHisanobu Shinya === Subject: ask for helpHow to prove : if a function f de'ned on [a.b] is l.s.c., then there is a monotone increasing sequence of l.s.c. step functions on [a,b] s.t. for each x in [a,b], we have f(x) = lim g_n(x) === Subject: Re: ask for helpX-DMCA-Noti'cations: -0800, ah_kit_ah_kit@yahoo.com.hk (anthony)>How to prove : if a function f de'ned on [a.b] is l.s.c., then there is a >monotone increasing sequence of l.s.c. step functions on [a,b] s.t. >for each x in [a,b], we have f(x) = lim g_n(x)Take [a,b] = [0,1] to make this easier to type.You could de'ne g_n to be constant on each open interval(j/2^n, (j+1)/2^n); the value of g on that interval should bethe largest number which will make g_n <= f (so that valueshould be the ____ of f.) De'ne g_n to be as large aspossible at the endpoints of those intervals, such thatg_n be lsc. Now use the fact that f is lsc to show thatg_n increases to f.************************David C. Ullrich === Subject: Involutionary CalculusConsider the following de'nitions: 1)The hyperderivative of a function f at a point x is de'ned as the limit as h approaches 0,of (f(x+h)/f(x))^(1/h),provided it exists. 2)The hyperintegral(or more properly the productal) of a function f over an interval [a,b] over which it is positive is de'ned as the limit as n approaches in'nity,of the continued product as k varies from 0 to n-1,of [f(a+(b-a)k/n)^1/n],provided it exists. It turns out to be that the two operations are related through following theorem: Suppose that F has f as its hyperderivative over some interval (a,b) and continuous over [a,b] then then the hyperintegral of f over [a,b] is F(b)/F(a). The calculus thus cerated has a structure analogous to the known calculus and many theorems have their analogues. It remains true,however,that the hyperderivative and the hyperintegral can be expressed in terms of the known derivatives and known integrals. If this idea is so much beautiful why is not it popular in mathematical circles? === Subject: Re: Involutionary Calculus> Consider the following de'nitions:> > 1)The hyperderivative of a function f at a point x is de'ned as the > > limit as h approaches 0,of (f(x+h)/f(x))^(1/h),provided it exists.> > 2)The hyperintegral(or more properly the productal) of a function f over an > interval [a,b] over which it is positive is de'ned as the > > limit as n approaches in'nity,of the continued product as k varies from 0 > to > n-1,of [f(a+(b-a)k/n)^1/n],provided it exists.> > It turns out to be that the two operations are related through following > theorem:> Suppose that F has f as its hyperderivative over some interval (a,b) and > continuous over [a,b] then> then the hyperintegral of f over [a,b] is F(b)/F(a).> > The calculus thus cerated has a structure analogous to the known calculus > and many theorems have their analogues.> It remains true,however,that the hyperderivative and the hyperintegral can > be expressed in terms of the known derivatives and known integrals.> If this idea is so much beautiful why is not it popular in mathematical > circles?We call it taking logs. === Subject: Re: Involutionary Calculus> > Consider the following de'nitions:> > 1)The hyperderivative of a function f at a point x is de'ned as the > > limit as h approaches 0,of (f(x+h)/f(x))^(1/h),provided it exists.> > 2)The hyperintegral(or more properly the productal) of a function f over an > interval [a,b] over which it is positive is de'ned as the > > limit as n approaches in'nity,of the continued product as k varies from 0 > to > n-1,of [f(a+(b-a)k/n)^1/n],provided it exists.> > It turns out to be that the two operations are related through following > theorem:> Suppose that F has f as its hyperderivative over some interval (a,b) and > continuous over [a,b] then> then the hyperintegral of f over [a,b] is F(b)/F(a).> > The calculus thus cerated has a structure analogous to the known calculus > and many theorems have their analogues.> It remains true,however,that the hyperderivative and the hyperintegral can > be expressed in terms of the known derivatives and known integrals.> If this idea is so much beautiful why is not it popular in mathematical > circles?> > We call it taking logs. I give a damn to what you call it or in what manner can the result be established.If you cannot see the inherent aesthetic appeal behind the results you better keep your comments with you. === Subject: Re: Involutionary Calculus> We call it taking logs.> > I give a damn to what you call it or in what manner can the result be > established.If you cannot see the inherent aesthetic appeal behind the> results you better keep your comments with you.Oh no. What will happen if I don't keep my comments with me? === Subject: Re: Involutionary CalculusX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>> >> Consider the following de'nitions:>> >> 1)The hyperderivative of a function f at a point x is de'ned as the >> >> limit as h approaches 0,of (f(x+h)/f(x))^(1/h),provided it exists.>> >> 2)The hyperintegral(or more properly the productal) of a function f over an >> interval [a,b] over which it is positive is de'ned as the >> >> limit as n approaches in'nity,of the continued product as k varies from 0 >> to >> n-1,of [f(a+(b-a)k/n)^1/n],provided it exists.>> >> It turns out to be that the two operations are related through following >> theorem:>> Suppose that F has f as its hyperderivative over some interval (a,b) and >> continuous over [a,b] then>> then the hyperintegral of f over [a,b] is F(b)/F(a).>> >> The calculus thus cerated has a structure analogous to the known calculus >> and many theorems have their analogues.>> It remains true,however,that the hyperderivative and the hyperintegral can >> be expressed in terms of the known derivatives and known integrals.>> If this idea is so much beautiful why is not it popular in mathematical >> circles?>> We call it taking logs.>> I give a damn to what you call it or in what manner can the result be > established.If you cannot see the inherent aesthetic appeal behind the> results you better keep your comments with you.Uh, right. All hail Ashurosh, the discoverer of the amazinghypercalculus.The point is that your assumption that hypercalculus is notpopular is simply incorrect. It's just calculus plus logarithms,and logarithms are _very_ popular...************************David C. Ullrich === Subject: Re: Please help prove! about modules. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, studying algebra by myself.>> >> I cannot 'gure out two proofs about modules.>> >> 1. S-module M is projective <=> each short exact sequence 0->A->B->M->0>> splits.>> >> 2. Q, the abelian group of rational numbers, is an injective Z-module.>> >> If anyone could help me, it would be greatly appreciated!>> these two results are pretty>standard and can be found in most Algebra texts - I recommend>Hungerford. (They are exercises in Lang.) >>If you can't locate a suitable text, ask again.>>-- >Paul Sperry>Columbia, SC (USA) === Subject: weigh with 5 weights by support1.mathforum.org (8.11.6/8.11.6/The Math book written by Theodore Shifrin:A druggist has the 've weights of 1, 3, 9, 27, and 81 ounces, and a two-pan balance. Show that he can weigh any integral amount up to and including 121 ounces. How can you generalize this result?while i see how this is possible, i have no clue how to possibly generalize the result. can someone give me a clue or tell me the === Subject: Re: weigh with 5 weights> This is a problem from my abstract algebra book written by Theodore Shifrin:> > A druggist has the 've weights of 1, 3, 9, 27, and 81 ounces, and a two-pan > balance. Show that he can weigh any integral amount up to and including 121 > ounces. How can you generalize this result?> > while i see how this is possible, i have no clue how to possibly generalize > the result. can someone give me a clue or tell me arbitrary positive integer, with weights 1, 3, 9, ..., 3^n, and a two pan balance scale of suf'cient strength one can weigh any any integral amount from 1 through (3^(n+1) -1)/2 inclusive. === Subject: Re: weigh with 5 weights>Subject: Re: weigh with 5 weights>Message-id: is a problem from my abstract algebra book written by Theodore>Shifrin:>> >> A druggist has the 've weights of 1, 3, 9, 27, and 81 ounces, and a>two-pan >> balance. Show that he can weigh any integral amount up to and including 121> ounces. How can you generalize this result?>> >> while i see how this is possible, i have no clue how to possibly generalize> the result. can someone give me a clue or tell an arbitrary positive integer, with weights 1, 3, 9, ..., >3^n, and a two pan balance scale of suf'cient strength one can weigh >any any integral amount from 1 through (3^(n+1) -1)/2 inclusive.>Suppose you generalized it to weights of powers of an arbitrarybase b. If your weights were 1, 5, 25 you can't form 2 or 3. You could form any number if you had b-1 copies of each weightusing addition simply by creating the weight as a radix b number.But since the two pan balance allows bothe addition and subtraction,the set of weights 1, 1, 5, 5, 25 allows any integer up to 37.So you could generalize the question to: given weights as powersof b, what is the minimum number of each do you need to form every integer up to n?--MensanatorAce of Clubs === Subject: Re: weigh with 5 weights> This is a problem from my abstract algebra book written by Theodore> Shifrin:> > A druggist has the 've weights of 1, 3, 9, 27, and 81 ounces, and a> two-pan balance. Show that he can weigh any integral amount up to and> including 121 ounces. How can you generalize this result?> > while i see how this is possible, i have no clue how to possibly> generalize the result. can someone give me a clue or tell me the answer?Look at the sequence 1, 3, 9, 27, 81.How is each term related to its predecessor?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Surjective Schroeder-BernsteinConsider the followingProposition. If there exist surjections from A to B and from B to A, then there exists a bijection from A to B.Can this propostion be proved in ZF without Choice? If not, is it equivalent to the Axiom of Choice?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Surjective Schroeder-Bernstein>Consider the following>Proposition. If there exist surjections from A to B and from B >to A, then there exists a bijection from A to B.>Can this propostion be proved in ZF without Choice? If not, is it >equivalent to the Axiom of Choice?We already have had postings of counterexamples withoutchoice.AFAIK, it is not even known that the stronger hypothesis:if there is a surjection from A to B there is an injectionfrom B to A; implies choice.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Interesting numerical effect I cannot explain...> > Hi Jeroen,> > well, what you are suggesting is indeed quite simple. It seems like it> does not even depend on the particular calculations that I am> performing. Does that mean that any numerical operations will suffer> from this effect?> Any calculation of which the result cannot be represented accuratly enough. So if you you have amantissa of 53 bits, then using integers up to those 53 bits is OK, even integers up to 54 bits alsoetcettera, and many ¤oats also. As soon as the calculation result is truncated in order to 't itin the available number of bits of the mantissa, you'll get the error. Instead of seeing it as an absolute error, it might be better to see it as a relative error. Forepsilon 2.2e-16, this means that your calculation result is always accurate to approx 15 decimals ifwritten as, for was very helpful.> > Nik> > >An explanation seems to be straightforward. Every numerical result R is represented by a mantissa M>and an exponent E. The value of M must be interpreted between -1 and 1 which can be achieved by>dividing M by max(M), then by de'nition:>> R = (M/max(M))*2^E>>I'm not sure which IEEE standard is used on the PC, but for both M and E a constant number of bits>is reserved. Because calculations have an uncertainty of +/- 1 in M, you 'nd:>> R' = ([M+/-1]/max(M))*2^E = R +/- 2^E / max(M).>>If you de'ne 1/max(M) as your epsilon Oe' (2.2e-16, this corresponds exactly with a 53 bit>mantissa!), then:>> R' = R +/- e*2^E>>this con'rms your 'ndings. This error will come back in all calculations, so the 'nal>error-result may depend on the particular order of the intermediate calculation you used to get the>'nal result...>>It should be easy now to 'll in the rest of your questions.>> Jeroen> > -- Nikita A. Visnevski> Adaptive Systems Laboratory> CRL, McMaster University> Phone : (905) 525-9140 x 27282> Web : http://soma.crl.mcmaster.ca> Interesting numerical effect I cannot explain...I suppose that is exactly why the largest N I have seen on win32 platform was 2^15 = 32768. In MSVC++ there is a constant in <¤oat.h> de'ned as follows:#de'ne DBL_DIG 15 /* # of decimal digits of precision */That's what I am using now to check for the result in the range +/- 2^15 * epsilon.Anyhow, thatnks averybody for your great help.Nik> > Any calculation of which the result cannot be represented accuratly enough. So if you you have a> mantissa of 53 bits, then using integers up to those 53 bits is OK, even integers up to 54 bits also> etcettera, and many ¤oats also. As soon as the calculation result is truncated in order to 't it> in the available number of bits of the mantissa, you'll get the error. > > Instead of seeing it as an absolute error, it might be better to see it as a relative error. For> epsilon 2.2e-16, this means that your calculation result is always accurate to approx 15 decimals if> written as, for example, 1.23e-8.> > Jeroen> -- Nikita A. VisnevskiAdaptive Systems LaboratoryCRL, McMaster UniversityPhone : (905) 525-9140 x 27282Web : http://soma.crl.mcmaster.ca === Subject: Re: help please>can you help 'nd me all the positive integer solutions (x,y) to the equation>2y^2=x^4+8x^3+8x^2-32x+15>any help is much appreciated.>choogu> >Notice to All:A bit too late, but I suggest you submit your answer to Mathematics Magazine rather than help the OP here. The problem is posed in the most recent issue thereof.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: help please> >can you help 'nd me all the positive integer solutions (x,y) to the equation>2y^2=x^4+8x^3+8x^2-32x+15>any help is much appreciated.>choogu> >> Notice to All:> > A bit too late, but I suggest you submit your answer to Mathematics > Magazine rather than help the OP here. The problem is posed in the most > recent issue thereof.dear chooguit has not answer by formols but you can plot it and calcolate it whit areacan you plot it? === Subject: Re: help please> >can you help 'nd me all the positive integer solutions (x,y) to the equation>2y^2=x^4+8x^3+8x^2-32x+15>any help is much appreciated.>choogu> >> Notice to All:> > A bit too late, but I suggest you submit your answer to Mathematics > Magazine rather than help the OP here. The problem is posed in the most > recent issue thereof.oops, I am sorry ..I did not realize that. Someone had asked me to doit. Please do not post anymore help.Choogu === Subject: Re: help pleasechoogu@yahoo.com integer solutions (x,y) to the> equation 2y^2=x^4+8x^3+8x^2-32x+15> any help is much appreciated.> chooguAdd one to both sides and replace x by v-2 and theequation becomes2y^2 +1= (v^2-8)^2Ahhhh...much prettier....Bart === Subject: Re: help please> can you help 'nd me all the positive integer solutions (x,y) to the equation> 2y^2=x^4+8x^3+8x^2-32x+15> any help is much appreciated.> choogudear choogu is your means that you want integrate from it, please say more thank you hupo === Subject: Re: Graph Theory: Number of maximal MatchingsHi Felix,thank you for your hints. Unfortunately my bipartite graphs are not regular.With best wishesbjj> >>Hi Robert,>>you are right, I should more precisely specify my problem.>>Let G = (V1, V2, E) be a bipartite graph with>>1) Edges connects vertices from V1 with V2 only>>2) |V1| = |V2|>>3) There exists a perfect matching covering all vertices>>My Problem: Is there a tighter bound on the maximal number of perfect >>matchings than n!?>>For example if |V1| = |V2| = 3 and each vertex has degree one, then I >>conclude the number of perfect matchings is one. But is there a general >>formula as a function on the number |E| of edges , degree sequence, etc, >>which gives a good bound on the number of perfect matchings?> [...]> > Well, there is something along these lines [1,p.173]:> > Theorem:> Let G be a k-regular bipartite graph on n vertices. Then the number of> perfect matchings is at least n!((k/n)^n).> > The proof is based on the observation that the permanent of the> adjacency matrix counts perfect matchings and on Van der Waerden's> conjecture (that had been proved by Egorychev and Falikman).> > On the same page there is a mention of another result, due to Bregman> [2]:> > Theorem:> Let G be a k-regular bipartite graph on n vertices. Then the number of> perfect matchings is at most ((k!)^n)/k.> > HTH,> Felix.> > [1] Asratian, Denley & Haggkvist, Bipartite Graphs and Their> Applications, Cambridge University Press, 1998.> > [2] L.M. Bregman, Some properties of nonnegative matrices and their> permanents, Soviet Math. Dokl. 14 (1973), 945-949.> === Subject: Re: JSH: Howard Aiken quote, my situation> Herman then, is Girard Desargues.>> He really was discarded by the botanical style.>> Can you explain what you mean by a botanical style?>> Sorry, i don't have more details. For years, i've tried to> get my hands on an original text, but failed, so far.> As i understand it, he used terms from botany to denote> mathematical objects (say, ¤ower instead of line, or so.)>> His main work is> Brouillon project d'une atteinte aux evenemens des> rencontres d'un cone avec un plan (1639)> This title in itself was already considered whimsical in his days:> Sketchy investigation of what happens when a cone shakes> hands with a plane.> It didn't sound like serious math, therefore it wasn't... :-)>> But, as said, this is all from hear-say. O, could i> only see the original text... :-)>> The brief biography at> > mentions that he explain[ed] matters in the vernacular, but nothing> Rough draft for an essay on the results of taking plane sections of a> cone, but I guess that's a liberal translation.My main source was Egmont Colerus. (Perhaps not the greatest ofauthorities, but in stories like these usually pretty trustworthy, imo.And of course one should take _all_ history with a grain of salt.)Online, i've found this: http://www.bib.ulb.ac.be/coursmath/bio/desargue.htmQuote: Ce ne fut pas tres apprecie par ses contemporains a cause du systemede termes mathematiques derives de noms botaniques et du manque de notationscartesiennes.Apart from that:Desargues said things like parallel lines _do_ have a point o'ntersection, only it is a point at in'nity. In the age of cartesiancoordinates and Euclidean space conceived as the only reasonablemathematics, people must have thought he was nuts.But that the minor minds of his days didn't follow him is notsurprising. What is surprising is that, despite appreciationby a few of his great contemporaries, he was really forgotten,for a long time, and nearly ended up totally forgotten.BTW, i'm quite surprised at the kind of opposition, on this.The history of mathematics (as of anything else) really _does_contain much Ohowling of Boetians'. The mainstream has noother choice, that's what makes it mainstream.What is surprising is that even in mathematics, this mainstreamsometimes has greater in¤uence on the course of things than thereal geniuses of the time.Or rather, it is not surprising. It's just annoying, for usmathematicians, because we like to think so much thatthe math community is different. It's not. Math is doneby humans (and let's hope it stays that way :-) ).> Evidently, you're not likely to 'nd any of his other original> writing.Well, i've had an opportunity to buy it, for 29500 euro.But i was too narrow-minded to do it. === Subject: Re: JSH: Howard Aiken Kronecker was certainly vociferous in his|rejection, but then again Hilbert was equally eloquent in his|defense.I've read that most of Kronecker's known comments on the philosophyof mathematics are from one talk he gave, and that generally he wasmuch nicer about the whole matter than his reputation suggests.Hilbert, years later, certainly was emphatic, but I'm not sure howeloquent one could say he was. One of his main attempts at defendingthe status quo failed on account of Goedel's second incompletenesstheorem.|Cantor caused a real problem in mathematics, but it was not|so much his results that were criticized, as much as some|consequences from those results (and let's be fair: Cantor's approach|->was<- formally wrong and led to paradoxes; the idea that any|speci'cation led to a set is certainly problematical, as Cantor|himself recognized with Cantor's Paradox) which were rejected. No, let's actually be fair: Cantor's approach was consistent anddidn't lead to any paradoxes. Cantor didn't assert the unrestrictedseparation axiom; he made a special point of distinguishingbetween cases where the things satisfying a property constituteda set and when they didn't. A somewhat polished-up version ofhis explanation why some failed to do so (there are too many ofthem), known as the limitation of size, is still favored by sometoday.Frege's system had the contradiction in it, and he never reallysettled on an answer to the problem.By the way, Zermelo was identifying his axioms for the sake offormalizing the proof of the well-ordering theorem, not as anattempt at patching over paradoxes either.Keith Ramsay === Subject: Re: irrationality of sqrt(2): easy question>> How to prove a^2 even => a even without using irrationality of sqrt(2)?>> a is not even>> ==> there is a n such that a = 2n+1>> ==> there is a n such that a^2 = 4n^2 + 4n + 1>> ==> there is a m such that a^2 = 2m + 1, namely m = 2n^2+2n>> ==> a^2 is not even>> So at the end you need to know that no integer is both even and odd;> at the beginning you need to know an integer is either even or odd.>> Odd integers are de'ned as having the form a = 2n+1>> Integers which are not odd, are de'ned as even ;-)> > Which now leaves the question: why is an even integer divisible by 2?> (There's a little work behind this. You need to know the Euclidean> division algorithm, at least for dividing by two.)All overkill. Proving sqrt(2) irrational needs onlyPrime Divisor Property of 2: 2|MN => 2|M or 2|Nwhich has a trivial proof by induction on N, namely:Base case N = 0,1 is clear, as is the inductive step 2|MN => 2|M(N-2) => 2|M or 2|N-2 via inductive hypothesis => 2|M or 2|N-Bill Dubuque === Subject: Re: irrationality of sqrt(2): easy question>> How to prove a^2 even => a even without using irrationality of sqrt(2)?Doesn't it follow easily from unique factorization into primes? interested in scienti'c online workshops, please visitsitehttp://de.geocities.com/scienceworkshops/Its goal is to organize study groups on scienti'c topics like quantum'eld theory,probabilistic inference or neural nets, to name a few topics I ampersonally interested in(of course, arbitray topics may be suggested).The idea is to study some text (which is freely available over theinternet, by setting upa study plan for 12 weeks and open a discussion groups whereparticipants can post questions,or solutions to exercises.The text must be of academic level, ranging from introductory texts topost-graduate level.Participation is free in these groups.I would like to start the 'rst workshop in fall, so if you areinterested you might eithersubscribe to our mailing list or === Subject: Re: Study groups in scienceJochen online workshops, please visit> site> http://de.geocities.com/scienceworkshops/> Its goal is to organize study groups on scienti'c topics like quantum> 'eld theory,> probabilistic inference or neural nets, to name a few topics I am> personally interested in> (of course, arbitray topics may be suggested).What positive precautions are you taking to prevent the idiots morons andkooks from taking it === Subject: Re: Study groups in science: What positive precautions are you taking to prevent the idiots morons and: kooks from taking it over, as has happened in sci.physics?Perhaps the textbook and the equations will keep people like you away?-- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea: prankster, fablist, magician, liar === Subject: Re: JSH: how you can prove to the mathematicians you are rightX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>> Consider this project, which aims to provide a formal proof of the Kepler>> Conjecture:>> >> http://www.math.pitt.edu/~thales/¤yspeck/>> Why not produce a formal proof of your results? Such a proof would be>> undeniable. Even those mathematicians who don't understand what you are>> saying would have to accept that you are correct. Third parties, whether>> mathematicians or not, could check your claims, by simply running your proof>> through the proof checker themselves.>>Good suggestion! I've actually threatened a couple of times to come>up with a formal proof that can be run through a computer, but haven't>really put any effort into it as of yet, as I didn't really have a>clue how to do the entire thing (and no links Keith Ramsay PLEASE).>>That's why I said threatened above as while I was serious when I said>I'd go out and do it, somehow I found myself never actually working at>it very hard.Of course not. If you ever did correctly formalize your argument soit could be run through a proof checker the machine would reportthat it was wrong.>I'll check the links I saw on your reference though to see if I can>start getting a better feel for how it can be done.>James Harris************************David C. Ullrich === Subject: Re: JSH: how you can prove to the mathematicians you are right>Consider this project, which aims to provide a formal proof of the Kepler>>Conjecture:>> http://www.math.pitt.edu/~thales/¤yspeck/>>Why not produce a formal proof of your results? Such a proof would be>>undeniable. Even those mathematicians who don't understand what you are>>saying would have to accept that you are correct. Third parties, whether>>mathematicians or not, could check your claims, by simply running your proof>>through the proof checker themselves.> > > Good suggestion! I've actually threatened a couple of times to come> up with a formal proof that can be run through a computer, but haven't> really put any effort into it as of yet, as I didn't really have a> clue how to do the entire thing (and no links Keith Ramsay PLEASE).> > That's why I said threatened above as while I was serious when I said> I'd go out and do it, somehow I found myself never actually working at> it very hard.Somehow I 'nd that I'm not surprised.Gib === Subject: Re: JSH: how you can prove to the mathematicians you are right> norabaron@hotmail.com (Nora provide a formal proof of the Kepler>> Conjecture:>> >> http://www.math.pitt.edu/~thales/¤yspeck/>> >> Why not produce a formal proof of your results? Such a proof would be>> undeniable. > Implying that proof-checking programs are infallible? How do> you know that?> > Maybe not infallible, so such a proof might not be undeniable. But> you must admit that a successful formalization and veri'cation would> add considerable evidence to James's claims. > I was asking out of genuine curiosity in this case. Maybe itsuf'ces to show that the proof-checker works on a 'nite subsetof logical statements. I would have guessed however that it wouldbe something like a compiler for a higher-level language: you couldthrow an exceedingly long and complex compound statement at itthat it might not parse correctly, even if it did just 'ne withshort ones. > A successful veri'cation of any of James's most dubious claims would> require a careful explanation of how the proof-checker failed. That's a safe assumption. The other side is the following: i't rejects a JSH proof, what is HE going to say about it?I have now proven not only that there is an error in Core Mathematics, but also there is a Core Error in the theory ofproof-checkers!!! Elsewhere in this thread he launches a pre-emptive strikewhich amounts to saying that at least one of his concepts cannot be expressed in the current language of mathematics: too advanced even fora proof checker. The response there should be, If your proofcannot be stated in the accepted language of mathematical logic,then you don't have a proof by anyone's standards.> Later JSH heard about some proofchecking software that was> touted as being easy to run and described his plan for transcribing> his proof into machine-readable language. I don't think that went> anywhere. It is not as easy as it sounds, whether for Wiles' proof> or the Kepler conjecture or even for JSH's short non-proofs. You are quite> right that third parties could check it if he translated what he > has done into acceptable language, etc., but I very much > doubt he will put in the effort. My view is there is no point in> doing it because we already have solid counterexamples to his> claims and we know exactly where he is making his mistakes. > Nevertheless I would agree with you in encouraging him to try this> route. He might learn something from it, and certainly he is not > learning anything now.> > Of course, I agree that it's unlikely James will do anything that> requires actual effort. He does make efforts and actually thinks hard about some things.But he rarely makes a serious attempt to learn from math books,courses, or people. If it isn't summarized succinctly on some webpage he ignores it. He *has* learned a few things: (1) Fermat's littletheorem; (2) the Barlow-Abel relations; (3) what algebraic integersand algebraic numbers are; (4) the theorem that roots of non-monicprimitive irreducible polynomials with integer coef'cients cannotbe algebraic integers. I don't think he could reliably give thede'nition of a ring, certainly not an ideal, probably not a group.Not much to show for 8 years of effort. Learning how to use aproof-checker, even the most user-friendly one around, is not goingto be trivial. Nora B. === Subject: Re: JSH: how you can prove to the mathematicians you are right <87znbpo5p5.fsf@phiwumbda.org> sha1:UQ7OGlEGd+G7cjl8mt3dEdLqhnk=norabaron@hotmail.com (Nora James's most dubious claims would>> require a careful explanation of how the proof-checker failed. > That's a safe assumption. The other side is the following: if> it rejects a JSH proof, what is HE going to say about it?> I have now proven not only that there is an error in Core > Mathematics, but also there is a Core Error in the theory of> proof-checkers!!!No. In my experience, if one's trying to formalize a proof, but theresult keeps getting spit out by the checker, one always 'gures thathis formalization is off, not that his basic argument is wrong.This is one reason that suggestions James formalizes his argumentaren't so helpful. When you're in the thick of things, it's hard totell the difference between a bad argument and problems turning a goodargument into a formal proof.> Elsewhere in this thread he launches a pre-emptive strike which> amounts to saying that at least one of his concepts cannot be> expressed in the current language of mathematics: too advanced even> for a proof checker. The response there should be, If your proof> cannot be stated in the accepted language of mathematical logic,> then you don't have a proof by anyone's standards.Of course that's correct, but James has no idea what formal languagesare. >> >> Of course, I agree that it's unlikely James will do anything that>> requires actual effort.> He does make efforts and actually thinks hard about some things.> But he rarely makes a serious attempt to learn from math books,> courses, or people. If it isn't summarized succinctly on some web> page he ignores it. He *has* learned a few things: (1) Fermat's little> theorem; (2) the Barlow-Abel relations; (3) what algebraic integers> and algebraic numbers are; (4) the theorem that roots of non-monic> primitive irreducible polynomials with integer coef'cients cannot> be algebraic integers. I don't think he could reliably give the> de'nition of a ring, certainly not an ideal, probably not a group.> Not much to show for 8 years of effort. Learning how to use a> proof-checker, even the most user-friendly one around, is not going> to be trivial.Well, you're more impressed by his efforts than I am.-- Jesse HughesLike the ski resort full of girls hunting for husbands and husbands hunting for girls, the situation is not assymmetrical as it might seem. -- Alan MacKay === Subject: Re: JSH: how you can prove to the mathematicians you are right> norabaron@hotmail.com (Nora James's most dubious claims would>> require a careful explanation of how the proof-checker failed. > That's a safe assumption. The other side is the following: if> it rejects a JSH proof, what is HE going to say about it?> I have now proven not only that there is an error in Core > Mathematics, but also there is a Core Error in the theory of> proof-checkers!!!> > No. In my experience, if one's trying to formalize a proof, but the> result keeps getting spit out by the checker, one always 'gures that> his formalization is off, not that his basic argument is wrong.> No??? I mean, yes, most people would conclude they had screwedup the translation and the formalism, but JSH is not most people. Or do you disagree on that also?> This is one reason that suggestions James formalizes his argument> aren't so helpful. When you're in the thick of things, it's hard to> tell the difference between a bad argument and problems turning a good> argument into a formal proof.> Look: he is not going to put in the kind of effort it would taketo do this. Why? (1) Because he resists learning from anyone else;(2) I think he knows he is wrong. Nobody can come as close as hehas to understanding our arguments without actually understandingthem. Hence his refusal to EVER even quote Decker's main point.If he really believed he is right, the motivation to have his proof validated by machine would be enormous. It is quite possiblethat he would become famous [Letterman, Leno, Larry King, etc.], even rich. He knows this. Thus if he really believed his own claims,he would be breaking his butt to learn how to use a proof-checker.Clearly he isn't. This is not the 'rst time this has come up. ButI think, beneath his patina of arrogant con'dence, he knows he iswrong and wasting time with a proof-checker would be an exercise in futility. He would rather prolong the inevitable by interminable squabbling with people here.> Elsewhere in this thread he launches a pre-emptive strike which> amounts to saying that at least one of his concepts cannot be> expressed in the current language of mathematics: too advanced even> for a proof checker. The response there should be, If your proof> cannot be stated in the accepted language of mathematical logic,> then you don't have a proof by anyone's standards.> > Of course that's correct, but James has no idea what formal languages> are. > >> >> Of course, I agree that it's unlikely James will do anything that>> requires actual effort.> He does make efforts and actually thinks hard about some things.> But he rarely makes a serious attempt to learn from math books,> courses, or people. If it isn't summarized succinctly on some web> page he ignores it. He *has* learned a few things: (1) Fermat's little> theorem; (2) the Barlow-Abel relations; (3) what algebraic integers> and algebraic numbers are; (4) the theorem that roots of non-monic> primitive irreducible polynomials with integer coef'cients cannot> be algebraic integers. I don't think he could reliably give the> de'nition of a ring, certainly not an ideal, probably not a group.> Not much to show for 8 years of effort. Learning how to use a> proof-checker, even the most user-friendly one around, is not going> to be trivial.> > Well, you're more impressed by his efforts than I am. Just look at a typical JSH non-rant posting: lots of algebra [often carelessly done to be sure], some ideas [usually unsound], some evidenceof thinking and calculating and head-scratching: evidence of effort, I would say, but not evidence of a worthwhile result, andnever evidence of an effort to understand much beyond high-school algebra. He agonizes over things, thinks about them, works atthem. But when it comes to learning any of the existing body ofmathematical theory, he is extremely lazy. No, I am not impressed. Nora B. === Subject: Re: JSH: how you can prove to the mathematicians you are right <87znbpo5p5.fsf@phiwumbda.org> <87smhgokhq.fsf@phiwumbda.org> sha1:l1I+upjt0yUMJDCFUZyh5Kk/0to=norabaron@hotmail.com (Nora formalize a proof, but the>> result keeps getting spit out by the checker, one always 'gures that>> his formalization is off, not that his basic argument is wrong.>> >> No??? I mean, yes, most people would conclude they had screwed> up the translation and the formalism, but JSH is not most people. > Or do you disagree on that also?Oh, no, no, no, I do *not* disagree that JSH isn't most people. Idon't quote most people, but I quote him regularly.But nothing about JSH indicates that he's more likely to decide thathis argument is wrong rather than his attempt to formalize it. On thecontrary, he's already shown a certain stubbornness regarding hisargument (maybe you've even noticed this). Why would he decide thathis formalization is correct and his argument wrong?>> This is one reason that suggestions James formalizes his argument>> aren't so helpful. When you're in the thick of things, it's hard to>> tell the difference between a bad argument and problems turning a good>> argument into a formal proof.>> >> Look: he is not going to put in the kind of effort it would take> to do this. Why? (1) Because he resists learning from anyone else;> (2) I think he knows he is wrong. I agree on (1) and (2) is not implausible.> Nobody can come as close as he has to understanding our arguments> without actually understanding them. Hence his refusal to EVER even> quote Decker's main point. If he really believed he is right, the> motivation to have his proof validated by machine would be enormous.> It is quite possible that he would become famous [Letterman, Leno,> Larry King, etc.], even rich.And that's just the letter L! When he gets to O, there's Oprah, andat R he gets Rose (or does he get Charlie Rose at C?). >> Well, you're more impressed by his efforts than I am.>> Just look at a typical JSH non-rant posting: lots of algebra [often > carelessly done to be sure], some ideas [usually unsound], some evidence> of thinking and calculating and head-scratching: evidence of > effort, I would say, but not evidence of a worthwhile result, and> never evidence of an effort to understand much beyond high-school > algebra. He agonizes over things, thinks about them, works at> them. But when it comes to learning any of the existing body of> mathematical theory, he is extremely lazy. No, I am not impressed.It's the theory that would keep him from learning to formalize hisargument. Sure, he diddles about with computations and twistedarguments, but he rarely learns much or puts in any real effort tofurther his understanding.-- Now I'm informing all of you that the people arguing against me are EVIL,yes they are real, live EVIL people as mathematics is that important, soit's important enough for Evil itself to send minions like them. -- James Harris on Evil's interest in Algebraic Number Theory === Subject: Re: JSH: how you can prove to the mathematicians you are rightthe website is certainly a tacit acknowledgment thatHales' proof was un'nished. I read about it in an expository book, andthe reason has nothing to do with computers per se (althoughI seem to recall that he was aiming to make some sortof metaproof ... a-ha -- new insightinto the James Harris Experience !-)> Consider this project, which aims to provide a formal proof of the Kepler> Conjecture:> > http://www.math.pitt.edu/~thales/¤yspeck/ as for productivity, forget it. we have a *huge* de'citwith China e.g., and the funding for the mortgagesof the housing bubble are investments from abroad (as wellas a lot of the bigger houses being those of wealthy foreigners). of course, a lot of this is moot, if we allow Cheenyto grind us into Tony's McCrusade (Usama's MacJihad). see my sig.--Give the World a Trickier Dick Cheeny -- out of of'ce after GIGA years.http://www.benfranklinbooks.com/http://larouchepub.com/ www.rand.org/publications/randreview/issues/rr.12.00/http:// members.tripod.com/~american_almanachttp://www.wlym.com/PDF-68 soros.html === Subject: Re: how you can prove to the mathematicians you are rightTim Smith this project, which aims to provide a formal proof of the Kepler> Conjecture:>> http://www.math.pitt.edu/~thales/¤yspeck/>> Why not produce a formal proof of your results? Such a proof would be> undeniable. Even those mathematicians who don't understand what you are> saying would have to accept that you are correct. Third parties, whether> mathematicians or not, could check your claims, by simply running yourproof> through the proof checker themselves.>That is until he discovers Ocore error in ¤yspeck'.l8r, Mike N. Christoff === Subject: Re: JSH: how you can prove to the mathematicians you are right> Consider this project, which aims to provide a formal proof of the Kepler> Conjecture:> > http://www.math.pitt.edu/~thales/¤yspeck/> > Why not produce a formal proof of your results? Such a proof would be> undeniable. Even those mathematicians who don't understand what you are> saying would have to accept that you are correct. Third parties, whether> mathematicians or not, could check your claims, by simply running your proof> through the proof checker themselves.I checked the link and it's fascinating, so I recommend others toconsider it as well.Now on to my second reply to your question. I realize that theproblem I'm facing may be translating *one* concept into today'smathematics, and I can give it to you in a couple of lines:The expression(f_1(x) + 1)(f_2(x) + 1) = 7P(x)where f_1(x), and f_2(x) are algebraic integer functions, meaning thatfor an algebraic integer x they give an algebraic integer, wheref_1(0) = f_2(0) = 0,and P(x) is a polynomial that is also an algebraic integer function,cannot be true in the ring of algebraic integers.If you know how to state a proof that mathematicians will accept thenplease do so!!!Unfortunately I don't know if there's an automated system out therewhich can handle such a statement, and I'm con'dent that I can proveit true, using basic things like the distributive property. But I'vehad arguments that have gone on for months, and months, and months onthe newsgroup here that lead me to think that mathematicians can'tcomprehend the methods in this area.Any ideas from *ANYONE* out there? I'll even read replies by peoplelike Nora Baron or David Ullrich, and I won't curse anyone in thisparticular thread.(Still might in others though.)James Harris === Subject: Re: JSH: how you can prove to the mathematicians you are right> > Any ideas from *ANYONE* out there? I'll even read replies by people> like Nora Baron or David Ullrich, and I won't curse anyone in this> particular thread.> > (Still might in others though.)Considering his attitude, I hope that neither of them (nor anyone else)will attempt to help the little weasel in any way whatsoever.-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: how you can prove to the mathematicians you are right> > The expression> > (f_1(x) + 1)(f_2(x) + 1) = 7P(x)> > where f_1(x), and f_2(x) are algebraic integer functions, meaning that> for an algebraic integer x they give an algebraic integer, where> > f_1(0) = f_2(0) = 0,> > and P(x) is a polynomial that is also an algebraic integer function,> > cannot be true in the ring of algebraic integers.> Well plugging in x=0 we get P(0)=1/7. As this is not analgebraic integer we are done. However, I suspect that thisis not exactly what you were looking for.Based on previous posts, I think what you want to prove is: The expression (f_1(x) + 7)(f_2(x) + 1) = 7P(x) where f_1(x), and f_2(x) are algebraic integer functions, meaning that for an algebraic integer x they give an algebraic integer, where f_1(0) = f_2(0) = 0, and P(x) is a polynomial that is also an algebraic integer function, and P(0)=1 can only be true in the ring of algebraic integers if f_1(x)/7 is an algebraic integer for all x.Unfortunately, a proof of this will be hard, because it is false. (it is true if we add the condition that f_1 and f_2 are polynomials [1]) - William Hughes[1] Actually, though I have made this claim before, I recently tried to prove it and failed. The claim is true for the ordinary integers, but there the proof involves prime factorization of the polynomial coef'cients (this is not going to work in the algebraic integers). I thought I had a way of substituting GCD's, but this fell apart when I tried to 'll in the details. I guess the statement it is true if we add the condition that f_1 and f_2 are polynomials should be classed as a conjecture (HnFnLC [2]). Can anyone provide a proof?[2] Hughes' neither First nor Last Conjecture === Subject: Re: JSH: how you can prove to the mathematicians you are right Adjunct Assistant Professor at the University of Montana.>Based on previous posts, I think what you want to prove is:> The expression> > (f_1(x) + 7)(f_2(x) + 1) = 7P(x)> > where f_1(x), and f_2(x) are algebraic integer functions, meaning that> for an algebraic integer x they give an algebraic integer, where> > f_1(0) = f_2(0) = 0,> > and P(x) is a polynomial that is also an algebraic integer function,> and P(0)=1> > can only be true in the ring of algebraic integers if f_1(x)/7> is an algebraic integer for all x.>>Unfortunately, a proof of this will be hard, because it is false. >(it is true if we add the condition that f_1 and f_2 are polynomials [1])>> - William Hughes>>[1] Actually, though I have made this claim before, I recently> tried to prove it and failed. The claim is true for the ordinary> integers, but there the proof involves prime factorization of> the polynomial coef'cients (this is not going to work in the> algebraic integers). I thought I had a way of substituting > GCD's, but this fell apart when I tried to 'll in the details.> I guess the statement it is true if we add the condition that> f_1 and f_2 are polynomials should be classed as a conjecture> (HnFnLC [2]). Can anyone provide a proof?>>[2] Hughes' neither First nor Last ConjectureYes. It follows from Dedekind's Prague Theorem, though I am sure BillDubuque will shortly post explaining that I am doing things the hardway (as usual):DEDEKIND'S PRAGUE THEOREM (Lemma 2 in Chapter 2, Section 5, of DavidHilbert's Zahlbericht) If the coef'cients a1,a2,....,b1,b2,... of twopolynomials in one variable F(x) = a1*x^r + a2*x^{r-1} + ... G(x) = b1*x^s + b2*x^{s-2} + ...are algebraic integers, and the coef'cients of the product F(x)*G(x) = c1*x^{r+s} + c2*x^{r+s-1} + ...are all divisible by the rational integer w, then each of the numbersa1*b1, a1*b2, a1*b3,...,a2*b1, a2*b2,... is also divisible by w.(Also proven Kronocker, Mertens, and Hurwitz).So assume we have (f_1(x) + 7)(f_2(x) + 1) = 7P(x)and f_1, f_2, and P are polynomials with algebraic integercoef'cients. Since 7 divides the coef'cients of the product, andf_2(x)=0, then for each coef'cient of f_1(x), 1 times thatcoef'cient must be a multiple of 7.-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu === Subject: Re: JSH: how you can prove to the mathematicians you are right> Consider this project, which aims to provide a formal proof of the Kepler> Conjecture:> > http://www.math.pitt.edu/~thales/¤yspeck/> > Why not produce a formal proof of your results? Such a proof would be> undeniable. Even those mathematicians who don't understand what you are> saying would have to accept that you are correct. Third parties, whether> mathematicians or not, could check your claims, by simply running your proof> through the proof checker themselves.> > I checked the link and it's fascinating, so I recommend others to> consider it as well.> > Now on to my second reply to your question. I realize that the> problem I'm facing may be translating *one* concept into today's> mathematics, and I can give it to you in a couple of lines:> > The expression> > (f_1(x) + 1)(f_2(x) + 1) = 7P(x) where f_1(x), and f_2(x) are algebraic integer functions, meaning that> for an algebraic integer x they give an algebraic integer, where> > f_1(0) = f_2(0) = 0,> > and P(x) is a polynomial that is also an algebraic integer function,> > cannot be true in the ring of algebraic integers.> This makes no sense at all.> If you know how to state a proof that mathematicians will accept then> please do so!!!> > Unfortunately I don't know if there's an automated system out there> which can handle such a statement, and I'm con'dent that I can prove> it true, using basic things like the distributive property. But I've> had arguments that have gone on for months, and months, and months on> the newsgroup here that lead me to think that mathematicians can't> comprehend the methods in this area.> > Any ideas from *ANYONE* out there? Try learning some math.> I'll even read replies by people> like Nora Baron or David Ullrich, and I won't curse anyone in this> particular thread.> If you cannot translate your ideas into language that a proof-checker can understand, you do not have a proof. Nora B.> (Still might in others though.)> > > James Harris === Subject: Re: JSH: how you can prove to the mathematicians you are right > Now on to my second reply to your question. I realize that the > problem I'm facing may be translating *one* concept into today's > mathematics, and I can give it to you in a couple of lines: The expression > > (f_1(x) + 1)(f_2(x) + 1) = 7P(x) > > where f_1(x), and f_2(x) are algebraic integer functions, meaning that > for an algebraic integer x they give an algebraic integer, where > f_1(0) = f_2(0) = 0, > > and P(x) is a polynomial that is also an algebraic integer function, cannot be true in the ring of algebraic integers. > > If you know how to state a proof that mathematicians will accept then > please do so!!!That is trivial. Set x = 0, on the left hand side we have 1, on theright hand side we have 7P(0), so P(0) = 1/7, which is not an algebraicinteger. But that is *quite* different from what you are attemptingto prove.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: how you can prove (f_1(x) + 1)(f_2(x) + 1) = 7P(x)>> where f_1(x), and f_2(x) are algebraic integer functions, meaning that> for an algebraic integer x they give an algebraic integer, where>> f_1(0) = f_2(0) = 0,>> and P(x) is a polynomial that is also an algebraic integer function,>> cannot be true in the ring of algebraic integers.Since this reduces to P(0) = 1/7, which is not an algebraic integer, the problem you have is to defend the assertion that P(x)*is* an algebraic integer function, given the evidence that it is *not*. However, even if you succeed in 'nding some what aroundthis contradiction it still won't shed any light on the *real* problem you have been grappling (unsuccessfully) with.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Re: JSH: how you can prove to the mathematicians you are right>> Consider this project, which aims to provide a formal proof of the Kepler>> Conjecture:>> >> http://www.math.pitt.edu/~thales/¤yspeck/>> Why not produce a formal proof of your results? Such a proof would be>> undeniable. Even those mathematicians who don't understand what you are>> saying would have to accept that you are correct. Third parties, whether>> mathematicians or not, could check your claims, by simply running your proof>> through the proof checker themselves.>>I checked the link and it's fascinating, so I recommend others to>consider it as well.>>Now on to my second reply to your question. I realize that the>problem I'm facing may be translating *one* concept into today's>mathematics, and I can give it to you in a couple of lines:>>The expression>>(f_1(x) + 1)(f_2(x) + 1) = 7P(x)>>where f_1(x), and f_2(x) are algebraic integer functions, meaning that>for an algebraic integer x they give an algebraic integer, where>>f_1(0) = f_2(0) = 0,>>and P(x) is a polynomial that is also an algebraic integer function,>>cannot be true in the ring of algebraic integers.>>If you know how to state a proof that mathematicians will accept then>please do so!!!>for x=0 you get 7P(0) = 1, so P(0) can't be and algebraic integer, soP(x) isn't an algebraic integer function-- Wim Benthem === Subject: Re: JSH: how you can prove to the mathematicians you are right>> Consider this project, which aims to provide a formal proof of the Kepler>> Conjecture:>> >> http://www.math.pitt.edu/~thales/¤yspeck/>> >> Why not produce a formal proof of your results? Such a proof would be>> undeniable. Even those mathematicians who don't understand what you are>> saying would have to accept that you are correct. Third parties, whether>> mathematicians or not, could check your claims, by simply running your proof>> through the proof checker themselves.>>I checked the link and it's fascinating, so I recommend others to>consider it as well.>>Now on to my second reply to your question. I realize that the>problem I'm facing may be translating *one* concept into today's>mathematics, and I can give it to you in a couple of lines:>>The expression>>(f_1(x) + 1)(f_2(x) + 1) = 7P(x)>>where f_1(x), and f_2(x) are algebraic integer functions, meaning that>for an algebraic integer x they give an algebraic integer, where>>f_1(0) = f_2(0) = 0,>>and P(x) is a polynomial that is also an algebraic integer function,>>cannot be true in the ring of algebraic integers.>>If you know how to state a proof that mathematicians will accept then>please do so!!!>You seem to be asking how to prove that P(0) = 1/7 is not analgebraic integer. Are you so far gone that you can't even do that for yourself?John Roberts-Jones === Subject: Re: JSH: how you can prove to the mathematicians you are right> Tim Smith this project, which aims to provide a formal proof of the Kepler> Conjecture:> > http://www.math.pitt.edu/~thales/¤yspeck/> > Why not produce a formal proof of your results? Such a proof would be> undeniable. Even those mathematicians who don't understand what you are> saying would have to accept that you are correct. Third parties, whether> mathematicians or not, could check your claims, by simply running your > proof> through the proof checker themselves.> > Good suggestion! I've actually threatened a couple of times to come> up with a formal proof that can be run through a computer, but haven't> really put any effort into it as of yet, as I didn't really have a> clue how to do the entire thing (and no links Keith Ramsay PLEASE).> > That's why I said threatened above as while I was serious when I said> I'd go out and do it, somehow I found myself never actually working at> it very hard.You have also threatened, but never followed through on, to prove all sorts of wild claims, from providing simple proofs of FLT onward and downward.> > I'll check the links I saw on your reference though to see if I can> start getting a better feel for how it can be done.> > > James Harris === Subject: Re: JSH: how you can prove to the mathematicians you are right>> Why not produce a formal proof of your results? Such a proof would be>> undeniable. > > Implying that proof-checking programs are infallible? How do you know> that?I beleive that the work of such programs is checkable by humans.-- --Tim Smith === Subject: because I've always thought> of the geometric (and negative binomial) as being the distribution of> the number of failures before the 'rst (some 'xed number of> successes). When I looked for a reference they are de'ned in terms of> the number of trials until some 'xed number of sucesses. Is it just> me (and my ancient Mood, Graybill and Boes) who have this view?I'm happy to report thathttp://planetmath.org/encyclopedia/ GeometricRandomVariable.html hasthe proper de'nition!Ian Smith === Subject: Columbia University uses False Arrest, Document Destruction!For explanation of the message title please see wesbiteshttp://www.cuspeech.org and === Subject: Re: Columbia University uses False Arrest, Document far to the Left that it will admitanything even vaguely human (short of a credentialed White male) andgive it a PhD for emulating Bush the Lesser's National Guardattendence record. Columbia Emeritus Professor Barton Sholod is mycousin. He routinely taught Spanish to spics who could not learnSpanish, and they graduated on time fully degreed without havingsuffered the hate burden of acquiring detectable education or ofpaying tuition out of pocket. They had compassionate scholarships.Anybody who cannot succeed in such a milieu is a phenomenon.Try CUNY instead - the standards are lower.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === Subject: Re: Columbia University uses False Arrest, Document Destruction!Distribution: worldCan you formulate in a few sentences what exactly is the problem?That would be helpful for those who don't take the effort of reading wholesites.Gr,Michiel Borkent === Subject: Re: Columbia University uses False Arrest, Document Destruction!Distribution: world> Can you formulate in a few sentences what exactly is the problem?> That would be helpful for those who don't take the effort of reading whole> sites.Essentially, he says that he was arrested because of email he sent toseveral people about a couple of members of the Columbia administration,including accusations that evidence from his student 'le had beendestroyed to prevent him from using it in an upcoming legal action.In addition to his arrest court orders were issued forbidding him tohave any contact with two of the people who received the email, as wellas the two administrators he named (and to whom he says he did not sendthe email). Now he also says that Columbia has contacted the INS andis trying to have him deported.I'm not taking any position here on the merits of his claims, justsummarizing them for you. He has posted copies of the emails, courtorders, and other documents on the web site so you can read them anddecide for yourself.-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Web Sites for Complexity Theory Complexity Classes,ftp://cs.nyu.edu/pub/local/yap/complexity-bk/Oded Goldreich, Introduction to Complexity Theory - Lecture Notesftp://ftp.eccc.uni-trier.de/pub/eccc/lectures/ IntroComplTh/all.ps.gzhttp://www.eccc.uni-trier.de/eccc/ Mendoza Rodriguezlet me=josejuanmr in let privacy=lycos in let net=es in me@privacy.net === Subject: Re: Web Sites for in message> Are there any good books in this 'eld, or good web sites that provide> an introduction or lecture notes with de'nitions and examples? I am> really trying to teach myself this stuff before I take the class next> semester.>> > While not a site to use to actually learn complexity theory, it does make a> great reference on known complexity classes.> > http://www.cs.berkeley.edu/~aaronson/zoo.html> > It contains literally hundreds of classes.> > Also, you may want to check out> > http://fortnow.com/lance/complog/> > a blog site by Lance Fortnow - a heavyweight in complexity theory research.> It contains more than a few tutorials on the subject.> > > > l8r, Mike N. ChristoffDo you know such good websites about Numbertheory ? === Subject: Re: Web Sites for Complexity Theory>Are there any good books in this 'eld, or good web sites that provide>an introduction or lecture notes with de'nitions and examples? I am>really trying to teach myself this stuff before I take the class next>semester.Try the course notes for 6.045J and 6.840J on MIT's OCW site(ocw.mit.edu). Follow the links to the Electrical Engineering andComputer Science course listings and from there to the course notes.--gregbogds at best dot com === Subject: Re: easy complex analysis problem......> > complex function f(z) = 1 / [2*{x^(1/2)}]> > int f(z) dz = ??> |z| = 1Is there supposed to be some connection between z and x? If so, what? === Subject: Questions - a bit weird1. Have any of you guys seen any paper in analysis which uses the variableiterations in a way similar to the way time is used in standardanalysis?2. If so, have (any of) the paper(s) used iterations in a === Subject: Re: 'led seven patents? You should have gotten reference> > numbers for this, right? What are they?> > (8) Unknown> Btw, the bbc does have a bureau in Delhi.Why don't you give them a call and show them yourdevice sticking off the ceiling?Also, you could report it to http://www.cnn.com/feedback/tips/.Lots === Subject: Sequence of Diophantine Equations by support1.mathforum.org (8.11.6/8.11.6/The Math with the followingproperty: for any z>M, there exist positive integers x,y_1,y_2,...,y_z such thatx^x=(y_1)^(y_1)+(y_2)^(y_2)+...+(y_z)^(y_z) ?(Please remark that the y's are >=1 and need not to benecessarily === Subject: Re: The role of in'nity in math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id constructivists regard as genuine mathematics only what can be >> obtained by a 'nite construction. The set of real numbers, or any >> other in'nite set, cannot so be obtained.>> This is denounced as pure mysticism by those who want to actually>be able to obtain things in mathematics by a 'nite construction.There are not only 'nitely many numbers in mathematics.The ultra'nitists simply pursue a model of the 'nite consideration of things in'nite.The limit approximation is a useful tool. It characterizes 'nitely some aspects of functional relations that imply their tractability in mainline calculist systems.The circle's not a regular polygon. A regular polygon can only approximate a circle, and not be one. One third can not be the solution of summing 'nitely many positive integer powers of 1/2, yet it can in'nitely.No classes in set theory.Consider the empty set, {}. Consider if that conceptualization is {;seogsergsgsbs; s...}, or that the empty set is the set of all sets, or not. If the set of all sets that are the set of all sets is the empty set, would not the empty set be the set of all sets?Ross F. === Subject: Re: Involutionary Calculus by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id following de'nitions:> > 1)The hyperderivative of a function f at a point x is de'ned as the > > limit as h approaches 0,of (f(x+h)/f(x))^(1/h),provided it exists.>> 2)The hyperintegral(or more properly the productal) of a function f over an > interval [a,b] over which it is positive is de'ned as the >> limit as n approaches in'nity,of the continued product as k varies from 0 to > n-1,of [f(a+(b-a)k/n)^1/n],provided it exists.>> It turns out to be that the two operations are related through following > theorem:> Suppose that F has f as its hyperderivative over some interval (a,b) and > continuous over [a,b] then> then the hyperintegral of f over [a,b] is F(b)/F(a).>> The calculus thus cerated has a structure analogous to the known calculus > and many theorems have their analogues.> It remains true,however,that the hyperderivative and the hyperintegral can > be expressed in terms of the known derivatives and known integrals.> If this idea is so much beautiful why is not it popular in mathematical > circles?Does it have any use? What good is it? What problems can it solve that are not otherwise solveable, if any? In what ways may it offer simpler explanations of phenomena explained by more convoluted expression?Solve a problem with it, and challenge others to solve the problem any other way.Ross F. === Subject: Integrating over open and closed ballsHi. I have a function in L^1(R^n). I'll use BC to denote the closedball of radius 1 centred at 0 and use BO to be the open ball of radiusone centred at 0. Is int_BC f = int_B0 f ? My intuition says yes,and came up with the following:int_B0 f = int_(0,1) int_S f(r,y)r^(n-1) ds(y) dr (changing topolar coords) = int_[0,1] int_S f(r,y)r^(n-1) ds(y) dr (since {0} and{1} have measure zero) = int_BC f (changing back to usual coords)Does this look correct? Is === Subject: Re: Integrating over open and closed balls>Hi. I have a function in L^1(R^n). I'll use BC to denote the closed>ball of radius 1 centred at 0 and use BO to be the open ball of radius>one centred at 0. Is int_BC f = int_B0 f ? My intuition says yes,>and came up with the following:>>int_B0 f = int_(0,1) int_S f(r,y)r^(n-1) ds(y) dr (changing to>polar coords)> = int_[0,1] int_S f(r,y)r^(n-1) ds(y) dr (since {0} and>{1} have measure zero)> = int_BC f (changing back to usual coords)>>Does this look correct? Is there a more obvious way to see Rodrigues === Subject: Re: complex problem...reloaded>> thus>> int [1/{2*e^(i@/2)}]*ie^i@ d@>> 0~2pi>> >> No; it's not 0~2pi, it's -pi~pi.>> >> >> ******0~2pi => -pi~pi correct??>> No; the function that you are interating is not periodic with period>> 2pi.>> >> = (i/2)*int e^(i@/2) d@>> >> = -2>> >> Yes, this is correct.> > I'm getting 2i for the answer. In the domain C (-oo,0], z^(1/2) is a > (holomorphic) antiderivative for f(z). So the integral should be> > lim_{z -> -1 from above} z^(1/2)> > - lim_{z -> -1 from below} z^(1/2)> > = i - (-i) = 2i.You're right, of course. By the method used by the OP, the result shouldhave been(i/2)*int(e^(i@/2),-pi <= @ <= Santos === Subject: prime relative to involutionDe'ne a natural number n to be hypercomposite if there exist two natural numbers a and b such that a^b=n (b is not 1),and hyperprime if no such numbers exist. Then, Every natural number is either a hyperprime or can be represented uniquely as the result of a left to right involution with a hyperprime base and normal prime exponents;e.g:81=(3^2)^2,256=((2^2)^2)^2. Every natural number is either a hyperprime or can be represented uniquely as the result of a right to left involution with hyperprime exponents;e.g: 81=3^(2^2),256=2^(2^3). Does any one 'nd this representation theorem important? The fact that there are two representation theorems is due to the fact that involution is non commutative. === Subject: Re: prime relative to involutionX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>De'ne a natural number n to be hypercomposite if there exist two natural numbers> a and b such that a^b=n (b is not 1),A hyperprime would usually just be called a power.>and hyperprime if no such numbers exist.> Then,> Every natural number is either a hyperprime or can be represented uniquely > as the result of a left to right involution with a hyperprime base and normal> prime exponents;e.g:81=(3^2)^2,256=((2^2)^2)^2.> Every natural number is either a hyperprime or can be represented uniquely> as the result of a right to left involution with hyperprime exponents;e.g:> 81=3^(2^2),256=2^(2^3).> > Does any one 'nd this representation theorem important?Who knows? At least the 'rst one is pretty obvious - it's just theprime factorization of the greatest common divisor of the exponentsappearing in the prime factorization of the original number. Haven'tthought about the second one...> The fact that there are two representation theorems is due to the fact that > involution is non commutative.************************David C. Ullrich === Subject: Re: Limit Of # Of Coprime Integers by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 10:08:29 -0500Leroy,For the original n=2 case, the multiplier out front should be1/m^3 rather than 1/m^2, of course.But also the numerical value seems wrong; for the limit I getproduct ( (1-1/p) + (1/p)*(1-1/p)^2 ) = 0.428...(the product being taken over all primes)while for the expression with zeta(3) I get12/pi^2 - 1/zeta(3) = 0.383...The factor in the in'nite product simpli'es to1 - 2/p^2 + 1/p^3which is unfortunate. If it could be expressed as products offactors of the form1 - p^kthen we could write the in'nite product in terms of1/zeta(k)but it can't be so expressed, so I'm at a loss.For general n, the in'nite product would be insteadproduct ( (1-1/p) + (1/p)*(1-1/p)^n )Don Coppersmith>Let H(m;k,j) = the number of positive integers <= m which are coprime>to *both* k and j.>>(So, for instance, H(m;k,j) = H(m;kj,1).) >>(And H(kj;j,k) = phi(jk), the Euler phi function.) > >>I believe that>>limit{m-> oo}>> m m > --- ---> 1 --- > > H(m;k,j)>m^2 / /> --- ---> k=1 j=1> 12 1>= ---- - ------- ,> pi^2 zeta(3)>which is, in linear-mode,>limit{m-> oo}>>(1/m^2) *sum{k=1 to m} sum{j=1 to m} H(m;k,j)>= 12/pi^2 - 1/zeta(3).>>(zeta(r) is, generally, sum{j=1 to oo} 1/k^r.)>>And, more generally,>limit{m-> oo} >> m m m > --- --- ---> 1 ------- > > ... > H(m;k_1,k_2,..,k_n)>m^(n+1) / / /> --- --- ---> k_1=1 k_2=1 k_n=1>> n+1> ---> 6*n 1>= ---- - / ------- ,> pi^2 --- zeta(j)> j=3>which is, in linear-mode:>>limit{m-> oo} >>(1/m^(1+n))*> sum{k_1=1 to m} sum{k_2=1 to m}...sum{k_n=1 to m} H(m;k_1,k_2,..,k_n)>= 6*n/pi^2 - sum{j=3 to n+1} 1/zeta(j) .>And H(m;k_1,k_2,..,k_n) =>>the number of integers, j, coprime with (k_1)*(k_2)*(k_3)*...*(k_n)>and such that 1 <= j <= m.>Am I Quet === Subject: Re: Riemann Hypothesis and P vs NP> from Chapter 10 of The Music of the Primes by Marcus du Sautoy:> > If the Riemann Hypothesis is true, then there is a fast way to> discover the primes used to build the RSA codes on which the security> of e-business currently relies.This sentence is true, strictly speaking, but it suggests to me that theauthor has misunderstood some aspects of this:We have had, for some time, fast ways to discover the primes used tobuild RSA codes. Namely, fast randomized primality testing algorithmssuch as that of Solovay and Strassen [4] or Rabin [3].I guess that du Sautoy is thinking of Miller's proof that if theExtended Riemann hypothesis is true, then there are polynomial timedeterministic tests for primality [2]. This is fascinating from atheoretical point of view, but in practice it would make no difference:the randomized algorithms are good enough.In any case, there is a polynomial-time primality test that doesn'tdepend on the Riemann Hypothesis [1]. But perhaps this is too recent(2002) to have made it into the book (published doesn't affect the security of RSA:you need a faster factorization algorithm for that.[1] M. Agrawal, N. Kayal and N. Saxena, PRIMES is in P, 2002. http://citeseer.nj.nec.com/agrawal02primes.html[2] L. Miller. Riemann's hypothesis and tests for primality. J. Comput. Sys. Sci., 13:300--317, 1976.[3] M. O. Rabin. Probabilistic algorithm for testing primality. J. Number Theory, 12:128--138, 1980.[4] R. Solovay and V. Strassen. A Computing, 6:84--86, 1977.-- Gareth Rees === Subject: Re: e is transcendental (was: classes of transcendental numbers ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 10:28:50 -0500>panamars@otenet.gr (Eur Ing Panagiotis relationship e^[ipi]+1=0 , or>>e^[ipi]=-1 >My comments>>--------------->>Since e^[iPi]=cosPi+isinPi>>or , e^[iPi]=-1+i[0]>>then there are two solutions here, to the given equatio:>A) e^[ipi]=-1 the real part solution and >B) e^[ipi]=i[0] , or e^[ipi]=0 the imaginary part solutio.>>The reasoning behind A and B is not evident to me. Please explain>where B comes from. e^[ipi]=0 is The solution if the imaginary part. Just think that angle is 179 degthe real component is cos179 = -0999847695and the imaginary is i*sin179=i*0.017452406e^[iPi179/180]= -0999847695 +i*0.017452406in polar form this will be e^[iPi179/180]=mod 1, arg 179 e^[i pi] = -1 is right of course, but the last >step that you mentioned in your derivation looks completely >unjusti'ed.>My question :>>What is the implication of this second value of e^[ipi]=0 ?>>No implication. It is not true. Panagiotis Stefanides http://www.stefanides.gr/why_logarithm.htm >>David McAnally>> know about the subject. There are lots of compact>minimal surfaces with that are Euclidean without being ¤at planes. >There are fully *eight* simply-connected geometric 3-manifolds with>compact quotients,You're right, I don't, you tactless crybaby! This is a bull courseto ful'll my science requirement, and Seeds' analogies are almost aspoor as your exposition. You have the worst case of professor'ssyndrome I have ever seen.The author was obviously comparing the area of circles on his planarexamples to some other area, but HE DIDN'T SPECIFY in what type ofdimension. The surface area enclosed by a circle will obviously besome pi * r^2, since it can be calculated by taking an equation of itscross-section to the perimeter of the circle and then rotating itabout an axis. Basic calculus. To what area is Seeds comparing this? Idon't understand why this is such a hard question. === Subject: easy...analysis problem....hello.......genius teacher........let function f :E -> R is differentiable on open set E in R1. let any a in E , f''(a) exist.show that f''(a) = lim {f(a+h) - 2f(a) + f(a-h) / (h^2)} , h->02. let any a in E , n-derivative f^(n) (a) exist.express f^(n) (a) by limit of f(x)-------------------------um.....i solved 'rst problem.lim {f(a+h) - 2f(a) + f(a-h) / (h^2)}h->0= lim [{f(a+h)-f(a)} - {f(a-h+h)-f(a-h)} / (h^2)]= lim {f'(a) - f'(a-h)} / h , let h=(-t)= lim {f'(a) - f'(a+t)} / (-t)t->0= lim {f'(a+t) - f'(a)} / t= f''(a)it's right??but i can't solve second problem....let me advice, please.....sir~ === Subject: radiation view factor would be between a verticalplate and the ground. Assume that the plate is resting on the ground andthere are no obstructions, that is, the plate sees only the ground and sky.It seems like it should be 0.5 for the plate-to-ground VF and 0.5 for theplate-to-sky VF, doesn't it? But a program that I use to compute viewfactors tells me that the plate-to-ground VF is only 0.4337. The programcomputes other view factors (like perpendicular equal-area plates joined atan edge) correctly. I know because I can check those sci.physics === Subject: Transcendental or algebraic?If x^x=2 then x is transcendental or algebraic? === Subject: Re: Transcendental or algebraic?> If x^x=2 then x is transcendental or algebraic?Let's examine three possibilities:x is rational(p/q)^(p/q) = 2 leads to (p^p)/(q^p) = 2^qUsing just the fact that 2 is an integer implies that the lowest-termsp/q has q=1. There is no solution this way.(I would go a little further: If x and x^x are both rational, x must bean integer.)x is algebraic and irrationalLook up Hilbert's Seventh Problem. x^x would end up transcendentalrather than 2.x is transcendentalThis is the only possibility left.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Transcendental or algebraic?>If x^x=2 then x is transcendental or algebraic?>My best guess is transcendental because the equation directly transforms to :LOG(2) = x, using base of x for the logarithm function. [would this be properly typed in text form as LOG_x (2) = x ?]G C === Subject: Re: Transcendental or algebraic? then x is transcendental or algebraic?>> > My best guess is transcendental because the equation directly transforms to :> LOG(2) = x, using base of x for the logarithm function. and this tells us...?J> > [would this be properly typed in text form as LOG_x (2) = x ?]> > G C> -- Jim Nastos, B.Math, B.Ed | Of'ce: 117 Athabasca HallMSc Candidate | Of'ce Phone: (780) 492-5046University Of Alberta | Edmonton, AlbertaDepartment of Computing Science | T6G 2H1nastos@cs.ualberta.ca |http://www.cs.ualberta.ca/people/grad/nastos.html === Subject: Re: My Childhood and In'nity> I ask you what is 2 times in'nity? In'nity, you might say, but that> doesn't make any sense and you know it. I am claiming that there is no> evidence for the Peano axioms that say that every number has a> successor! It seems clear to me that no one has come close to proving> this. On the other hand, there is abundant evidence for the number M> existing, i.e., Asimov's number. And 2M cannot exist by de'nition.If every number doesn't necessarily have a successor, can there only beone such number? And would that number be M? Is it possible that therecould be numbers without successors and yet not be the largest number?For instance, if 6 did not have a successor, would that preclude the existence of 8?Suppose I told you there were 3000000 people living in Chicago and toprove it, I gave you the address of Mr. 3000000. You could then driveover to his house and verify that he exists. But if I understand yourargument, I would need to drive over to the houses of the other 2999999people to prove that Mr. 3000000 is actually the 3000000th person.So if I gave you the driving instructions to the house of Mr. FirstSixth Generation Type 21211 Mersenne Hailstone, you could get in yourCollatzmobile, put in reverse and drive to his house (the address ofwhich has over 53000 digits, so it's a bit larger than Asimov's number).Now you certainly can't deny the existence of Mr. FSGT21211MH (as he's known to his friends) because you have explicit instructions on how toget there (the Collatz sequence). But if every number doesn't have a successor, I really don't know how many neighbors he has? Is that thenub of your gist?> > > Dr. Ben Zona === Subject: Re: My Childhood and In'nity> the largest number possible which I now call M.Suppose that your M exists. If M > 1, then M^2 > M, but that wouldimply the existence of a number (M^2) which is greater than thelargest number (M).So M has to be less or equal to 1. But 1 is the greatest number lessor equal to 1, therefore M=1. You don't to search === anymore.Subject: Re: My Childhood and Under NO circumstances will postings containing illegal or copyrighted material through this service be tolerated!! And I always thought the largest number was G! You say it's M? Damn! How could I have been wrong all these years?DaveROn 9 Feinstein)>When I was a young three and a half year old lad, I used to write>numbers on paper and line them across the room. My goal was to get to>the largest number possible which I now call M. Then my mother told>me that the largest number is in'nity. I asked her what things are>in'nite. She could not answer me. Then I asked her what do you mean>goes on for ever? How can something go on for ever? She could not>answer me and told me to ask my teacher, as that is her job.>>So the next day I went to nursery school and was playing on the swing>and I bumped my nose and started crying. The nursery school teacher>saw me crying and tried to make me feel better. I decided at that>point that it would be a good opportunity to ask her what in'nity is.>So I did. But she didn't understand me because I was still crying.>>For years, I blocked this out of my memory, this disturbing incident.>And I went to college and eventually got a PhD from the University of>San Moritz, a non-accredited but well-respected university, writing my>thesis on the role of life experience in scienti'c thought. partial>differential equations and started to shake. What I was doing was>meaningless! What my teachers told me was a lie! Partial differential>equations presuppose the existence of in'nity! And how do I know that>there is an in'nity? No one knows in fact. Then I remembered myself>as a child lining up numbers trying to 'nd M, the largest number in>the universe. I decided that I would convert my partial differential>equation paper about difference equations instead. Then my life>changed. I decided that I would pick up the pieces of where I left off>as a child and renew my search for M. I am currently using my computer>to 'nd this number, as I have concluded that this is not a job for a>human being.>>I am looking for collaborators for this big project. Anyone who wants>to participate in this revolution, literally, in mathematics and>science should please contact me. I can be found at my brother-in-law>Craig Feinstein's email address. Just do not make it known to him that>I am using it, as he thinks it is not healthy for me to continue this>project, and that I should just do what my doctor says.>>Dr. Ben Zona === Subject: Re: My Childhood and In'nity>> You were ingorant as a child. You are stupid as an adult. How big is>> 2M, git?>You missed the point of my argument entirely. 2M cannot exist.> > I feel this is a good point; as, if M exists, M is the largest number,> then 2M is indeed nonsense.> > Hisanobu ShinyaWell, suppose this M is an integer (the same argument works with therationals and the reals). Since the integers form a group undermultiplication, the product ab must be in Z if a and b are in Z. Alittle substitution gives... 2M in Z. If we put the standard orderon Z, then M<2M, assuming M is positive. Since the OP does not wantthis, we can only conclude that he intends M to be something otherthan an integer, in which case M does not have the same logical statusas elements of Z. Perhaps he's thinking of non-archimedean models ofarithmetic? (Doubtful. He obviously doesn't have the mathematicalsophistication necessary.)'cid Oooh === Subject: Re: The Clearest by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id proof of Goldbach's conjecture does not work.The heart of the proposed proof is the assertion (Lemma 2.5, second part) that for each large odd q,there is no setting of residues r_i corresponding to the primes p_i < q/2,satisfying the following two conditions:(1) for each i, there are at least two odd integers less than q in the residue class (x = r_i mod p_i);(2) each odd prime p 13 + 0.245 * pi(q).Consider the 'rst six odd primes: p_1=3, p_2=5, p_3=7, p_4=11,p_5=13, p_6=17, which I will call the small primes.Each odd prime p_i between 18 and q is not divisible by any of thesesix small pimes; these p_i will be the large primes.The primes p_i between 18 and (q/4 -2) will be called medium primes; some primes are both medium and large. Consider a choice of residue classes r_i unequal to 0 modulo p_i for these six small primes.The number of choices of such residue classes is (3-1)*(5-1)*(7-1)*(11-1)*(13-1)*(17-1)=92160.For each such choice of (r_1,...,r_6), some fraction of the largeprimes p_i fall outside all six residue classes.The average value of these fractions (averaged over the 92160 choices) is (3-2)*(5-2)*(7-2)*(11-2)*(13-2)*(17-2)/92160 = 22275/92160 < 0.2417.So we can select one setting of (r_1,...,r_6) for which the fraction of large primes p_i falling outside all six residue classes is less than 0.2417.(In fact for q suf'ciently large, each choice of (r_1,...,r_6)gives a fraction suf'ciently close to 22275/92160.)So these six odd primes, and their associated residue classes r_i mod p_i, hit a fraction at least 1-0.2417=0.7583 of the large primes p_i.For each of the remaining large primes p_k (fewer than 0.245(q/log q) of them), and the six small primes,assign one of the medium primes p_i, and select r_i so that p_k = r_i mod p_i.Since p_i 6+0.245*pi(q), so there are enough medium primes to assign to theleftover large primes and the six small primes.The upshot is that we have chosen residue classes (r_i mod p_i) for the small and medium primes, such that each (small or large) prime falls into one of the residue classes; this is the F whose existence Lemma 2.5 (part 2) denies.Don Coppersmith >In http:// www.geocities.com/erdosfan/solution.html the new Let>me ask the questions here. ...>>Hisanobu Shinya === Subject: Re: Sine without approximating> Is it possible to solve:> sin x = x/2> Without graphing, and without approximating?? Someone already mentioned the connection with Kepler's equation andfunctions related to it. My remark: The inquiry rejects two approaches (graph and approximation)but does not postulate acceptable options. (Presumably, a 'nitecomposition of arithmetical operations and functions from a list includingelementary functions and perhaps some other, unspeci'ed ones.) So: (1) How do we obtain values of sin(x)? By a well-known in'nite series,or in'nite product, with shortcuts and improvements that take advantageof periodicity, symmetry and addition formulas. In essence, it's anin'nite process truncated after required precision is achieved. Can yousay approximation? (2) If we do accept programmable in'nite series or products with errorcontrol, where do we draw the line? Why stop there? What would be wrongwith in'nite function composition? That's another name for processes thatinclude in'nite continued fractions and iteratively formed sequences. Ad (2): the positive solution of sin(x)=x/2 can be obtained, as mostreaders know anyway, by the in'nite composition of the functiong(x)=2*sin(x) with itself, applied to, say, x=2. Or for the less patient,you compose h(x) = x - (x - === Subject: Re: Sine without approximating> Is it possible to solve:> sin x = x/2> Without graphing, and without approximating??Of course, Fishfry is correct about one value of x at least.x = 0 IS a solution.;)But there is also a *positive* (and negative) value === Subject: Re: Sine without approximating> Is it possible to solve:> sin x = x/2> Without graphing, and without approximating??What are the units of Ox', degrees, radians, etc? === Subject: Re: Sine without message> Is it possible to solve:> sin x = x/2> Without graphing, and without approximating??>> What are the units of Ox', degrees, radians, etc?The units are radians. === Subject: Re: Sine without approximating> Is it possible to solve:> sin x = x/2> Without graphing, and without approximating??Only semi-seriously:x = -i*W(exp(2ix)-1),where W() is the Lambert W function (one value of W() here, more precisely).But, alas, x Quet === Subject: Re: Sine without approximating> >> Is it possible to solve:>> sin x = x/2>> Without graphing, and without approximating??>> >> > > Well, x=0, 2pi, 4pi, 6pi, ... come to mind.> > Are you saying that sin(2pi) = pi?Oops. Late night posting. === Subject: Oh, how I wonder...To me, integration makes perfectly sense but,Where is the irrasionality of irrasional numbers?They seem to be fully rational.Where is the inductive logic in mathematical induction?It seems to be deduction.Whats primary with prime numbers?1 seems to be more primary.Whats imaginary with imaginary numbers?-1 seems to be the only imaginary number.What is platonic with platonic 'gures?They seem to be of this world.Whats defferentiated during differensiation?No difference is attempted found. === Subject: Re: Oh, how I wonder...>Where is the irrasionality of irrasional numbers?>They seem to be fully rational.They're not ratios (of whole numbers). Hence ir-ratio-nal.>Where is the inductive logic in mathematical induction?>It seems to be deduction.>>Whats primary with prime numbers?>1 seems to be more primary.>>Whats imaginary with imaginary numbers?>-1 seems to be the only imaginary number.>>What is platonic with platonic 'gures?>They seem to be of this world.>>Whats defferentiated during differensiation?>No difference is attempted found.It is a mistake to think that the common de'nitions of thewords used in mathematics will tell us something about thestructure of the mathematical objects. They have a structure,independent of our language(s). We make clear what structurewe wish to study by giving a formal de'nition. Now, we havethe OPTION of using a long name for these things; for examplewe can refer to convex,face-congruent,face-regular polyhedronif we wish. (That would be the chemists' approach.) But wedon't like that mouthful, and we have a record that Plato mentioned them, so we OPT to call them Platonic solids.That doesn't imbue them with any additional property beyondwhat is already a consequence of their de'nition.Sometimes the choices we have collectively made turn out tobe good ones, sometimes not so good. Imaginary number wasprobably a poor choice, in retrospect. Likewise transcendental.Regular topological space was uninspired. Perfect number was great PR for a pretty unimportant idea; I might put chaos in this category too. Spectral sequence would play well in the popular press if they could only get a sense of what to do with it.But they're all just names. The trick is to learn the actualde'nition and see the consequences of that de'nition. A simple group by any other name would be as cool.dave === Subject: Re: Riemann Hypothesis and P vs NP by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, Somwhere in the>> chapter the author mentions P versus NP since factorising numbers is a>> dif'cult problem (personally, I don't know if it is NP-hard or>> NP-complete).It is not known to be in either set. I think it unlikely thatit is NP-Complete or NP-Hard. If it were, it would mean that thereare no truly exponential algorithms in NP, since factoring is alreadyknown to be sub-exponential. I have insuf'cient knowledge to saywhether factoring might be in P. I think it unlikely, but Riemann Hypothesis is true, then there is a fast way to>> discover the primes used to build the RSA codes on which the security>> of e-business currently relies.This is true, but vacuous. It is vacuous in the sense thatthere is a fast way to compute the primes to build an RSA key*regardless* of whether GRH is true or false. This has been truefor many years. Both Maurer's algorithm and the Shawe-Tayloralgorithm give fast methods for generating fully proved primes.(and Preda Mihailescu and I developed yet a faster method.)Note that I said GRH and not RH. RH alone does not suf'ce. Weneed a bound on the smallest primitive root of a prime and thisrequires information about general characters of Z/pZ*, thusrequring info about L-series in general and not just RH.Please note that the recent Agrawal et.al. method (AKS) is ALSOirrelevent to whether we can quickly generate primes. AKS gives aP-Time algorithm for *testing* any randomly chosen integer forprimality. But this is a different problem from *generating* primes. The latter is in P and does not depend on AKS or RH. Its runtime does depend on the PNT for A.P.'s. >> My question is: If the RH is proved to be true does it automatically>> follow that P=NP?No. AFAIK there is no established connect between the two problems.Such a connection, if it exists, would be truly amazing.>>There seems (to me) to be a ¤aw in the author's argument.>>Either the Riemann Hypothesis is true or it is not true.Correct.>If one were to assume that it were true, from which it automatically>follows that there is a fast way to discover the primes (from the>author's argument).The truth of GRH (not RH by itself) gives an algorithm whose runtime is O(log^3 N * M(N)) for proving that an integer N is prime.M(N) is the time needed to multiply integers mod N. Work by EricBach gives an effective constant of 2 for the O() estimate.Maurer and Shawe-Taylor can generated fully proved primes inO(log^2 N M(N)). [i.e. faster!]The complexity of AKS seems to be O(log^4 N M(N))Note that the complexity of M(N) depends upon whether one uses aclassical algorithm, giving M(N) = O(log^2 N) or convolution methodsgiving M(N) = O(log N loglog N logloglog N).>>For any given number that needs to be factored, this fast way could be>used, and then the results quickly veri'ed. You are confusing the problem of generating the primes for an RSA modulus with the problem of *factoring* that modulus. >If the results are>incorrect (or if it is not fast), and the author's assumption is>true, then the RH is false.Huh? This makes no sense to me. Please explain further.>>So, you can either crack every code fairly quickly or you can disprove>the RH. Either one sounds like a huge advance.Once again. Generating primes has NOTHING to do with the problemof breaking an RSA key. We have fast algorithms for building anRSA key. We always have had them. The algorithms we have for factoring the modulus have nothing to do with whether RH/GRH istrue or false. (AFAAK)A proof of GRH would give us yet another (in addition to AKS) proofthat PRIMES is in P. It would give us a faster method than AKS.But it would still be slower than methods currently in use. And it isunlikely that it would have an impact on the security of RSA keys.[it is always possible that some technique discovered while provingGRH might lead to a fast factoring algorithm. But it seems unlikely].>>Perhaps the author is intimating that the proof of the RH should>reveal a fast way to discover the primes, but I don't see how this>follows.A proof of GRH *would* give a faster (than AKS) P-Time algorithmfor proving random integers prime. But since we can already computeprimes faster it is IRRELEVANT.I have discussed this many times in this newsgroups, as well as insci.crypt. Why do we keep hearing the same confused claims?Bob SilvermanC.S. Draper Laboratory === Subject: Re: Resolution to Decker Quadratic Issue> Turns out there's another approach to prove a problem with the old> concepts about the ring of algebraic integers.> > Decker put forward the quadratic> > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > > where his a's are roots of > > a^2 - (x - 1)a + 7(x^2 + x).> > Letting x=2, you have> > a_1(2)^2 - a_1(2) + 42 = 0, which gives > > a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.> > Now consider the quadratic > > y^2 - by - 7 = 0, which has as one of its roots > > (b + sqrt(b^2 + 28))/2.> > Note that root is an algebraic integer factor of 7 for all algebraic> integers y, and b.> > Now consider> > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2> > which is> > (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z> > > and I can divide both sides by 4 to 'nally get> > 49 z^4 + 7b z^3 + (42b^2 - 581) z^2 - 125bz + 1764 = 0.>Not quite. Not that it affects your argument, but you actually get 7z^4 + bz^3 +(6b^2 + 83)z^2 - 6bz + 252 = 0 > Importantly, for any integer b, such that it is irreducible over Q,> *none* of the solutions for z can be an algebraic integer!> Okay. So for the b's that make the polynomial above irreducible,you've shown that (b + sqrt(b^2 + 28))/2doesn't divide (1 + sqrt(-167))/2in the algebraic integers. Not much of a surprise, in spite ofthe fact that the former divides 7 and the latter divides 42.> Now then, imagine that there exists some algebraic integer b for which> it is reducible over Q, then the root will be a fraction with a 7 or> 49 in the denominator.> > > So consider the root c/7, which gives> > (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))c/7,> > which would force (1 + sqrt(-167)) to be coprime to 7.>Why would that be? > Therefore, (1 + sqrt(-167)) has no factors in common with 7 in the> ring of algebraic integers!!!> But that's simply not true, as Dik reminded us in an earlier response.> === Subject: Elliptic curvesIf we set X = {(x,y) in C^2 | y^2 = x^3 + ax + b} and Y = {(x,z) inC^2 | z^2 = bx^4 + ax^3 + x} (z = x^2y) and de'ne X' and Y' byremoving points with x = 0. One can create a bijection X' --> Y' by(x,y) |-> (1/x,x^2y) and if one glues together X and Y along X' and Y'one gets something like when one completes C by the Riemann sphere.Question is now, how do I visualize this one? I have no background intopology. === Subject: Re: JSH: Open letter to Jim Ferry> I'm intrigued by the questions raised by your recent posts, as for> years you were this guy who came up with rather creative ways to> insult me, and now I 'nd it hard not to 'gure you're just doing so> again.Is it okay to ask you questions, talk in any familiar way, or in any wayact as if I wish you to reply or am addressing you?Those proscriptions apply only to David Ullrich, Virgil, and anyone whoposts under a palindromic pseudonym, right?Okay then. You ask a good question. Just what is my intent? I'm notsure precisely why I'm doing what I'm doing, but I'll try to answer youearnestly and respectfully.> Now I'll embarrass you a bit as from what I've read on the web you> have one of the highest IQ's out there, so it seems to me there's> probably some reason to what you're doing, and possibly I'm wrong> about what it is.This does embarrass me because I'm certainly not one of the big 'sh onsci.math. You must be basing this statement on the fact that I oncejoined something called Mega, which purports to be a high-IQ society forthose of 1-in-a-million intelligence. I now realize that by joining, Iwas implicitly making this arrogant claim about myself, but I reject thatclaim. The fact that I was able to ace the math part of their test justindicates that it wasn't hard enough, because it's easy to 'nd peoplebetter at math than I. Indeed, you can 'nd lots of them on sci.math.And some of them even take the time to analyze your work, James.> Therefore, I'm going to give you the opportunity that I feel *I* don't> get, which is the bene't of the doubt.You're asking me to clarify me position, which I am about to do. I thinkit's fair to say, however, that others have given you the same opportunity,i.e., that they've asked you to clarify your position.> Tell me succinctly and in a way that will minimize potential> embarrassment for both of us, what it is that your up to, and no, none> of this wild stuff about how great I supposedly am, or how I've proven> FLT or any of that, as I just want you to say something that 'ts into> a worldview that makes sense.> > What's your intent?You think that I'm mocking you, but you're not entirely sure. DavidUllrich thinks it's incredible that it's not obvious to you that I'mmocking you.Well let me clear things up. Yes, I'm mocking you. I've made a seriesof posts over the last six months in which I've appeared to be convertedto a religion in which you are the Messiah of Mathematical Truth beingcruci'ed my the benighted masses. Most people consider that absurd andtherefore conclude that I must have been being sarcastic. You, however,do not consider the idea that you are the Messiah of Mathematical Truthto be absurd. You consider it to be essentially correct -- a little offsomehow: a little over the top, or emotionally overblown, but basicallythe correct attitude to take.Long ago I decided that creative mockery is the best response to yourpresence on sci.math. So I have tried to 'nd new and better waysto mock you. My latest effort has been the most elaborate, as wellas the most successful (in my opinion). But it has forced me toput myself in your mindset -- a form of Method acting. And in thismindset, I perceived how abusively this newsgroup treats you, and itmade me angry. The abuse that people heap upon you is sickening andwrong.I know that this makes me sound like a hypocrite, but just because Imock you does not mean that I'm against you. You may not be able tofathom this, but I honestly feel that I'm on your side. I want youto succeed, and the way you can succeed is this: to join me in mockingyou!Learn to laugh at yourself, James! Your attitude is really quiteamusing, quite ridiculous! Normal people get continuous feedback abouttheir strengths and weaknesses and adjust their self-image accordingly,but you've decided to suppress any negatives, and exaggerate the positivesto such a ridiculous extent that you've turned yourself into a runawayself-esteem engine. But you don't perceive the self-esteem: what itfeels like to you is constant self-doubt because the pressure of allthe negative feedback you suppress is constantly leaking in. So youpump the bellows faster and faster! Yee-hah! Here comes the NarcissisticPersonality Disorder Express! Look at Oer go! Ridiculous! Laughable!Don't you see how funny it is?In fact, it's so humorous, it would be worth you getting treatment foryour NPD just so you can see how magni'cently ridiculous you've been!There's no point in trying to hide from your ridiculousness, and there'sno need either. There is a certain 'nite amount of ridiculousness whereshame is maximized. You've gone far, far beyond that -- you've traveledinto realms of ridiculousness where shame is no longer a factor, or hasachieved some negative value called pride. Be proud of your absurdity!Take a bow, James. You're spectacular!But I'm getting ahead of myself. First you need to stop the absurditybefore you can look back and laugh at it. Go to a mental healthprofessional. Tell them you have NPD, and if they don't believe youtell them to give you a test. Then get treatment. And then, acceptwhat you've been up until now: spectacularly ridiculous! fantasticallyabsurd! Laugh at yourself for about six weeks straight. And then startbuilding yourself a new and better life as a normal person who's notconstantly obsessed with being better or worse than everyone else.Peace be with you,-Jim Ferry === Subject: Re: JSH: Open letter to Jim FerryX-DMCA-Noti'cations: http://www.giganews.com/info/dmca.html>I'm intrigued by the questions raised by your recent posts, as for>years you were this guy who came up with rather creative ways to>insult me, and now I 'nd it hard not to 'gure you're just doing so>again.>>Now I'll embarrass you a bit as from what I've read on the web you>have one of the highest IQ's out there, so it seems to me there's>probably some reason to what you're doing, and possibly I'm wrong>about what it is.>>Therefore, I'm going to give you the opportunity that I feel *I* don't>get, which is the bene't of the doubt.>>Tell me succinctly and in a way that will minimize potential>embarrassment for both of us, what it is that your up to, and no, none>of this wild stuff about how great I supposedly am, or how I've proven>FLT or any of that, as I just want you to say something that 'ts into>a worldview that makes sense.>>What's your intent?Sorry to break this to you, but he's just making fun of you withhis posts. The fact that this is not obvious to you seems simplyincredible.>James Harris************************David C. Ullrich === Subject: Re: JSH: Open letter to Jim Ferry> > Sorry to break this to you, but he's just making fun of you with> his posts. The fact that this is not obvious to you seems simply> incredible.Ahh, don't spoil our fun by letting him in on the joke! Some of usmight have money riding on how long it takes him to catch on...Of course, that raises the question, How do you collect on a bet o'n'nite length?-- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Open letter to Jim Ferry> I'm intrigued by the questions raised by your recent posts, as for> years you were this guy who came up with rather creative ways to> insult me, and now I 'nd it hard not to 'gure you're just doing so> again.> > Now I'll embarrass you a bit as from what I've read on the web you> have one of the highest IQ's out there, so it seems to me there's> probably some reason to what you're doing, and possibly I'm wrong> about what it is.> > Therefore, I'm going to give you the opportunity that I feel *I* don't> get, which is the bene't of the doubt.> > Tell me succinctly and in a way that will minimize potential> embarrassment for both of us, what it is that your up to, and no, none> of this wild stuff about how great I supposedly am, or how I've proven> FLT or any of that, as I just want you to say something that 'ts into> a worldview that makes sense.A worldview that makes sense and a worldview that JSH will accept are incompossible. I'm having a bit of trouble solving a recursion relation (or evengetting a reasonable lower limit)The relation is:S(1,m)=mS(n+1,m)=sum (i=1 to m-1) of S(n,i)I am speci'cally interested in S(n,n+1) and believe that this will beexponential in n but am unable to prove it :(Any suggestions? === Subject: trouble solving a recursion relation (or even> getting a reasonable lower limit)> > The relation is:> > S(1,m)=m> S(n+1,m)=sum (i=1 to m-1) of S(n,i)> > I am speci'cally interested in S(n,n+1) and believe that this will be> exponential in n but am unable to prove it :(> > Any suggestions?> > How about binomial coef'cient?? S(i,m) = m choose iThen S(n,n+1) = n+1 choose n = n+1, not exponential. .... > S(1,m)=m> S(n+1,m)=sum (i=1 to m-1) of S(n,i)> .... > > How about binomial coef'cient?? S(i,m) = m choose i> .... Perhaps Chris could try proving by induction on i that this is in fact the unique solution. Ken Pledger. === Subject: Re: there positive in'nity = 2147483647 and negative in'nity =-2147483648. Weird thing is 2*(positive in'nity) = 0. Where should I publish my 'ndings? === Subject: Re: there is no such 00001 n=1> 00002 1 n=n+1> 00003 print(3,4)n> 00004 if(n.eq.M) then print(3,4)M> 00005 else go to 1> 00006 end if> 00007 end It has currently reached about 2.0 x 10^18.You LIE! No output device in the world could keep up with the paceyou imply. If it output a million lines per second, your currentvalue would have === taken about 60,000 years.SocksSubject: Re: there is no such IMPORTANT: Under NO circumstances will postings containing illegal or copyrighted material through this service be tolerated!! in message>> It has currently reached about 2.0 x 10^18. Just as Einstein proved>> that there is no aether, I am convinced that I will prove that there>> is no in'nity and then write a book or two.>> >> Hey, you could be on to something! Just one question. What happens when you>> multiply the highest number by 2?>> >> ;-Peter>>The Great Computer in the sky gives an over¤ow error, of course.>And to answer other people's questions about the foundations of>Calculus, dx=1/M, so in fact, we can do away with derivatives and>replace them with difference equations. lim_{n goes to>in'nity}f(n)=f(M).>>So there you have it. Good wholesome math with no paradoxes. I feel>sorry for you people as you believe in stuff that doesn't even exist.>No one has ever measured anything beyond Asimov's constant, A, that I>gave before! My hypothesis is that M=A*c^2.Hmmm. I think I get it. Difference equations where lim_{n goes toin'nity.... Hey! Wait a cotton pickin' minute. In'nity doesn'texist! You said so yourself. Almost had me === fooled there Dr. Ben. >>Dr. Ben ZonaSubject: Re: there is no postings containing illegal or copyrighted material through this service be tolerated!! You think you can reach the largest number with a FORTAN program? Howcrude! Obviously, you need to write this program in C++. Then youmight get somewhere. Although, you might have to wait for Intel tocome out with a 64 cafeinst@msn.com (Craig Feinstein)>I've thought really hard about this one and came to the conclusion>that there is no scienti'c evidence of in'nity existing. The highest>number that anyone has ever measured to according to Isaac Asimov in>his book Science and Human Thought is only about 5.0 x 10^48. No one>has ever gotten past that number. Doesn't this sound weird?> >What's to say that eventually there is a number where it is impossible>to count higher than? If someone were to 'nd this number and prove>that it is in fact the highest number, then that person would>undoubtably be rich and famous.>>I am currently running a computer program that will eventually 'nd>this magic number (I hope and pray) that I call M for short. It>counts and counts and counts and my theory is that it will eventually>stop at M. I am looking for collaborators in this experiment so that I>can use their computer time. The program in FORTRAN is simple:>>00001 n=1>00002 1 n=n+1>00003 print(3,4)n>00004 if(n.eq.M) then print(3,4)M>00005 else go to 1>00006 end if>00007 end>>It has currently reached about 2.0 x 10^18. Just as Einstein proved>that there is no aether, I am convinced that I will prove that there>is no in'nity and then write a book or two.>>Dr. Ben Zona === Subject: Re: there is no such thing as Cantor was a rusha. He confused generations of mathematicians by> telling them there are lots of in'nities. It is impossible to get a> BS in math without admitting that his theories are correct.> > The diagonal argument was a sham because it presupposes the existence> of in'nity without proving its existence. His theorem is proof that> in'nity does not in fact exist, because if it did, then there would> be only one type, not aleph null, aleph one etc., because in'nity by> de'nition is as far as one can go. He even de'led the Hebrew> alphabet by putting subscripts next to aleph when the gematria of> aleph is not in'nity or even M but one.> > > Okay, let's assume there is no in'nity, which is your whole point, no? > And now you have this magic number M.> > People have estimated that there are about 6x10^79 atoms in the entire > universe > (http://www.sunspot.noao.edu/sunspot/pr/answerbook/ universe.html#q70). > > I suppose this number 6x10^79 is your magic M. If you tried to write > down all the numbers 1 through 6x10^79, you would run out of space to > write them. (I won't even go into counting them, since even counting > one number every picosecond means this would take ~2x10^61 years) Now, you might be wondering why I am requiring you to write all these > umbers down (and why I think your computer program might be cheating)> > I can write 1 6x10^90, but that does not mean this number exists. It > might just be too big of a number, much larger than your M. So, unless > you have proof that all numbers up to your number M exists, proving M > exists is all for naught.> > Okay, maybe that number is too big. Afterall, people guess the universe > is only 13.7 billion years old > (http://wiener.math.csi.cuny.edu/UsingR/Data/age.universe.html ), or > roughly 4.3x10^42 yocto-seconds (10^-24). Maybe this should be M, since > the age all of existence is less than this number.> > Personally, I prefer to think of 42 as in'nity. Until you convince me > otherwise, I will just continue believing this way. 43? Doesn't even > exist.> Close, but not quite. Gematria tells that M is 40... === Subject: integration by substitution questionHello. I'm a college freshman currently taking calc 3 but thisquestion is really more about stuff I was supposed to know at thebeginning of calc 2. The problem is, when my textbook describessubstitution it starts with backwards chain rule. In fact, I lookedat some other texts and some even say that integration by chain ruleand integration by substitution is the same thing. However I don'tsee why chain rule is even necessary. Say you have integralthere are 2 ways of looking at this. One way is to say, this is justchain rule done in reverse, so the answer is U^2/2 + C. However, theway it's done in class and in all the later chapters in the bookpretty much 100% of the time is through manipulating differentials; inthis case, dU/dx*dx = dU, so the original integral is U*dU. But thiscan be done without any mention of chain rule whatsoever! So what amI missing?I'm not even going to get into the meaning of differentials in bothde'nite and inde'nite integral notation and all that stuff Icompletely don't understand, either. Any enlightenment would begreatly appreciated though :-) === Subject: Re: integration by substitution > .... Say you have integral> there are 2 ways of looking at this. One way is to say, this is just> chain rule done in reverse, so the answer is U^2/2 + C. However, the> way it's done in class and in all the later chapters in the book> pretty much 100% of the time is through manipulating differentials ....> So what am I missing?.... Leibniz tried and discarded several ideas for good calculus notation, before settling on dy/dx etc. His notation has some 'ne mnemonics built into it. For example, the chain rule is du from simple algebraic fractions. Of course it's really more subtle than that, but the simple appearance of that equation makes it easy to remember and use. His notation is helpful also in integration by substitution, where you can think of That's not a thorough explanation of what's going on, but it makes life easier for all of us. So when you casually use differential notation to skim over the depths of integration by substitution, spare a thought for Leibniz who took the trouble to make it look so easy. Ken Pledger. === Subject: Re: integration by substitution question> Hello. I'm a college freshman currently taking calc 3 but this> question is really more about stuff I was supposed to know at the> beginning of calc 2. The problem is, when my textbook describes> substitution it starts with backwards chain rule. In fact, I looked> at some other texts and some even say that integration by chain rule> and integration by substitution is the same thing. However I don't> see why chain rule is even necessary. Say you have integral> there are 2 ways of looking at this. One way is to say, this is just> chain rule done in reverse, so the answer is U^2/2 + C. However, the> way it's done in class and in all the later chapters in the book> pretty much 100% of the time is through manipulating differentials; in> this case, dU/dx*dx = dU, so the original integral is U*dU. But this> can be done without any mention of chain rule whatsoever!Right, and we can just give you the Fundamental Theorem of Calculus without any justi'cation. What's behind manipulating differentials? Why does it work? What are differentials anyway? At the beginning calculus level, they're pretty much just a notational device - a trick really (although a useful one). The chain rule justi'es this device. Without understanding this, you're just pushing buttons and barking to your teacher's commands. (By the way,it's good that you're asking this question!) === Subject: Need help with proving that Z[i] is EuclideanHi everyoneI'm a high school student that has a little problem.In a project I'm doing in matheamtics, I'm trying to prove that Z[i]is Euclidean (thus, Euclide's algorithm can be apllied to Z[i] aswell). I'm stuck and I need help to complete this proof. It would bemost helpful if someone would give me a clue on how to proceed toprove this since the way that I started out on doesn't seem to get meanywhere. I tried to show that everytime the divisions algorithm isused for two Gaussian integer, the rest is also a Gaussian integer. Byshowing that if the divisions algorithm is apllied on this newGaussian integer, the rest is once again a Gaussian integer, I thoughtthat you could prove with induction that so is the case for every stepunitl you don't get any rest.Am I completely wrong or at least a bit on the way?After proving that Euclide's algorithm can be apllied to Gaussianintegers, I have to prove that the last rest is the gcd for these twoGaussian integers. How to proceed here? Once again, I do not want acomplete proof or anything, just a few guidelines on === Subject: Re: Need help with proving that Z[i] is Euclidean Adjunct Assistant Professor at the University of Montana.>Hi everyone>I'm a high school student that has a little problem.>In a project I'm doing in matheamtics, I'm trying to prove that Z[i]>is Euclidean (thus, Euclide's algorithm can be apllied to Z[i] as>well). To be Euclidean, there must be some way of measuring the size of theelements. What you want to show is that given two elements a and b ofZ[i], with b different from 0, then it is possible to divide a by b,which means 'nding a ->unique<- quotient q and a ->unique<- remainderr subject to the following two conditions: (1) a = b*q + r; and (2) either r=0, or the size of r is strictly smaller than the size of b.In the integers, the size is measured by the absolute value. Whatare you using to measure size in Z[i]?The usual thing to use is the norm, which takes values N(a+bi) = a^2 + b^2.>I'm stuck and I need help to complete this proof. It would be>most helpful if someone would give me a clue on how to proceed to>prove this since the way that I started out on doesn't seem to get me>anywhere. I tried to show that everytime the divisions algorithm is>used for two Gaussian integer, the rest is also a Gaussian integer.The remainder/residue? Well, that's the de'nition of divisionalgorithm, so you must actually be trying to prove that you can->de'ne<- a division algorithm for the Gaussian integers; that is,that you can 'nd q and r as described above.> By>showing that if the divisions algorithm is apllied on this new>Gaussian integer, the rest is once again a Gaussian integer, I thought>that you could prove with induction that so is the case for every step>unitl you don't get any rest.>Am I completely wrong or at least a bit on the way?What you are describing is how you would use the division algorithm toimplement Euclid's algorithm to 'nd a greatest common divisor betweentwo given algebraic integers. For the induction to work, you need toshow that each time you divide, the remainders get smaller, and thatthey cannot get in'nitely smaller. In the integers, you do thatbecause at each step you get a positive integer strictly smaller thanthe previous remainder, or you get 0, so you must eventually get tozero.But 'rst you must show that it is possible to pick q and r as above,and that they are unique (or at least, unique enough; maybesometimes you could pick a+bi, or a-bi, or b+ai, or b-ai, or -a+bi,etc, so you have to specify how you are going to pick them...)>After proving that Euclide's algorithm can be apllied to Gaussian>integers, I have to prove that the last rest is the gcd for these two>Gaussian integers. How to proceed here?You can do this the same way as you do it for the integers.Say you are given a and b; and a=b*q + r1, as above. You want to showthat gcd(a,b) = gcd(b,r1). Then, when you divide b by r1, you will getgcd(r1,b) = gcd(r1,r2), where s is the remainder you get from thatdivision. Continuing that way until you get to the last nonzeroremainder, call it d, you will have gcd(a,b) = gcd(b,r1) = gcd(r1,r2) =gcd(r2,r3) = ... = gcd(rn,d) = d, and that will give you what youwant.So, 'rst: let d be the gcd of a and b. That means that (i) d divides a and divides b; and (ii) if e divides a and also divides b, then e divides d.We want to show that d is the gcd of b and r1. That is, we need toshow that: (i') d divides b and divides r1; and (ii') if e divides b and also divides r1, then e divides d.Since a - b*q = r1, and d divides both a and b, it follows that ddivides r1. So d divides b and r1.Now assume that e divides both b and r1. Then it divides b*q + r1, soit must divide a. So e divides b and a, and by (ii), it must dividesd. Thus, d satis'es (i') and (ii'), so it has to be the gcd of b andr1. Therefore, gcd(a,b) = gcd(b,r1).-- =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu === Subject: what is the signi'cance of left relative residual and right relative appoximation B. De'ne left relative residual tobe L=(A-B)*A^(-1), right relative residual to be R=A^(-1)*(A-B). What arethe signi'cance and relationship of their norms? Would lot,-Joenyim === Subject: matrix problem: what can you say about the inf-norm(E*A^(-1)) and inf-norm(A^(-1)*E)?Here inf-norm is the maximal absolute row sum.A is a square matrix, E is a O'small'' matrix of the same order.I wonder if there exists some theorem/conclusion/hypothesis on therelationship betweenINF-NORM(E*A^(-1)) and INF-NORM(A^(-1)*E) ?Can anybody === Subject: open set aware of a construction (or a reference to a construction)of a (nontrivial) open set containing all rational points on aninterval with a proof of its exact measure. For example, aconstruction of an open set on [0,1] with measure exactly 1/2,containing every rational point in the interval. (Of course, it's aneasy exercise to construct a set with these properties with a measurewithin an arbitrarily good === Subject: optimal solutionGroup, I hope someone will give me advice, or point me in the right directionso that I can solve the following type of problem. Given:N THINGS that produce GOOD stuff, but at a COST as follows;COST(i) = exp(slope(i)*(GOOD(i)-offset(i))); i=1,Nwhich implies GOOD(i) = log(COST(i))/slope(i) + offset(i); i=1,NI have a target amount of GOOD I need to get out of my THINGS, but I want tominimize the COSTso SUM(GOOD(i)) = Target SUM(COST(i)) is minimumIt seems to me that the minimum is reached when the slopes of the GOOD vsCOST curves of each THING are equal, or 1/(COST(i) * slope(i)) = SOME_CONST.Whats the best way to go