mm-24 How can I create a unique number from the two that will not be>produced by another pair?The classic method for this is through use of the chinese remainder> theorem. [example deleted]Who uses the classic method, and why?Any computer programmer given the task would most likely justinterlace the digits of the two numbers (probably in binary)e.g. f(77777, 333) = 7070737373I don't really know how to go about starting this proof. Does anyone have anyideas how this should go?Question: For any integer n>2, show that there are at least 2 elements in U(n)that satisfy x^2=1. I know it should be by induction but I don't know where to go after 3 and I'mnot really sure how to set it up. U(n) is the set of all positive integersless than n and relatively prime to n.Is this proof correct?Question: Let a and b be elements of an abelian group and let n be anyinteger. Show that (ab)^n=a^nb^n. Proof: When n=1, (ab)^1=ab=a^1b^1Assume (ab)^n=a^nb^n. Now (ab)^(n+1)=(ab)^n(ab)=a^nb^n(ab)=(a^n)a(b^n)b(since it is an abeliangroup). So (a^n*a)(b^n*b)=a^(n+1)b^(n+1). Thus, (ab)^n=a^nb^n. > I don't really know how to go about starting this proof. Does anyone have > any> ideas how this should go?Question: For any integer n>2, show that there are at least 2 elements in > U(n)> that satisfy x^2=1. I know it should be by induction but I don't know where to go after 3 and I'm> not really sure how to set it up. U(n) is the set of all positive integers> less than n and relatively prime to n.It is easy to see that the residue classes, mod n, of 1 and n-1 are both in U(n), and both square to give the residue class of 1, mod n.I see no need of induction. >It is easy to see that the residue classes, mod n, of 1 and n-1 are both >in U(n), and both square to give the residue class of 1, mod n.I see no need of induction. > I don't really know how to go about starting this proof. Does anyone have any> ideas how this should go? Question: For any integer n>2, show that there are at least 2 elements> in U(n) that satisfy x^2=1. I know it should be by induction but I don't know where to go after 3 and I'm> not really sure how to set it up. U(n) is the set of all positive integers> less than n and relatively prime to n.>There are no integers x in U(n) with x^2 = 1, since 1 coprime to n.> Is this proof correct? Question: Let a and b be elements of an abelian group and let n be any> integer. Show that (ab)^n=a^nb^n. Proof: When n=1, (ab)^1=ab=a^1b^1> Assume (ab)^n=a^nb^n. Now (ab)^(n+1)=(ab)^n(ab)=a^nb^n(ab)=(a^n)a(b^n)b(since it is an abelian> group). So (a^n*a)(b^n*b)=a^(n+1)b^(n+1). Thus, (ab)^n=a^nb^n.>I suppose so if one presumes you're using induction.If all the spaces weren't crammed out the equationsthey'd be easier to read and check for accuracy.So I'll just suppose they're correct. > I am trying to write a piece of software quite like Granular Synthesis> Demo 1 in http://www.dcs.gla.ac.uk/~jhw/audioclouds/ . Hmm, that looks like a very interesting project! It looks difficultbut well worth solving -- that's what I call a good problem.> I was wondering if anyone could explain the difference between a> probabilty distribution and a probabilty density?Well, a probability distribution is essentially a function that tells you how much stuff is within a given region. It turns outthat a very convenient way to construct such functions is to workinstead with the probability density -- that's a function that sayshere the stuff is piled up so deep, and there it's so deep, andstuff in a region -- voila, the probability distribution.> And if you have a probabilty distribution can a density be obtained> from it and if so how?Yes, you can get the density by differentiating the distribution.Whether it's easier to go from density to distribution or vv dependson the details of the problem. > Also the main bit i am stuck on is how a density can be conditioned> (in the example given it is conditioned by the x and y positions of a> mouse) how would you go about doing such a thing.By conditional on the mouse position they just mean that the density is a function of the mouse position -- conceptually there'sa different density for every mouse position. Of course, for position that are close together, the densities will be very similar;the trick is to figure out just how the density should vary.Robert Dodier PaulI just read Ohanian & Ruffini's 1.9 and it completely supports myposition and not yours IMHO.PZ: My position on what? On our ability to measure tidal effects within anarbitrarily small region, down to a spacetime point?JS: The following statements are true:1. To a good approximation the non-tensor connection field g-force on a where m(passive) = m(active)The approximation is two-foldA. Not near a space-time singularity, i.e. not falling behind the event horizon of a black hole.B. The scale is large enough so that quantum gravity metric fluctuations are ignorable.2. To the same approximation the connection field g-force (which reduces to Newton's gravity force) is eliminated on a timelike geodesic. The connection field is non-zero only on time-like non-geodesics. Rotation 3. The curvature tensor is also a local observable.IF you define a true gravity field as one in which the tidal curvature tensor is NOT zero, then one can always locally, in principle, distinguish a true gravity field from a fake gravity field (where the tidal tensor vanishes in all frames geodesic LIF & non-geodesic LNIF) by definition.On the other hand, if you define a gravity field as the connection field, then you cannot make such a distinction!Therefore, the problem here is only a semantic problem from wavering between the two different definitions of gravity field.principle locally see tidal effects as a torque around the center of mass or a precession of a small spinning gyroscope of course. Also you can see deformations of shape of a geodesic droplet if the surface tension is small enough.5. Since Einstein's GR comes from locally gauging ONLY the 4-parameter translation subgroup of the Poincare group, of course Einstein's early formulation of the equivalence principle was only an approximate rotational degrees of freedom. Einstein's guv field of curved space-time with the symmetric connection force field is simply the compensating gauge force field needed to restore the now local gauge symmetry which is equivalent to local general coordinate transformations. If, in addition, you locally gauge the 6-parameter Lorentz subgroup, then you get an additional torsion force field, i.e. an anti-symmetric piece of the connection field. This will obviously modify the predictions of GR for the tidal torques on extended test 6. Hagen Kleinet shows that tidal curvature, the basis for gravity waves, comes from stringy (vortex core if one puts in a vacuum coherence O(1) ODLRO parameter) disclination defects in a Planck lattice in 4D. There is no torsion field in 1915 GR, but if there was one in Nature, as Akimov & Shipov claim in Moscow, then it would correspond to dislocation lattice defects. The different 4D discrete world crystal symmetry groups of the tiled unit cell are different physical vacuum structures quite obviously. This is 4D (and maybe 11D) crystallography with additional supersymmetry matrix dimensions).7. There are still two more subgroups left in the 15 parameter Conformal Group. You will get new compensating gauge force field physics if you locally gauge the 1-parameter dilation subgroup and also locally gauge the 4-parameter special conformal transformation of constant acceleration boosts for hyperbolic motion in Special Relativity e.g. MTW has a chapter on this.elegant fashion show of weaves of the fabric of reality from spin networks to pre-geometric spin foams to quantized Area operators and world holograms et-al I find Penrose saying:The algebra I have used to treat linear displacements and rotations together, or linear and angular momentum together, I call the algebra of twistors. I have used the term 'twistor' to denote a 'spinor' for the six-dimensional (++----) pseudo-orthogonal group O(2,4).9. It is obvious to any School Boy that O(2,4) is simply Lorentz SR O(1,3) with string world sheet O(1,1) fiber. Therefore, string theory is inherent here. Penrose continues:The twistor group is the (++--) pseudo-unitary group SU(2,2) which is locally isomorphic with O(2,4). In turn O(2,4) is locally isomorphic with the fifteen-parameter (local) conformal group of space-time. Under a conformal transformation of space-time the twistors will transform linearly according to a representation of the group SU(2,2). p. 175 Combinatorial Space-Time Quantum Theory and Beyond (Cambridge, 1971).Suddenly the connections among1. Quantum loop gravity of spin networks --> foams with quantized area etc. operators2. Local gauge invariance3. String theory4. Twistors & Conformal Groupare becoming clearer. >what I think you are missing is that .999... (a zero, a dot and an>infinite amount of 9s) is equal to 1, so what you are asking is>whether 1^2, 1^3, 1^4, etc. is 'tending to' 1.i understand the premise.> Look at it this way: We can write a number like 0.9 as 1 - 0.1, and a>number like 0.99 as 1 - 0.01. In general we can write it down as:>f(n) = 1 - (1/(10^n)) for n>0 where n is the number of nines behind>the dot.>So the number 0.999... can be found by looking at:>lim{n->infinity}{f(n)} = >lim{n->infinity}{1 - (1/(10^n))} =>lim{n->infinity}{1} - lim{n->infinity}{1/(10^n)}but infinity means unlimited... limit{n->unlimited}{f(n)} = > limit{n->unlimited}{1 - (1/(10^n))} => limit{n->unlimited}{1} - limit{n->unlimited}{1/(10^n)}...which contradicts the premise.What premise does that contradict? If it is your premeise that 0.999... is not equal to 1, then good.>Since the limit of the fraction-part goes to zero, the whole thing>becomes equal to 1. In other words 0.999... is equal to 1.only if the premise the unlimited has a limit is believed.If it does not, then unlimited decimals have no meaning in teh first place, and your 0.999... is NaN.But in every mathematician's world, every Cauchy sequence, of which (1 - 1/10^n) is an example, has a real number limit, however unlimited you seem to think it should be. in teh case of (1 - 1/10^n), that limit is 1, and 1 is the only possible numerical value that 0.999... can possibly mean, to anyone who pretends to any mathematical skills.If 0.999... represents a number the that number is 1.If 0.999... does not represent a number, then this whole discussion is moot.If you can't stand that heat, get out of the mathematical kitchen.garry denke, geologist> denoco inc. of texas What is a double limit? lim(m,n->+oo) (1 - 1/10^m)^nDoes the above have an assigned value? lim(m->+oo) (1-1/10^m)^n = 1Why do you prefer the above as opposed to the below? lim(n->+oo) (1-1/10^m)^n = 0Garry Denke, GeologistDenoco Inc. of Texas > What is a double limit? lim(m,n->+oo) (1 - 1/10^m)^nDoes the above have an assigned value? lim(m->+oo) (1-1/10^m)^n = 1Why do you prefer the above as opposed to the below? lim(n->+oo) (1-1/10^m)^n = 0> Garry Denke, Geologist> Denoco Inc. of TexasThe double limit, lim_{(m,n)->(+oo,+oo)} (1 - 1/10^m)^n, does not exist.For it to exist it would be necessary at least for both the single limits above to have the same value, which they do not.The limit as m goes to +oo is relevant the meaning of 0.999..., whereas the limit as n goes to +oo is not.Just telling. Your welcome. > What is a double limit?> lim(m,n->+oo) (1 - 1/10^m)^n> Does the above have an assigned value?No. If it had a value, say L, then L would have to satisfy thecondition that fore every epsilon > 0, there exists M, N > 0 such that |(1-1/10^m)^n - L| < epsilonfor every m > M and for every n > N. The fact that no such L existsfollows from the fact that the iterated limits below have differentvalues.> lim(m->+oo) (1-1/10^m)^n = 1This holds for each n > 0, and therefore lim(n->+oo) lim(m->+oo) (1-1/10^m)^n = 1 (1)> Why do you prefer the above as opposed to the below? > lim(n->+oo) (1-1/10^m)^n = 0This holds for each m > 0, and therefore lim(m->+oo) lim(n->+oo) (1-1/10^m)^n = 0 (2)Comparing (1) and (2), we see that the limits do not commute.It's not a matter of prefering one limit to the other. It's simply amatter of understanding what the notation 0.999... means. Since0.999..., by definition, means lim (m->+oo) (1-1/10^m), it follows thatlim (n->+oo) 0.999...^n = lim (n->+oo) lim (m->+oo) (1-1/10^m)^n, whichis the limit that appears in (1).-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. How to integrate sinc(x) (without Fourier transformation) ??> That integral is not an elementary function. It is known as the Sineintegral function. >How to integrate sinc(x) (without Fourier transformation) ?? That integral is not an elementary function. It is known as the Sine> integral function.> Nice link. Beautiful images. :-)1. first of all, thank you for having confirmed me that it wasn't trivial2. my purpose was to find the Fourier transformation of sinc, and not justto take it from a book. Is it possible by an easy process, or else... no way?.. >How to integrate sinc(x) (without Fourier transformation) ??> That integral is not an elementary function. It is known as the Sine>integral function.> Nice link. Beautiful images. :-)1. first of all, thank you for having confirmed me that it wasn't trivial2. my purpose was to find the Fourier transformation of sinc, and not just> to take it from a book. Is it possible by an easy process, or else... no way> ?..> The infinite Fourier transform is fairly trivial. If you use theintegral representation sinc(x) = {{sin(pi x)} over {pi x}} = (2pi )^{-1} int _{-pi} ^{pi) {dt e^{ixt} }it is easy to see that the Fourier transform (multiply by e^{-ikx} andintegrate from -infty to infty ) is theta left({ pi ^2 - k^2}right)where the theta function is 1 for positive arguments, 0 for negative,and defined to be 1/2 for argument 0.-- Julian V. NobleProfessor Emeritus of ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything and thereby take away our free will and that share of glory that rightfully belongs to us. -- N. Machiavelli, The Prince. X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In PM, Joona I Palaste said:>At least in the University of Helsinki, if there is a function>f:X->Y, and U is a subset of X, then fU means {f(x) | x in U}.Are you sure that you mean fU and not f_U (f subscript U) orf|U?>Could there be any functions f:X->Y and subsets U of X so that fU>and f(U) would both be defined but would not be equal?If you're using standard notation then the two are identical bydefinition. If you're using fU to mean f|U then the two aredifferent by definition.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to Shmuel (Seymour J.) Metz following:> In , on said:>>At least in the University of Helsinki, if there is a function>>f:X->Y, and U is a subset of X, then fU means {f(x) | x in U}.> Are you sure that you mean fU and not f_U (f subscript U) or> f|U?Yes, I am sure. f_U would mean a function f associated with the set U,not iterated over it. f|U would mean the function f constrained intothe set U.>>Could there be any functions f:X->Y and subsets U of X so that fU>>and f(U) would both be defined but would not be equal?> If you're using standard notation then the two are identical by> definition. If you're using fU to mean f|U then the two are> different by definition.I can't see how using the notation I defined above, fU = {f(x) | x inU}, fU and f(U) could be identical by definition.-- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland ---------- http://www.helsinki.fi/~palaste --------------------- rules! --------/A friend of mine is into Voodoo Acupuncture. You don't have to go into heroffice. You'll just be walking down the street and... ohh, that's much better! - Stephen Wright > Shmuel (Seymour J.) Metz following:>> In , on said:>At least in the University of Helsinki, if there is a function>f:X->Y, and U is a subset of X, then fU means {f(x) | x in U}.>> Are you sure that you mean fU and not f_U (f subscript U) or>> f|U?> Yes, I am sure. f_U would mean a function f associated with the set U,> not iterated over it. f|U would mean the function f constrained into> the set U.>Could there be any functions f:X->Y and subsets U of X so that fU>and f(U) would both be defined but would not be equal?>> If you're using standard notation then the two are identical by>> definition. If you're using fU to mean f|U then the two are>> different by definition.> I can't see how using the notation I defined above, fU = {f(x) | x in> U}, fU and f(U) could be identical by definition.Because f(U) = { f(x) : x in U } is perfectly standard notation when U is asubset of the domain of f.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. Dave Seaman scribbled the following:>> Shmuel (Seymour J.) Metz following:> In , on said:>>At least in the University of Helsinki, if there is a function>>f:X->Y, and U is a subset of X, then fU means {f(x) | x in U}.> Are you sure that you mean fU and not f_U (f subscript U) or> f|U?>> Yes, I am sure. f_U would mean a function f associated with the set U,>> not iterated over it. f|U would mean the function f constrained into>> the set U.>>Could there be any functions f:X->Y and subsets U of X so that fU>>and f(U) would both be defined but would not be equal?> If you're using standard notation then the two are identical by> definition. If you're using fU to mean f|U then the two are> different by definition.>> I can't see how using the notation I defined above, fU = {f(x) | x in>> U}, fU and f(U) could be identical by definition.> Because f(U) = { f(x) : x in U } is perfectly standard notation when U is a> subset of the domain of f.I see. However, in the question I asked, I use fU for that concept andf(U) very specifically for f applied to U itself, not to every elementof U. I think David Ulrich got the right idea.-- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland ---------- http://www.helsinki.fi/~palaste --------------------- rules! --------/Shh! The maestro is decomposing! - Gary Larson Joona I Palaste scribbled the following:> I see. However, in the question I asked, I use fU for that concept and> f(U) very specifically for f applied to U itself, not to every element> of U. I think David Ulrich got the right idea.Typo, should be David Ullrich with two l's.-- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland ---------- http://www.helsinki.fi/~palaste --------------------- rules! --------/He said: 'I'm not Elvis'. Who else but Elvis could have said that? - ALFX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In feminine plural of the foreign loan word, nilpotent,>when written without vowels or other punctation, is written>nylpw_tn_tywt>where _t is teth.That last t would have to stand for a Tav, not a Tet. I assume thatthe w stands for a Vav. Also, are you sure that what you are lookingat is intended to be the plural of nilpotents (nilpotetioth) and notnilpotency (nilpotentiuth)?>What I would like to know is how it would be written>if one did write the vowels and other punctation.Well, transliterations of loan words, and transformations oftransliterations, are always dicey. >(0) which of the consonants is dotted?I would Expect a Dagesh where you have p (otherwise it would benilfotentioth>(1) is the y following the first n retained and the n endowed with a>chireq?ITYM hiriq. Yes to both.>(2) is there a schwa under the l?I would expect a schwa nach.>(3) is the w following the p replaced by a cholem or is it replaced>by a qamats qatan?I would expect a cholem in a transliteration of a worde with an o.>(5) is there a schwa under the second n?I would expect a schwa na'.>(6) is the y following the second teth retained and the teth endowed>with a chireq?>I would also like to know how I might figure out examples like this,>involving foreign loan words, on my own, instead of having to ask>people in every instance.Try to find niqud that matches the pronunciation of the original. Beglad that you recognized it as a transliteration; I once spent 5minutes trying to figure out the shoresh of vatergate before realizingthat there was none - it was a transliteration of Watergate.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot > In feminine plural of the foreign loan word, nilpotent,>when written without vowels or other punctation, is written> >nylpw_tn_tywt>where _t is teth. That last t would have to stand for a Tav, not a Tet.I had thought that words transliterated into Hebrew strictly used tet.YS---Checked by AVG anti-virus system (http://www.grisoft.com). >In the feminine plural of the foreign loan word, nilpotent,>>when written without vowels or other punctation, is written>>nylpw_tn_tywt>>where _t is teth.> That last t would have to stand for a Tav, not a Tet.> I had thought that words transliterated into Hebrew strictly used tet.> Certainly not. For words of English origin, th becomes Tav and tbecomes Teth. Case in point: Mathematics -> mathematika,mem-tav-mem-teth-yod-qoph-heh.Alan Alan Greenwood schreef in bericht[ ... ]>I had thought that words transliterated into Hebrew strictly used tet.Certainly not. For words of English origin, th becomes Tav and t> becomes Teth. Case in point: Mathematics -> mathematika,> mem-tav-mem-teth-yod-qoph-heh. AlanBut the old English of Euclid and Ptolemy is hard to read. .a6[IDoubleDot] [EDoubleDot] [IDoubleDot] [Divide]I would have expected becoming teth and becoming tav instead? in>> In the feminine plural of the foreign loan word, nilpotent,>> >when written without vowels or other punctation, is written>> >nylpw_tn_tywt>>where _t is teth.>That last t would have to stand for a Tav, not a Tet.> I had thought that words transliterated into Hebrew strictly used tet.Certainly not. For words of English origin, th becomes Tav and t> becomes Teth. Case in point: Mathematics -> mathematika,> mem-tav-mem-teth-yod-qoph-heh. AlanAre there any other examples of this occurance?YS---Checked by AVG anti-virus sha1:SHDIYfVy+L0N0jSittzBcKXLiKo =>> That last t would have to stand for a Tav, not a Tet. I had thought that words transliterated into Hebrew strictly used> tet.They do--for the t sounds found in the borrowed word. The final Tafis part of the Hebrew feminine plural, so it is not changed to Tet.Len. > That last t would have to stand for a Tav, not a Tet.> I had thought that words transliterated into Hebrew strictly used>tet. They do--for the t sounds found in the borrowed word. The final Taf> is part of the Hebrew feminine plural, so it is not changed to Tet. Len.>YS <3ffd484c$2$fuzhry+tra$mr2ice@news.patriot.net> tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In PM, David W. Cantrell said:> Or maybe dx post facto. e^x, dy, dx e^x, dx (Aristophanes)or perhaps d(hi/ho)-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply <3ffa32dd$8$fuzhry+tra$mr2ice@news.patriot.net> <8765fp8z1v.fsf@becket.becket.net> <3ffc9d5e$9$fuzhry+tra$mr2ice@news.patriot.net> <87lloj9qj9.fsf@becket.becket.net> <87hdz79ppk.fsf@becket.becket.net>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In tb+usenet@becket.net (Thomas Bushnell, BSG) said:>Mathematics in antiquity and the middle ages (and even surprisingly>late) was thought in fairly concrete terms; consider Euclid's rule>that you can extend a line indefinitely. A line, for Euclid, is not>an infinite thing; it's a *finite* thing, but you can extend it as>much as you like--with it always remaining finite. And yet Euclid published a proof that there is no largest primenumber. For that matter, the proof that 2^.5 is irrational would seemto be an acknowledgement of the infinite in another sense.-- Shmuel (Seymour J.) Metz, SysProg > In PM, tb+usenet@becket.net (Thomas Bushnell, BSG) said:>Mathematics in antiquity and the middle ages (and even surprisingly>late) was thought in fairly concrete terms; consider Euclid's rule>that you can extend a line indefinitely. A line, for Euclid, is not>an infinite thing; it's a *finite* thing, but you can extend it as>much as you like--with it always remaining finite. And yet Euclid published a proof that there is no largest prime> number. Oh, indeed! For us, that means there are an infinite number ofprimes. But for Euclid, it means that you can produce as manyprimes as you like. It's easy to miss the difference, because we have come to see suchdifferences as (mathematically) pointless. > For that matter, the proof that 2^.5 is irrational would seem> to be an acknowledgement of the infinite in another sense.Um, no. The Greeks didn't have anything like decimal expansions.The proof was taken to establish that two particular lines are notcommensurate, that is, there is no line segment which is an integraldivisor of each. The usual Greek commentary on this discovery amounted to there beingno common measure for all lines or something like that, not anythingabout infinity.Thomas <3ffa32dd$8$fuzhry+tra$mr2ice@news.patriot.net> <8765fp8z1v.fsf@becket.becket.net> <3ffc9d5e$9$fuzhry+tra$mr2ice@news.patriot.net> <87lloj9qj9.fsf@becket.becket.net>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In tb+usenet@becket.net (Thomas Bushnell, BSG) said:>Of course they did; it means a thing without limit; in some>contexts, more exactly, without any limit in some regard.That's hand waving, not a definition. What did they mean by withoutlimits?>Finite means whatever can be limited. Anything can be limited.>Well, it's my stock and trade, yes. I have a definition: the>infinite is whatever is not limited Again, that begs the question.>A very easy thing to say; a very hard thing to proveCount your wife's teeth.>But yes, he did. Then please explain what he meant by completed infinite and as atonce, without handwaving or circular definitions.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to > In PM, tb+usenet@becket.net (Thomas Bushnell, BSG) said:>Of course they did; it means a thing without limit; in some> >contexts, more exactly, without any limit in some regard.That's hand waving, not a definition. What did they mean by without> limits? >Finite means whatever can be limited. Anything can be limited.I'm not sure how you find it to be vague to say without limits atthe same time as you assert that anything can be limited. >A very easy thing to say; a very hard thing to proveCount your wife's teeth.So, Aristotle said something incorrect in one area, he must be crazyabout math? (And everyone else too?) By this measure, I don't thinkNewton fares too well, not to mention Euler, Gauss...>But yes, he did. Then please explain what he meant by completed infinite and as at> once, without handwaving or circular definitions.I gave several examples: lines infinitely extended, for example, andas at once in terms of co-existence of physical things at the samemoment in time. ThomasX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In Andres AlvarezMarin) said:>I think that you should teach her what are the mathematics useful>forfew years I may have left, is an an understanding and love for thebeauty of Mathematics. But I don't know how to impart that. If hewanted to impart an understanding and love for the beauty of art,would you tell him to teach how to use paintings to cover cracks inthe wall?>Don't use the mathematician point of view, maybe the engineer point>of view... application.That's fine if he wants her to be an engineer who is blind to beauty.>Also, have you considered teach her to program?.That might be a good thing in it's own right, but has nothing to dowith teaching an appreciation of Mathematics.Children are born with boundless curiosity and sense of play. Thepublic school system carefully and systematically tries to destroythose. It is the job of parents and grandparents to protect theirchildren as much as they can from the stultifying atmosphere of theschools. There will be time in later life for her to learn thatMathematics can be useful; now is the time for her to learn thatMathematics is fascinating and fun.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to >In Andres Alvarez>Marin) said:>>I think that you should teach her what are the mathematics useful>>for>few years I may have left, is an an understanding and love for the>beauty of Mathematics. But I don't know how to impart that. If he>wanted to impart an understanding and love for the beauty of art,>would you tell him to teach how to use paintings to cover cracks in>the wall?Considering that mathematics was (and is) used to gain insight intopractical problems (e.g. engineering), I don't see why it is wrong toteach applications. IMHO, applications reinforce the theory.>>Don't use the mathematician point of view, maybe the engineer point>>of view... application.>That's fine if he wants her to be an engineer who is blind to beauty.I strongly disagree. Being an engineer does not necessarily blind oneto the beauty of mathematics. It may be the case, as I said above,that the engineer prides herself on the practical application ofmathematical principles. Not everyone needs to be a theorist. Noteveryone wants to be a theorist.>>Also, have you considered teach her to program?.>That might be a good thing in it's own right, but has nothing to do>with teaching an appreciation of Mathematics.Again, this is an opportunity for practical application of theory.What bothered me a bit about the OP's post was that it implied thatthere was something wrong with not studying higher math and studyingmolecular biology instead. There's nothing wrong with that, imho, andcertainly any degree from Caltech is a considerable accomplishment.>Children are born with boundless curiosity and sense of play. The>public school system carefully and systematically tries to destroy>those. It is the job of parents and grandparents to protect their>children as much as they can from the stultifying atmosphere of the>schools. There will be time in later life for her to learn that>Mathematics can be useful; now is the time for her to learn that>Mathematics is fascinating and fun.Why not useful, fascinating, and fun?--gregbogds at best dot comX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In , on said:>Is it just me, or does this happen to everyone?I find that when I read things that I believe I have completelyforgotten, it all starts coming back to me. Also, I find thatsometimes the complex things are easy but the simple things aredifficult. YMMV.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Terminate: SPA(GIS) In said:>Point 3:>As the one-to-one correspondence N <-> R is not possible (Point 1), >the criterion settled by Cantor (Point 2) is not valid to resolve >whether R is countable or not.No. If you stipulate that R is uncountable, then it immediatelyfollows that R is uncountable. If you don't stipulate it, a reductioad absurdum is perfectly reasonable, although not necessary. Areductio ad absurdum does not assume a false statement: it proves thatthe statement must be false by proving that it leads to acontradiction.>Point 4: >If a proof (any) comes to the conclusion that R is not countable >because it is not possible to accomplish the criterion established>in Point 2, then the proof is uselessIncorrect.>because that criterion is not valid for that purpose (Point 3). No, it is your Point 3 that is not valid.>DEFINITION>Useless proof: A correct proof that proves nothing.What do you mean by proves nothing?>QUESTION>Does somebody know any other criteria (different from the one >established in Point 2) to resolve whether R is countable or not?Do you know a definition of frequency other than the number of cyclesper unit of time? The definition is the definition, and any proofultimately comes down to showing that it satisfies the definition.>NOTE: >Someone could interpret that if N <-> R is not possible, then R is >uncountable. This conclusion is true, but in this case, it will>imply that R is uncountable because we havent got a suitable tool>to count its elements,No. It will imply that no such tool exists.>since the properties of N and R are incompatible.You still haven't defined what you mean by that statement inMathematical terms.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply <3ffcc389$17$fuzhry+tra$mr2ice@news.patriot.net> tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In PM, David W. Cantrell said:>(Or maybe you meant to say properly, rather than poperly. ;-)That depends on whether Gauss was speaking ex cathedra.Yes, I have a tendencey towards dropped, doubled and transposedletters; I meant properly, and I was also thinking of the issue ofwhether he allowed the one-point compactification.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In Wolfram Research said:What was the upshot of your dispute with CRC a few years ago? -- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to >|> Infinitesimals are totally unintuitive and correspond to no physical>|> objects. Recall the history of infinitesimals in calculus/analysis.>|> They were widely dismissed and ridiculed (e.g. Bishop Berkeley).>|>|There will always be foot-draggers.I think Bishop Berkeley was using his criticism of infinitesimal> calculus as an indirect defense of theology. He who can digest> a second or third fluxion, a second or third difference, need not,> we think, be squeamish about any point of divinity. -- The Analyst It sounds like he was saying I know you are but what am I?-- http://hertzlinger.blogspot.com >|> Infinitesimals are totally unintuitive and correspond to no physical>|> objects. Recall the history of infinitesimals in calculus/analysis.>|> They were widely dismissed and ridiculed (e.g. Bishop Berkeley).>|>|There will always be foot-draggers.>I think Bishop Berkeley was using his criticism of infinitesimal>calculus as an indirect defense of theology. He who can digest>a second or third fluxion, a second or third difference, need not,>we think, be squeamish about any point of divinity. -- The Analyst It sounds like he was saying I know you are but what am I?No, he was saying that you can't appeal to but I can't see it! intheology if you don't admit the same argument in physics.Thomas > : o the answer appears to be > : 1/[(999/1024) + 1] = .5061789... > This doesn't have the right limiting behavior.It has. > If I toss the coin 1e6 > times and it still comes up heads every time, then I can be (almost) > certain that I have the two-headed coin. Right? Yes. Make n the number of tosses, the answer is: 2^n / (999 + 2^n)which goes to 1 as n goes to infinitity.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ >Your start is already wrong. P(two headed) = 1/1000 by the definition >of the problem. Coming from the recesses of my memory I find Bayes' >theorem: > P(A|B) = P(A and B)/P(B) >Let's have A = two headed B = 10 heads in a row. >You want P(A|B) i.e. the probability of A (two headed) assuming that >B (10 heads in a row) did occur. Now P(A and B) = 1 > How can P(A and B) be greater than P(A) = P(two headed) = 1/1000?mind. P(A and B) = P(A) - P(A and not B) = P(A) - 0 = 1/1000.Revise the remainder accordingly and I get: P(B) = P(A and B) + P(not A and B) = 1/1000 + 999/1024000.So P(A|B) = (1/1000)/(1/1000 + 999/1024000) = 1024/2023 ~ 50.61%.-The recesses of my mind are indeed pretty deep. The last time Iused Bayes was about 30 years ago...-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ : One apparent way of avoiding the paradoxes of naive set theory is to: turn set-defining characteristic functions into partial functions from: sets to the three-valued logic {T,F,bot}. This three-valued logic: extends classical logic in the obvious way with: and(T,bot)=or(F,bot)=bot, and(F,x)=F, or(T,x)=T, not(bot)=bot, so that: every truth function has a fixed point.Very much Lukasiewicz' trivalent logic, then. That is good, and it allowseasy extension to polyvalent logics.: Obviously the law of excluded middle does not hold: or(a,not(a)) only: implies that a is T or bot. This gives the logic a constructive: character.Constructivism has been one of the driving reasons behind the introductionof alternate logics. Removing the constraint of excluded middlegeneralises one into the realm of Heyting algebras, where constructivetheories roam, while allowing the avoidance of the antinomies found in theBoolean.: In my formulation, I identify each set with its characteristic: function from sets to {T,F,bot}. Thus given s:set, the membership of: an element x can be tested with s(x). In the general case, this is a: partial function, returning bot for some values. I call such sets: partial sets, and sets whose characteristic function is in {T,F}: total sets. Every set of ZF and NF is a total set, with this theory: admitting a strictly larger class of sets than either.In some ways, this is the structure of a topoi, or at least a functor from atopoi category to something similar. Is the approach meant to be categorial(which, by the way, is a good thing in my eyes -- I'm just curious)? Then,your partial classification appears to mainly distinguish the topoi of setsfrom some of the many other topoi.: Russell's set R={s:set|not(s(s))} is then a partial set. All ZF sets: are elements of R, while some elements of NF are not elements of R,: while some new sets such as R itself are of undecidable membership.:: I've translated the ZF axioms to this set theory, rephrasing them in: terms of characteristic functions and new for-all and there-exists: logic operators performing logical conjunction and disjunctions across: all elements of a characteristic functions. Everything appears to be: sound and avoids known paradoxes.The categorial study of paradox is becoming a large field these days, and itappears you may be repeating some of the work already done (which can besoooo frustrating sometimes!). I don't mean to assume any level of study,but perhaps I might suggest that, if you haven't, you should check out someonline I can suggest.: With the new axioms, it is easy to construct a bijection from the: universal set to its power set. Cantor's proof that |P(x)|>|x| for: all non-empty sets x proceeds by constructing C={a:x|not(P(x)(a)} and: using the law of excluded middle to derive a contradiction on its: membership in P(x). This goes away for lack of excluded middle,: leaving C a partial set which appears not to be constructively: contradictory.:: The one worrying aspect of this approach is that it identifies sets: with characteristic functions from sets to logic values:: Set=Set->{T,F,bot}. I have only been able to develop an intuition of: such sets in a purely constructive way, by writing down a finite list: of possibly self-referential equations defining sets, and convincing: myself that a unique solution exists. This is much in the style of: NF's axiom that every (possibly cyclic) graph corresponds to a set,: but I allow unlimited comprehension.:: Are there any known problems with this approach to set theory? Any: pointers to research on the topic?No known problems that I am aware of. In fact, it seems to me to be one ofthe more successful modern approaches for classifying paradox. However, ifI could make one suggestion, it would be to not restrict yourself to yourtrivalent logic. Any Heyting algebra is possible, and expands your researchinto the much more fruitful world that all topoi present. In fact, becauseof natural distinctions that present themselves between propositions thattake on the middle value, trivalent theories are often looked at only assummarisations of a more natural infinitely valent theory. Good resourcesfor this can be found in intuitionist discussions, but it is more general.Also, may I ask why you posted to comp.lang.functional? This intrigues mebecause some of my own research has been around the evaluation of the lambdacalculus and proof / evaluation theory in the context of non standardlogics, but I do not see this approach explicitly stated in your message.Good luck with your researche!-- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea: prankster, fablist, magician, liar > I just read a section in a real analysis book that states that up to a set of> measure zero any L.measurable function on the line is equal to a function of> Baire class two.Really??? Where can one find a proof of this? Is it true for b.c. 1? I'm sure the answer is no, but I can't think of a counterexample. Yeah, is that really true? Can anybody suggest any books, or (if itsnot too much to ask) an outline of a proof? That's really cool. Can't think of a counterexample for class one either...>I just read a section in a real analysis book that states that up to a set of>measure zero any L.measurable function on the line is equal to a function of>Baire class two. > hi> thank you very much for your posting, but seem not to be able to find> in it the answer to my question - how they are the same 'entities' if> they are coded by different numbers?thank you> maximThey are not really the same entities. Signs of type >1 are variablesof the same type number but as a sequence of one number. Just as 0 isnot the same as {0}, the number X is not the same as the sequencecontaining exactly one number X. Signs of type >1 are then joinedtogether to form elementary formulas. Godel defines the operation ofjoining sequences (such as signs) to form a new sequence. So you haveto have sequences, not individual numbers, to join them to form a newsequence, an elementary formula, in definition (20) in his paper.Charlie VolkstorfCambridge, MA>> hi,>> i have very technical and boring question, but may be somebody out>> there read the english translation of the Godel's original paper and>> could help me.>Boring? Quite the contrary. It is refreshing to see someone discuss>something of significance. (Too much time is spent quibbling over>exercises in elementary arithmetic!)>> My question is, in what sense 'sign of type n' and 'variable of type>> n' are the same ('dasselbe')?>Let represent the sequence of numbers a, b, c, . . . >Godel's system P (PM plus Peano's Axioms) contains:>variables of type 1: X1, Y1, Z1, . . .>variables of type 2: X2, Y2, Z2, . . .>variables of type 3: X3, Y3, Z3, . . .>etc.>signs of type 1: 0, S0, SS0, . . . , X1, SX1, SSX1, . . . , Y1, SY1,>SSY1, . . . , Z1, SZ1, SSZ1, . . .>signs of type 2: , , , . . >signs of type 3: , < , . . .>etc.>As you point out, the signs of type >1 are just the variables of type>>1 seen as sequences of length 1. The reason Godel defines signs of>type >1 is so that he can juxtapose these sequences to form elementary>formulas in (20) as: sign of type n+1(sign of type n). That is:>elementary formulas: X2(0), X2(S0), X2(SS0), . . . , X2(X1), X2(SX1),>X2(SSX1), . . . , Y2(0), Y2(S0), . . ., X3(X2), X3(Y2), . . .>Elementary formulas are relation variables with arguments. Subsequent>definitions include logical operators negation, disjunction and>quantification in order to build Godel numbers for all of (higher>order) Predicate Calculus theorem proving and show that the>corresponding relationships are recursive.>BTW: In (19) Godel includes v=R(x)=2^x so that v=also occurs in 11.) Why does he include this redundant clause? IMHO:>This shows that these functions and relations are indeed recursive>(see theorem IV.)>I use the same technique in my Program Synthesis system>(http://www.mathpreprints.com/math/Preprint/ CharlieVolkstorf/20021008.1/1>section IV Proof 1 Step 2 clause ~LT(J,A) in my synthesis of programs>that determine if one number is a factor of another.) Godel is just>using Logic as a programming language, just as Peano did before him to>write a program to demonstrate that the universal set of the natural>numbers is recursively enumerable, which is axiom 1 in Section VI of>http://www.arxiv.org/html/cs.lo/0003071 .>In fact, many of Godel's axioms and definitions of recursive functions>and relations are formal rules and theorems in this paper. This is>because a computer program is simply a constructive proof that a>particular set is recursive or recursively enumerable. Mathematicians>such as Peano, Godel and Turing for generations have been showing that>various sets and functions are, or are not, recursive or recursively>enumerable by constructing programs in various bases of computing. >Today we call that process Computer Programming!>Charlie Volkstorf>Cambridge, MA>> thank you>>maxim so, when he says that they are the same (dasselbe), he is wrong?>hi>thank you very much for your posting, but seem not to be able to find>in it the answer to my question - how they are the same 'entities' if>they are coded by different numbers?>thank you>maximThey are not really the same entities. Signs of type >1 are variables> of the same type number but as a sequence of one number. Just as 0 is> not the same as {0}, the number X is not the same as the sequence> containing exactly one number X. Signs of type >1 are then joined> together to form elementary formulas. Godel defines the operation of> joining sequences (such as signs) to form a new sequence. So you have> to have sequences, not individual numbers, to join them to form a new> sequence, an elementary formula, in definition (20) in his paper.Charlie Volkstorf> Cambridge, MA>> hi,>>i have very technical and boring question, but may be somebody out>>there read the english translation of the Godel's original paper and>>could help me.>>Boring? Quite the contrary. It is refreshing to see someone discuss>> something of significance. (Too much time is spent quibbling over>> exercises in elementary arithmetic!)> My question is, in what sense 'sign of type n' and 'variable of type>>n' are the same ('dasselbe')?>>Let represent the sequence of numbers a, b, c, . . . >> Godel's system P (PM plus Peano's Axioms) contains:>>variables of type 1: X1, Y1, Z1, . . .>> variables of type 2: X2, Y2, Z2, . . .>> variables of type 3: X3, Y3, Z3, . . .>> etc.>> signs of type 1: 0, S0, SS0, . . . , X1, SX1, SSX1, . . . , Y1, SY1,>> SSY1, . . . , Z1, SZ1, SSZ1, . . .>> signs of type 2: , , , . . >> signs of type 3: , < , . . .>> etc.>>As you point out, the signs of type >1 are just the variables of type>>1 seen as sequences of length 1. The reason Godel defines signs of>> type >1 is so that he can juxtapose these sequences to form elementary>> formulas in (20) as: sign of type n+1(sign of type n). That is:>>elementary formulas: X2(0), X2(S0), X2(SS0), . . . , X2(X1), X2(SX1),>> X2(SSX1), . . . , Y2(0), Y2(S0), . . ., X3(X2), X3(Y2), . . .>>Elementary formulas are relation variables with arguments. Subsequent>> definitions include logical operators negation, disjunction and>> quantification in order to build Godel numbers for all of (higher>> order) Predicate Calculus theorem proving and show that the>> corresponding relationships are recursive.>>BTW: In (19) Godel includes v=> R(x)=2^x so that v=> also occurs in 11.) Why does he include this redundant clause? IMHO:>> This shows that these functions and relations are indeed recursive>> (see theorem IV.)>>I use the same technique in my Program Synthesis system>> (http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ 20021008.1/1>> section IV Proof 1 Step 2 clause ~LT(J,A) in my synthesis of programs>> that determine if one number is a factor of another.) Godel is just>> using Logic as a programming language, just as Peano did before him to>> write a program to demonstrate that the universal set of the natural>> numbers is recursively enumerable, which is axiom 1 in Section VI of>> http://www.arxiv.org/html/cs.lo/0003071 .>>In fact, many of Godel's axioms and definitions of recursive functions>> and relations are formal rules and theorems in this paper. This is>> because a computer program is simply a constructive proof that a>> particular set is recursive or recursively enumerable. Mathematicians>> such as Peano, Godel and Turing for generations have been showing that>> various sets and functions are, or are not, recursive or recursively>> enumerable by constructing programs in various bases of computing. >> Today we call that process Computer Programming!>>Charlie Volkstorf>> Cambridge, MA> thank you>maxim ... >> If c and 7 are coprime in A then there are elements p and q in A >> such that cp + 7q = 1. > >Be careful here. In James' all-inclusive ring of two complex conjugates, >one can be a unit while the other is not a unit. A bit strange, and I >do not know whether his ring really does exists, but the proof >fails for that. Because if c is a unit, q = 0 and p = inv(c). But >p' = inv(c)' does not necessarily belong to the ring (conjugation is >not a ring operation). >> So 1 = 1' = (cp + 7q)' = (cp)' + (7q)' = c'p' + 7'q' = c'p' + 7q', >> and since p' and q' are in A, c' and 7 are coprime in A. >Indeed, a valid proof is conjugation is a valid operation in the ring. > Actually, I thought I was careful to restrict my argument to the > algebraic integers, the ring in which James alleged coprimality.I do not know whether James alleges coprimality in that ring. He ispretty unclear about that. > I agree that such an argument might not go through in the Harris > ring, but might it not fail for a more basic reason? It is not > obvious that his ring would be a B'ezout domain.That is something I am also wondering about. But I think James' ringdoes not even need to be really be Bezout, but what he utters abouthis ring is not even sufficient to clear this. If I remember right,it has been shown that you can get rings with only one of two complexconjugates adjoined to the integers. > James has on several occasions lowered his shield of vagueness > sufficiently to allow identification of a specific element of his > ring that is not an algebraic integer.I have done a few times too, and each time I could not get throughthe conjugation. In James' case you need many complex conjugatesof which only one is included in the ring. Does that lead to acontradiction? I have no idea. It may be that it is possible tomake a choice in each case such that you get a complete ring. Onthe other hand, that might be impossible. > Each time, people have > jumped up and down pointing out the unintended consequences if that > number's (algebraic) conjugates were also included.Yup. > The usually > implicit appeal to the symmetry rendering conjugates algebraically > indistinguishable probably went way over James' head. Certainly, > he has always run away from the questions raised, but his silence > might be explained as withdrawal of his examples after realising > their inapplicability, perhaps due to some ineffable subtlety > visible only to himself.I think this is *not* the case. Note his current insistance where(I presume in his ring) one of two complex conjugate numbers isdivisible by 7 and the other is coprime to 7, where you can not tellwhich is which due to the duality of the sqrt operator. > I am therefore reluctant to deduce any > properties of his ring from those examples.The problem with James' proposition is that he needs a ring whereonly two of three factors are divisible by 7 and the third isco-prime to it. It is my opinion that *if* such a ring doesexist (i.e. where complex conjugates are not necessarily bothelement of the ring), that might be a proof of FLT for n=3 (I shouldread through James' proof to see whether that is true, but I amreluctant to do so until the existance of such a ring is established).The sad thing is that the inverse is not true (I think). So provingthe existence of such a ring is harder than proving FLT for n = 3.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ >...>>If c and 7 are coprime in A then there are elements p and q in A>>such that cp + 7q = 1.>>Be careful here. In James' all-inclusive ring of two complex conjugates,>>one can be a unit while the other is not a unit. A bit strange, and I>>do not know whether his ring really does exists, but the proof>>fails for that. Because if c is a unit, q = 0 and p = inv(c). But>>p' = inv(c)' does not necessarily belong to the ring (conjugation is>>not a ring operation).>> So 1 = 1' = (cp + 7q)' = (cp)' + (7q)' = c'p' + 7'q' = c'p' + 7q',>>and since p' and q' are in A, c' and 7 are coprime in A.>>Indeed, a valid proof is conjugation is a valid operation in the ring.>Actually, I thought I was careful to restrict my argument to the>algebraic integers, the ring in which James alleged coprimality.I do not know whether James alleges coprimality in that ring. He is>pretty unclear about that.>For once, he was reasonably explicit, in the thread title and inthe antepenultimate paragraph of the post starting this thread:>>Of course the problem with the ring of algebraic integers is the>>implication from previous interpretations (here I rely on what I've>>heard primarily from sci.math posters like Arturo Magidin, Nora>>Baron, and Dik winter) is that *both* roots have non-unit factors in>>common with 7 in the ring of algebraic integers.>I agree that such an argument might not go through in the Harris>ring, but might it not fail for a more basic reason? It is not>obvious that his ring would be a B'ezout domain.That is something I am also wondering about. But I think James' ring>does not even need to be really be Bezout, but what he utters about>his ring is not even sufficient to clear this. If I remember right,>it has been shown that you can get rings with only one of two complex>conjugates adjoined to the integers. > James has on several occasions lowered his shield of vagueness>sufficiently to allow identification of a specific element of his>ring that is not an algebraic integer.I have done a few times too, and each time I could not get through>the conjugation. In James' case you need many complex conjugates>of which only one is included in the ring. Does that lead to a>contradiction? I have no idea. It may be that it is possible to>make a choice in each case such that you get a complete ring. On>the other hand, that might be impossible. > Each time, people have>jumped up and down pointing out the unintended consequences if that>number's (algebraic) conjugates were also included.Yup. > The usually>implicit appeal to the symmetry rendering conjugates algebraically>indistinguishable probably went way over James' head. Certainly,>he has always run away from the questions raised, but his silence>might be explained as withdrawal of his examples after realising>their inapplicability, perhaps due to some ineffable subtlety>visible only to himself.I think this is *not* the case. Note his current insistance where>(I presume in his ring) one of two complex conjugate numbers is>divisible by 7 and the other is coprime to 7, where you can not tell>which is which due to the duality of the sqrt operator.I think I see what you mean. Perhaps James accepts that neither ofthe zeros, a and a', of x^2 - x + 42 is coprime to 7 in the ring Aof algebraic integers, but is asserting that in his ring, H, a isdivisible by 7 and a' is coprime to 7.In A, choose v = gcd {6, a} and w = gcd {7, a} so that a = vw.Then a' = v'w' and ww'=7u for some unit u in A. If A is a subring of H then a' = v'w' in H, and if a' is coprime to 7 in H then w' must be a unit in H and w/7 = u/w' is in H. It is then clear that 7 is not a unit in H, since 1 = a + a' = vw + v'w' = 7v(u/w') + v'w' and v'w' is coprime to 7 in H.At first sight, James would have no problems in working in one ofthe rings formed by adjoining a or a', but not both, to A.However, if he also adjoins quotients arising from equations otherthan the only one he is really interested in, all he achieves is toincrease the potential for undesirable interactions. But it is typical of James that, instead of adjoining just theelements he needs, he throws in everything but the kitchen sink andno one can say what are the properties of the ensuing monstrosity.> I am therefore reluctant to deduce any>properties of his ring from those examples.The problem with James' proposition is that he needs a ring where>only two of three factors are divisible by 7 and the third is>co-prime to it. It is my opinion that *if* such a ring does>exist (i.e. where complex conjugates are not necessarily both>element of the ring), that might be a proof of FLT for n=3 (I should>read through James' proof to see whether that is true, but I am>reluctant to do so until the existance of such a ring is established).>The sad thing is that the inverse is not true (I think). So proving>the existence of such a ring is harder than proving FLT for n = 3. I have not had time to look at any cubic examples.John Roberts-Jones >Message-id: Collatz tree? >Maybe this is known so just disregard!>I couldn't find anything about it on web searches about the> >Collatz.>Probably of little value for the overall conjecture but still a>curiosity!>The Collatz tree level count is odd up to level 5 where level>1,2,3,4,5 [1,2,4,8,16] has a count of (1) for each level. >At level 6 the count becomes even and fall one for one on each>side of the tree up too and including level 16.>This changes @ level 17 where the left side of the tree adds one>more path.>This happens also for certain levels higher than 17 where the>left side adds 1 more path over the right side.>I have denoted the 2 sides of the tree by their highest-ranking>sequence.>Where [1,2,4,8,16] is the trunk and the first main branch>assigned to the RIGHT side of the tree is --> >32,64,128,256,512,1024,2048,4096...and with all applicable nodes>branching from it creating all branches on the RIGHT side of>the Collatz tree.>Where [1,2,4,8,16] is the trunk and the first main branch>assigned to the LEFT side of the tree is -->5,10,20,40,80,160,320,640,1280,2560,...and with all applicable>nodes branching from it creating all branches on the LEFT side of>the Collatz tree.> 8) 20 3 21 128 > | | /> 7) 10 64> /> 6) 5 32> /> 5) 16> 4) 8> 3) 4 > 2) 2> Levels 1) 1>Why does the left side have this property of adding (1) more>branch at certain levels =>17 then the right side?> >Will this mean, at a very high level there will be many more> >branches on the left side of the tree than on the right side> >of the tree making the tree lopsided?>Putting the above question in another perspective, at some higher>level will the right side branching number eventually catch up and> >equal the left side branching number?>It could be that the left side of the tree is facing the sun giving>that side of the tree a more vigorous growth. ;-)>DanPersonally, I don't like the way you draw the Collatz tree. My preference> is for vertical placement to always represent x*2 and for the diagonal> placement to always represent (x-1)/3: 3 20 128 21> | | /> 10 64> | |> 5 32> |> 16> |> 8> |> 4> |> 2> |> 1By your method, 32 is part of the right branch whereas I consider the right> branch > single left branch and right branch. Rather, there is the central trunk from> which> sprout an infinite number of branches on either side, so I don't see any> problem.Note that in my view each branch starts with an odd number and extends> vertically> to infinity, all of which are even. Does this mean there are infinitely more> even> numbers than odd?Infinity always catches up in the end.Mansenator,I know the way I did the tree was different but I used each sequenceas a one to one correspondence (left,right) with each other.Where the (2) sequences, [3,6,12,24...]is the last sequence left ofcenter and [21,42,84,168...] is the last sequence right of center.This contains the inner limits and actually separates the tree intotwo halfs because these two sequences are 0 (mod 3)'s and willcontinue to double and continue on vertically ----->oo.The other two sequences, [5,10,20,40,80,160...]is the last sequenceor outer branch limit of the left side of tree and [32,64,128,256,...]is the last sequence or outer branch limit of the right side of treewhere both sequences double ----->oo.Presenting the tree in this why, it just seemed neat, up to level16 that you have this even looking tree and with boundry limitseven after level 16.Using this representation of the tree it is obvious the level seedcounts are the same as the standard tree layout.Right now it appears after level 16 there are increasingly more sequences with a (5) in its path other than a (32) in its path.True or false? This can be checked with the standard tree also proving or disprovingmy claim.Dan >Message-id: >Message-id: Collatz tree? >Maybe this is known so just disregard!>>I couldn't find anything about it on web searches about the>> >Collatz.>Probably of little value for the overall conjecture but still a>>curiosity!>>>The Collatz tree level count is odd up to level 5 where level>>1,2,3,4,5 [1,2,4,8,16] has a count of (1) for each level. >>At level 6 the count becomes even and fall one for one on each>>side of the tree up too and including level 16.>>This changes @ level 17 where the left side of the tree adds one>>more path.>>This happens also for certain levels higher than 17 where the>>left side adds 1 more path over the right side.>I have denoted the 2 sides of the tree by their highest-ranking>>sequence.>Where [1,2,4,8,16] is the trunk and the first main branch>>assigned to the RIGHT side of the tree is -->>32,64,128,256,512,1024,2048,4096...and with all applicable nodes>>branching from it creating all branches on the RIGHT side of>>the Collatz tree.>Where [1,2,4,8,16] is the trunk and the first main branch>>assigned to the LEFT side of the tree is -->> >5,10,20,40,80,160,320,640,1280,2560,...and with all applicable>>nodes branching from it creating all branches on the LEFT side of>>the Collatz tree.>>> 8) 20 3 21 128 >> | | />> 7) 10 64>> />> 6) 5 32>> />> 5) 16>> 4) 8>> 3) 4 >> 2) 2>> Levels 1) 1>Why does the left side have this property of adding (1) more>>branch at certain levels =>17 then the right side?>Will this mean, at a very high level there will be many more>>branches on the left side of the tree than on the right side>>of the tree making the tree lopsided?>Putting the above question in another perspective, at some higher>>level will the right side branching number eventually catch up and>> >equal the left side branching number?>It could be that the left side of the tree is facing the sun giving>>that side of the tree a more vigorous growth. ;-)>Dan>>> Personally, I don't like the way you draw the Collatz tree. My preference>> is for vertical placement to always represent x*2 and for the diagonal>> placement to always represent (x-1)/3:>>> 3 20 128 21>> | | />> 10 64>> | |>> 5 32>> |>> 16>> |>> 8>> |>> 4>> |>> 2>> |>> 1>>> By your method, 32 is part of the right branch whereas I consider the right>> branch >no>> single left branch and right branch. Rather, there is the central trunk>from>> which>> sprout an infinite number of branches on either side, so I don't see any>> problem.>>> Note that in my view each branch starts with an odd number and extends>> vertically>> to infinity, all of which are even. Does this mean there are infinitely>more>> even>> numbers than odd?>>> Infinity always catches up in the end.Mansenator,I know the way I did the tree was different but I used each sequence>as a one to one correspondence (left,right) with each other.>Where the (2) sequences, [3,6,12,24...]is the last sequence left of>center and [21,42,84,168...] is the last sequence right of center.>This contains the inner limits and actually separates the tree into>two halfs because these two sequences are 0 (mod 3)'s and will>continue to double and continue on vertically ----->oo.>The other two sequences, [5,10,20,40,80,160...]is the last sequence>or outer branch limit of the left side of tree and [32,64,128,256,...]>is the last sequence or outer branch limit of the right side of tree>where both sequences double ----->oo.>Presenting the tree in this why, it just seemed neat, up to level>16 that you have this even looking tree and with boundry limits>even after level 16.>Using this representation of the tree it is obvious the level seed>counts are the same as the standard tree layout.Ok, my diagram is biased towards the way I use the tree, so Ishouldn't complain if you're trying to prove a point.Right now it appears after level 16 there are increasingly more >sequences with a (5) in its path other than a (32) in its path.True or false?Well, I would say it's probably true. On my diagram, 5 is lower onthe trunk than the next live branch (85, 21 being 0 mod 3 has nosub-branches). For any given level, we would then expect thereto be more sub-branches leading to 5 since it got a head startin sprouting sub-branches.>This can be checked with the standard tree also proving or disproving>my claim.>I don't know if you saw that message I posted about sci.mathappreciation. If not, I want to thank you because that thread abouttwin primes in Collatz sequences led me to discover the solution toa problem that's had me stumped for 4 years. Sometimes thethinking out loud that takes place in these newsgroups is beneficialto those who are replying to other's queries.Dan--MensanatorAce of Clubs I am obssesively addicted to working on hard problems, especially likethe 3x+1(see http://www.cecm.sfu.ca/organics/papers/lagarias/.) Everytime, I look at any mathematical text, or do anything else I get drawnback to hard mathematical problems. I could easily spend hours, days,weeks, or even years working on these problems.I am currently a Sophomore in College, but I did very poorly lastsumester, and I don't know what to do. Peole have always though Ishould straight A's, but my grades have always muich worse. Many timesthis obsession is so strong I have trouble even reading or studyingmathematics as soon a some new idea possibly applicable to some hardproblem comes up.If I make a bit of an effort to stay away from math problems for alittle while, I easily get addicted to internet poker, and get a bitof a gambling problem.Another issue I face is that I have trouble writing, especially if itrelates novels or other fiction. I can do outlines, research, etc. butwhen it comes down to actually writing I get stuck, and can sit forhours and never be able to get started. I can write internet posts,but I can't write full papers.I talked to a psychiatrist, and I think I may have A.D.D. I guess Icould get some drugs. Does anyone have any ideas of things I could do?I don't really this obssesion to go away, but I also don't really wantto flunk out of college.Has anyone else experienced similar problems? Anyone have any advice,or know something I should do? >I am obssesively addicted to working on hard problems, especially like>the 3x+1(see http://www.cecm.sfu.ca/organics/papers/lagarias/.) Every>time, I look at any mathematical text, or do anything else I get drawn>back to hard mathematical problems. I could easily spend hours, days,>weeks, or even years working on these problems.I am currently a Sophomore in College, but I did very poorly last>sumester, and I don't know what to do. Peole have always though I>should straight A's, but my grades have always muich worse. Many times>this obsession is so strong I have trouble even reading or studying>mathematics as soon a some new idea possibly applicable to some hard>problem comes up.If I make a bit of an effort to stay away from math problems for a>little while, I easily get addicted to internet poker, and get a bit>of a gambling problem.Another issue I face is that I have trouble writing, especially if it>relates novels or other fiction. I can do outlines, research, etc. but>when it comes down to actually writing I get stuck, and can sit for>hours and never be able to get started. I can write internet posts,>but I can't write full papers. I am a 1967 Putnam fellow who did poorly in most English classes.I aced the grammar rules in 8th grade, but detested fiction.A book review which criticized the characters for double negativesand other grammatical errors was not what the teachers wanted. Go to the college library and find journals such as American Mathematical Monthly, College Mathematics Journal,Mathematics Magazine, Fibonacci Quarterly with problem sections.Do as many problems as you can, but be sure to write up your solutions.Install (free) Tex (probably latex2e) on your systemso you can format mathematical text. Review your solution a fewdays later to see whether it is clear and accurate, and to lookfor simplifications. When you are happy, submit it.In another year, look for some of your solutions to be published,Don't forget to try the Putnam in December, even if those solutionsare written up by hand. Peter Montgomery-- Intelligence agents foresee foreign flights crashing into landmarks.Those flights are from earth to mars. pmontgom@cwi.nl Home: San Rafael, California Microsoft Research and CWI > I am obssesively addicted to working on hard problems, especially like> the 3x+1(see .> I am currently a Sophomore in College, but I did very poorly last> this obsession is so strong I have trouble even reading or studying> mathematics as soon a some new idea possibly applicable to some hard> problem comes up.> If I make a bit of an effort to stay away from math problems for a> little while, I easily get addicted to internet poker, and get a bit> of a gambling problem.> Another issue I face is that I have trouble writing, especially if it> relates novels or other fiction. I can do outlines, research, etc. but> when it comes down to actually writing I get stuck, and can sit for> hours and never be able to get started. I can write internet posts,> but I can't write full papers.> or know something I should do?Just take courses that suit you and don't worry about a degree ?Or get into an engineering school where things like psychology-basedanalysis of literature is not so important ?Or just take one course per quarter ?Gambling ? Try trading options or futures. This high risk investing isactually less risk than gambling and thus can be reliable endeavor. >I am obssesively addicted to working on hard problems, especially like>the 3x+1(see .>I am currently a Sophomore in College, but I did very poorly last>this obsession is so strong I have trouble even reading or studying>mathematics as soon a some new idea possibly applicable to some hard>problem comes up.>If I make a bit of an effort to stay away from math problems for a>little while, I easily get addicted to internet poker, and get a bit>of a gambling problem.>Another issue I face is that I have trouble writing, especially if it>relates novels or other fiction. I can do outlines, research, etc. but>when it comes down to actually writing I get stuck, and can sit for>hours and never be able to get started. I can write internet posts,>but I can't write full papers.>or know something I should do?Just take courses that suit you and don't worry about a degree ?Or get into an engineering school where things like psychology-based> analysis of literature is not so important ?Or just take one course per quarter ?Gambling ? Try trading options or futures. This high risk investing is> actually less risk than gambling and thus can be reliable endeavor.Math,I can't tell if your ADD, though your grades history indicates theposability.When you speak of uncontrolled urges that is usually called ObsessiveCompulsive. I beleive I got my ADD from my mother. She had bothgambling and alchohol problems as well. It is not unusual to havemore than one condition.In any case I think seeking a professional opinion is an excellentidea for you.One man's opinion. Next!Ray > I can write internet posts, but I can't write full papers.>.. know something I should do?Better to resolve emotional conflicts by oneself, taking out time forit. Else BM or Behaviour Modification methods can be tried on bytrained psychologists. optimization software/solvers: it turns outto me in NEOS there are two solvers for integer optimzation: I'd like to trybut just don't know there capacity and where are they running... Ipreviously had some experiences on running some large scale it was just too slow...So I want to ask what are normally the capacity of the NEOS integeroptimization solvers? Are they running fast, for hundreds variables integeroptimization problems?I have also seen a number of solvers/softwares on Internet... it was just aquestion that which ones are better before I try one by one... I have onlyone computer to use, so if I spend several days on one solver, then turn toanother it is bad, it will need me several months to at least get a littleidea...Also there are a number of modeling languages, one must learn in order to do my hundredsvariable integer optimization problem?We have CPLEX 8.1 in our school, what is the relationship between this CPLEX8.1 and other softwares on the Internet?Oooops, after several days diving on Internet, these stuff really confusedme... Please help me to choose the What is the smallest number of dimensions that can support the largestnumber (e.g >500,000)of objects meeting the following two conditions? Condition One: circle radius = xCondition Two: circles boundaries touch (and do not overlap) What is the smallest number of dimensions that can support the largest> number (e.g >500,000)of objects meeting the following two conditions?Condition One: circle radius = x> Condition Two: circles boundaries touch (and do not overlap)If I understand your question correctly you are looking forY. Edel, E. M. Rains and N. J. A. Sloane, On Kissing Numbers inDimensions 32 to 128, Elect. J. Combinatorics 5(1), paper R22 (1998).http://www.combinatorics.org/Volume_5/PDF/v5i1r22. pdfFor n=36 a kissing number of 438872 is obtained, and for n=40 991792.Hugo Pfoertner Suppose A is a real square matrix whose entries are all in (0,1) andsuch that the sum of each of its columns is 1. It can be seen as amatrix whose entries are probabilities. Then, it's easy to see thatA^2 satisfies these same properties and, therefore, so does A^n forevery natural n. Using a spreadsheet I concluded that the sequence(A^n) converges to a matrix whose colums are identical, have sum 1 andcorrespond to an eigenvector of A associated with the eigenvalue 1.However, I couldn't give a formal proof for this fact. I tried usingfixed points theorems (maybe Brower's could do), but got confused. Ialso tried using the minimal polynomial of A. Could anyone give anidea?In the convergence processe, I defined the norm of A as ||A|| =Sqrt(Sum(i=1 to n Sum (j=1 to n) (a_i_j)^2) (just and extension of thenorm of a Euclidean vector - is this the usual definition of the normof a matrix?) and, based on Cauchy criterion, tried to make ||A^m -A^n|| < eps for m and n sufficiently large.I noticed that if the entries a_i_j are allowed to be in [0,1] thenthe above conclusion may fail, and I get a permutation of A if one ofthe a_i_j's is 1 (which implies the others in the same column are 0)Artur >Suppose A is a real square matrix whose entries are all in (0,1) and>such that the sum of each of its columns is 1. It can be seen as a>matrix whose entries are probabilities. Then, it's easy to see that>A^2 satisfies these same properties and, therefore, so does A^n for>every natural n. Using a spreadsheet I concluded that the sequence>(A^n) converges to a matrix whose colums are identical, have sum 1 and>correspond to an eigenvector of A associated with the eigenvalue 1.>However, I couldn't give a formal proof for this fact. I tried using>fixed points theorems (maybe Brower's could do), but got confused. I>also tried using the minimal polynomial of A. Could anyone give an>idea?The transpose of your matrix is an irreducible aperiodic stochastic matrix. Look up Markov chain or Perron-Frobenius Theorem.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In said you thought>the set theory you had in mind was from the 1950s.What I saw was a text book, not one of the original papers. Now I'mgoing to have to see whether I can track down a copy. I should havebought it when I had the chance :-(I'll get back to you if I can find a copy and check whether I got anydetails wrong.-- Shmuel (Seymour J.) Metz, SysProg and <3ffc9a68$7$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In doing things like arbitrarily banning self-membership of sets>or postulating some sort of infinite hierarchy of sets. What's wrong with an infinite hierarchy of sets? And how do you doAnalysis without infinite sets?>To me, it all seems so unnecessary and so inelegant. De gustibus non diputandem est.>One way -- and I don't>know if it is really anything new -- is to have a simplified set>theory that does not postulate the existence of any particular set>or sets.That somewhat restricts its utility, doesn't it?>It seems to work and neatly avoid problems like RP that have plagued naive set theory.Problems that were dealt with nearly a century ago. Certainly GBN andZF don't have those problems.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to > In by doing things like arbitrarily banning self-membership of sets>or postulating some sort of infinite hierarchy of sets. What's wrong with an infinite hierarchy of sets?Nothing. It just seems to be unnecessary. That's all.And how do you do> Analysis without infinite sets?>Who said anything about doing without infinite sets? If you want to developnumber theory in my system, for example, you simply start with a premisecorresponding to Peano's Axioms which includes an axiom of infinity. What doyou do in ZF if you want to develop group theory? It's axioms are not builtinto ZF. Wouldn't you simply start by assuming the group axioms?>To me, it all seems so unnecessary and so inelegant. De gustibus non diputandem est. >One way -- and I don't>know if it is really anything new -- is to have a simplified set>theory that does not postulate the existence of any particular set>or sets. That somewhat restricts its utility, doesn't it?>Not really.>It seems to work and neatly avoid problems like RP that have plaguednaive set theory. Problems that were dealt with nearly a century ago. Certainly GBN and> ZF don't have those problems.>Indeed. The only advantage my system offers is its simplicity. My purpose,however, is not to supplant ZF or GBN, but to provide a learning tool forthe non-specialist taking introductory courses in courses in logic or puremath. While it may suit your purposes, for them, ZF is needlesslycomplicated and overly abstract.DanVisit DC Proof Online at http://www.dcproof.com -- FREE download > De gustibus non diputandem est.You should get your Latin right, if you are going to post it.Thomas <1001nipi5rcmc28@corp.supernews.com> How about base -10. You get the 1's place, the -10's place, the 100's place, the -1000 place, and> so on. With this scheme you can write any number positive or negative> without using a minus sign. And in base -2, N bits can represent all the numbers in the range[-2/3*((2^N)-1), 1/3*((2^N)-1)], inclusive. Interesting. If onlythere were a way to do calculations with these numbers... :) Anyone have any references for this system? Looks interesting.[Crossposted to sci.math.]-Arthur(I guess that also implies that 3 | (2^N)-1, which looked interestingat first until I saw that 2^2k = 4^k = 1 (mod 3), so duh.) , Arthur> How about base -10.> You get the 1's place, the -10's place, the 100's place, the -1000 place,>and>so on. With this scheme you can write any number positive or negative>without using a minus sign.You do need 10 digits, though.Continue with decimals, places worth -1/10, 1/100, -1/1000, etc. And in base -2, N bits can represent all the numbers in the range> [-2/3*((2^N)-1), 1/3*((2^N)-1)], inclusive. Interesting. Another interesting one for COMPLEX numbers uses base -1+i and digits{0,1}. Represent all complex numbers that way.> If only> there were a way to do calculations with these numbers... :)> Anyone have any references for this system?Knuth, The Art of Computer Programming. Vol 3 I think.> Looks interesting.-Arthur(I guess that also implies that 3 | (2^N)-1, which looked interesting> at first until I saw that 2^2k = 4^k = 1 (mod 3), so duh.)-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ I was wondering, what are some of the criteria that mathmaticians usein order to figure ouf if there is (an) anlytical solution(s) to agiven ODE or ODE system?Is there any textbook that will list such criteria if there is suchthing?any thoughts are welcome. john I was wondering, what are some of the criteria that mathmaticians use> in order to figure ouf if there is (an) anlytical solution(s) to a> given ODE or ODE system?analytical? Do you mean analytic in the sense the solution isgiven by a power series? Or do you mean closed form in some sense?The theory of integration in closed form has some things to say aboutsolution of DEs in closed form. See M. Bronstein, _Symbolic Integration I: Transcendental Functions_ (Springer-Verlag 1997)-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ >> I was wondering, what are some of the criteria that mathmaticians use>> in order to figure ouf if there is (an) anlytical solution(s) to a>> given ODE or ODE system?>analytical? Do you mean analytic in the sense the solution is>given by a power series? Or do you mean closed form in some sense?>The theory of integration in closed form has some things to say about>solution of DEs in closed form. See> M. Bronstein, _Symbolic Integration I: Transcendental Functions_> (Springer-Verlag 1997)Closed-form solutions can often be found using symmetries. These methodsare implemented in Maple's dsolve command. See e.g. Symmetry and Integration Methods for Differential Equations by George W. Bluman, Springer-Verlag2002, ISBN 0387986545.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 >Closed-form solutions can often be found using symmetries. These methods>are implemented in Maple's dsolve command. See e.g. Symmetry and Integration >Methods for Differential Equations by George W. Bluman, Springer-Verlag>2002, ISBN 0387986545.Sorry, that should be George W. Bluman and Stephen C. Anco. Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 > I was wondering, what are some of the criteria that mathmaticians use> in order to figure ouf if there is (an) anlytical solution(s) to a> given ODE or ODE system?If the coefficients are analytical, then there is an analyticalsolution. This is Cauchy-Kovalevskaya's theorem; seehttp://www-gap.dcs.st-and.ac.uk/~history/Projects/Ellison/ Chapters/Ch5.html> Is there any textbook that will list such criteria if there is such> thing?I suggest the first chapter of A primer of real analytic functions, bySteven G. Krantz and Harold R. Parks.Jose Carlos Santos So this is where George Cox hangs out.It actually seems as though he is reasonable and helpful with people inhere.You must have a split personality George - you make an arse of yourselfelsewhere with your trolling!!LOLLOLRussell--You had better stop fighting by the time I get back,oryou're ALL grounded.-God >> How can i proove that equation x^3+2y^3+4z^3-34xyz=0 is true>> then x=y=z=0, if x,y,z are whole numbers?that x, y, and z are even, so x/2, y/2, z/2 is also a solution.Sorry, i made a mistake writing question, i need to prove that ONLY ONE solutionexists (0,0,0) in whole numbers------------------------------------------------------- -------------------->John R Ramsden (jr@adslate.com)>--------------------------------------------- ------------------------------>Eternity is a long time, especially towards the end.> Woody Allen En el escribi.97:> Ramsden says...>> On > How can i proove that equation x^3+2y^3+4z^3-34xyz=0 is true> then x=y=z=0, if x,y,z are whole numbers?>> that x, y, and z are even, so x/2, y/2, z/2 is also a solution.> Sorry, i made a mistake writing question, i need to prove that ONLY> ONE solution exists (0,0,0) in whole numbersIt is the same yo has said before, no?Suppose that (x, y, z) is a solution.Then,x^3+2y^3+4z^3 = 34xyz > x must be even. Let x = 2x'8x'^3 + 2y^3 + 4z^3 = 68x'yz >4x'^3 + y^3 + 2z^3 = 34x'yz 68x'y'z = 34x'y'z' > (x', y', z') also is a solution!Then, if you have a solution (x, y, z), then (x/2, y/2, z/2) is a integersolution, then ((x/2)/2, (y/2)/2, (z/2)/2)) = (x/4, y/4, z/4) also is ainteger solution, ...., as (x/2^k, y/2^k, z/2^k). You have infinitely manyinteger solutions, each one of them with less absolute value that theprevious. But it is impossible, except that x/2 = x, y/2 = y and z/2 = z > x = y = z = 0.This method of prove, a particular case of reasoning by contradiction, iscalled 'infinite descend', ant it is due to Fermat.-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com thanx to both of you, now it's clear for me:)This method of prove, a particular case of reasoning by contradiction, is>called 'infinite descend', ant it is due to Fermat.>-- Ignacio Larrosa Ca.96estro>A Coru.96a (Espa.96a)>ilarrosaQUITARMAYUSCULAS@mundo-r.com>-- Ignacio Larrosa Ca.96estro>A Coru.96a (Espa.96a)>ilarrosaQUITARMAYUSCULAS@mundo-r.com Aahh !I got it !!> if lim [p(x)] = lim [q(x)] when x-> c, then lim[p'(x)] = lim> [q'(x)] when x-> c. when x-> c.[/quote:0fc2fa9f81]When c = oo ( oo = infinite ), if lim p(x) =oo, lim q(x) =oo whenx->oo, then lim [p(x) /q(x)] = lim[p'(x) / q'(x)] ( that isL'Hopital rule )let h(x) = ax + b, lim [f(x)/h(x)] = lim [f'(x)/h'(x)] = 1 when x->oo ( because h(x) is asymptote of f(x).So lim f'(x) = lim h'(x) ( x-> oo ), or f'(oo) = h'(oo) = aSince g''(x) >0, we have g'(x) is an increasing function forx>0,so g'(x) < g'(oo) = f'(x) - a < f'(oo) - a = 0 ( because f''(x)> 0 for x>0 )Thus g'(x) < 0 for all x>0And so, g(x) is decreasing functioc for x>0, so g(x) > g(oo) =lim [f(x) - h(x)] = 0 when x>0[/b:0fc2fa9f81]Is my solution exact ?---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- > The matter now is that I don't know this is right or wrong :> if lim [p(x)] = lim [q(x)] when x-> c, then lim[p'(x)] = lim> [q'(x)] when x-> c.> May anyome confirm that for me, plzzz.> If it's right, I can solve the problem above.News==----> http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 > ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption > =---Let c = 0, p(x) = x and q(x) = x^2, to show your conjecture false. Given a nonzero cardinal k, what is known about the maximal cardinality of all sets of pairwise non-homeomorphic topologies on k? For infinite k, is it equal to (2^k)^k? Is there a known combinatorial formula for finite k? How about the maximal cardinality of sets of non-homeomorphic T_x topologies (x in {0, 1, 2, 3, 3.5, 4, 5})? Of metric spaces?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu > Given a nonzero cardinal k, what is known about the maximal> cardinality of all sets of pairwise non-homeomorphic topologies on> k? For infinite k, is it equal to (2^k)^k?You probably mean 2^(2^k), seeing as (2^k)^k = 2^k. Anyway, that isthe answer to your question: there are 2^(2^k) non-homeomorphictopologies on a set of cardinality k. There are 2^(2^k) topologies ona set of (infinite) cardinality k, and each homeomorphism classcontains at most 2^k of those topologies.> Is there a known combinatorial formula for finite k?Beats me. See A001930 at topologies (x in {0, 1, 2, 3, 3.5, 4, 5})? Of metric spaces? >Given a nonzero cardinal k, what is known about the maximal>>cardinality of all sets of pairwise non-homeomorphic topologies on>>k? For infinite k, is it equal to (2^k)^k?>>>You probably mean 2^(2^k), seeing as (2^k)^k = 2^k. Anyway, that is>the answer to your question: there are 2^(2^k) non-homeomorphic>topologies on a set of cardinality k. There are 2^(2^k) topologies on>a set of (infinite) cardinality k, and each homeomorphism class>contains at most 2^k of those topologies.>Yes, sorry - I meant 2^(2^k), the cardinality of P(P(k)). Is the proof relatively easy? If so, hints? If not, reference? >Is there a known combinatorial formula for finite k?>>>Beats me. See A001930 at> scribbled the following:>> Given a nonzero cardinal k, what is known about the maximal>> cardinality of all sets of pairwise non-homeomorphic topologies on>> k? For infinite k, is it equal to (2^k)^k?> You probably mean 2^(2^k), seeing as (2^k)^k = 2^k.Don't you mean (2^k)^k = 2^2k, assuming that ^ means exponentation?-- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland ---------- http://www.helsinki.fi/~palaste --------------------- rules! --------/I said 'play as you've never played before', not 'play as IF you've neverplayed before'! - Andy Capp > Fred Galvin scribbled the following:>> Given a nonzero cardinal k, what is known about the maximal>> cardinality of all sets of pairwise non-homeomorphic topologies on>> k? For infinite k, is it equal to (2^k)^k? > You probably mean 2^(2^k), seeing as (2^k)^k = 2^k. Don't you mean (2^k)^k = 2^2k, assuming that ^ means exponentation?I meant (2^k)^k = 2^(k^2), but it's all the same since k is infinite(and I'm assuming the axiom of choice), k^2 = k = 2k. A variable X follows an Ito Process if:dx=a(x,t)dt+b(x,t)dz Where dz is a wiener process.Ito's lemma shows that if a variable G is a function of x and t:G=g(x,t), it will follow the process described in Ito's differential.If G is a function of x, t, and G itself: G=g(x,t,G), can I use Itolemma? if not, is there any alternative way to solve the problem? Consider a simpler situation first. Suppose X_t = f(X_t). If this isto hold almost surely, then P(X_t in S) = 1, where S = {x: f(x) = x}.But then again, if X_t is in S almost surely, then we don't care aboutthe values of f(x) outside of S, so we may as well assume f(x) = x.> A variable X follows an Ito Process if:> dx=a(x,t)dt+b(x,t)dz > Where dz is a wiener process.> Ito's lemma shows that if a variable G is a function of x and t:> G=g(x,t), it will follow the process described in Ito's differential.> If G is a function of x, t, and G itself: G=g(x,t,G), can I use Ito> lemma? if not, is there any alternative way to solve the problem? I didn't see this thread till after I posted. I find thatmany of my questions have ongoing or recent threads.NSA was what I was thinking of but had forgotten what they called it.(Don't know how my message got posted twice.)Van > Say we are working in the ring of differential operators...How would you go about simplifying the following...> [Using * for the ring multiplication][cos(t) * d/dt * sin(t) * d/dt]Basically I am not sure if you can 'forget' the rightmost d/dt since> that is just to differentiate the function - i.e it does not> simplify...so would it then become [cos(t) * [d/dt * sin(t)] * d/dt]> [cos(t) * [cos(t) + Sin(t) * d/dt] * d/dt]> [cos^2(t) * d/dt + cos(t)sin(t) *d/dt ] etc...Or am I forgetting something ?> - SandosPlease this has me quite stuck... I am unsure how to =>>> [cos(t) * d/dt * sin(t) * d/dt]>>> Basically I am not sure if you can 'forget' the rightmost d/dt>> since that is just to differentiate the function - i.e it does not>> simplify... Please this has me quite stuck... I am unsure how to proceedYou can't throw a differential operator away. Somebody else shouldpipe up if I'm wrong, but this looks like a chain rule. Put D ford/dt, and you get: cos(t) * D * sin(t) * D= cos(t) * D * (sin(t) * D) (by associativity)= cos(t) * (D*sin(t) + DD) (by the chain rule)= cos(t) * (cos(t) + D^2)= cos^2(t) + cos(t)*D^2Len. > I have to build a context free grammar that generates the following >language : >{ a^i b^j c^k | i,j,k >= 0 and k <= i+j }> [his attempt deleted]It looks like you're on the wrong track.First solve the problem of generating b^j c^k with k<=j,That's what was supposed to do S5.> and then try to generate a^i B c^k with k<=i, where> B is something you have already generated.DGA >I have to build a context free grammar that generates the following language :>{ a^i b^j c^k | i,j,k >= 0 and k <= i+j }I take it you are unaware of the pumping lemma for CFL's?Yes, I am. I have been learning grammars for a week . > Perhaps you should do some of the exercises that appear in the> first hit at Can You agree to the following?The important part of a function is the way from the independentvariable in the domain to the dependent in the range (the arrowin f:R ->R )When i monitor the weather in my place, i 'll geta function, i attribute temperature to time. Temperature is not a term calculated from time.In math You have to be precise, so if You have a function, which is not term-defined, You have to attribute to every element of thedomain an element of the range, the domain must be precise.Attribute to every nunber the amount of digits of this number indecimal notation, f.e. 3 ->1, 2.114 ->4,...This function is not term-defined, at least not for a simple mind like me. Choose the domain R, then You are not able to draw a graph,still You can choose a domain at Your choice.f(x)=1/x is a term-defined function - You can say it's a functionR -> R, as everybody knows the rules of calculation ofdivision (x must not be zero) and by these rules can determinethe exact domain. The same with the function f:R -> R: x -> ln (x^3+x+3), as ln is defined for values greater zero.The algebraic structure in which You are calculating isimportant. f: R2 -> R2 : z=(x,y) -> (1/x, 0) is not term-defined.f(z) = ( 1/ ( z dot(1,0)) ) real scalar multiplication (1,0)is term defined and obviously in (R2, +, r.s.m., dot ), theeuclidian vectorspace. And the term requires, that the firstcomponent x must not be zero.This function can not be defined by a term in (R2, +, * ),the commutative field, which is known under the nameof complex numbers.Have fun Hero Can You agree to the following?The important part of a function is the way from the independentvariable in the domain to the dependent in the range (the arrowin f:R ->R )When i monitor the weather in my place, i 'll geta function, i attribute temperature to time. Temperature is not a term calculated from time.In math You have to be precise, so if You have a function, which is not term-defined, You have to attribute to every element of thedomain an element of the range, the domain must be precise.Attribute to every nunber the amount of digits of this number indecimal notation, f.e. 3 ->1, 2.114 ->4,...This function is not term-defined, at least not for a simple mind like me. Choose the domain R, then You are not able to draw a graph,still You can choose a domain at Your choice.f(x)=1/x is a term-defined function - You can say it's a functionR -> R, as everybody knows the rules of calculation ofdivision (x must not be zero) and by these rules can determinethe exact domain. The same with the function f:R -> R: x -> ln (x^3+x+3), as ln is defined for values greater zero.The algebraic structure in which You are calculating isimportant. f: R2 -> R2 : z=(x,y) -> (1/x, 0) is not term-defined.f(z) = ( 1/ ( z dot(1,0)) ) real scalar multiplication (1,0)is term defined and obviously in (R2, +, r.s.m., dot ), theeuclidian vectorspace. And the term requires, that the firstcomponent x must not be zero.This function can not be defined by a term in (R2, +, * ),the commutative field, which is known under the nameof complex numbers.Have fun Hero mogensn@mogensn.dk (Mogens Nielsen)>litterature on the real case.Actually come to think of it I'm not nearly as sure of what I said asI was yesterday... let's review:>>If you want me to be more precise, here is the problem:>>Let K be a compact subset of R^k. Let T:C(K) -> C(K) be an operator,>> >where C(K) is the set of continuous functions from K to R. In the>>of T, without defining it.>>> He's probably referring to the complex spectral radius, ie the sup >> of |lambda| for lambda in the (complex) spectrum. Because that's>> the _standard_ definition of the spectral radius, even for real >> Banach algebras.Come to think of it, if B is a real Banach algebra then it really makes no sense to talk about the complex spectrum: L is inthe spectrum of x if x - Le is not invertible, but if L is complexthere's no such thing as x - Le in the first place.What seems most likely to me is that C(K) is actually the spaceof continuous functions from K to C; that's the standard meaning,it seems to me. Are you sure he's talking about functions fromK to R?Another possibility is that the standard definition of the spectrumof an element of a real Banach algebra is the spectrum in thecomplexification of the algebra. And another possibilityis that he's really talking about the real spectrum (this lastseems least likely to me).Just realized I'm not certain what's the standard definitionof the spectrum in a real Banach algebra, how embarassing(never thought about real Banach algebras). But I betit's supposed to be the spectrum in the complexificationof the algebra - that would be consistent with the standarddefinition of the spectrum of a real matrix, for example.>Mogens************************David C. Ullrich the smallest non-definable ordinal is definable.>why is this argument wrong?When you post the same message to two different groups youshould _cross-post_ instead of posting the message separately!You would do that by putting sci.math,sci.logic in thenewsgroups field (how you do that depends on whatsoftware you're using.)The reason is this: If you crosspost then readers in sci.logicwill see replies from sci.math (I would not have bothered with giving an explanation in sci.logic if I'd known there werealready correct explanations in sci.math. More important,if you're posting a question you assume that the readerswill be interested in the answer - so you should assume thatreaders of sci.logic will benefit from replies made in sci.math.)************************David C. Ullrich > the smallest non-definable ordinal is definable.> why is this argument wrong?> I don't think that it is definable, because it seems to me that definability is not a first order property. You presumably say a set x is definable if there is some first order statement p such that x = { y : p(y) } ?. If you try to turn this into a first order property then you end up trying to get your quantifiers to range over predicates, which dosn't work.So, as you can't talk about the property 'is definable' in a first order way, there certainly doesn't have to be a least such element with that property in a first order theory.Or did you mean something else by definability?The thing to be careful about in ZF and other axiomatic set theories is that, while most of the time intuitive and verbal reasoning about the axioms will do the job, they really are first order theories and you need to stay within the first order reasoning if you expect them to hold together.David > the smallest non-definable ordinal is definable.> why is this argument wrong?> It's nothing to do with ordinals. The same this happens with:The least natural number not definable in fewer than twentyEnglish words > the smallest non-definable ordinal is definable.> why is this argument wrong?Because you cannot express the smallest non-definable ordinal using the language of first order logic within ZF. I think that what it boils down to is that it is not possible, within ZF, to show that the class of all sets is a model for set theory.Let me give more details. ZF is consistent if and only if NBG is consistent. NBG is a theory in which the basic objects are classes rather than sets. In NBG a class is defined to be a set if it is an element of another class. Amongst NBG's axioms is: if a is a set, and if F(x) is a well formed formula in which all quantifiers are over sets, then {x in a: F(x)} is also a set.What you would like to do is to show that the class of all sets forms a model for ZF. But if you could do that, this would contradict Goedel's Theorem. It turns out that in order to go through the usual construction to show that the class of all sets is a model of ZF that one would need to work in a richer theory, that is, with this axiom: if a is a set, and if F(x) is a well formed formula, then {x in a: F(x)} is also a set. This theory has to be richer, because it is possible to prove that ZF is consistent within this theory.In order to construct the smallest non-definable ordinal within your set theory, you would need this richer set theory. This is not a paradox - all you have done is to define the smallest ordinal not definable in ZF, which will certainly not be the smallest ordinal definable in this richer theory.I hope this helps.Stephen >the smallest non-definable ordinal is definable.>why is this argument wrong?Because you cannot express the smallest non-definable ordinal using> the language of first order logic within ZF. I think that what it> boils down to is that it is not possible, within ZF, to show that the> class of all sets is a model for set theory.I think you have pulled the wool over giovanni's eyes (or tried to).Nothing in his question mentioned ZFC. Presumably he means something like definable in English. Of course,there is still an answer, and it proceeds similar to yours, butdifferently, in some interesting ways.Thomas Thomas Bushnell, BSG ha scritto nel messaggio> Presumably he means something> like definable in English.no, I mean definable in the sense of the descriptive set theory!!!! > Thomas Bushnell, BSG ha scritto nel messaggio>Presumably he means something>like definable in English.no, I mean definable in the sense of the descriptive set theory!!!!Oh, then the original answer seems entirely correct to me. Now I have just solved the problem, I have understood. > I think that what it boils down to> is that it is not possible, within ZF,> to show that the class of all sets> is a model for set theory.scuse me, can you explain to me what has that to do with it?> What you would like to do is to> show that the class of all sets> forms a model for ZF.why is it necessary for my argument?> It turns out that in order to go> through the usual construction to> show that the class of all sets is> a model of ZF that one would> need to work in a richer theoryBernays-Morse, for example.but in my opinion I don't want (and don't have) to show that the class of allsets is a model of ZF.> In order to construct the smallest> non-definable ordinal within your> set theory, you would need this> richer set theory.why?I think I can construct the smallest non-definable ordinal also in the ordinalc of the continuum.I hope you can explain to me why I'm wrong... Now I have just solved the problem, I have understood. > .9^2 = .81> .99^2 = .9801> .999^2 = .998001> .9999^2 = .99980001> .99999^2 = .9999800001> .999999^2 = .999998000001> .9999999^2 = .99999980000001> .99999999^2 = .9999999800000001> .999999999^2 = .999999998000000001> .9999999999^2 = .99999999980000000001> .99999999999^2 = .9999999999800000000001> .999999999999^2 = .999999999998000000000001> .9999999999999^2 = .99999999999980000000000001> .99999999999999^2 = .9999999999999800000000000001> .999999999999999^2 = .999999999999998000000000000001> .9999999999999999^2 = .99999999999999980000000000000001> .99999999999999999^2 = .9999999999999999800000000000000001> .999999999999999999^2 = .999999999999999998000000000000000001> .9999999999999999999^2 = .99999999999999999980000000000000000001> .99999999999999999999^2 = .9999999999999999999800000000000000000001Each position after the decimal point eventually becomes a 9 and stays> that way.Daniel W. Johnson has proven that 0.999... will never equal 1.I knew someone here could do it.Garry Denke, GeologistDenoco Inc. of Texas > Daniel W. Johnson has proven that 0.999... will never equal 1.No, I've proven that the 8's, 0's, and 1's in your blitherings areirrelevant. Because no matter what number you choose less than 1, thesquares of the original Cauchy sequence will eventually exceed it (andremain above it).In other words, the squares of 0.999... tend toward 1.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W In sci.math, Garry = .81>> .99^2 = .9801>> .999^2 = .998001>> .9999^2 = .99980001>> .99999^2 = .9999800001>> .999999^2 = .999998000001>> .9999999^2 = .99999980000001>> .99999999^2 = .9999999800000001>> .999999999^2 = .999999998000000001>> .9999999999^2 = .99999999980000000001>> .99999999999^2 = .9999999999800000000001>> .999999999999^2 = .999999999998000000000001>> .9999999999999^2 = .99999999999980000000000001>> .99999999999999^2 = .9999999999999800000000000001>> .999999999999999^2 = .999999999999998000000000000001>> .9999999999999999^2 = .99999999999999980000000000000001>> .99999999999999999^2 = .9999999999999999800000000000000001>> .999999999999999999^2 = .999999999999999998000000000000000001>> .9999999999999999999^2 = .99999999999999999980000000000000000001>> .99999999999999999999^2 = .9999999999999999999800000000000000000001>>> Each position after the decimal point eventually becomes a 9 and stays>> that way.Daniel W. Johnson has proven that 0.999... will never equal 1.> I knew someone here could do it.All he's proven is that (.999...9)^2 leads to a Cauchy sequence.This is a good start, as Cauchy sequences inherently definereal numbers.http://mathworld.wolfram.com/CauchySequence.htmlSo what is the limit of that sequence?We know .999...9 = 1 - 10^(-n) for any finite expansion.Pick an epsilon > 0, and I can find an M = ceil(log(1/epsilon))which is such that, for any N > M, 0 < (1 - 10^(-N) ) - 1 < epsilon.So I can get as close to 1 as desired. Squaring, cubing, fourthing,fifthing -- it doesn't matter too much as the only adjustmentis that I might have to use a slightly bigger M:M = ceil(log(5/epsilon)) would probably work for fifth powers,for example.Of course things get a little weird if one uses variable powers.(1 - 1/n)^n = 1/e, for example, as n tends towards positive infinity.(In your case, that might equally easily be expressed as(1 - 10^(-n))^(10^n), leading to .9^10, .99^100, .999^1000, etc.Try it -- that series, at least, does *not* tend to 1.)Garry Denke, Geologist> Denoco Inc. of Texas-- #191, ewill3@earthlink.netIt's still legal to go .sigless. > In sci.logic, Dave Seaman>> Omega+1 has the same cardinality as omega, but they are different>> ordinals. If the (w+1)-strings of digits are ordered lexicographically,>> then there is no order-preserving map of such strings to the reals.> Interesting. So expressions such as omega+1 and 2*omega actually do> make sense? If so, mea culpa.>> Yes. In fact, w is identical to N, the set of natural numbers (including 0),>> and w+1 = w U {w} = { 0, 1, 2, 3, ..., w }. >> Addition and multiplication are noncommutative: >> 1+w = w < w+1, (w+1 has a last element but 1+w does not)>> 2*w = w < w*2 = w+w. (an w of pairs vs. a pair of w's)> Ugh. My brain's beginning to hurt now... :-)> So which one of these applies to such Garry-Denke-isque decimal> expansions as> (.999....)^2 = .999...99800...001 ?None of them, as far as I can see.> It feels like 8 is in the w+2 or w+3 position but I want to make sure.> 1 is presumably in the position w*2 + 2.That seems to be implied by the notation, but I'll resist any urge to tryto make sense of it.> Presumably 1+w refers to things like> 9,(9,9,9,9,...)> where one tacks things onto the left side of an infinite sequence.> If one computes .999... / 10 + .8 = .8999..., one gets a similar> sequence.> However, if we do allow transfinites and infinitesimals into our> number system a la Garry Denke, I'm not sure how the rest of mathematics> is going to deal with it (the concept of limit in particular may have> to be redefined).> Hopefully it's a simple retrofit but I wonder.>> There's nothing wrong with considering transfinite ordinal sequences of>> digits in general, but as I said, there is no order-preserving map of>> such objects to the reals for ordinals > w.> That might be good enough. :-) Of course that's probably related> to the impossibility of reliably defining things such as> .999... - .999...99800...001 = .000...?!?!?!...>:-)I wouldn't go so far as to say it's impossible. After all, nonstandardanalysis makes sense of infinitely large integers. But then, thehyperintegers of NSA are not the same as the transfinite ordinals, and Iam not aware that anyone has tried to use digit strings indexed by thehyperintegers to define values in the hyperreals.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. In sci.logic, Dave Seaman> Omega+1 has the same cardinality as omega, but they are different> ordinals. If the (w+1)-strings of digits are ordered lexicographically,> then there is no order-preserving map of such strings to the reals.>> Interesting. So expressions such as omega+1 and 2*omega actually do>> make sense? If so, mea culpa.> Yes. In fact, w is identical to N, the set of natural numbers (including 0),> and w+1 = w U {w} = { 0, 1, 2, 3, ..., w }. > Addition and multiplication are noncommutative: > 1+w = w < w+1, (w+1 has a last element but 1+w does not)> 2*w = w < w*2 = w+w. (an w of pairs vs. a pair of w's)>> Ugh. My brain's beginning to hurt now... :-)>> So which one of these applies to such Garry-Denke-isque decimal>> expansions as>> (.999....)^2 = .999...99800...001 ?None of them, as far as I can see.>> It feels like 8 is in the w+2 or w+3 position but I want to make sure.>> 1 is presumably in the position w*2 + 2.That seems to be implied by the notation, but I'll resist any urge to try> to make sense of it.Yeah, well, it doesn't make much sense anyway in light of thesubtraction below. :-) As Barb has pointed out, 10^(-omega) = 0by any reasonable metric, so if one states that 8 is in the w+2position then the 8's value equals 10^(-(omega+2)) = 10^(-omega) * 10^2= 0 * .01 = 0.Personally, I think .999... is a great example of a Cauchy sequence,with limit 1. :-)>> Presumably 1+w refers to things like>> 9,(9,9,9,9,...)>> where one tacks things onto the left side of an infinite sequence.>> If one computes .999... / 10 + .8 = .8999..., one gets a similar>> sequence.>> However, if we do allow transfinites and infinitesimals into our>> number system a la Garry Denke, I'm not sure how the rest of mathematics>> is going to deal with it (the concept of limit in particular may have>> to be redefined).>> Hopefully it's a simple retrofit but I wonder.> There's nothing wrong with considering transfinite ordinal sequences of> digits in general, but as I said, there is no order-preserving map of> such objects to the reals for ordinals > w.>> That might be good enough. :-) Of course that's probably related>> to the impossibility of reliably defining things such as>> .999... - .999...99800...001 = .000...?!?!?!...>>:-)I wouldn't go so far as to say it's impossible. After all, nonstandard> analysis makes sense of infinitely large integers. But then, the> hyperintegers of NSA are not the same as the transfinite ordinals, and I> am not aware that anyone has tried to use digit strings indexed by the> hyperintegers to define values in the hyperreals.> NSA presumably routinely works with very large (1,024-bit)integers, but they *are* integers; the challenge is mostlyin defining operations therewith in terms of arraysof smaller integers, sort of like defining multidigitarithmetic in terms of the times table, except thatinstead of a 10 [Times] 10 matrix one can memorize by rote,one has a 65536 [Times] 65536 affair that is implemented by themicroprocessor's multiply instruction.It's not that difficult in principle but it's tricky.I have some ideas on how to compute M^e mod N, where allof M, N, and e are very large 1,024-bit integers, buthave no idea how well they'd work. (The simplest methodI can think of is to keep squaring M in an accumulator,multiplying the accumulator with M whenever there's a '1'.It would still take 1,024 to 2,048 multiplications of twomultiprecision numbers. There are probably more elegantmethods.)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. > ... The sum of _many_ beta distributions converges to a gaussian distribution, > but neither my beloved book (and some hard core integral solvining) or extensive search on the internet has given me> anything on what the distribution of the sum of two beta functions is. ... The distribution of a sum of independent Beta random variables is close to normal if no variance is large compared to the sum, and the parameters do not get too extreme;The pdf of the sum of 2 Beta rv's (the question posed above) will generally not be close to Normal. One can derive the characteristic function of the sum, and then invert it numerically to derive the pdf,for given parameter values. For example, here is a quick 2-line derivation and pdf plot usingmathStatica: http://www.mathstatica.com/Sumof2Betas/ Colin ______________________________Dr Colin RosemathStatica Pty LtdWeb: www.mathStatica.com______________________________ >... The sum of _many_ beta distributions converges to a gaussian distribution,>but neither my beloved book (and some hard core integral solvining) or extensive search on the internet has given me>anything on what the distribution of the sum of two beta functions is. ...> The distribution of a sum of independent Beta random variables> is close to normal if no variance is large compared to the sum,> and the parameters do not get too extreme;The pdf of the sum of 2 Beta rv's (the question posed above) will> generally not be close to Normal. One can derive the characteristic> function of the sum, and then invert it numerically to derive the pdf,> for given parameter values.For example, here is a quick 2-line derivation and pdf plot using> mathStatica: http://www.mathstatica.com/Sumof2Betas/> ColinBTW: is there a 'standardized' way to measure how fara pdf is from being normal? I mean, hm, a practicalone and may be only for some classes of pdf. Concerning the results of the Gelfound-Schneider Theorem(http://mathworld.wolfram.com/GelfondsTheorem.html), which states thatwhen A is an algebraic number, and B is an irrational and algebraicnumber, A^B is a transcendental number.So, given the contrapositive, which would state that if C is not atranscendental number, then the log(Base A) of C, where A is analgebraic number, would not be irrational and algebraic.And thus this leads me to believe, and please correct me if I've madea mistake in my logic, that a logarithm with any integer base, ofanother integer, is rational or transcendental. Does this mean thatmost logarithms with integer bases of integers are transcendental? Orcould they all be rational? >Concerning the results of the Gelfound-Schneider Theorem>(http://mathworld.wolfram.com/GelfondsTheorem.html), which states that>when A is an algebraic number, and B is an irrational and algebraic>number, A^B is a transcendental number.>So, given the contrapositive, which would state that if C is not a>transcendental number, then the log(Base A) of C, where A is an>algebraic number, would not be irrational and algebraic.>And thus this leads me to believe, and please correct me if I've made>a mistake in my logic, that a logarithm with any integer base, of>another integer, is rational or transcendental. Does this mean that>most logarithms with integer bases of integers are transcendental? Or>could they all be rational?Most such logarithms are transcendental. The lograithm of one integerwith respect to another integer is only rational if they are both powersof the same integer.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. >Concerning the results of the Gelfound-Schneider Theorem>(http://mathworld.wolfram.com/GelfondsTheorem.html), which states that>when A is an algebraic number, and B is an irrational and algebraic>number, A^B is a transcendental number.So, given the contrapositive, which would state that if C is not a>transcendental number, then the log(Base A) of C, where A is an>algebraic number, would not be irrational and algebraic.Yes. If C is algebraic then either A is transcendental or B is eitherrational or transcendental.>And thus this leads me to believe, and please correct me if I've made>a mistake in my logic, that a logarithm with any integer base, of>another integer, is rational or transcendental.Yes.>Does this mean that>most logarithms with integer bases of integers are transcendental? Or>could they all be rational?Let log_a(c) = n/m where n and m are relatively prime. Then c isinteger only if a^(n/m) is integer which is true only if a = d^(km)where d and k are integers. If a is prime then the only solution is k= 1, d = a and m = 1. So any logarithm with a prime integer base willonly have integer and transcendental values. Download beta 1 - www.master-graph.com/mgraph20b1.exe (only 477KB).All who will help me to improve this program will get a free registrationkey.You can do the following:-find bugs-feedback about you impression by the program-advice me to change or add some new features-translate it to the other language (seewww.master-graph.com/instructions/interface.html)Feel free to contact with me - >2. Why can symmetry be used in the solving procedure ?If an equation has a certain symmetry, one can show that the *set* of> its solutions has to have the same symmetry....Do you know the name of the person who proved that ? >One see people>-expand the the assumed solution in a set of basis functions that have>the same symmetry as the system considered,Well, usually the basis functions have *not* the same symmetry as the> system... (for example, not all wave functions for the H atom are> spherically symmetric!)>and solve for the>coefficients as system of linear equations.(Gauss-elimination)> The basis functions of the one-dimesional representaion of the group>that the system belongs to, are the basis functions that are used. Why?Sorry, I'm not very familiar with this...>Because the symmtry operators and the hamiltonian operator commute>(?????????)Help !When one can find operators for certain symmetries which commute with> the Hamiltonian, one usually chooses the wave functions to be> eigenfunctions to these operators, too. Again, this is done because it> usually simplifies finding the solutions (example: for solving the H> atom, it is *very* helpful to try to find a wave function which is an> eigenstate for angular momentum, too - because then one already knows> that the wave function has to be a multiple of a spherical harmonic!)Any references about the theory of the commutation ? >3. How does variational theory (and what is it ?) come into the>solving theory of Schr.9adinger eqn. ?AFAIK,AFAIK ?? You've lost me. the Schroedinger equation usually can't be solved by the> variational theory - one can only find *approximate* solutions. What it> is? Well, the idea (in the case of the Schroedinger equation) is to use> some test functions and calculate the expectation value for the energy> for them. It is easy to prove that these expectation values have to be> greater or equal to the ground state energy, hence by looking for the> lowest possible energy among all of the test functions, one gets an> upper bound for the ground state energy. (this is only the tip of the> iceberg, but I think you get the general idea...)> Easy introductory text?> 4. Perturbation theory in quantum mechanics. Is it one of these>none-proven things ?>Well, I think in many mechanics classes, this is considered to be an> advanced topic.>When you suggest books, consider that I have the education of an>electronics engineer. No fancy mathematics courses like functional>analysis, etc.But apparently you have some experience with differential equations and> matrices, don't you? What about the general definition of vector> space? What about the notion of differential operators?> Sure Bjoern, I've read some intriductory quantum mechanics books, andthey often have a section of functions whicj are orthogonal to eachthat I can writea diff. eqn. like (d^2/dt^2)y + y=0, like (D^2 +1)*y=0.I've calculated these type of equations in the calculus courses. But Ihave no clue of the behindlying theory. It is the theory I want to diginto....> Lasse Then the> branches depend more immediately on these cuts which define a domain> for them than on the branch points. Discussion of the consequences of> this requires a few more paragraphs and will be for another time. The following supplies what was left out of my Jan6 post. In order for branches to be fully defined a domain has to beprovided for them. The use of cuts to produce a simply-connectedregion where the monodromy theorem applies is familiar and sosingle-valuedness of each branch on the domain is obtained. A branchis just a function from the cut plane to a region of the w-plane andis distinguished from its fellows by its initial value. If the given relation is linear in the independent variable z thenthe branches have disjoint ranges. For a proof let an ordinary point whave the single counter-image z and be in the ranges of the brancheswhich have the initial values wh and wk. An arc in the cut plane say Afrom z0 to z is imaged by (among others) an arc from wh to w and anarc in the cut plane say B from z0 to z is imaged by an arc from wk tow. Then the circuit AinverseB is imaged by the arc wh to w to wk. Bythe cut construction and the monodromy theorem the circuit must beimaged by a circuit so wh=wk and so w cannot belong to two ranges.When f(w,z) is non-linear in z overlapping ranges can be expected. Permutations on branches are intrinsic to closed circuits but onewants to associate them with crossings of cuts. NB: A permutationcannot generally be associated consistently with just a branch-point.Put an arrow (orientation counts) across each cut. Now take a circuitfrom and to the initial point which crosses one cut only in thedirection of its arrow. The permutation produced by the circuit isgiven to the arrow. Crossing the opposite way gives the inversepermutation. A general circuit crosses more than one cut. The permutation itproduces is the same as the product of the assigned permutations takenin the order in which the cuts are crossed. This is shown bydistorting the circuit back to the initial point in between crossings, Now translatability, to get permutations for a system of cuts fromthose assumed already known for another system. A circuit crossing onecut only in the system for determination may cross more than one cutin the known system. But its permutation is determinable by theprevious paragraph and this permutation just has to be given to thecrossed cut. It only needs another step to define a Riemann surface. But perhapsimplicit functions can be studied just as well staying with branchesand using the above theory. One advantage is that the z-plane and thew-plane can be represented on different parts of the blackboard andthe details filled in without any strain on the visual imagination. Copyright Jan04. Quotation with acknowlegement permitted. ...> Permutations on branches are intrinsic to closed circuits but one>wants to associate them with crossings of cuts. NB: A permutation>cannot generally be associated consistently with just a branch-point.>Put an arrow (orientation counts) across each cut. Now take a circuit>from and to the initial point which crosses one cut only in the>direction of its arrow. The permutation produced by the circuit is>given to the arrow. Crossing the opposite way gives the inverse>permutation.> A general circuit crosses more than one cut. The permutation it>produces is the same as the product of the assigned permutations taken>in the order in which the cuts are crossed. This is shown by>distorting the circuit back to the initial point in between crossings,...I don't know whether you (or anyone here) took the (not inconsiderable)trouble to look up the references I gave, to several papers (two orthree by Orevkov, one by Zoladek), in an earlier post. Within the a full-scale, at least a more detailed exposition of the point of view which is (somewhat sketchily) described in their papers (and, aboutas sketchily, in some earlier publications of mine). For the timebeing, I will simply point out here that (in the case where yourRiemann surface is sitting inside C^2, as the set where someholomorphic function f(z,w) of two complex variables vanishes)you can do better than putting permutations on branch cuts:you can (if you chose the cuts canonically, to be where two distinct branches have equal real part) put *braids* onbranch cuts. (And in this case the arrows are really important;in the generic situation where the permutation for each cut isjust a transposition, and therefore self-inverse, the arrowsare insignificant. One way to think of the braid group on nstrings is, in fact, to start with the presentation of thepermutation group on n letters which has has generators thetranspositions of adjacent letters from 1 to n, and asrelations the rules (1) (i i+1)(i+1 i+2)(i i+1) =(i+1 i+2)(i i+1)(i+1 i+2) for i from 1 to n-2, (2)(i i+1) commutes with (j j+1) for j>i+1, (3) the squareof each generator is the identity, and throw away therelations (3).) When you do so you (potentially, andactually) learn more about f(z,w) than you do withoutbraids. (And you learn some interesting things aboutbraids, too.)Lee Rudolph > Here is a simplified reformulation of my previous problem:> How to solve this integer matrix equation?> [ a1^2*u1, a1*a2*u2, a1*a2*u3, a2^2*u4]> [ a1*a3*u1, a1*a4*u2, a3*a2*u3, a2*a4*u4]> [ a1*a3*u1, a3*a2*u2, a1*a4*u3, a2*a4*u4]> [ a3^2*u1, a3*a4*u2, a3*a4*u3, a4^2*u4]> => [ 16 16 -16 -4]> [ 4 -16 -4 4]> [ 4 4 16 4]> [ 1 -4 4 -4]a1, a2, a3, a4, u1, u2, u3, u4 should be 2's positive integer power, such> as -2, +1, etc....> Since there are more equations than unknowns, sometimes there are> conflicting assignment to the variables, so only approximated assignment can> be made to minimize the mean square error between the left hand side and> right hand side in the above equation......> The only problem is that one the matrix gets larger, with hundreds of> unknowns, [exhaustive search] cannot be used any more...Can anybody give me more ideas and pointers?I haven't followed your other posts so don't know how typicalthe above example is of the structures of the real problem; ifit is at all typical, you would be better off treating it as aset of independent equations. Making obvious number of equations: -jiw Hey,We all know the 3X+1 conjecture is unprooved to this moment,But I was wondering if we know by proof that f.i. the 5X+1 ( or 7X+1, ...)analogue problem has divergent sequences.My guess is that such a proof could be 'simpler' and maybe is 'already outthere' but i don't ...Does someone know of any proof of these generalizations or can help me withsome links for more informatioon about this generalizations?Thx in advance Hey,We all know the 3X+1 conjecture is unprooved to this moment,But I was wondering if we know by proof that f.i. the 5X+1 ( or 7X+1, ...)analogue problem has divergent sequences.My guess is that such a proof could be 'simpler' and maybe is 'already outthere' but i don't ...Does someone know of any proof of these generalizations or can help me withsome links for more informatioon about this generalizations?Thx in advanceKristof Clevers >Last moon>right over my head, I tossed one rupee coin.>It was Head.>Today morning, I tossed again one rupee coin in blue sky.>Again, it was Head.>So God is telling me to file one more patent application. I have>already filed 7 patent applications.> No, God is telling you that Rosencrantz and Guildenstern are dead.-----------------------------In the end, R&G resign themselves to their fate, although Guildensternsays, There must have been a moment, at the beginning, when we couldhave said -- no. But somehow we missed it. Perhaps.-----------------------------No? To what?Is it beginning? Or is it end?In the beginning, we are never aware of what is going to happen infuture. By this time, I am sure that absolutely everything is setup.What is happening now, it could have happened when I was in Mumbai forabout 50 days and it could have happened when I had about Rs.50,000/-I ran to Mumbai for eight times, filed 7 patent application. And nowwhen, I am completely devastated, running Rs.40,000/- negative, I havegot complete understanding of this mechanism of gravity.Destiny. It is already written, well scripted. This entire Universe isa stage and we are just Actors. Somekind of drama is taking place andit has been shown to me that, yes, everything is setup. But He willnever tell me what is scripted in future.It is up to God now, I said it and I have dropped envelope containingmy patent application no.8 in post box just short time ago. It isgoing by ordinary post. Patent office may never receive it or even ifthey receive, those people may never issue any receipt. I will neverknow.For my beloved Gravity...------------------------------------tadap tadapke is dil se Aah nikalti rahimujhko saza di pyaar Ki aisa kya gunaah kiyato lut gaye haan lut gayeto lut gaye Hum teri mohabbat mein-----------------------------------Tomorrow morning, I will start execution sequence. As soon as I seethis Action Device hanging in air, I will disclose it and mechanism tolocal media, people. I am going to follow my deadline 14 January 04.14PM(IST).I just can't hang anymore...Apocalypse NOW!!-Abhi. >I ran to Mumbai for eight times, filed 7 patent application. And now>when, I am completely devastated, running Rs.40,000/- negative, I have>got complete understanding of this mechanism of gravity.Rs.? Are you talking about Indian Rupees? I hope not, because thatworks out to about US$880 that you're in debt. Not exactly the Enroncollapse, Dude.--V.G.People are more violently opposed to fur than leather, because it is easier to harrass rich women than it is motorcycle gangs. - Bumper Sticker(This sig file contains not shield. I met the following problem i one book,which can be solved either bymeasure theory or by Lebesque Dominated Convergence theorem,but theauthor says it is too difficult to handle without these means .I thinkI need help on it.Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each xin [0,1].Then Int(fn(x)dx,0,1)--->0!Mladen I met the following problem i one book,which can be solved either by> measure theory or by Lebesque Dominated Convergence theorem,but the> author says it is too difficult to handle without these means .I think> I need help on it.Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x> in [0,1].Then Int(fn(x)dx,0,1)--->0!> I remember seeing a problem like this in one of the early chapters in Real and Complex Analysis by Rudin. My memory is that it also contained a reference to a paper which had the solution to this problem. But it was about 20 years ago when I saw this, so my memory might be faulty. > I met the following problem i one book,which can be solved either by>measure theory or by Lebesque Dominated Convergence theorem,but the>author says it is too difficult to handle without these means .I think>I need help on it.Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x>in [0,1].Then Int(fn(x)dx,0,1)--->0!Define f_n for n >= 2 by 2 f (x) = n x for x in [0,1/n] n = n(2 - nx) for x in [1/n,2/n] = 0 for x in [2/n,1]Each f_n is continuous on [0,1] and f_n(x) -> 0 for each x in [0,1],yet, for all n >= 2, |1 | f (x) dx = 0 | 0 nAll the f_n are dominated by 1/x, but 1/x is not in L^1[0,1]. Thereis no function in L^1[0,1] that dominates all the f_n.Rob Johnson take out the trash before replying >> I met the following problem i one book,which can be solved either by>>measure theory or by Lebesque Dominated Convergence theorem,but the>>author says it is too difficult to handle without these means .I think>>I need help on it.>>Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x>>in [0,1].Then Int(fn(x)dx,0,1)--->0!>Define f_n for n >= 2 by> 2> f (x) = n x for x in [0,1/n]> n> = n(2 - nx) for x in [1/n,2/n] = 0 for x in [2/n,1]He did say fn:[0,1]--->[0,1].Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 I met the following problem i one book,which can be solved either by> measure theory or by Lebesque Dominated Convergence theorem,but the> author says it is too difficult to handle without these means .I think> I need help on it.Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x> in [0,1].Then Int(fn(x)dx,0,1)--->0!It is false as stated. See if you can find a counterexample.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ >> I met the following problem i one book,which can be solved either by>> measure theory or by Lebesque Dominated Convergence theorem,but the>> author says it is too difficult to handle without these means .I think>> I need help on it.Help on what? The measure theory, the Lebesgue Dominated Convergence Theorem,or the too difficult way? >> Let fn:[0,1]--->[0,1] are continuous functions and fn--->0 for each x>> in [0,1].Then Int(fn(x)dx,0,1)--->0!>It is false as stated. See if you can find a counterexample.I see nothing to complain about here except the grammar, unless you're reading ! as factorial, or taking fn as a net instead of a sequence.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 The page with links to mathematical departments in the worldhttp://www.math.hr/~duje/mathdept.htmlnow contains links to 1003 departments from 93 countries.Any comments and suggestions are very welcomed.Andrej Dujella Is there an efficient way to solve the following problem:Given a,b,eps>0 find all positive n of being integer.For example, let a=sqrt(2), b=sqrt(3) and eps=10^-k. I've compiled a table of> n that work for various small values of k. Is it complete?[The numbers are shortened: 297..072(26) is the 26 digit number starting with> 296 and ending with 072]k=13> 297..072(26)> 617..409(26)> 915..158(26)k=14> 394..552(28)> 483..617(28)> 685..472(28)> 774..537(28)k=15> 135..884(30)k=16> 218..750(32)k=17> 173..971(34)k=18> 506..352(36)> 546..517(36)> 972..568(36)k=19> 131..132(38)k=20> 239..576(40)> 296..519(40)> 904..493(40)> 961..436(40)richJust an idea: Use continued fractions, seeBest Rational Approximations to Real Numbers by R. Knotthttp://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/ cfINTRO.html#convergents Hugo Pfoertner Jack Zamat> In the development of the calculus, notation was an important factor. The> typical notation of d/dt and the integral sign are from Liebntiz, and> because Newton had such jacked up notation, England fell behind the restof> continental Europe in analysis for over 100 yrs. My question is this: Iknow> that Newton used the dot notation for differentiation, but what type of> notation did he use for the integral/integration? If anybody knows theanser> or can point me to a source, I would be most appreciative.A translation of Newton's Principia would tell the tale, assuming that thetranslator has not modernized the notation. If I remember correctly, he hadno explicit notation for integrals. If he wanted to say (for example) thatx^3 / 3 is the integral (inverse fluxion) of x^2, he would phrase it likex^2 will be the fluxion of x^3 / 3. I do know that he used a whole longseries of ingenious but rather arbitrary synthetic constructions, in thestyle of Euclid and Archimedes, in his problems on integration.LH <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <87brpe1wtx.fsf@phiwumbda.org> <878ykhzlss.fsf@phiwumbda.org> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In flaw in this argumentWhat is obvious to you and what is true are two very different things.>is that every real number has a definition.You're making a pun, not presenting a logical argument. The statementthat there are only a finite number of definable real numbers refersto a specific definition of definable, and it is not the definitionthat you are using.>It is hard to imagineThat is a flaw in your imagination.>Doesn't this mean that every real number is computable?No. Again, you're making a pun instead of an argument.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to <3ffd3edf$21$fuzhry+tra$mr2ice@news.patriot.net> tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Terminate: SPA(GIS) In said:>I enjoy Shmuel's postings -- usually he sets people straight. In>this case however I think he misremembers what Goedel did.G.9adel did several things. One of them was to translate staements aboutapparently stronger systems into statements about naturals. That lead,among other things, to his incompleteness result, but I'm pretty surethat he also proved con(PA) => con(ZF).-- Shmuel (Seymour J.) Metz, SysProg <3ffd3edf$21$fuzhry+tra$mr2ice@news.patriot.net> <40019929$48$fuzhry+tra$mr2ice@news.patriot.net> > In said:>I enjoy Shmuel's postings -- usually he sets people straight. In>>this case however I think he misremembers what Goedel did. G.9adel did several things. One of them was to translate staements about> apparently stronger systems into statements about naturals. That lead,> among other things, to his incompleteness result, but I'm pretty sure> that he also proved con(PA) => con(ZF).Pretty sure? As a result of pretty sure, you advised me thus?,----| Well, it takes more machinery. Read a Scientific American papaer| called Goedel's Proof for background, then read On the Consistency| of the Axiom of Choice and the Generalized Continuum Hypotheses.`----This after saying (in the same post), [...] but it's oldnews. Google for G.9adel. I suppose it's good that you're stating your opinions with a bit ofhumility now, but it would've been better if you had expressed thatyou're pretty sure all along. I made the mistake of taking your word for this claim in a couple ofposts (although, I always made clear it was hearsay for me). I didthis despite your annoying arrogance and condescension just becauseyou seemed confident of the result of which you're now pretty sure. Isuppose I learned my lesson.references unless you are certain that they contain the result ofinterest. For extra points, it would be swell if you were certainthat *some* paper contain the result of interest. If I hadn't been too damn lazy to look up Goedel's original, I'd be atouch pissed.Aside: You write, One of them was to translate staements aboutapparently stronger systems into statements about naturals. That lead,among other things, to his incompleteness result. What apparentlystronger system was relevant for his proof of the incompletenesstheorems? Unless you take his proof as involving a translation ofmetamathematical statements into arithmetic, I don't know what youmean here. Personally, I wouldn't use the term translation there,since the metamathematics were not treated as a formal theory as faras I recall. Maybe you have something else in mind.-- Jesse F. Hughes[Mathematical] society has evolved far enough away from mainstreamsociety that it has become rogue, and now is willing to push its needsagainst that of the majority. -- James S. Harris >>Rationals are Uncountable> Then how do you explain the numerous _injective_ mappings one may > construct from Q to N?For example:An arbitary rational may be represented uniquely as p/q, where, WLOG, we > may require that: > p is an integer, and > q is a positive integer, and> p and q have no common factors except 1 and > if p = 0 then q = 1Define f : Q -> N by> f(p,q) = 2^p*5^q if p is positive or zero, and> f(p/q) = 3^(-p)*5^q if p is negative.It is trivial to show that f is injective, thus the cardinality of the > rationals cannot be larger than that of the naturals.And, of course, the trivial injective mapping g:N -> Q : n -> n/1> completes the issue.We already know the rationals are countable. The idea is to find the defect(s) in his proof. >Also, thank you to R3769, Michel Hack and Jesse F. Hughes for correcting>>my sloppy thinking concerning the notion of inconsistency with regard to>>mathematics. I now understand that an inconsistency in ZFC does not>>mean an inconsistency in mathematics. >Well, now we must be a bit careful. Shmuel Metz informs me that>Goedel proved the relative consistency of ZFC to PA --- so that *if*>PA is consistent, *then* ZFC is consistent. Hence, an inconsistency>in ZFC would in fact yield an inconsistency in PA --- not due to the>simple fact that PA has a model in ZFC, but to the more surprising>(reported) fact that ZFC has a model in PA.>>Umm. Are you sure you're remembering that correctly? > I'm fairly sure that I'm remember what Shmuel said and am stating it> correctly. I don't know anything about the claim than I've said> above.My apologies to Shmuel if I've misrepresented things.> Having discussed this with a friend who knows more about the subject than me, we've concluded that the statement is wrong if one assumes the consistency of ZF(C)Here's a proof:Suppose you can construct a model of ZFC in PA. There is a set in ZFC whic his a model of PA, so in that set we can then construct a model of ZFC. This means we have some set in ZFC which is itself a model of ZFC, so ZFC must be consistent. This contradicts one of Goedel's theorems, as a system strong enough to prove it's own consistency must be inconsistent (if it has a model of PA in it).David >> In jesse@phiwumbda.org (Jesse F. Hughes) said:> I don't see that an inconsistency in ZFC would necessarily yield an> inconsistency in, say, Peano arithmetic.>> Because, having modelled ZFC in PA, an inconsistency in ZFC can be>> translated back into an inconsistency in PA.>> ZF with the axiom of infinity replaced by its negation is equivalent> to PA.You are helpful too!!Russell--Luck is when the paths of opportunity and preparation cross. : Its the same machine.> : Assume I have a TM that always finds a natural number larger than> : any natural number on any given finite tape.> : This same TM will try to find the largest natural on an infinite tape.> : The TM has no way of knowing the tape is infinitely long.> : This TM will always say there is a natural number larger than> : any natural it has found on the tape. What is the output of the following program if it never reaches> the end of input?I am assuming the computer can perform an infinite number ofoperations in finite time.> while (!end_of_input)> {> input x;> if (x>max)> max=x;> }> print max;It will print a natural number.We are assuming that every input is a natural number.max must also be a natural number.> What do you think the output of this program would be? while (true)> {> n=n+1;> }> print done;>It will print an infinitely long string of numbers followed bythe word done.Russell- 2 many 2 count I am assuming the computer can perform an infinite number of>operations in finite time.Yes. I suppose there is nothing *wrong* with making this assumption providedyou are willing to live with it's consequences. Clearly, you will need to bemore careful about what infinite number of operations means. It's unlikelythat you will find many followers of your beliefs without first demonstratingsome very interesting (and useful) consequences of your assumption. It's alsopossible you will deduce from your assumption something that is untenable, andthus have to abandon it. In any event you will have learned something, andthat is always a very good thing.When I last asked about whether there were any models of computation thatallowed an infinite number of operations in finite time, the answer was: no,but there are for oracle is the best (however lame) advice I can offer. Good Luck!rich > : Its the same machine.>: Assume I have a TM that always finds a natural number larger than>: any natural number on any given finite tape.>: This same TM will try to find the largest natural on an infinite tape.>: The TM has no way of knowing the tape is infinitely long.>: This TM will always say there is a natural number larger than>: any natural it has found on the tape.> What is the output of the following program if it never reaches>the end of input?I am assuming the computer can perform an infinite number of> operations in finite time.> while (!end_of_input)>{>input x;>if (x>max)>max=x;>}>print max;It will print a natural number.> We are assuming that every input is a natural number.> max must also be a natural number.But, from a tape containing all natural numbers, it will never reach an end of input condition, so it will never print anything.> What do you think the output of this program would be?> while (true)>{>n=n+1;>}>print done;> It will print an infinitely long string of numbers followed by> the word done.(1) It will not print any numbers, since there is no instruction to do so, and (2) it will never print done since it will be locked in an endless loop not containing the print command.> Russell> - 2 many 2 countyou mean - 2 dumb 2 count <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> > What is the output of the following program if it never reaches>the end of input? I am assuming the computer can perform an infinite number of> operations in finite time. You assume wrongly. PLEASE PLEASE PLEASE think before you post!-Arthur <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <1g75yr4.1si6z861lg1554N%panoptes@iquest.net> <97adneZXNdeRbWCi4p2dnA@comcast.com>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In PM, Russell Easterly said:>There is no theoretical limit on how fast something can be computed.There is, however, a theoretical limit one whether something thatdoesn't satisfy the definition of a TM is a TM.In PM, Russell Easterly said:>I am assuming that a TM can perform an infinite>number of operations in finite time.Such a device might have a role in the theory of computation, but itwould not be a turing machine and theorems about Turing Machines wouldnot apply to it.>It appears that induction and infinite computation come to different>conclusions on some problems.No. The different conclusions come from applying things to contextswhere they are inapplicable.-- Shmuel (Seymour J.) Metz, SysProg and tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In at 07:04 PM, r3769@aol.com (R3769) said:>So every theorem of ZFC is also a theorem of ZF. But not the same theorem.>At this point the only real problem is that we all>know the C in ZFC stands for Choice, and not, say, Cohen. The flaw is that you are assuming that the corresponding theorems arein fact identical, which they are not. Theorems in ZFC translate intovery different, and more complex, theorems in ZF.>However, as near as I can figure, yes it is true that you can encode>statements about ZFC as statements about the naturals but NOT in the>way Shmuel Metz suggests. That there is an interpretation, f, from>the theory cn(PA) to the theory ZFC is a fact. You've got it backwards. What is at issue is translating statementsabout ZFC into statements about PA. You need to be very careful inlooking at this, because the translations are not what you wouldnaively expect.>Because interpretations are, by definition, 1-1. No. All that is required is that the interpretation satisfies the sameaxioms. For a more recent, and perhaps more understandable, example Metz, SysProg and JOATnot reply to > at 07:04 PM, r3769@aol.com (R3769) said:>So every theorem of ZFC is also a theorem of ZF. But not the same theorem.>At this point the only real problem is that we all>>know the C in ZFC stands for Choice, and not, say, Cohen. The flaw is that you are assuming that the corresponding theorems are>in fact identical, which they are not. Theorems in ZFC translate into>very different, and more complex, theorems in ZF.>Yes, I agree the argument is flawed. What I can't figure out is how theoremsin ZFC translate into theorems of ZF, yet axioms can't be so translated.>>However, as near as I can figure, yes it is true that you can encode>>statements about ZFC as statements about the naturals but NOT in the>>way Shmuel Metz suggests. That there is an interpretation, f, from>>the theory cn(PA) to the theory ZFC is a fact. You've got it backwards. What is at issue is translating statements>about ZFC into statements about PA. You need to be very careful in>looking at this, because the translations are not what you would>naively expect.>True enough. I would add that translating theorems in ZFC to theorems in PAwas also an issue.>>Because interpretations are, by definition, 1-1. No. All that is required is that the interpretation satisfies the same>axioms. Ok. I clearly need to think about this some more. . richX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Terminate: SPA(GIS) question: Do you mean to say that ZFC is inconsistent implies>ZF is inconsistent or that ZF can't be modelled in PA?Kurt G.9adel proved that. Likewise if ZF plus the Generalized ContinuumHypothesis is inconsistent then ZF is inconsistent.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In at 11:02 PM, r3769@aol.com (R3769) said:>However, the *relative* inconsistency of>PA isn't really what we are concerned about. For mathematics to be>doomed, you still need to show the inconsistency of ALL other set>theories, ST, where ST can be modelled in PA. If PA is inconsistent, then anything more powerful than PA isinconsistent. So proving the inconsistency of anything modeled in PAwould indeed be a fatal blow. -- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to <87smiudtrj.fsf@phiwumbda.org> <3FF9DD0F.CC42A336@mdli.com> <87isjmyzmw.fsf@phiwumbda.org>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In jesse@phiwumbda.org (Jesse F. Hughes) said:>I won't get into that can of worms.I resolve that by claiming to be a Formalist on odd numbered days anda Platonist on even numbered days; I'm not really happy with eitherposition.-- Shmuel (Seymour J.) <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In jesse@phiwumbda.org (Jesse F. Hughes) said:>I know who Goedel that you read abouthis work, not his identity.>What does the consistency of the axiom of choice have to do with>this?Nothing. Why do you assume that a paper will contain only the resultsummarized in its title?>Is this where he models ZFC in PA?papers, Paul Cohen's book may still be in print.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> > In jesse@phiwumbda.org (Jesse F. Hughes) said:>What does the consistency of the axiom of choice have to do with>>this? Nothing. Why do you assume that a paper will contain only the result> summarized in its title?I assume that every result in the paper has some relation to the maintopics, even if it is only illustrative of the methods. Or does thispaper also have a section on the mating habits of aardvarks?>>Is this where he models ZFC in PA? papers, Paul Cohen's book may still be in print.How is it that, of all the participants in this thread, you are theonly one who claims Goedel proved Con(PA) -> Con(ZFC)? This is anhonest question. This result, if true, is surely *at least* asimportant and interesting as the consistency of the axiom of choice.Maybe you should be less condescending and ask yourself if you'reabsolutely sure that this result is found in Goedel's work. If it is,it would be very enlightening if you could sketch the development, asI surely cannot spend time on researching it myself at present. Evenbarring that, a reference to any other work (preferably academic)which claims Con(PA) -> Con(ZFC) is contained, even implicitly, in thepublications of Goedel would be useful. As it is, I think I can't take your assertions at face value, althoughI did at first. I did so because I'm admittedly not familiar with theGoedel paper to which you refer, and you seemed pretty certain of theresult. However, given that others have questioned this claim, I'llsimply wait until more evidence comes in. A more ambitious lad than Iam, or one with more time or for whom these claims are close to hiscurrent research interests, would look up the original sources, as yousuggest, but I won't at present.-- All intelligent men are cowards. The Chinese are the world's worstfighters because they are an intelligent race[...] An average Chinesechild knows what the European gray-haired statesmen do not know, thatby fighting one gets killed or maimed. -- Lin Yutang jesse@phiwumbda.org (Jesse F. Hughes) said:>What does the consistency of the axiom of choice have to do with>this?>> Nothing. Why do you assume that a paper will contain only the result>> summarized in its title?I assume that every result in the paper has some relation to the main>topics, even if it is only illustrative of the methods. Or does this>paper also have a section on the mating habits of aardvarks?>>Is this where he models ZFC in PA?>> papers, Paul Cohen's book may still be in print.How is it that, of all the participants in this thread, you are the>only one who claims Goedel proved Con(PA) -> Con(ZFC)? This is an>honest question. This result, if true, is surely *at least* as>important and interesting as the consistency of the axiom of choice.Maybe you should be less condescending and ask yourself if you're>absolutely sure that this result is found in Goedel's work. If it is,>it would be very enlightening if you could sketch the development, as>I surely cannot spend time on researching it myself at present. Even>barring that, a reference to any other work (preferably academic)>which claims Con(PA) -> Con(ZFC) is contained, even implicitly, in the>publications of Goedel would be useful. I've been waiting for someone who really knows this stuff to speakup, but nobody has (or if anyone has, he's also included disclaimersabout how he could be wrong...) So, although I'm not an expert,I'll say what I think happened here. Note that anything I say couldbe wrong, but I think I can explain how it could be thatShmuel came to believe that Goedel proved Con(PA) impliesCon(ZF), even though Godel proved no such thing.Note that the last clause is supposed to be within the scope of the it could be that - I'm not _asserting_ that Goedel did nosuch thing, although I'm pretty sure he didn't. It seems clearhe didn't, since Con(PA) _is_ a theorem of ZF (I think,in a suitable sense), while Godel _did_ prove that Con(ZF)is not a theorem of ZF... but there are subtleties in allthis sort of thing so I could be wrong.One can give formulas of ZF Godel numbers, formalize thenotion of proof, etc. So the statement that a given sequenceof formulas is a valid proof in ZF is equivalent to a certainstatement about the natural numbers.Ie, although of course it doesn't work out this way:Could be that n is the Godel number of an axiomof ZF if and only if n is congruent to 6 mod 10, whileA follows from B and C by modus ponens if andonly if the corresponding Goedel numbers satisfyb = a*c. So saying that a sequence of formulas isa valid proof in ZF is the same as saying that acertain sequence of integers has the property that[*] each one is either congruent to 6 mod 10 or isthe quotient of two preceding integers on the list.Of course that's not how it works, but there _is_ atrue statement that's conceptually the same as thepreceding paragraph, just _much_ more complicated.So in some sense we've reduced statements aboutsets to statements about natural numbers. If oneis not careful about various things one could easilyjump from here to the statement that if PA isconsistent then so is ZF, since in some sensewe're reduced ZF to arithmetic.But Con(PA) implies Con(ZF) certainly does notfollow from the above arithmetization of ZF.Because exactly the same arithmetization as abovecan be performed with _any_ theory in place ofZF, consistent or not!For those of us who still don't get it: Suppose forthe sake of argument that the Godel number ofnot P is 7 times the Goedel number of P. IfZF is inconsistent then there is a _valid_ proofin ZF of P and also a valid proof of not P, forsome P. After arithmetization as above, thatmeans there is a valid sequence of Goedelnumbers that ends with 42, and another validsequence of Godel numbers that ends with7*42. The existence of those two sequencesof numbers, both satisfying [*], shows thatZF is inconsistent, but it does not show thatPA is inconsistent - they _are_ sequencesof integers which satisfy [*], so what? Thefact that we have two sequences satisfying[*], one of which ends in 42 and one of whichends in 7*42, is not a proof of 0 = 1 from theaxioms of PA.>As it is, I think I can't take your assertions at face value, although>I did at first. I did so because I'm admittedly not familiar with the>Goedel paper to which you refer, and you seemed pretty certain of the>result. However, given that others have questioned this claim, I'll>simply wait until more evidence comes in. A more ambitious lad than I>am, or one with more time or for whom these claims are close to his>current research interests, would look up the original sources, as you>suggest, but I won't at present.************************David C. Ullrich 10:35:20 +0100, jesse@phiwumbda.org (Jesse F.>> In jesse@phiwumbda.org (Jesse F. Hughes) said:>What does the consistency of the axiom of choice have to do with>>this?>> Nothing. Why do you assume that a paper will contain only the result> summarized in its title?>>I assume that every result in the paper has some relation to the main>>topics, even if it is only illustrative of the methods. Or does this>>paper also have a section on the mating habits of aardvarks?>>Is this where he models ZFC in PA?>> papers, Paul Cohen's book may still be in print.>>How is it that, of all the participants in this thread, you are the>>only one who claims Goedel proved Con(PA) -> Con(ZFC)? This is an>>honest question. This result, if true, is surely *at least* as>>important and interesting as the consistency of the axiom of choice.>>Maybe you should be less condescending and ask yourself if you're>>absolutely sure that this result is found in Goedel's work. If it is,>>it would be very enlightening if you could sketch the development, as>>I surely cannot spend time on researching it myself at present. Even>>barring that, a reference to any other work (preferably academic)>>which claims Con(PA) -> Con(ZFC) is contained, even implicitly, in the>>publications of Goedel would be useful. I've been waiting for someone who really knows this stuff to speak>up, but nobody has (or if anyone has, he's also included disclaimers>about how he could be wrong...) So, although I'm not an expert,>I'll say what I think happened here. Note that anything I say could>be wrong, but I think I can explain how it could be that>Shmuel came to believe that Goedel proved Con(PA) implies>Con(ZF), even though Godel proved no such thing.the it could be that - I'm not _asserting_ that Goedel did no>such thing, although I'm pretty sure he didn't. It seems clear>he didn't, since Con(PA) _is_ a theorem of ZF (I think,>in a suitable sense), while Godel _did_ prove that Con(ZF)>is not a theorem of ZF... but there are subtleties in all>this sort of thing so I could be wrong.>One can give formulas of ZF Godel numbers, formalize the>notion of proof, etc. So the statement that a given sequence>of formulas is a valid proof in ZF is equivalent to a certain>statement about the natural numbers.Ie, although of course it doesn't work out this way:>Could be that n is the Godel number of an axiom>of ZF if and only if n is congruent to 6 mod 10, while>A follows from B and C by modus ponens if and>only if the corresponding Goedel numbers satisfy>b = a*c. So saying that a sequence of formulas is>a valid proof in ZF is the same as saying that a>certain sequence of integers has the property that[*] each one is either congruent to 6 mod 10 or is>the quotient of two preceding integers on the list.Of course that's not how it works, but there _is_ a>true statement that's conceptually the same as the>preceding paragraph, just _much_ more complicated.>So in some sense we've reduced statements about>sets to statements about natural numbers. If one>is not careful about various things one could easily>jump from here to the statement that if PA is>consistent then so is ZF, since in some sense>we're reduced ZF to arithmetic.But Con(PA) implies Con(ZF) certainly does not>follow from the above arithmetization of ZF.>Because exactly the same arithmetization as above>can be performed with _any_ theory in place of>ZF, consistent or not!For those of us who still don't get it: Suppose for>the sake of argument that the Godel number of>not P is 7 times the Goedel number of P. If>ZF is inconsistent then there is a _valid_ proof>in ZF of P and also a valid proof of not P, for>some P. After arithmetization as above, that>means there is a valid sequence of Goedel>numbers that ends with 42, and another valid>sequence of Godel numbers that ends with>7*42. The existence of those two sequences>of numbers, both satisfying [*], shows that>ZF is inconsistent, but it does not show that>PA is inconsistent - they _are_ sequences>of integers which satisfy [*], so what? The>fact that we have two sequences satisfying>[*], one of which ends in 42 and one of which>ends in 7*42, is not a proof of 0 = 1 from the>axioms of PA.And then come to think of it, carrying this a step fartherleads to something which someone might even moreplausibly misunderstand or misremember as saying thatCon(PA) implies Con(ZF): Encoding sequences of integersas integers, the statement ZF is not consistent is equivalent to[**] There exists a positive integer n such that [some complicated condition],_where_ the condition only involves existentialquantifiers - that's not precisely stated, butin fact that condition satisfies some technicalcondition, so that in fact if there is an integersatisfying that condition then this fact can beproved in PA just by exhibiting n and verifyingthat it satisfies the condition. So it's correctto say There is no proof in PA of not Con(ZF)implies Con(ZF).>>As it is, I think I can't take your assertions at face value, although>>I did at first. I did so because I'm admittedly not familiar with the>>Goedel paper to which you refer, and you seemed pretty certain of the>>result. However, given that others have questioned this claim, I'll>>simply wait until more evidence comes in. A more ambitious lad than I>>am, or one with more time or for whom these claims are close to his>>current research interests, would look up the original sources, as you>>suggest, but I won't at present.>************************David C. Ullrich************************David C. Ullrich <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> <87smim7rqf.fsf@phiwumbda.org> > I've been waiting for someone who really knows this stuff to speak> up, but nobody has (or if anyone has, he's also included disclaimers> about how he could be wrong...) So, although I'm not an expert,> I'll say what I think happened here. Note that anything I say could> be wrong, but I think I can explain how it could be that> Shmuel came to believe that Goedel proved Con(PA) implies> Con(ZF), even though Godel proved no such thing.[snip rest]I think that you're exactly right about how Shmuel may havemisunderstood matters, assuming he did so. I was fairly explicitabout what I think is necessary[1] for the translation of proofs in ZF tofollowing:,----[ <87u135nv7i.fsf@phiwumbda.org> ]| PA, presumably in a natural way, so that any proof of P & ~P in ZFC| would yield a proof of Q & ~Q for some formula Q of PA.`----Unless a proof of P & ~P in ZFC translates to a proof of Q & ~Q (orsomething similar) in PA, the mere existence of a translation ofproofs in ZFC to proofs in PA does not appear to yield Con(PA) -> Con(ZFC).Shmuel referred satisfied this condition, but I was explicit aboutthat assumption in using presumably. I don't assume any such thingany longer.Footnotes: [1] Well, not *really* necessary in the technical sense, but(a) sufficient and (b) the most likely way that such a proof wouldproceed. Another way it could proceed, of course, is by proving firstthat Con(PA) is false, but I think I would've heard of that proof.-- Jesse HughesTime and again, history has shown that people who think their beliefstrump reality lose, and lose badly. Luckily, I don't have to listento you. -- James Harris on reality avoidance <87smiudtrj.fsf@phiwumbda.org> <3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net> <87smisrg0b.fsf@phiwumbda.org> <3ffd41a6$22$fuzhry+tra$mr2ice@news.patriot.net> <871xqa1n9s.fsf@phiwumbda.org> <40009ae6$11$fuzhry+tra$mr2ice@news.patriot.net> > In jesse@phiwumbda.org (Jesse F. Hughes) said:>>I know who Goedel that you read about> his work, not his identity.Son, you shouldn't be insulting. I am familiar with some of his work,too, but nothing that supports your claim.-- [I]t's good for the economy to charge for intellectual property, soopen source software cannot be good, while Microsoft is the mostfar-thinking company around and is doing it all for the good of thepublic. -- Linus Torvalds paraphrases Microsoft VP Craig Mundie <3ffadc52$20$fuzhry+tra$mr2ice@news.patriot.net> <1g75muz.1d66n4i14gjvdlN%panoptes@iquest.net> <3ffd4579$1$fuzhry+tra$mr2ice@news.patriot.net> <1g78quy.1a1bzig1e5uy44N%panoptes@iquest.net>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In described a bijection between the rationals in [0,1) and the>naturals. He claimed to. Hew was wrong.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to > He claimed to. Hew was wrong.While there are several inconsistencies among his claims, I don't thinkthat was one of them.Any non-negative integer can be written as a finite suma1*1! + a2*2! + a3*3! + ... + an*n!where each ak is an integer such that 0 <= ak <= k. If trailing zeroesare removed from the sum, this representation is unique. (Note that thecase of 0 ends up with zero terms.)One way to prove the necessary things about this representation is byinduction on the maximum number of terms; the numbers from 0 to (n+1)!-1can be represented uniquely (up to trailing zeroes) with at most nterms.Any rational in [0,1) can be written as a finite sumb1/2! + b2/3! + b3/4! + ... + bn/(n+1)!where each bk is an integer such that 0 <= bk <= k. If trailing zeroesare removed from the sum, this representation is unique. (Note that thecase of 0 ends up with zero terms.)This proof is trickier. It would probably be necessary to use inductionon n to show that all multiples of 1/(n+1)! in [0,1) can be representeduniquely (up to trailing zeroes) with at most n terms, and note that anyrational in [0,1) has some factorial that can be used as a denominator.(Uniqueness would require either induction on the trailing terms or acombinatorial argument.)And then the bijection consists of transferring the coefficients fromone sum to the other.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W In particular, you simply> misstate Turing's definition here. Turing defines a computable number as any number that can be> represented by a computable sequence. A computable sequence is the> output tape of a TM that has an infinite number of 0's and/or 1's. Turing did *not* allow an infinite number of 1's on the output tape,> so this is simply not his definition.I recently read Turing's ON COMPUTABLE NUMBERS,WITH AN APPLICATION TO THEENTSCHEIDUNGSPROBLEMhttp://www.cs.umass.edu/~immerman/cs601 /turingReference.html#section-2I may have mis-read it, but it appears this is how he defined computablenumbers in this paper. Computable numbers may have been defineddifferently in later papers.2. Definitions.Automatic machines.If at each stage the motion of a machine (in the sense of 1) is completelydetermined by the configuration, we shall call the machine an automaticmachine (or a-machine). For some purposes we might use machines (choicemachines or c-machines) whose motion is only partially determined by theconfiguration (hence the use of the word possible in 1). When such amachine reaches one of these ambiguous configurations, it cannot go on untilsome arbitrary choice has been made by an external operator. This would bethe case if we were using machines to deal with axiomatic systems. In thispaper I deal only with automatic machines, and will therefore often omit theprefix a-.Computing machines.If an a-machine prints two kinds of symbols, of which the first kind (calledfigures) consists entirely of 0 and 1 (the others being called symbols ofthe second kind), then the machine will be called a computing machine. Ifthe machine is supplied with a blank tape and set in motion, starting fromthe correct initial m-configuration, the subsequence of the symbols printedby it which are of the first kind will be called the sequence computed bythe machine. The real number whose expression as a binary decimal isobtained by prefacing this sequence by a decimal point is called the numbercomputed by the machine.At any stage of the motion of the machine, the number of the scanned square,the complete sequence of all symbols on the tape, and the m-configurationwill be said to describe the complete configuration at that stage. Thechanges of the machine and tape between successive complete configurationswill be called the moves of the machine.{233}Circular and circle-free machines.symbols of the first kind it will be called circular. Otherwise it is saidto be circle-free.A machine will be circular if it reaches a configuration from which there isno possible move, or if it goes on moving, and possibly printing symbols ofthe second kind, but cannot print any more symbols of the first kind. Thesignificance of the term circular will be explained in 8.Computable sequences and numbers.A sequence is said to be computable if it can be computed by a circle-freemachine. A number is computable if it differs by an integer from the numbercomputed by a circle-free machine.We shall avoid confusion by speaking more often of computable sequences thanof computable numbers.I understand this to mean that a circle free machine must writean infinite number of symbols of the first type.> By the way, induction over all natural numbers has nothing to do with> feasibility and everything to do with the definition of N and with the> meaning of universal quantification. Induction does not require that> one actual go through all of the natural numbers and ensure that the> next one satisfies the proposition.The system I am describing is similar to hypercomputation.http://www.alanturing.net/turing_archive/ pages/Reference%20Articles/hypercomputation/ hypercomputation.htmlDo we need induction if we assume we can actually go through allof the natural numbers and ensure that they all satisfy the proposition?Would induction and hypercomputation always give the same results?Russell- 2 many 2 countX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft at 03:26 PM, Tom Adams said:>He then shows how the metric can define a topological space.Presumably he defines the topology by taking the d-open sets as abasis. >That sounds right, but I don't see why it must be so. The>definitions of T-limit points and d-limit points are so different. If p is a d-limit point of S, p in U and U in T, then there is anepsilon>0 such that B(p,epsilon) C U. By the definition of d-limit,that ball intersects S. But then U intersects S.Conversely, is p is a T-limit, then for every epsilon>0, B(p,epsilon)in T, and therefor intersects S.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to > I'm reading an introductory book on topology and the author defines a> d-limit point p of a set S in a pseudometric space as a point having the> property that for every eps>0 there is in S a point s different than p> such that the metric d(p,s) property that d(x,y)=0 only if x=y.) In the topological space (X, T) he defines a T-limit point p of a set S> as a point having the property that every element of T containing p> meets S in a point other than p. He then shows how the metric can define a topological space. He then says when we begin with a pseudometric d on a set X, we obtain> a collection of d-open sets that is a topology T, and the T-limit points> of a set coincide with the d-limit points of a set. That sounds right, but I don't see why it must be so. The definitions> of T-limit points and d-limit points are so different. We might argue> that every open set S is the union of some collection of open sets> (which should be the same for the metric and topological spaces), but I> don't see how that shows that the T-limit points and d-limit points of S> are the same.>If x is a T-limit pt of S, then for all open U nhood x, Ux / S isnonnul. As any d-ball containing x is open nhood of x in T, it quicklyfollows x is a d-limit pt of S.Conversely if x is a d-limit point of S and U an open nhood x, thenthere's some d-ball B with x in B subset U. As x is d-limit pt of S,Bx / S is nonnul, so also Ux / S is nonnul. >>I'm reading an introductory book on topology and the author defines a>>d-limit point p of a set S in a pseudometric space as a point having the>>property that for every eps>0 there is in S a point s different than p>>such that the metric d(p,s)>property that d(x,y)=0 only if x=y.)>>In the topological space (X, T) he defines a T-limit point p of a set S>>as a point having the property that every element of T containing p>>meets S in a point other than p.>>He then shows how the metric can define a topological space.>>He then says when we begin with a pseudometric d on a set X, we obtain>>a collection of d-open sets that is a topology T, and the T-limit points>>of a set coincide with the d-limit points of a set.>>That sounds right, but I don't see why it must be so. The definitions>>of T-limit points and d-limit points are so different. We might argue>>that every open set S is the union of some collection of open sets>>(which should be the same for the metric and topological spaces), but I>>don't see how that shows that the T-limit points and d-limit points of S>>are the same.If x is a T-limit pt of S, then for all open U nhood x, Ux / S is> nonnul. As any d-ball containing x is open nhood of x in T, it quickly> follows x is a d-limit pt of S.> It seems the key point is switching between the T and d point of view by noticing that any d-ball is T-open (because the metric d determines openness in the metric space and in the derived topology T).> Conversely if x is a d-limit point of S and U an open nhood x, then> there's some d-ball B with x in B subset U. As x is d-limit pt of S,> Bx / S is nonnul, so also Ux / S is nonnul.Here the key point seems to be that a d-open set is a T-open set (is an element of T) so the d-type operations on d-open sets give us information about T-limit pts.explain this. I looked at another text and this interesting topic wasn't even discussed. Maybe I need a better book.Tom Adams <40014E86.4060308@comcast.net> >>I'm reading an introductory book on topology and the author defines a> >>d-limit point p of a set S in a pseudo-metric space as a point having the>>property that for every eps>0 there is in S a point s different than p>>such that the metric d(p,s)>property that d(x,y)=0 only if x=y.)>In the topological space (X, T) he defines a T-limit point p of a set S>>as a point having the property that every element of T containing p>>meets S in a point other than p.>He then shows how the metric can define a topological space.>>He then says when we begin with a pseudometric d on a set X, we obtain>>a collection of d-open sets that is a topology T, and the T-limit points>>of a set coincide with the d-limit points of a set.>That sounds right, but I don't see why it must be so. The definitions> >>of T-limit points and d-limit points are so different. We might argue>>that every open set S is the union of some collection of open sets>>(which should be the same for the metric and topological spaces), but I>>don't see how that shows that the T-limit points and d-limit points of S>>are the same.> If x is a T-limit pt of S, then for all open U nhood x, Ux / S is>nonnul. As any d-ball containing x is open nhood of x in T, it quickly>follows x is a d-limit pt of S. It seems the key point is switching between the T and d point of> view by noticing that any d-ball is T-open (because the metric d> determines openness in the metric space and in the derived topology T). > Conversely if x is a d-limit point of S and U an open nhood x, then>there's some d-ball B with x in B subset U. As x is d-limit pt of S,>Bx / S is nonnul, so also Ux / S is nonnul. Here the key point seems to be that a d-open set is a T-open set (is an> element of T) so the d-type operations on d-open sets give us> information about T-limit pts. explain this. I looked at another text and this interesting topic> wasn't even discussed. Maybe I need a better book.>You're welcome. Don't despair quite yet, did he leave such proofs asexercises for the student in the problem section?Did the author explain why the d-open balls for metric or pseudo-metric isa topology? The hard part is to show the intersection of two balls isopen. >>I'm reading an introductory book on topology and the author defines a>>d-limit point p of a set S in a pseudo-metric space as a point having the>>property that for every eps>0 there is in S a point s different than p>>such that the metric d(p,s)>property that d(x,y)=0 only if x=y.)>>In the topological space (X, T) he defines a T-limit point p of a set S>>as a point having the property that every element of T containing p>>meets S in a point other than p.>>He then shows how the metric can define a topological space.>>He then says when we begin with a pseudometric d on a set X, we obtain>>a collection of d-open sets that is a topology T, and the T-limit points>>of a set coincide with the d-limit points of a set.>>That sounds right, but I don't see why it must be so. The definitions>>of T-limit points and d-limit points are so different. We might argue>>that every open set S is the union of some collection of open sets>>(which should be the same for the metric and topological spaces), but I>>don't see how that shows that the T-limit points and d-limit points of S>>are the same.>>If x is a T-limit pt of S, then for all open U nhood x, Ux / S is>nonnul. As any d-ball containing x is open nhood of x in T, it quickly>follows x is a d-limit pt of S.>>It seems the key point is switching between the T and d point of>>view by noticing that any d-ball is T-open (because the metric d>>determines openness in the metric space and in the derived topology T).>Conversely if x is a d-limit point of S and U an open nhood x, then>there's some d-ball B with x in B subset U. As x is d-limit pt of S,>Bx / S is nonnul, so also Ux / S is nonnul.>>Here the key point seems to be that a d-open set is a T-open set (is an>>element of T) so the d-type operations on d-open sets give us>>information about T-limit pts.>>explain this. I looked at another text and this interesting topic>>wasn't even discussed. Maybe I need a better book.You're welcome. Don't despair quite yet, did he leave such proofs as> exercises for the student in the problem section?> Don't see it in the exercises for this and the next section. He may get to it later. I'll be OK with the generous help I get from this group.> Did the author explain why the d-open balls for metric or pseudo-metric is> a topology? Yup, very interesting, too. > The hard part is to show the intersection of two balls is> open.He does that indirectly by showing every set is T-open iff it is an element of the topology T. So d-open balls are T-open, and are members of the topology T; and the intersection of two of them is still a member of the topology T and therefore open.Tom Adams >>the form Adot = f(x,p)..where x and p are the states of this dynamical>>system and f is a nonlinear function in x, p. If everything is real you might want to write A as tilde B B , namely as> the product of a matrix and its transpose. Whatever the resulting equations,> A will be guranteed symmetric positive-definite. Perhaps you could make> B upper-triangular for simplicity.> And most likely, you can rewite your differential equation for A intoone for B; then you can work with B only, never need any checks,and form A at the very end...Arnold Neumaier > Let me try to pose the problem, I am trying to solve.I have a matrix, A (for simplicity let us assume it is a 2 X 2 matrix), that> is supposed to be symmetric and positive definite. There is a differential> equation the numerical solution of which gives me A at a current instant, in> the form Adot = f(x,p)..where x and p are the states of this dynamical> system and f is a nonlinear function in x, p. Since A has only three unique> parameters for this case, I actually obtain, theta_dot = g(x,p), where theta> A(t) thereby enforcing symmetry..... To enforce positive definiteness, I> need to do the following,theta(1) > 0, theta(2) > 0 and -sqrt(theta(1)*theta(2)) < theta(3) <> sqrt(theta(1)*theta(2))...Then by using some sort of a projection algorithm, I could enforce positive> definiteness in this way, however, this doesnt scale well for higher> dimensioned matrices.If the indefiniteness is simply caused by roundoff, multiplying the diagonalafter each integration step by 1.00000001 should cure the problem (assumingdouble precision). If not, something is likely to be wrong with your modeling...Arnold Neumaier > En el escribi.97:>I'm doing som math and came to this>question about graphs.> The Graph y= f(x) has a symmetry line x=3> a) decide f(7), if f(-1) = 0>b) decide f(5), if f(1) = 3>c) Great the Graph> I have calculated that a) should be 0>and b) should be 3.> what I wonder is how do I create a graph? (question (c))>How do I create the equation like (y= -2x^2 +2x +1)?>I do need that one I think to calculate all the Points in the graph.> Please advice a newbe from Sweden!I think that in some way it is implicit that you can seek a quadratic> function. Then, as x = 3 is a simetry line, the equation must bey = a(x - 3)^2 + bReplace x by 7 or -1 (it is the same) and y by 0, and you get a eqution in a> and b. Do the same with x = 5 or 1 and y = 3, and you get a system with two> equations with two unknowns, a and b.I'm still a bit blur about how to precede.You say that I get two unknowns, a and b.what I wanted was the equation to put intomy calculator to get the Graph written.I dont understand this high match. Can youexplain a bit more?However I did do what you suggestedand calculatet a bit with your equation.y = a(x - 3)^2 + bdunno how to get it right however. En el escribi.97:> Ignacio Larrosa Ca.96estro En el escribi.97:> I'm doing som math and came to this> question about graphs.>> The Graph y= f(x) has a symmetry line x=3>> a) decide f(7), if f(-1) = 0> b) decide f(5), if f(1) = 3> c) Great the Graph> I have calculated that a) should be 0> and b) should be 3.>> what I wonder is how do I create a graph? (question (c))> How do I create the equation like (y= -2x^2 +2x +1)?> I do need that one I think to calculate all the Points in the> graph.>> Please advice a newbe from Sweden!>> I think that in some way it is implicit that you can seek a quadratic>> function. Then, as x = 3 is a simetry line, the equation must be>> y = a(x - 3)^2 + b>> Replace x by 7 or -1 (it is the same) and y by 0, and you get a>> eqution in a and b. Do the same with x = 5 or 1 and y = 3, and you>> get a system with two equations with two unknowns, a and b.> I'm still a bit blur about how to precede. You say that I get two unknowns, a and b.> what I wanted was the equation to put into> my calculator to get the Graph written. I dont understand this high match. Can you> explain a bit more? However I did do what you suggested> and calculatet a bit with your equation.> y = a(x - 3)^2 + b> dunno how to get it right however.If you must to look for a quadratic function f(x) = p*x^2 + q*x + r, and itmust be simmetric with respect to the line x = 3, it must bef(x) = a(x - 3)^2 + bTo see that, equate f(3 - t) = f(3 + t), as x = 3 is an axis of simmetry.f(3 - t) = p(3 - t)^2 + q(3 - t) + r = p*t^2 - (6p + q)t + 9p + 3q + rf(3 + t) = p(3 + t)^2 + q(3 + t) + r = p*t^2 + (6p + q)t + 9p + 3q + rf( 3 - t) = f(3 + t) for all t > -(6p + q) = 6p + q ==> q = - 6p ==>f(x) = p*x^2 - 6p*x + r = p*x^2 - 6p*x + 9p + r - 9p = p(x - 3)^2 + (r - 9p)= a*(x - 3)^2 + bThen,As f(-1) = 0 = -16aAs f(1) = 3 a = -1/4 , b =4f(x) = (-1/4)(x - 3)^2 + 4-- Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com Imagine yourself in space and look at this third rock from Sun, Earth.Known most intelligent living being named human being rules thisPlanet, 6 billion in number.For millions of years, we lived in jungle, caves. We had to go throughendless sufferings. When we gained consciousness, we began to askourselves, where from we came, what is going on around us. Our questto understand nature began.We have come a long way in this quest of understanding our universe,God. When I am typing this, Spirit is exploring Mars. But ifradius of our Universe is 1 trillion Kilometer, we can not go a singlemillimeter in our entire lifetime using Newton's Third Law. IfEinstein is right, we can not travel FTL. Then why creator of thisuniverse has created those billions of Galaxies?Where things have gone wrong? You are trying to understand quantum mechanics, Newtonian mechanics,Relativity rather than trying to understand our Nature and God. Thisis the biggest mistake you are doing. You are caught in jungle ofknowledge and you have abandoned that supreme force, God. You arephysics theory, you must not talk about God! Why not? Because there isno proof of such supernatural thing.No proof? Things are hidden, life is created and we are forced to findthose hidden things. This is happened thousands of times, in everydiscovery, every invention. We can find something only when someonehides it or lost by ourselves. Certainly every discovery or inventionwas not lost by us in past. So definitely, someone has hidden itlong before creation of Life and Universe. Why don't you peopleunderstand such simple logic.entire knowledge built over thousands of years comes down crashing andyou will be forced to spend your days and nights on streets withoutshelter, like those people surviving Earthquake. Like those homelesspeople, you will be theoryless. You are going to see awesome power ofGravity. Through this Action Device, you are going to see none otherthan that Supreme Force, Gravity, God in Action.Watch Out...-Abhi.(BTW, why don't you see a thread, Action Device Working Model, PartII in sci.math NG. I learned about this thread in the night ofJanuary 06. I really never see any message in sci.math NG. Some textfrom this thread is given below.)------------------------------------------------------ This is what has happened until now...> called action device, I decided to give it a try in all fairness.> After all, everyone is saying it should be so easy to build. So I got> myself some wood, steel angles, springs and bolts and built it. And> guess what, it really works!> I climbed on the roof (I didn't want the thing sticking to the> ceiling and then having to do a lot of explaining), I stretched my> arms, and let go. Wow, it imediately started moving in an orthogonal> angle to the plain, as had been described. The stupid thing is I must> have been holding it upside down or something, because instead of> going off into space, it shot straight downwards and landed on the> driveway. At that moment my mom's truck drove over it and reduced it> to a miserable heap of splinters... So I'm going to have to wait to> get new materials and give it a new try. I'll give you the news as> soon as I get to it.PART IIaccident, I got fresh materials and fresh insights. I built a newaction device with a few changes however. First the angle of thedevice is a lot sharper, about 45 degrees, this should provide forgreater push, then I painted marks on the thing to make sure whichpart is up and which one is down (figuring that out wasn't easythough). I must tell you of something I had already observed the firsttime around which I had kept to myself because it was an unconfirmedobservation: When I tried to install the springs, for some strangereason, they resisted being stretched! I felt there where undiscoveredforces in action even before completion of the device! After climbingto the roof I gave the thing a nice shove, just to make sure it landedon the lawn if something went wrong this time. Again, it worked ! thething started moving up and away from me. Then fate struck again, Ihaven't figured out what exactly happened, my guess is that point ddetacthed itself and continued the trip without the rest of thedevice, or something. I'll tell you when I find out.However there is something else I wan't to share with you, I carefullyobserved the trajectory of the device from the time I released it tothe time it landed and found it was a parable. I'm going to try andfigure the implicatins this has, but I already consider it a veryinteresting and revolutionary fact in ------------------------------ our entire lifetime using Newton's Third Law. If> Einstein is right, we can not travel FTL. Then why creator of this> universe has created those billions of Galaxies? Where things have gone wrong?The mobile over a crib is out of reach too...Maybe it is a way of teaching us to extend our reach.David A. Smith It actually wasn't a homework problem, in case anybody was wondering.I'm past the geometry stage of math (currently in second yearcalculus), but my geometry has gotten a little rusty. Actually, I'mnot sure we ever learned that in Geometry class. Typical AmericanPublic Schools, I suppose. (On the other hand, I wasn't the bestGeometry student. My teacher gave liberal amounts of extra credit fordoing math team, and so I got A's all year without doing thehomework...)The question arose when I calculated (I was bored, away from home,with nothing to do) that the koch snowflake occupies exactly 4/5 ofthe bounding hexagon, if that makes any sense. So then I was wonderingif that was the only way to occupy exactly 4/5 of a regular hexagon,and then I was wondering if it was actually possible to draw a kochsnowflake with only compass adn straightedge, and the only step I wasunsure of was trisecting a line.Anyway, today I realised that you can occupy 4/5 of a regular hexagonwith simpler, finite triangles. Split the hexagon into 6 triangles,mark off 20 equal segments on each side of each of the triangles, drawsmall equilateral triangle with that as the side length, which willfill the entire hexagon up with 2400 small triangles. Then just slect4/5 of them.So then I felt a bit stupid. But anyway...Jonathan ChristensenIs there an easy way to trisect a line segment, i.e. split it into> three equal portions, with just straightedge and compass?If not an easy way, is there any way at all?Jonathan Christensen > Is there an easy way to trisect a line segment, i.e. split it into> three equal portions, with just straightedge and compass?If not an easy way, is there any way at all?Jonathan ChristensenIn 1995 two highschool freshmen discovered a construction different from the usual onedating back to Euclid. They named itthe GLaD Construction -- Goldenheim, Litchfield and Dietrich. Look attheir web page: http://www.gfacademy.org/GLaD/glad.htmlAt the bottom of this page is a link to a MS Powerpoint presentationFibonacci Sketchpad Euclidhttp://www.gfacademy.org/GLaD/presentation/ Presentation.pptexplaining their method.Hugo Pfoertner > Is there an easy way to trisect a line segment, i.e. split it into> three equal portions, with just straightedge and compass?Define easy.Here's one approach that avoids the tedium of constructing a coupleparallel lines:Call the endpoints of the segment P1 and P2. Draw a line through P2(preferably close to perpendicular). Use the compass with the center atP2 to mark two points P3 and P4 on the new line which are equidistantfrom P2 (preferably about as far away as P1). Draw the line P1P3.Bisect the segment P1P3 and call its midpoint P5. Draw the segmentP4P5. Call the intersection of P4P5 and P1P2 P8. Draw a circle withits center at P8 and passing through P2; call its other intersectionwith P1P2 P9. P8 and P9 are the desired trisection points.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W Is there an easy way to trisect a line segment, i.e. split it into> three equal portions, with just straightedge and compass?>Yes. Draw a ray such that it shares an endpoint (A) with the segment. Makethree points on the ray B, C, and D such that AB = BC = CD. Draw a linefrom D to the other endpoint of the segment (point E). Now draw a linethrough B parallel to DE. Repeat at point C.The details can be filled in by any high school geometry book.-Tralfaz > Is there an easy way to trisect a line segment, i.e. split it into>three equal portions, with just straightedge and compass?> Yes. Draw a ray such that it shares an endpoint (A) with the segment. Make> three points on the ray B, C, and D such that AB = BC = CD. Draw a line> from D to the other endpoint of the segment (point E). Now draw a line> through B parallel to DE. Repeat at point C.The details can be filled in by any high school geometry book.... or from the fountainhead: Euclid VI.9. Ken Pledger. > Is there an easy way to trisect a line segment, i.e. split it into> three equal portions, with just straightedge and compass?Yes. In fact, there is an easy way of dividing a segmentinto N congruent sub-segments. (Assuming this is homework)Suppose you had a different segment, already divided into three equal parts. Can you see how to use that other segment together with some facts about parallels and aboutsimilar triangles to trisect the original segment? Canyou easily construct such a helper segment, in a positionthat's convenient (say, one side of a triangle ...)? Yes, each of the sequences except for 0.9999....^2 tends toward 0. It justtakes longer to do so as you add more 9's. It is what happens at infinitythat makes this a special case....Take a look.Lets look at 0.99999....Let's just prove that is 1.Now, in an earlier post you defined infinity, so you completely agree thatit goes on forever. That is important to understand for this small proof.Let x = 0.999999999......Then 10x = 9.9999999....Now lets take 10x - x.10x = 9.999999....- x = 0.999999....Remember allthe way to infinity.So, 9x = 9.So, x=1.But we said x was equal to 0.999999.... That means 0.999999.... = 1.Since 0.9999.... = 1 then...0.999999...^2 = 1^2 = 1.Hope that explains it.Markis>> Squares of 0.999... tend away from 1> There is only one square of 0.999...; perhaps you can tell me why you>used the plural? Square, cube, etc. > But your numerical evidence below supports that the squares of>0.9, 0.99, 0.999, ... tend TOWARD 1, not AWAY. To see this, all you>need to do is arithmetic. Sequence 0.9^2, 0.9^3, 0.9^4, ... tends toward 0> Sequence 0.99^2, 0.99^3, 0.99^4, ... tends toward 0> Sequence 0.999^2, 0.999^3, 0.999^4, ... tends toward 0> etc., etc., etc., Sequence 0.999...^2, 0.999...^3, 0.999...^4, ... tends toward 0> Garry Denke, Geologist> Denoco Inc. of Texas >> Squares of 0.999... tend away from 1>There is only one square of 0.999...; perhaps you can tell me why you>used the plural?Square, cube, etc. > Ok. You should have said powers instead of squares. But now I understand your statement.>But your numerical evidence below supports that the squares of>0.9, 0.99, 0.999, ... tend TOWARD 1, not AWAY. To see this, all you>need to do is arithmetic. Sequence 0.9^2, 0.9^3, 0.9^4, ... tends toward 0agreed.> Sequence 0.99^2, 0.99^3, 0.99^4, ... tends toward 0agreed.> Sequence 0.999^2, 0.999^3, 0.999^4, ... tends toward 0agreed.> etc., etc., etc.,agreed.Sequence 0.999...^2, 0.999...^3, 0.999...^4, ... tends toward 0> I do not agree. Why do you think this is true? Yes, I can. Proof: 0.999... = 9/10 + 9/100 + 9/1000 + ...= (9/10)/(1-1/10) = 1. So, 0.999...^n = 1^n = 1 for all positive integers n. So, the sequence 0.999...^2, 0.999...^3, 0.999...^4,...is the sequence 1, 1, 1, ... which tends toward 1. So this is > Garry Denke, Geologist> Denoco Inc. of Texas >Does sequence .999...^2, .999...^3, .999...^4, ... tend to 0 or to 1?Since .999... = 1 in your question, your sequence is 1,1,1,..., and > doesn't tend anywhere, it stays put at 1.Virgil assumes that the unlimited value .999... equals the limitedvalue 1, but fails to prove that the powers of the unlimited value.999... tend toward the limited value 1. For all values less than thelimited value 1 in the sequence a^2, a^3, a^4, ... the tendancy istoward 0, yet Virgil hangs on to the misconception that theunlimited is limited ...Main Entry: unlimited Pronunciation: -'li-m&-t&dFunction: adjective1 : lacking any controls : UNRESTRICTED2 : BOUNDLESS, INFINITE3 : not bounded by exceptions : UNDEFINED- unlimitedly adverbMain Entry: limitedFunction: adjective1 a : confined within limits : RESTRICTED b of atrain : offering faster service especially by making a limited numberof stops2 : characterized by enforceable limitations prescribed (as by aconstitution) upon the scope or exercise of powers 3 : lacking breadth and originality - limitedly adverb- limitedness noun... Virgil's logic is limited.Garry Denke, GeologistDenoco Inc. of Texas >>Does sequence .999...^2, .999...^3, .999...^4, ... tend to 0 or to 1?>Since .999... = 1 in your question, your sequence is 1,1,1,..., and >doesn't tend anywhere, it stays put at 1.Virgil assumes that the unlimited value .999... equals the limited> value 1, but fails to prove that the powers of the unlimited value> .999... tend toward the limited value 1.On the contrary, provided the unlimited value of .999... means the limit lim_{m -> oo} (1-1/10^m), the continuity of f(x) = x^n and the fact of lim_{m -> oo} (1-1/10^m) = 1, both of which I presented in this thread before but which Denke then ignored, proves it.> For all values less than the> limited value 1 in the sequence a^2, a^3, a^4, ... the tendancy is> toward 0, yet Virgil hangs on to the misconception that the> unlimited is limited ...The only misconception is Denke's of what others post.Main Entry: unlimited > Pronunciation: -'li-m&-t&d> Function: adjective> 1 : lacking any controls : UNRESTRICTED> 2 : BOUNDLESS, INFINITE> 3 : not bounded by exceptions : UNDEFINED> - unlimitedly adverbMain Entry: limited> Function: adjective> 1 a : confined within limits : RESTRICTED b of a> train : offering faster service especially by making a limited number> of stops> 2 : characterized by enforceable limitations prescribed (as by a> constitution) upon the scope or exercise of powers monarchy 3 : lacking breadth and originality head -- Virginia Woolf - limitedly adverb> - limitedness nounI do not recall ever hearing of Virginia Woolf's expertize in mathematics so I cannot accept such definitions as relevant to my meanings.The only dictionaries relevant in a discussion of mathematical meanings are mathematical dictionaries. ... Virgil's logic is limited.So is the logic of every sane person limited by the laws of logic, but Denke's willful misunderstandings appear unlimited.A sequence may be unlimited in its number of terms and simultaneously limited by being Cauchy sequence, as is (1-1/10^m), in which case the sequence converges to a limit, as does (1-1/10^m).Garry Denke, Geologist> Denoco Inc. of Texas > Let a_n = 1 - 10^(-n). Then a_0 = 0, a_1 = .9, a_2 = .99, etc. If we> take the sequence {(a_0)^2, (a_1)^2, (a_2)^2, (a_3)^2...} we can> examine what happens to the squares of .9999...999 as n grows to> infinity.We have that lim (n -> infinity) (a_n)^2 = lim (n -> infinity) 1 - 2 *> 10^(-n) + 10^(-2n) = 1 - lim [2 * 10^(-n) + 10^(-2n)] = 1 - 0 = 1.Notably, you don't seem to be interested in only the squares of the> numbers, since you include also the fourth power. Maybe you should> state more clearly precisely what you want.I am interested in the sequence a^2, a^3, a^4, ...for a = .9, a = .99, a = .999, through a = .999... Garry Denke, GeologistDenoco Inc. of Texas In sci.math, Garry a_n = 1 - 10^(-n). Then a_0 = 0, a_1 = .9, a_2 = .99, etc. If we>> take the sequence {(a_0)^2, (a_1)^2, (a_2)^2, (a_3)^2...} we can>> examine what happens to the squares of .9999...999 as n grows to>> infinity.>>> We have that lim (n -> infinity) (a_n)^2 = lim (n -> infinity) 1 - 2 *>> 10^(-n) + 10^(-2n) = 1 - lim [2 * 10^(-n) + 10^(-2n)] = 1 - 0 = 1.>>> Notably, you don't seem to be interested in only the squares of the>> numbers, since you include also the fourth power. Maybe you should>> state more clearly precisely what you want.I am interested in the sequence a^2, a^3, a^4, ...> for a = .9, a = .99, a = .999, through a = .999... lim(n->+oo) (1 - 10^(-n))^n= lim(n->+oo) (1 - exp(-n * log(10))^n= lim(n->+oo) (1 - n*exp(-n * log(10)) + (n)(n-1)/2!*exp(-2*n*log(10)) - (n)(n-1)(n-2)/3!*exp(-3*n*log(10)) ...)lim(n->+oo) n*exp(-n * log(10))= n - n^2*log(10)/1! + n^3*log(10)^2/2! - n^4*log(10)^3/3! ... + (-1)^k*n^(k+1)/k! ...It's a bit difficult to prove but this series is in fact absolutelyconvergent (since k! > n^(k+1) for sufficiently large k) and infact the limit is 0. Therefore, lim(n->+oo) (1 - 10^(-n))^n = 1.Or one can look atlim(n->+oo) log(n * exp(-n * log(10)) = lim(n->+oo) log(n) - n*log(10)If one sets f(x) = x and g(x) = log(n), then f'(x) = 1 and g'(x) < 1for all x > 1. (g'(x) = 1/x, in fact.)Therefore h(x) = log(x) - x*log(10) is such thath'(x) = (1/x) - log(10). For x > log(10), h'(x) is negative;therefore for sufficiently large x, h(x) < K for some K.In fact, for sufficiently large x, h(x) < 0 -- certainly forx > 10, h(x) < h(10) = -9*log(10) < 0, as is easy to see.It turns out thatlim(n->+oo) log(n * exp(-n * log(10)) = -oothough I am not entirely sure of the easiest method of proof.> Garry Denke, Geologist> Denoco Inc. of Texas-- #191, ewill3@earthlink.netIt's still legal to go .sigless. >Let a_n = 1 - 10^(-n). Then a_0 = 0, a_1 = .9, a_2 = .99, etc. If we>take the sequence {(a_0)^2, (a_1)^2, (a_2)^2, (a_3)^2...} we can>examine what happens to the squares of .9999...999 as n grows to>infinity.>We have that lim (n -> infinity) (a_n)^2 = lim (n -> infinity) 1 - 2 *>10^(-n) + 10^(-2n) = 1 - lim [2 * 10^(-n) + 10^(-2n)] = 1 - 0 = 1.>Notably, you don't seem to be interested in only the squares of the>numbers, since you include also the fourth power. Maybe you should>state more clearly precisely what you want.I am interested in the sequence a^2, a^3, a^4, ...> for a = .9, a = .99, a = .999, through a = .999... > Garry Denke, Geologist> Denoco Inc. of TexasBeen there, done that, and have the t-shirt.To repeat for the terminally illiterate, let f(n) = (1-1/10^m)^n, for fixed m, giving the series you cite an interest in above, and let g(m) = (1-1/10^m)^n, for fixed n, giving the series that everyone else is interested in.Then lim_{n -> oo} f(n) = 0 but lim_{m -> oo} g(m) = 1and for suitable relations between m and n, lim_{(m,n) ->(oo,oo)} (1-1/10^m)^n can be any value in [0,1]X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In about is notsquares but arbitray powers. > .9^4 = .6561Is not a square of .99 . . .-- Shmuel (Seymour J.) Metz, SysProg and Let's use w as a stand-in for the Greek character omega here. As many here know, we have the following recursive definition for w_a, also called aleph_a, for all ordinals a.1) w_0 = w2) w_S(a) = the least cardinal greater than w_a, where S(a) = a+1 is the successor of a.3) w_a = sup {w_b: b < a} when a is a limit ordinal.It is easy to show that a <= w_a for all a. Can one show in ZF that the inequality is always strict? How about in ZFC? In ZF+GCH?It is clear that if a = w_a, then a is a limit cardinal.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu >Let's use w as a stand-in for the Greek character omega here. As many >here know, we have the following recursive definition for w_a, also >called aleph_a, for all ordinals a.>1) w_0 = w>2) w_S(a) = the least cardinal greater than w_a, where S(a) = a+1 is >the successor of a.>3) w_a = sup {w_b: b < a} when a is a limit ordinal.>It is easy to show that a <= w_a for all a. Can one show in ZF >that the inequality is always strict? How about in ZFC? In ZF+GCH?>It is clear that if a = w_a, then a is a limit cardinal.Define the sequence a_0 = w and a_S(n) = w_{a_n} for all natural numbers n, and let b = sup {a_n : n < w}, then b = w_b.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. > Let's use w as a stand-in for the Greek character omega here. As many > here know, we have the following recursive definition for w_a, also > called aleph_a, for all ordinals a.> 1) w_0 = w> 2) w_S(a) = the least cardinal greater than w_a, where S(a) = a+1 is > the successor of a.> 3) w_a = sup {w_b: b < a} when a is a limit ordinal.It is easy to show that a <= w_a for all a. Can one show in ZF > that the inequality is always strict? How about in ZFC? In ZF+GCH?It is clear that if a = w_a, then a is a limit cardinal.> I believe that the inequality is not strict. Consider the ordinal defined recursively:a_0 = 0,a_n = w_{a_{n+1}}Then I believe that a = a_w satisfies a = w_a.I recall seeing somewhere some kind of fixed point theorem for ordinals, something to the effect that if f is an increasing function on the class of ordinals, perhaps satisfying some other axioms as well, then there exists an ordinal a = f(a). I recall seeing somewhere some kind of fixed point theorem for ordinals, > something to the effect that if f is an increasing function on the class > of ordinals, perhaps satisfying some other axioms as well, then there > exists an ordinal a = f(a).> Now I think about it, I think that the additional axiom would be something like f(a) = sup{f(b):bI came across 2 questions I had that maybe someone can answer :>1) If A is an n x n matrix, and ||A|| is defined as the smallest number c>such that ||Ax|| <= c||x|| for all x in C^n, the n-fold complex numbers,>then why is it true that a sequence of matrices A_m converges to a matrix A>if and only if ||A_m - A|| --> 0?by definition.> verify that ||.|| above is a norm.> 2) Consider an infinite series whose terms are matrices : A_0 + A_1 + A_ 2>+ ... If the sum (from m = 0 to infinity) of ||A_m|| < infinity, then>the infinite series of matrices is said to converge absolutely.this is an extension of a calculus theorem. it's proof is similar.> Why is it true that if a series converges absolutely, then the partial>sums of the series form a cauchy sequence?let space V possess a norm ||.||, and let epsilon > 0.we know that if S = Sum fj converges, then> ||Sn - S|| < epsilon/2 for some n large enough,> where Sn = Sum fj, j = 1 to infty.oops, j = 1 to n (NOT infty)> ||Sn - Sm|| loe ||Sn - S|| + ||Sm - S|| < epsilon> for m and n large enough.> Moshe >Well, ||A_m - A|| -> 0 is often the definition of the convergence of a>sequence of matrices. What's your definition?> Let A_n be a sequence of complex matrices. A_n converges to a matrix A if> each entry of A_n converges to the corresponding entry of A.So it's enough to show || (jth column of A_n) - (jth column of A) || -> 0, in the sense of vectors in C^m. But this can be written as || (A_n - A)(e_j) ||, where e_j is the jth standard basis vector. My latest brainchild, as minor as most of the others, is this mini-theorem.Let s_0 and s_1 be nonnegative integers, not both zero. Define inductivelys_{n+2} = | s_{n+1} - s_n |for n >= 0. Show that the sequence (s_n) eventually looks like0, x, x, 0, x, x, 0, x, x, ...and x = gcd(s_0,s_1).For example, if s_0 = 69 and s_1 = 39, the sequence runs69, 39, 30, 9, 21, 12, 9, 3, 6, 3, 3, 0, 3, 3, 0, ...LH with log(x) and log(y)...In an optimization problem, I basically need to seek the minization of thefunction of the form log(x+y)... but in order for further mathematicalmanipulation, I need to break log(x+y) up into log(x) and log(y),right?-Walala function that relates log(x+y) with log(x) and log(y)...In an optimization problem, I basically need to seek the minization of the>function of the form log(x+y)... but in order for further mathematical>manipulation, I need to break log(x+y) up into log(x) and log(y),>right?>-WalalaA simple-minded idea: minimizing x+y will minimize log(x+y).- Ken there any function that relates log(x+y) with log(x) and log(y)...In an optimization problem, I basically need to seek the minization of the> function of the form log(x+y)... but in order for further mathematical> manipulation, I need to break log(x+y) up into log(x) and log(y),> log(x+y)=log(x)+log(1+y/x) might help.But for minimization it is better to set the derivatives to zerorather than do such transformations...Arnold NeumaierX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft In over what appeared to be a slight variant of Zeno's>paradox, with him insisting that the passage of time cannot be>trully analog, since if time was continuous, the vector of timeThere is no such thing.>would have to have infinite speedThere is no such thing.>speed as it passed over ONE particular real second How does time pass over anything?>The above was quite interesting for me.It's all handwaving. Nothing that you quoted has any meaning,Mathematical or physical.>If the passing seconds are represented as real numbers, then indeed, >a _specific_ second in the future (which is a specific real number)>approaching the present and becoming past, How does a second approach a time?>passes from the state future to the state past in 0 time. Why? Why don't say that it takes future_time - past_time to pass?>Thus the speed vector of time as it passes over>a _specific_ (real) second, Again, that whole phrase has no meaning.>In other words, the rate of conversion How do you convert seconds, and how do you measure the rate?>In mathematical terms, No. The one thing that you have *NOT* done is to use anything vaguelyresembling Mathematical terms.>The above is related to Zeno's paradox,No.>The above smells fishy to me.It's total balderdash.>Indeed the quantifier for every, forces the brain to consider>mathematical continuity here, but in practical terms isn't always a>_specific_ epsilon that's being tested? What good would that do? If you want to prove continuity then you mustdeal with the universal quantifier.>So, although the quantifier intuitively (and>mathematically) satisfies this crave for continuity, in actual>practice, the available epsilons are discreetized in our mindsNo.>We certainly cannot>practically test EVERY available epsilon > 0.Irrelevant. Even if we could it would be a brute force proof and you'dbe asked to find a more elegant proof.>It appears though that the notion of philosophical continuity is>still somehwat elusive. (to me at least).It's a notion that has nothing to do with Mathematics and may evenhave nothing to do with Physics.>To top the cake, if one considers any possible action inside this>universe,One would not be considering Mathematics, but Physics.>every such action is quantized. That's an assumption. It is certainly *NOT* required by Quantum Mechanics.>Following this, does it make sense to talk>about the philosophical notion of having time being continuous?Sure, in a Philosophy news group. Here, time is off topic. Insci.physics, the issue of quantized time is very much an openquestion.>Time itself might well exist outside our existence continuously,ObMinkowsky If we take current physical theory seriously, time doesnot exist, any more than space does. Only spacetime exists.>But if consciousness is quantized,>then the term continuous time is nonsense.To a solipsist. Not to one who believes that there is a universebeyond the perceptions.>It appears as though the mathematician's main stance 1) above is>invalid.Irrelevant; it is a stance that relates to Biology, Psychology andPhysics, not to Mathematics.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot > It appears as though the mathematician's main stance 1) above is invalid.>Read Immanuel Kant's Critique of Pure Reason.Ignore Dedekind/Peano succession. The concept of number in mathematics isassociated with a closure property in the fundamental theorem of algebra.Geometric intuitions rule. Kant treated time and space in ideality ratherthan absolutely.Kant's notion of infinity precludes Cantor's use of the successor function toobtain transfinite numbers. It can only be interpreted in terms of single-pointcompactification, whence you get the Riemann sphere, stereoscopic projections,and the spherical encodings and quantizations discussed by Conway and Sloane inSphere Packings, Lattices, and Groups.Quantization is an artifact of serializing sense data into grammatical formsused to talk about sense data.You will get far better answers on a philosophy group.:-)mitch > Two days ago I had an interesting discussion with my coworker at the Math*snip*That was a long post & I am sure it was very interesting. To answer thequestion posed in the subject line of your post; read last month's issue ofScientific American. If loop quantum gravity is correct, then both space &time are discrete. > The above was quite interesting for me. It appears as though it is true. If>the passing seconds are represented as real numbers, then indeed, a>_specific_ second in the future (which is a specific real number)>approaching the present and becoming past, passes from the state future to>the state past in 0 time. Thus the speed vector of time as it passes over>a _specific_ (real) second, must be infinite. I do not know what a speed vector of time is supposed to be.Best measurements to date place it at about 1 billion nanoseconds per secondin the future direction. There are other directions, such as the pastdirection and dream directions.There are times when the magnitude of the time vector slows down by afactor of perhaps 100. Exact measurement of the factor is difficult todetermine as such times occur only in moments of unusual alert or duress.Sometimes of timelessness have also be noted.When awake, time is continuous. At other times, time is quantizedinto waking periods and dreams.> Do you view the present moment as the origin and specific moments in the> future as moving backward toward the origin? If so, I see no infinite.> I see the constant vector -1.>In no time at all, time had begun, for it's time had come. William Elliot [CapitalEth][EDoubleDot][Micro] .b3 time is supposed to be. Best measurements to date place it at about> 1 billion nanoseconds per second> in the future direction. There are other directions, such as the past> direction and dream directions. There are times when the magnitude of the time vector slows down by a> factor of perhaps 100. Exact measurement of the factor is difficult to> determine as such times occur only in moments of unusual alert or duress.> Sometimes of timelessness have also be noted. When awake, time is continuous. At other times, time is quantized> into waking periods and dreams. > Do you view the present moment as the origin and specific moments in the>future as moving backward toward the origin? If so, I see no infinite.>I see the constant vector -1.> In no time at all, time had begun, for it's time had come.--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------- -----------------------Eventually, _everything_ is understandable > Then we went over what appeared to be a slight variant of Zeno's paradox,> with him insisting that the passage of time cannot be trully analog, since> if time was continuous, the vector of time would have to have infinite> speed as it passed over ONE particular real second (represented as a real> number on the Real line of seconds).> Nonsense. >Then we went over what appeared to be a slight variant of Zeno'sparadox,>with him insisting that the passage of time cannot be trully analog,since>if time was continuous, the vector of time would have to have infinite>speed as it passed over ONE particular real second (represented as areal>number on the Real line of seconds).Nonsense.I know it probably is. You could have done us all the courtesy of showing usexactly where the nonsense lies though, genius.But, what should I expect with a name like fishfry?--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------- -----------------------Eventually, _everything_ is understandable <1073788832.691260@athnrd02.forthnet.gr> >Then we went over what appeared to be a slight variant of Zeno's>> paradox, with him insisting that the passage of time cannot be>> trully analog, since if time was continuous, the vector of time>> would have to have infinite speed as it passed over ONE particular>> real second (represented as a real number on the Real line of>> seconds).> Nonsense. I know it probably is. You could have done us all the courtesy of showing us> exactly where the nonsense lies though, genius.>Not before long, the number of Zeno steps would be infinitely increasingto swallow the same eternally unvarying constant quantum of time.> But, what should I expect with a name like fishfry?Sole food? 1073788832.691260@athnrd02.forthnet.gr...>Then we went over what appeared to be a slight variant of Zeno's> paradox,>> with him insisting that the passage of time cannot be trully analog,> since>> if time was continuous, the vector of time would have to haveinfinite>> speed as it passed over ONE particular real second (represented as a> real>> number on the Real line of seconds).> Nonsense. I know it probably is.So why publish it?You could have done us all the courtesy of showing us> exactly where the nonsense lies though, genius.Well, the same argument say that speed over any paricular point in anycontinuous motion is infinite. This is not the *definition* of speed But, what should I expect with a name like fishfry?> --> Ioannis Galidakis> http://users.forthnet.gr/ath/jgal/> ------------------------------------------> Eventually, _everything_ is understandable> Denis Feldmann probably is. So why publish it?I guess because I don't quite understand the notion of rate of continuoustime flow.After thinking about Wade's answer for a while, (constant vector of -1), Ithink maybe what he meant was:One could define rate of time flow as 1 sec/sec, which indeed is theopposite of what Wade suggested (because he assumed that I was looking attime backwards).I don't understand in what sense this can be a vector in the conventionalsense though. This vectorappears to be dimensionless.Correct me if I am wrong.> You could have done us all the courtesy of showing us>exactly where the nonsense lies though, genius. Well, the same argument say that speed over any paricular point in any> continuous motion is infinite. This is not the *definition* of speedQuite so.--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------- -----------------------Eventually, _everything_ is understandable The World Wide Wade >The above was quite interesting for me. It appears as though it is true.If>the passing seconds are represented as real numbers, then indeed, a>_specific_ second in the future (which is a specific real number)>approaching the present and becoming past, passes from the statefuture to>the state past in 0 time. Thus the speed vector of time as it passesover>a _specific_ (real) second, must be infinite. I do not know what a speed vector of time is supposed to be. Do you view> the present moment as the origin and specific moments in the future as> moving backward toward the origin? If so, I see no infinite. I see the> constant vector -1.I can't define it mathematically, cause I would need to define time in termsof time again. Think of it as follows:Imagine (your) consciousness (being a point) moving along the time line,forward, with time being continuously mapped onto the Real line. In a sense,you are continuously moving along the reals in the forward direction.A future point in time has an exact real representation somewhere in frontof you. It hasn't become present yet.Call this time-point t_0. Now it approaches and it transits in front of you,becoming past the moment your clock ticks t_0.What is the transition speed of t_0, as it passes from being a futuretime point to being a past time point RELATIVE to yourpoint-consciousness? If it was any finite number k, after, say, 2 wholeseconds, there would be 2*k points having been transitioned fromfuturepoints to past points. But, say, 2 whole seconds worth of time,have a finite measure, contradicting k finite, since 2*k time-points havemeasure 0. So k must be infinite (in fact uncountable), otherwise yourconsciousness couldn't possibly have moved any finite amount of timeforward.Can you see it?--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------- -----------------------Eventually, _everything_ is understandableLet p and q be distinct primes. Suppose that H is a proper subset of theintegers and H is a group under addition that contains exactly 3 elements ofthe set {p, p+q, pq, p^q, q^p}. Determine which fothe following are the 3elements in H.a) pq, p^q, q^pb) p+q, pq, p^qc) p, p+q, pqd) p, p^q, q^pe)p, pq, p^qThe answer turns out to be e) but it took me forever to figure this out. Sinceit was on a GRE math practice exam, it must be fair game for a regular exam. My question is this: Is there any fast way to know that the answer is e) asopposed to the others? I would have had to skip the question on a real exambecause it took me far too long. >Let p and q be distinct primes. Suppose that H is a proper subset of the>integers and H is a group under addition that contains exactly 3 elements of>the set {p, p+q, pq, p^q, q^p}. Determine which fothe following are the 3>elements in H.>a) pq, p^q, q^p>b) p+q, pq, p^q>c) p, p+q, pq>d) p, p^q, q^p>e)p, pq, p^q>The answer turns out to be e) but it took me forever to figure this out. Since>it was on a GRE math practice exam, it must be fair game for a regular exam. >My question is this: Is there any fast way to know that the answer is e) as>opposed to the others? I would have had to skip the question on a real exam>because it took me far too long.Since H is a group under addition, then H is the set of the integral multiples of some integer. Since H is a proper subset of the integers,then H is the set of integral multiples of some nonnegative integer distinct from 1. This means that the gcd of a set of elements of His a nonnegative integer distinct from 1. In all cases a)-d), the gcd of the three in the list is 1, thus eliminating them as possibilities.This leaves only e), where the gcd is p, and H is the set of integral multiples of p.David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. I want some of what he is smoking! of Prof Frank Tipler, David Deutsh, > Prof Paul Davies and Nick Bostrum that we are all living within a > 'computer programme of an ancestral simulation' created by an advanced > super-civilisation within the far-flung future!JS: Like The Matrix films. BTW I will be in London at least March 8 - 13.MD: What if the 'simulation' so unaware at first becomes aware of its > own predicament by becoming frustrated at not being able to 'break > through the wall of light' and therefore assumes true sentience by > understanding that it is being caged within a loop of an enfolding > programmed cave of information!Would this then make the 'Godmind' of the programmer very irate who then > decides to wipe the programme clean to then start again from scratch! > Perhaps this has happened many a time already and that each time of a > new Alpha of a begining when reaching an Omega point foreseen, the > programme contains elements of primary memories previously attained that > avoid being wiped clean, which survive unseen within our collective myths!Myths of previous worlds remembering giant birds that flew across the > oceans spitting fire at cities to therefore destroy and of arrows that > could reach the moon from a heroe's bow. Perhaps, even to become aware > of ones part to live all over again at certain points in ones own life > reenacting and to foresee future events about to happen one already has > lived!Perhaps such memories are left there to drive us forwards towards a > future pre-written looped back!The Mayans have 2012 and the Chinese also have the same mythic symbolic > time indicator that has now become as a collective focus pertaining to > the end of 'time.' This shared time between America and China may > indicate that the scientists of these two nations prove something that > shocks humanity into awakening only then to initiate a conflaguration of > an apocalypse as the Godmind wipes the simulation clean!Maybe this time indicator indicates our collective awareness of the > predicament and thereby the attainment of true sentience, which > invariably leads to the simulation being wiped again by the Godmind of > the programmer for it might be somewhat jealous and does not want > another to rival it.But, what would happen if the simulation finds a way of not being wiped > clean and that it breaks free through the wall of light to strike the > Godmind in the eye?Perhaps the simulation might then be welcomed as a prodigal child, but > would such a child stand the pace, and if not, only then to reel back > into an autohypnotic state of self induced trauma remembering fractured > memories becoming as mythic symbols to rise again when the child starts > to heal slowly from the future shock!Of course there is another perspective and that this reality is a > 'self-generated simulation' amongst a myriad number of others!Then we have the somewhat mundane dilemma of dealing with certain past > individuals who have molded human society by their influence and were in > fact influenced by a signal looping back from the future!Should such an influence be proved that their electron dreams were > merely spun by a charged signal from the future it would have far > reaching implications that might set the very world on fire.Were such individuals such as Christ, Muhammad and Abraham for example > aware of the nature of the charged signal's true origin that spun their > electron dreams?Perhaps not, since the signal was filtered through their culturally and > socially imprinted neural-nets and therefore saw only what they were > expecting to see pertaining to whatever internal dialogue they ascribed to!Of course the signal is on going as to imprint certain individuals at > certain times whom see whatever they are imprinted to see, such as UFO's > and Grays where as before long ago they perceived Angels and Demons for > whatever is seen is always one step ahead as an unknowable!Why an unknowable, which is masked by imprinted collective dialogue of a > symbol?It is because a prodigal child who reaches into 'breaking through the > wall of light' will invariably reel back from the shock of it all into > an autohypnotic state of trauma that will mask the truer experince that > rises as a black tidal wave into consuming a fragile mind!What be this truer experience?That we are all sustaining a self-created reality of a particular brane > and that we are all unconsciously shifting between alternate branes all > of the time without even realising it and the proof of it lies within > the nature of synchronicity subtle, for each synchronicity is a stargate > into an alternate brane next door!This particular brane is just one of a myriad number of others tuned > into and that we can consciously engineer the tuning into others should > we desire!But if this be true that a subtle synchronicity indicates a shift into > an alternate brane next door one would then invariably find that the > true nature of 'Magick' and its power is very much alive and that this > power resides within each and every one of us all!The super technology of a computer and of walking a myriad worlds is > here Schwannmany minds cohered by emotively charged symbols that spin the electron> dreams of those zapped by lightening winding back as in a loop!How many a Moses, Christ's, Abraham's, Muhammad's, Pacal Votan's, Buddha's> etc does it take to change a lightbulb as they all stand one legged on a> tightrope of a spun super-stringed lightening zap of a telephone line?What can we make of those who wander wind swept fields in the dead of night> dressed in black creating crop-circles and when caught out they say that > the> aliens in their heads got them to do it?Of course such individuals may just go and phone up certain people making> out that they are an alien or two as they speak like a computer and again> driven to do so by the voices strobing their brains!And when they are not about to get spun into the Wyrd of the web > trembled by> a possible consciously engineered directed signal of an emotive charge> certain predisposed individuals see dancing lights that hum a vortex into> opening up on a side of the hill as the eye of the moon hides its self away> behind a looming triangular cloud!As the lights do dance to hypnotise into spinning those into seeing and> without realising they are spun into an alternate brane next door!Perhaps a certain individual whose eyes becomes glazed over as if in trance> not themselves for a moment becomes as a window to look through by a time> traveller of a tourist who zaps their neural-net by spinning their electron> dream as a stargate to open to thereby perceive when not in trance a crop> circed mandala that suddenly appears within another brane verily spun into!The reasoning behind all of this perhaps being that to engineer fractal> realities of alternate branes next door to (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08GkkK10487; Sorry, I need to make yet more corrections. The property (1) B_n(r) subset B_{n+1}(r) listed below should read:If B_n(r) = (x_1, x_2, ...,x_n) and B_{n+1}(r) = (y_1, y_2, ...,y_{n+1}), then {x_1, x_2, ...,x_n } subset {y_1, y_2, ...,y_{n+1}}As for the sequence (which was supposed to be an 8-gon)(x, (x+y)/2, y, (y-x)/2, -x, (-x - y)/2, -y, (-y +x)/2)2 should be replaced with sqrt(2) to make all vectorsof the same norm.Note: Presumably, showing s_n -> 0 shouldn't be too hardsince every bounded sequence in a finite dimensional spacehas a convergent subsequence.>Three corrections to my first message:>1. dim X >= 2 should be prerequisited.>2. lim n -> infty (n*s_n/2r) >= 2 does not follow> from the triangle inequality as shown. It follows> from norm homogenity:> For x != 0, take the 2-gon (x, -x)> Then the distance from x to -x,and from -x back> to x, divided by twice the radius r = ||x|| is> (||x - (-x)|| + ||-x - x||)/(2||x||) = 4||x||/(2||x||) = 23. Not so much a correction, as (hopefully helpful) statement> I forgot to include: should be expected. If I remember correctly from my (lost)> notes, one can go from the square 4-gon (x, y, -x, -y)> where ||x+y|| = ||x-y|| to an 8-gon (x, (x+y)/2, y, (y-x)/2, -x, (-x - y)/2, -y, (-y +x)/2) Increasing the lower bound estimate of lim n -> infty (n*s_n/2r)> from ~2.6 to something closer to 3. Of course, one > first has to assume that the limit exists, i.e. (n*s_n/2r)> is a Cauchy-sequence. >>The kite equation thread inspired me to think back >>to another property Ive had in the back of my mind for a >>while. I think I can be fairly sure someone has thought of>>it before.>>Let X be a complex normed space. Let B_n(r) = (x_1, x_2, ...,x_n) >>(a finite sequence of elements of X) be an n-gon of radius >>r; this means>>(a) x_i != x_j for all 1 <= i,j <= n>>(b) dim (lin {x_1, x_2, ...,x_n}) = 2>>(c) ||x_1|| = ||x_2|| = ... ||x_n|| = r > 0>>>(d) ||x_1 - x_2|| = ||x_2 - x_3|| = ... = ||x_n - x_1|| = s_n > 0>>Call (B_n(r)) a sequence of n-gons if: >>for all n in naturals:>>(1) B_n(r) subset B_{n+1}(r) >>(2) B_n(r) != B_{n+1}(r)>>(3) B_{n+1}(r) is an n-gon of radius r>>For all n in naturals, define s_{n+1} to be as in (d), above,>>for B_{n+1}(r).>>>What if all we know about X is that it is a normed space and:>>lim n -> infty (n*s_n/2r) -> pi = 3,14...) for every n-gon sequence>>(or use for every two dimensional subspace, there exists an n-gon>>sequence... - which I believe should be equivilant to the above)>>>Notice that if X is a normed space and ||x|| = ||y|| > 0,>>then ||x-y||/||x|| <= (||x|| + ||y||)/||x|| = 2||x||/||x|| = 2>>Thus, by requiring X be normed space, we are actually>>stating that lim n -> infty (n*s_n/2r) >= 2 for X. It is also>>the property that there are nonzero vectors orthogonal to each >>other (-> ||x+y|| = ||x-y||) (this may be the case in every >>lim n -> infty (n*s_n/2r >= 2,6 just by using the parallelogram >>equation (hint: make a square). >>Note: actually, I forgot exactly what the number was (2,6 or >>something close) since its been a while and I dont know where my >>notes are on that anymore. However, it shouldnt take long to >>recheck.>>For me, it would (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08I3IK16839; In Hebrew, the feminine plural of the foreign loan word, nilpotent, when>written without vowels or other punctation, is writtennylpw_tn_tywtwhere _t is teth. What I would like to know is how it would be written>if one did write the vowels and other punctation. For example:>(0) which of the consonants is dotted?>(1) is the y following the first n retained and the n endowed with a chireq?>(2) is there a schwa under the l?>(3) is the w following the p replaced by a cholem or is it replaced by> a qamats qatan?>(4) what is under the first teth: a schwa, a segol or a chataf segol?>(5) is there a schwa under the second n?>(6) is the y following the second teth retained and the teth endowed with> a chireq?I would also like to know how I might figure out examples like this, involving>foreign loan words, on my own, instead of having to ask people in every>instance.Ignorantly,>Allan Adler>ara@zurich.ai.mit.edu*********************************** *****************************************>* *>* Intelligence Lab. My actions and comments do not reflect *>* in any way on MIT. Moreover, I am nowhere near the Boston *>* metropolitan area. *>* *>************************************************************ ****************That doesn't look like the feminine plural, butrather an abstraction, like nilpotency or support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08K3AS26439; What (if at all) the existence of higher ordinal infinities (such as strongly inaccessible cardinals and higher) imply in terms of CH or GCH ? > What (if at all) the existence of higher ordinal infinities (such as> strongly inaccessible cardinals and higher) imply in terms of CH or> GCH ?The general rule so far is that adding large cardinal axioms does notdecide CH one way or the other.Of course, some as-yet-unconsidered large cardinal support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08K3AG26434; >> Mathematicians are literate guys, so who ever first used homogeneous meant>> of the same form i.e. ax +by+cz+b isn't homogoneous since b isn't of the>> same form as ax,by,cz?According to Jeff Miller's nice webpage at http:// members.aol.com/jeff570/mathword.htmlthe first use is in:>HOMOGENEOUS EQUATIONS is found in 1815 in the second edition of Hutton's>mathematics dictionary: Homogeneous Equations ... in which the sum of the>dimensions of x and y... rise to the same degree in all the terms (OED2).> That definition of homogeneous is still used in many d.e. books where it causes enormous confusion with homogeneous linear d. e. which is completely different. That last use of the word homogenousis more in keeping with what, in modern linear (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i08K3Bw26471; > Let p(z) be polynomial of degree n such that >>> |p(z)|<= M for all z in the unit disc i.e. |z|<1>> >> Prove that |p(z)|<= M|z|^n for all |z|>1>this function f brings sex in the city.How can I prove now (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933el25569; > In general is there an agreed concensus as to naming rights in> mathematics, and/or is there a body that governs naming rights?>>> No. Most things are named after totally irrelevant people>> (e.g. Pell's equation) :-)And:during a representation theory conference some speaker talks of using>Brauer's trick. A hand goes up. 'What's Brauer's trick?' Asks Richard>Brauer.third hand anecdote, so open to corrections that actually it was Witten,>or someone else entirely.Herr Weyl you must explain me one thing-what is What about Chapman inequality?Is it named after Robin support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933dF25511; >> 1)The function f(x) has f ''(x) > 0 for all x 0 and the graph is asymptotic to y = ax + b as x. >> Show that f(x) - ax - b has negative derivative for all x > 0,>and that f(x) > ax + b for all x 0.[/quote:b29fe8cde7](1)Suppose f'(p) - a > 0 for some x. Then, since f'(x) isincreasing (f''(x) > 0), f'(x) > f'(p) for all x >= p andhence: f(x) > f(p) + (x - p) f'(p) = = f(p) - p f'(p) + f'(p) xfor all t >= x. Thus: lim f(x)/x >= f'(p) > a,which contradicts our assumption that f is assimptotic toy = ax + b.Hence f'(x) <= a. But f' is strictly increasing, sof'(x) < a.(2)Function f(x) - ax - bis strictly decreasing (since it has negative derivative)and has a limit 0 when x tends to infinity. Hence it is(strictly) positive.Hope this support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933dI25515; 2001 00:40:03 +0200, Alexander Pambook>Really ! Who? Who is inventor of multiplication table?>What exactly is the approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933dF25523; Could a proof of the Goldbach Conjecture be worth of a Fields prize? P.S. Since the proofs for higer dimensional cases of Poincare Conjecture could be worth the prize, I think the proof of Poincare Conjecture could also support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933de25519; >I had written a program to find Pythagorean Triples years ago. I have>>now restructured it to find quadruples with power 3 as well - there are>>a lot, even a few starting with 1^3.>>But there are no quintruples (is that word correct?) with power 4, no>>hexuples (is that word correct?) with power 5, etc. (Not even triples or>>quadruples with higher powers.)>>What I haven't tried yet is quadruples or quintruples with powers 2 and>>3...-- >http://www.tarpley.netCould you send me that program support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933e525581; Anyone know of any recent work on the odd perfect number question?Ciao, support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0933du25543; Wouldn't a dictionary without vowels be a dctnry (or sometimes, dctnr)?You said you are using a dictionary, right? What does it say?>If it too has no vowels, you probably ought to invest in one>that has vowels.>> In Hebrew, the feminine plural of the foreign loan word, nilpotent, when>> written without vowels or other punctation, is written>>> nylpw_tn_tywt>>> where _t is teth. What I would like to know is how it would be written>> if one did write the vowels and other punctation. For example:>> (0) which of the consonants is dotted?>> (1) is the y following the first n retained and the n endowed with a chireq?>> (2) is there a schwa under the l?>> (3) is the w following the p replaced by a cholem or is it replaced by>> a qamats qatan?>> (4) what is under the first teth: a schwa, a segol or a chataf segol?>> (5) is there a schwa under the second n?>> (6) is the y following the second teth retained and the teth endowed with>> a chireq?>>> I would also like to know how I might figure out examples like this, involving>> foreign loan words, on my own, instead of having to ask people in every>> instance.>>> Ignorantly,>> Allan Adler>> ara@zurich.ai.mit.edu>>> ************************************************************** **************>> * *>> * Intelligence Lab. My actions and comments do not reflect *>> * in any way on MIT. Moreover, I am nowhere near the Boston *>> * metropolitan area. *>> * *>> ************************************************************** support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09DLVu01322; >Squares of 0.999... tend toward 0, not toward 1 .9^2 = .81> .9^3 = .729> .9^4 = .6561> .99^2 = .9801> .99^3 = .970299> .99^4 = .96059601 .999^2 = .998001> .999^3 = .997002999> .999^4 = .996005996001> .9999^2 = .99980001> .9999^3 = .999700029999> .9999^4 = .9996000599960001 .99999^2 = .9999800001> .99999^3 = .999970000299999> .99999^4 = .99996000059999600001 .999999^2 = .999998000001> .999999^3 = .999997000002999999> .999999^4 = .999996000005999996000001 .9999999^2 = .99999980000001> .9999999^3 = .999999700000029999999> .9999999^4 = .9999996000000599999960000001> .99999999^2 = .9999999800000001> .99999999^3 = .999999970000000299999999> .99999999^4 = .99999998000000059999999600000001 .999999999^2 = .999999998000000001> .999999999^3 = .999999997000000002999999999> .999999999^4 = .999999996000000005999999996000000001 .9999999999^2 = .99999999980000000001> .9999999999^3 = .999999999700000000029999999999> .9999999999^4 = .9999999996000000000599999999960000000001 .99999999999^2 = .9999999999800000000001> .99999999999^3 = .999999999970000000000299999999999> .99999999999^4 = .99999999996000000000059999999999600000000001 .999999999999^2 = .999999999998000000000001> .999999999999^3 = .999999999997000000000002999999999999>.999999999999^4 = .999999999996000000000005999999999996000000000001[snip rest for brevity]Prove squares of 0.999... tend toward 1>or stop claiming that 0.999... = 1 > Garry Denke, Geologist>Denoco Inc. of Texas You're a geologist? Don't you need to know arithmetic for that? In the first place since 0.999... is a single number it makes no sense to talk about squares plural. What you (perhaps) mean is that squares of the partial sums, 0.9, 0.99, 0.999, etc. I say (perhaps) because you appear to be really looking at cubes, fourth powers, etc. The squares YOU give are: .9^2= 0.81, .99^2= 0.9801, .999^2= 0.998001 ... to .999999999999^2 = .999999999998000000000001. And you claim that these tend to 0??? What IS true is that if a< 1 then a^2, a^3, a^4 ... tends to 0- but that's NOT the same as saying approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09DLVV01334; support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09EwYE08366; >Rationals are UncountableLet S be the set of all rational numbers [0,1).>s is a member of S if 0.000... <= s < 1.000...>and s is rational.Assume s is represented in base factorial (!).>Base ! is used because every rational number>has a finite representation in base !.In base ! the allowable digits for>position k are (0,1,...,k).>(k starts at 1)Every position, k, represents 1 / (k+1)!.k>1 1/2! = 1/2>2 1/3! = 1/6>3 1/4! = 1/24.123 (base !) = 1/2 + 2/6 + 3/24 = 0.958333... (base 10).Every rational number has an unique finite base !>representation. Any finite base ! number is rational.We can create the set S defined above by taking the>set produced by counting in base !..0>.1>.01>.11>.02>.12>.001>.101>...There exists a rational number, x, not in S.If S(i) is of the form .111...1 and>its length is equal to or greater than x>then set x to a string of 1's one longer than S(i).>If I understand correctly, you seek an x of the form .111...1 one digit longer than the maximum x of this form. Since you list all of them (all numbers of this form are rationals in the [0,1) range) there's no such maximum, and therefore no such x.>x differs from every member of S.>x is a rational number because it has a finite number>of digits. The length of x is exactly one greater>than some member of S.0.0 <= x < 1x = 1/2! + 1/3! + ... + 1/k!>and equals the largest rational number>less than the fractional part of e.Actually, it is the smallest rational approximation of e that is not in set>S>where a rational approximation of e is any number of the form .111...111.>Russell>- Zeno is right. Motion is support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09EwZO08407; I have a problem to solve where I stuck with this subproblem:Say m(x) is a certain function, for which transformation T(m(x)) = k(x)can be said that: Sum(i=1 to (p-1)){(-1)^(i+1)*(diff^i(k(x))/(diff(x))^i))} is always larger than (-1)^(p+1)*( diff^p(k(x))/(diff(x))^p) ) could approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09FdvP11549; A MATHEMATICAL PLAY IN THREE ACTSPROLEGOMENON[An informal definition of naturals] an infinite set of whole numbers...., etc. A direct outcome of this definition is that * Point 1: There are not naturals with infinite nonzero digits.* Point 2. As a result of Point 1, it is not possible to assume the bijection N <-> R. - Proof (see below). # FIRST ACT[Cantor and Human Curiosity]HC: Will it be possible to count the elements of R?C: Perhaps.HC: But how?[Cantor thinks for a moment, and states a criterion]C: Well, I think it is easy. Without getting into extreme formalities, we would be able to say that an infinite set would be countable if and only if we can establish a bijection between K and N, i.e. N <-> K.HC: Thats great! It sounds elegant, logical and intuitive. However, I think we will have a little problem using it with R.C: Why?HC: Because Point 2 tells us that we cannot propose that bijection with R. So, your criterion doesnt work in this case. We will have to look for another criterion. Its a pity, because it was nice.C: Yes, thats true. # SECOND ACT[Mrs. Perfect Proof1, Mrs. Perfect Proof2, Mrs. Perfect Proof_n and Human Curiosity]answer to your curiosity is that the reals are uncountable, because it is no possible to achieve the bijecci.97n N <-> R.PP2: Yes I have reached the same conclusion.HC: Why?PP2: For the same reason that PP1.HC: And you?PP_n: Exactly the same that my but its not possible to conclude that R is not countable from that criterion, because it cannot be used with R.PP1: Oh, we are very sorry. We did our best. # FINAL ACT[Mr. Na.95ve and Human Curiosity]MR. NAIVE: Do you know, Mr. HC? If we admit Cantors criterion to count R, as we positively know that N <-> R is not possible (it doesnt matter why), it is clear that R will be necessarily uncountable. What do you think about it?HC: Well Mr. Na.95ve, we can do so, but it will not convince anybody. The problem is that we know that Point 2 invalidates Cantors criterion. Therefore, R would be certainly uncountable, but no because R be intrinsically uncountable, but because we dont have a valid criterion to count it.MR. NAIVE: Well then, perhaps there are other different criteria to count the reals.HC: I dont know it, but if we put a simple analogy perhaps we could clarify our thoughts. Lets try it. We may take the following equivalence table N <==> M = A set of linear units of measure R <==> A = The air that there is in a room Being our goal to find out whether A is measurable or not, we have the following analogy:POINT 1: A cannot be measured with linear unities, so we cannot assume M -> A.CRITERION: A set D will be lineally measurable if and only if it is possible M -> D.POINT 2: We cant use CRITERION to find out whether A is measurable or not, because POINT 1 does not allow it.PROOFS: We have found out that A is not measurable, because CRITERION case. NOSENSE: We can admit CRITERION to measure A, and consequently A is not measurable. Well, clearly it seems a foolish thing to say. In the first place because POINT 1 tells us that we cannot apply CRITERION to measure A, and in the second place because A would be able to be measured by other means.Good, Mr. Na.95ve, this is the end of our analogy.MR. NAIVE: Which is the result? Are there other criteria to find out whether R is countable or not?HC: Well, in our analogy A cannot be measured with linear units, but if we arrange orthogonally three of them, then we can. By analogy, using two or more bijections N <-> R at once perhaps we could count R. Think about this, MR. Na.95ve, think about this.MR. NAIVE: Ill do it.***************************Proof of Point 2INITIAL ASSERTIONS1) We can represent any real number using a given positional system of numeration. For ease we will use the decimal notation.2) Cantor uses a transfinite method (the diagonal argument) to prove that the premise N <-> R is false, and we will use another transfinite method to prove that the transfinite construction N <-> R is impossible, and therefore we cannot assume the bijection N <-> R.Proposition: The transfinite construction N <-> R is not possibleProof: Every real number within the interval (0, 1) has as first decimal digit one of the ten digits of the decimal system of numeration, i. e. it must be of the form 0.0, 0.1, 0.2, , 0.9. As you can see for this first digit there are 10 ^ 1 possibilities. For the second digit we have 0.00, 0.01, 0.02, , 0.99. Consequently, there are 10 ^2 possible combinations for the two first digits, 10 ^3 for the first three digits, and so on. When the number of digits is infinite we get R.Now, when we have only a digit of the decimal expansion of the reals, we make the one-to-one correspondence 0 <-> 0.0, 1 <-> 0.1, 2 <-> 0.2, , 9 <-> 0.9. Next, with two digits we begin the 1-1 correspondence 0 <-> 0.00, 1 <-> 0.01, , 10 <-> 0.10, 11 <-> 0.11, , 99 <-> 0.99. With three digits, we continue doing the same, and so on.And now is when the impossible transfinite construction appears. While we keep doing the 1-1 correspondence within the finite decimal expansion, everything will go fine. But, what natural would correspond to the decimal expansion of the number e? By induction it is obvious that it should be 71828182, i. e., a number with infinite nonzero digits, which doesnt belong to N (Point 1). Conclusion:The transfinite construction N <->R is not possible if we do not admit naturals with an infinite number of nonzero figures. If we admit that, then the reals would be countable.Nicolas de la support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09EwZA08393; okay the equations can be modified to1.(sig)^2 = E[a0^2] + E[a1^2] + E[a2^2] + ..... +E[aN^2]2.(sig)^2 = E[a0^2] + 2^2.E[a1^2] + 2^4.E[a2^2] + ..... +2^2N.E[aN^2].........................which reduces the problem to finding a closed-form expression for the matrix> -- -->| 1 1 1 . . . 1 |>| 1 2^2 2^4 . . . 2^2n |>| . . . . . . |>| 1 n+1^2 n+1^4 . . . n+1^2n | > -- --which is definitely a vandermonde matrix...i have tried the routines given in the numerical recipes in C book and the ways to invert i found on the net..i want to solve it for n as large as 1024..a routine written on the computer gets out of hand very fast as the intermediate steps in the matrix inversion entail finding factorials of numbers as large as n..is there a routine/method/algorithm that circumvents this problem of intermediate steps requiring huge numbers...please help!!!!!Ankit SeedherTexas Instruments, India>>Could you describe what problem you are trying to solve?The problem is the following: >An electronic circuit i am trying to analyze has an output of the form:>y(x) = a1.x + a2.x^2 + a3.x^3 + .... + aN. x^NThe inputs {x} are discrete points from the set {1,2,3,4,...N}The ideal output is y = x => a1 = 1, aj = 0 for j ~=1>However, the actual outputs are such that yj (output for x = j) is a gaussian random variable with mean j and variance j(sig)^2. All yj s are independent RVs. i.e. cov(yi,yj) = 0 for all i,j.>It can also be assumed that aj (for j ~=1) is a zero mean r.v. I basically want the variances of aj given that yj is known to be an r.v. with gaussian pdf N(j, j*(sig)^2). Also cov(ai, aj) = 0 for all i,j.So we have,>y1 = a1 + a2 + ....... + aN>y2 = 2.a1 + 2^2.a2 + ...... + 2^N.aN> ......................>yN = N.a1 + N^2.a2 + .......+ N^N.aNNow if we take the mean square expectation on this equation we get1.(sig)^2 = E[a1^2] + E[a2^2] + ..... +E[aN^2]>2.(sig)^2 = 2^2.E[a1^2] + 2^4.E[a2^2] + ..... +2^2N.E[aN^2]>.........................So we get a set of N linear equations. We wish to solve for E[a1^2], >E[a2^2],....,E[aN^2]. Basically, we wish to obtain a closed form expression for E[aj^2] which boils down to finding the inverse of the matrix:> -- -->| 1 1 . . . 1 |>| 2^2 2^4 . . . 2^2n |>| . . . . . . |>| n^2 n^4 . . . n^2n | > -- --I had expressed this matrix in a general from in my post yesterday.So we need to invert this matrix and get a closed form general expression for E[aj^2] in terms of (sig)^2. I shall try to dig up some info about Lagrange interpolation as u suggested to see if it can be applied here. Please let me know if you have some other suggestions now that I have described the problem.I would greatly appreciate any help.>Ankit Seedher>Texas Instruments, India.>> I am working on an engineering problem that has got stuck at a point where>>I need to find a closed form expression for the INVERSE of the following>>n-by-n matrix:> -- --> | 1 1 . . . 1 |> | x1 x1^2 . . . x1^n |> | x2 x2^2 . . . x2^n |> | . . . . . . |> | xn-1 xn-1^2 . . . xn-1^n|> -- -->>Could you describe what problem you are trying to solve?>>It looks like you are trying to interpolate a polynomial, perhaps finding>>coefficients of a polynomial through given points. Well, this can easily be>>accomplished using Lagrange interpolation.>>However, often you don't need the explicit values of the coefficients, but>>merely an efficient means of evaluating the polynonial at arbitrary points.>>In that case you might consider Chebyshev polynomials.>>I'll happily expand on these matters if you are interested.>>If you really want an analytic expression for the inverse of the matrix, I'd>>guess a simple formula could be given, but I don't have it here. I suggest>>considering a 3x3 matrix and see if you can determine approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09HoAO22250; > 1)The function f(x) has f ''(x) > 0 for all x > 0> and the graph is asymptotic to y = ax + b as x -> oo. > Show that f(x) - ax - b has negative derivative for all x > 0,> and that f(x) > ax + b for all x > 0.I have an eyeball approach to the problem . . . not much rigor.Restricting our discussion to x > 0, the curve is always concave UP.Since it approaches the line y = ax + b asymptotically,the distance f(x) - (ax + b) is constantly decreasing.Therefore, its derivative will be constantly negative.To be concave UP and approaching a line (slanted or horizontal),the curve must be ABOVE the line. (Make a sketch.)Therefore, f(x) > ax + support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09HoAq22246; >Let sigma(n) be the sum of the positive divisors of n.There is a term(Hardy ?; Ramanujan?) for sigma(n):sigma(n)=Pi^2n/6( 1+(-1)^n/4 + 2cos((2/3)nPi)/9 +> 2cos((1/2)nPi)/16 + 2(cos((2/5)nPi)+cos((4/5)nPi))/25 +> 2cos((1/3)nPi)/36 + ...)for which I have lost the reference.Is there a general expression for this term or how can I calculated the next approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i09KfO603630; >
>Sorry, I meant to ask: what day
of the year is the longest?>This question is usually given
a wrong or incomplete answer.> All the days are 24 hours
except when a leap second is added.> If you mean to ask
which day has the longest daylight at a given> location,
that would be the day of the summer solstice. >>That is true
if you define the day to be from sunrise to sunset.>>But if
you define the day to be from sunrise to sunrise, then>>the
day is longest when the Earth is closest to the Sun,
which>>happens around December 23 (a little after winter
solstice).I define the day to be from midnight to midnight.
Perhaps you meant to>ask about the solar day, which is the
time between transits of the>sun. However, your answer is
incorrect. Perihelion occurs in early>January (around January
8, IIRC), and therefore the answer is *not* the>perihelion
date. In general, the solar day is longer than average
near>the solstices (both of them) and shorter than average
near the>equinoxes.The length of the solar day is determined
by the sun's daily movement>in right ascension, which
determines the difference between the>sidereal day and the
solar day. What's confusing you is that the sun>moves fastest
*in longitude* at perihelion, but fastest movement
in>longitude does not equal fastest movement in right
ascension. The>obliquity of the ecliptic causes the fastest
movement in right>ascension to occur near the two solstices,
despite the fact that one of>them happens to be near
aphelion.>Why is Dec. 19 the longest solar day instead of
perihelion Jan. 4th?>The longest solar day is presently
achieved near the December solstice,>because that one is
closer to perihelion than the June solstice is.>Because of
precession of the equinoxes, there will come a time
when>perihelion is not particularly close to either solstice,
and therefore>nowhere near the longest solar day.-- >Dave
Seaman dseaman@purdue.edu> ++++ stop the execution of Mumia
Abu-Jamal ++++> ++++ if you agree copy these lines to your sig
++++>++++ see http://
www.xs4all.nl/~tank/spg-l/sigaction.htm
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i09JbDx30828;
>Okay I have never
been very good with math and hopfully someone out>there can
help. Here is the situation:I Have a total of five different
items: I am shipping out bowes with a>combination of these
different items and need to prepare all the>different
combinations before hand. So lets name the items:>1. tshirt>2.
glove>3. hat>4. ring>5. necklacenow a package can have either
all five items or a choice of lets say>number 1,2,4 or 1,2 or
1,4,5 or just 2 and so on... I think you probly>get the jist
of what im asking.What are the total number of combinations
for the items obviously>without just rearanging thier order.
and what is the formula to do so?Happy New YearAssume that
each item can be either in or out of the box that you
areworking on. Assume that the choices are independent. Assume
that theitems are unique (not diplicated).Then the number of
possible boxes is 2 x 2 x 2 x 2 x 2 = 32 , but oneis an empty
approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i09LKeV06757;
>It took about 22 seconds to print out
1,000,000 primes. How much faster is>it able to be?also the
program produces 10,000,000 primes in 22.27 secondsThe program
stores all the primes in a bit array and can be accessed to see
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0A3Bq119159;
I am working on
the program where is necessary to calculate huge>amount of
numbers in always the same sequence, before the
program>actually opens.In other words, these number have to be
calculated and saved into>memory before the program starts. I
was trying to use number pi, but to calculate 100 million
decimal>places of number pi would take forever and no one
would want to wait>that long. On the other hand I can't afford
to include the whole 100>million dec. places along with the
software, that would take too much>of the space and too much
time to download. Is there any other number like PI, which I
could calculate
faster?>Serge>------------------------------------------------
----------> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY
**>----------------------------------------------------------
> http://www.usenet.comWhat
do you mean when you say like pi? Transcendental?
approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i0A3Btg19224;
following probability problem as part of a>job-interview
pre-screen. After giving a totally wrong answer ('cause>I
wasn't thinking), I gave the following answer. However,
the>interviewer said that my answer was still incorrect.Fair
enough; I find probability unintuitive. But now my curiosity
is>peaked: What is the right answer? Question: You have a jar
with 1000 coins in it, 999 are normal, but>one is two-headed
(i.e both sides are heads). You choose a coin at>random and
toss it 10 times. With each toss the coin comes up heads,>so
you get 10 consecutive heads. What is probability that you
have>choosen the two-headed coin? My answer:P(two headed) = 1
- P(normal coin *and* 10 heads in a row)> = 1 - .999 *
(1/2)^10> = 1 - .999 * (1/1024)> = 1 - .000976> = 0.999024> =
99.9024%If this is wrong, what is the right answer? How do you
calculate it?StuartNice math, Stuart, but wrong.Instead of
saying there is one two-headed coin,say there is one red coin.
If you pick one coin at random out of 1000 (withoutlooking),
the probability that you picked the redcoin is 1/1000.You can
toss the coin all day long and this doesn'tchange.Get
approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i0A3BqJ19163;
I have just attended a course on scheduling
for project managers and there>the beta distribution comes up
in relation to estimation of task durations.I have looked up
in my old probability theory book regarding this and found>the
proof for the fact that the sum of _many_ beta distributions
converges>to a gaussian distribution, but neither my beloved
book (and some hard core>integral solvining) or extensive
search on the internet has given me>what I have found (this
includes characteristic function of the beta>distribution and
it's density function) I am beginning to believe that there>is
no distribution defined/named for the sum of two beta
distributions, but>I might be wrong.Can anyone answer this
one?/Torben>-- P.S. The views expressed above are my own.>Are
you sure you want the sum of two beta distributions?Perhaps
you want the pdf for the sum of tworandom variables, each
independently distributed accordingto beta distribution,
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0A3Bqd19155;
i have a rough
draft. the subjects/approach are these:1. using ternary logic
to2. axiomatize a universal set into existence while3.
avoiding russell's paradox by4. modifying the subsets and
foundation axioms in a way that theoretically shouldn't change
anything when those axioms are applied to crisp sets. in other
words, this is a strict extension of zfc. the main idea is
that in russell's paradox, if S={x in U: x is not an element
of x}, then the truth value of the statement S is in S is the
third truth value. the details are in my paper. as i alluded
to, when applied to crisp sets, the new subsets axioms and new
foundation axioms work the same way. i have only found two
fuzzy sets so far and they are both related to S. part of the
idea is to construct some new membership symbols; there are
four:1. a membership symbol to mean membership is ambiguous,
unknown, or general2. a membership symbol to mean membership
is true3. a membership symbol to mean membership has the third
truth value4. a memberhsip symbol to mean membership is false.
since symbols mentioned in 2-4 are crisp objects (eg, for 2,
either it is true that membership is true or it's false that
membership is true), standard deduction, contradiction, etc,
applies. in fact, if all other axioms are taken to mean 2 when
the usual symbol is written and 4 when the negation of the
usual symbol is written, then it should work out. in the
subsets axiom, i used a new biconditional that i defined by a
truth table. this biconditional (and the new conditional it's
based on) is an extension of the usual (bi)conditional, i.e.,
when the truth values are limited to T and F, it is the same.
here is a link to
it:http://www.physicsforums.com/attachment.php?s=&postid=
124723this is a zipped pdf located at
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0AD3T816500;
>
>Anyway, I
see no reason to stop there. Since>>the system already uses
deci- and centi-,>>why not use all metric fractional prefixes?
This>>would give: >decillion =3D 10^33>>centillion =3D
10^303>>millillion =3D 10^3003 =3D 10^(1000^1 + 3)
>>micrillion =3D 10^3000003 =3D 10^(1000^2 + 3)>>nanillion =3D
10^(1000^3 + 3) >>picillion =3D 10^(1000^4 + 3)>>femtillion =3D
10^(1000^5 + 3)>>attillion =3D 10^(1000^6 + 3)>>zeptillion =3D
10^(1000^7 + 3)>>yoctillion =3D 10^(1000^8 + 3) >Uh, oh... ran
out of prefixes! :-( =A0 BTW, the>>American version of that
last one has>>1,000,000,000,000,000,000,000,003 zeroes,>>and
the British version (10^(1,000,000^8))
has>>1,000,000,000,000,000,000,000,000,000,000>>
000,000,000,000,000,000 zeroes (somewhat>>difficult to write
out in full). Ooops... that should read:decillion =3D
10^33>centillion =3D 10^303>millillion =3D 10^3003 =3D
10^(3*1000^1 + 3) >micrillion =3D 10^3000003 =3D 10^(3*1000^2
+ 3)>nanillion =3D 10^(3*1000^3 + 3) >picillion =3D
10^(3*1000^4 + 3)>femtillion =3D 10^(3*1000^5 + 3)>attillion
=3D 10^(3*1000^6 + 3)>zeptillion =3D 10^(3*1000^7 +
3)>yoctillion =3D 10^(3*1000^8 + 3) where yoctillion has
3,000,000,000,000,000,000,000,003 zeroes>(American), or
6,000,000,000,000,000,000,000,000 zeroes (British).>
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0AD3V716544;
I'm writing
because a search showed that my infinite-precision calculator
(http://medialab.dyndns.org/bignum) was cited here by uncle
al.Unfortunately, dyndns.org have blocked our site without any
warning,making it point to 10.0.1.128 instead (whoch doesn't
exist, claiming that one of the 400 users on the site has
published something that is covered by copyright.Yeah,
right.Anyway, fortunately we have another name,
medialab.freaknet.org, for the same computer, so my calculator
can now only be reached as
http://medialab.freaknet.org/bignumPlease update your
approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i0AKqwU16572;
>>3.14159265358979323...>>3,1,4,5,?,?,....>
>How many digits of pi have you memorized? I did 100. The world
record> I knew about 1500 digits at one time. A. Say> I
HAVE A NUMBER > to remember PI as 3.1416 B. Desire> MAY I HAVE
A LARGE CONTAINER OF COFFEE > to remember PI as 3.1415926 C.
Chant> GOPEEBHAGYA MADHURVAATHA SHRADGI SHODHADHI SANDHIGA I>
KHALAJEEVITHA KHAATHAAVA GALAHAALAARA SANDHARA II> a sanskrit
verse from Atharvana veda,which is in praise of > Lord Vishnu
and Lord Shankara.If this is deciphered as per> language of
Vedic Mathematics,it denotes the value of PI/10 to > 32
decimal places as below:> .31415926535897932384626433832792>
You need to know/learn which of the alphabets represent each
of > the numerals 1,2,3,4 ....> Krsna Tirthaji
Maharaja(1884-1960)Sankaracharya of Govardhana>
Matha,Puri,Orissa State,INDIA) D. Professor Yasumasa Kanada
and nine other researchers at The > Information Technology
Centre at Tokyo University,calculated the > value of PI to
1.2411 trillion places,in September 2002.> which computed this
value working continuously for over > 400 hours.> - bsr A
approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i0AKr2B16618;
>> what I think you are missing is that
.999... (a zero, a dot and an>> infinite amount of 9s) is
equal to 1, so what you are asking is>> whether 1^2, 1^3, 1^4,
etc. is 'tending to' 1.i understand the premise.> Look at it
this way: We can write a number like 0.9 as 1 - 0.1, and a>>
number like 0.99 as 1 - 0.01. In general we can write it down
as:>>> f(n) = 1 - (1/(10^n)) for n>0 where n is the number
of nines behind>> the dot.>>> So the number 0.999... can be
found by looking at:>>> lim{n->infinity}{f(n)} = >>
lim{n->infinity}{1 - (1/(10^n))} =>> lim{n->infinity}{1} -
lim{n->infinity}{1/(10^n)}but infinity means unlimited...
limit{n->unlimited}{f(n)} = >limit{n->unlimited}{1 -
(1/(10^n))} =>limit{n->unlimited}{1} -
limit{n->unlimited}{1/(10^n)}...which contradicts the
premise.>> Since the limit of the fraction-part goes to
zero, the whole thing>> becomes equal to 1. In other words
0.999... is equal to 1.only if the premise the unlimited has a
limit is believed.garry denke, geologist>denoco inc. of texas
Now you seem to be playing games with words. The fact that n
is unlimited does not mean the sequence is. It is easy to show
that the limit limit{n->unlimited}{1} -
approve@localhost) by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i0AKr5116644;
If I post it here is anyone here willing to
look over an outline argument that there are no odd perfect
numbers and point out the flaw/s?Any flaw shouldn't be too
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i0AKrBi16725;
I have this
question here that I attempted and failed. some pointer
toward>this problem is greatly appreciated.The golden ratio is
the ratio b:a so that b:a = (a+b):b. The ratio of the>golden
ratio was considered by the ancient Greeks to be the most
perfect>ratio for the lengths of the sides of rectangles, such
as portraits. Show>that if b:a is the golden ratio, then b/a =
(1+ sqrt(5))/2.I guess the question is really saying to derive
b = (1+sqrt(5)), and a = 2>from b/a = (a+b)/b (*).So I tried to
manipulate with the algebra of (*), and failed.>Is there
anything i am doing wrong?> Gavin ,[ b+a]/b=b/a , or
ab+a^2=b^2 , or b^2-ab-a^2=0giving b=[a+(or-)sqrt{a^2+4a^2}]/2
=a[1+-sqrt(5)]/2 or b/a=[1+sqrt(5)] or b/a=[1-sqrt(5)]/2 But do
it simply (the Classical Greek-UNGRADUATED STRAIGHT EDGE AND
COMPASS mrthod): Draw sqrt(5),as the hypotenuse[AC] of the
orthogonal triangle[ABC] with horizontal[AB] side 2, and
vertical[BC] side 1. Etend AC by 1 unit to D to btain line
ACD=sqrt(5)+1 . Use compass and divide line AD, at point E so
that AE=(AC)/2 or AE = [sqrt(5)+1]/2.Panagiotis
Stefanideshttp://www.stefanides.grhttp://www.stefanides.gr/
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>> There is
another slightly subtle flaw with your argument. You know>>
that a_1(x,y) and a_2(x,y) are roots of >>> a^2 - (x - y) +
7(x^2 + xy).Typo, this should be a^2 - (x - y)a + 7(x^2 +
roots. Perhaps you choose as the 'negative' root,>>>
a_1(x,y) = ((x - y) - sqrt((x - y)^2 - 28*(x^2 + xy)))/2>>>
and the other as the 'positive' root,>>> a_2(x,y) = ((x - y)
+ sqrt((x - y)^2 - 28*(x^2 + xy)))/2.>>> The choice is
arbitrary. Your subsequent argument makes no>> use of whether
the negative or positive root is chosen.>>No. At x=0, one of
these roots ( the 'positive' root) is 0,>the other (the
'negative' root) is -y.>James chooses a_1(0,y) to equal 0 and
a_2(0,y) = -y. Values for>other x can be specified by chosing
a branch cut and insisting on>continuity. There is no
ambiguity as long as y is not equal to 0. > Yes, you're right
again. I knew this and had forgotten it.And similarly for the
cubic. Careless thinking on my part. Other arguments however
obviously still apply. I am wondering what error Harris
himself thinks he has found in his choice of adding the
variable y to the polynomial. Nora B.>>> The same problem
actually occurs with your argument about>> your cubic
polynomial, except there you are arbitrarily choosing>> one of
three roots a_1, a_2, and a_3 instead of one of two. You >>
conclude that two of the three must be divisible by 7, and the
>> other coprime to 7. Again James choses a_1 and a_2 to be 0
at 0. Again we can resolve the>ambiguity at other values of x
by using continuity and branch cuts.>The double root at x=0
(and possible other double or triple roots for>other values of
x) leads to a minor ambiguity, but not an important one. -
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>>How to integrate
sinc(x) (without Fourier transformation) ??>> That integral is
not an elementary function. It is known as the Sine>> integral
function.>> <http://
mathworld.wolfram.com/SineIntegral.html>Nice link.
Beautiful images. :-)1. first of all, thank you for having
confirmed me that it wasn't trivial2. my purpose was to find
the Fourier transformation of sinc, and not just>to take it
from a book. Is it possible by an easy process, or else... no
way>?..The Fourier transform of sinc(x) is a rectangle
function in thefrequency domain. You can derive this
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>If I post it here
is anyone here willing to look over an outline argument that
there are no odd perfect numbers and point out the flaw/s?Any
flaw shouldn't be too hard to spot: it's a mainly geometrical
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>>Message-id:
Collatz tree? >Maybe this is known so just disregard!>>I
couldn't find anything about it on web searches about the>>
>Collatz.>Probably of little value for the overall
conjecture but still a>>curiosity!>>>The Collatz tree
level count is odd up to level 5 where level>>1,2,3,4,5
[1,2,4,8,16] has a count of (1) for each level. >>At level 6
the count becomes even and fall one for one on each>>side of
the tree up too and including level 16.>>This changes @
level 17 where the left side of the tree adds one>>more
path.>>This happens also for certain levels higher than 17
where the>>left side adds 1 more path over the right
side.>I have denoted the 2 sides of the tree by their
highest-ranking>>sequence.>Where [1,2,4,8,16] is the trunk
and the first main branch>>assigned to the RIGHT side of the
tree is -->>32,64,128,256,512,1024,2048,4096...and with all
applicable nodes>>branching from it creating all branches on
the RIGHT side of>>the Collatz tree.>Where [1,2,4,8,16] is
the trunk and the first main branch>>assigned to the LEFT
side of the tree is -->>
>5,10,20,40,80,160,320,640,1280,2560,...and with all
applicable>>nodes branching from it creating all branches on
the LEFT side of>>the Collatz tree.>>> 8) 20 3 21 128 >> |
| />> 7) 10 64>> />> 6) 5 32>> />> 5) 16>> 4) 8>> 3) 4 >> 2)
2>> Levels 1) 1>Why does the left side have this property of
adding (1) more>>branch at certain levels =>17 then the right
side?>Will this mean, at a very high level there will be many
more>>branches on the left side of the tree than on the right
side>>of the tree making the tree lopsided?>Putting the
above question in another perspective, at some higher>>level
will the right side branching number eventually catch up and>>
>equal the left side branching number?>It could be that the
left side of the tree is facing the sun giving>>that side of
the tree a more vigorous growth. ;-)>Dan>>> Personally, I
don't like the way you draw the Collatz tree. My preference>>
is for vertical placement to always represent x*2 and for the
diagonal>> placement to always represent (x-1)/3:>>> 3 20
128 21>> | | />> 10 64>> | |>> 5 32>> |>> 16>> |>> 8>> |>> 4>>
|>> 2>> |>> 1>>> By your method, 32 is part of the right
branch whereas I consider the right>> branch >> single left
branch and right branch. Rather, there is the central trunk
from>> which>> sprout an infinite number of branches on either
side, so I don't see any>> problem.>>> Note that in my view
each branch starts with an odd number and extends>>
vertically>> to infinity, all of which are even. Does this
mean there are infinitely more>> even>> numbers than odd?>>>
Infinity always catches up in the end.Mansenator,I know the way
I did the tree was different but I used each sequence>as a one
to one correspondence (left,right) with each other.>Where the
(2) sequences, [3,6,12,24...]is the last sequence left
of>center and [21,42,84,168...] is the last sequence right of
center.>This contains the inner limits and actually separates
the tree into>two halfs because these two sequences are 0 (mod
3)'s and will>continue to double and continue on vertically
----->oo.>The other two sequences, [5,10,20,40,80,160...]is
the last sequence>or outer branch limit of the left side of
tree and [32,64,128,256,...]>is the last sequence or outer
branch limit of the right side of tree>where both sequences
double ----->oo.>Presenting the tree in this why, it just
seemed neat, up to level>16 that you have this even looking
tree and with boundry limits>even after level 16.>Using this
representation of the tree it is obvious the level seed>counts
are the same as the standard tree layout.Right now it appears
after level 16 there are increasingly more >sequences with a
(5) in its path other than a (32) in its path.True or false?>
>This can be checked with the standard tree also proving or
disproving>my claim.>DanWith further checking I believe the
statement I made above aboutmore sequenses with a (5) in them
instead of a (32) is false.It does appear to go back and forth
where sometimes the count for (32)sequences is higher than the
(5) sequences and visa versa or somtimesthe count is the same
at any given level of the Collatz tree.Checking higher levels,
what I thought was happening was not do to a programming
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what is inertia? is it
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I would like to
know if this is a mathematical proof of the Cantors goof, or
conversely it is just another mathematical goof about the
Cantors proof. A MATHEMATICAL PROOF IN THREE
ACTSPROLEGOMENON[The naturals]A direct outcome of the
definition of the natural numbers is that * Point 1: There are
not naturals with infinite nonzero digits.* Point 2. As a
result of Point 1, it is not possible to assume the bijection
N <-> R. - Proof: (see below). # FIRST ACT[Cantor and Human
Curiosity]HC: Will it be possible to count the elements of
R?C: Perhaps.HC: But how?[Cantor thinks for a while, and
states a criterion]C: Well, I think it is easy. Without
getting into extreme formalities, we would be able to say that
an infinite set K would be countable, if and only if we would
be able to establish a bijection between K and N, i.e. N <->
K.HC: Thats great! It sounds elegant, logical and intuitive.
However, I think we will have a little problem using it with
R.C: Why?HC: Because Point 2 tells us that we cannot propose
that bijection with R. So, your criterion doesnt work in this
case. We will have to look for another criterion. Its a pity,
because it was nice.C: Yes, thats true. # SECOND ACT[Mrs.
Perfect Proof1, Mrs. Perfect Proof2, Mrs. Perfect Proof_n and
Human Curiosity]answer to your curiosity is that the reals are
uncountable, because it is no possible to achieve the
bijecci.97n N <-> R.PP2: Yes I have reached the same
conclusion.HC: Why?PP2: For the same reason that PP1.HC: And
you?PP_n: Exactly the same that my colleagues.HC: Well ladies,
conclude that R is not countable from that criterion, because
it cannot be used with R.PP1: Oh, we are very sorry. We did
our best. # FINAL ACT[Mr. Mathman and Human Curiosity]MR.
MATHMAN: Do you know, Mr. HC? If we admit Cantors criterion to
count R, as we positively know that N <-> R is not possible (it
doesnt matter why), it is clear that R will be necessarily
uncountable. What do you think about it?HC: Well Mr. Mathman,
we can do so, but it will not convince anybody. The problem is
that we know that Point 2 invalidates Cantors criterion.
Therefore, R would be certainly uncountable, but no because R
be intrinsically uncountable, but because we dont have a valid
criterion to count it.MR. MATHMAN: Well then, perhaps there are
other different criteria to count the reals.HC: I dont know it,
but if we put a simple analogy perhaps we could clarify our
thoughts. Lets try it. We may take the following equivalence
table N <==> M = A set of linear units of measure R <==> A =
The air in a room Being our goal to find out whether A is
measurable or not, we have the following analogy:POINT 1: A
cannot be measured with linear unities, so we cannot assume M
-> A.CRITERION: A set D will be lineally measurable if and
only if it is possible M -> D.POINT 2: We cant use CRITERION
to find out whether A is measurable or not, because POINT 1
tells us that A cant be measured in this way.PROOFS: We have
found out that A is not measurable, because CRITERION
case.PROPOSAL: We can admit CRITERION to measure A, and
consequently A is not measurable. Well, no one will accept
this, because POINT 1 tells us that we cannot apply CRITERION
to measure A, and also because A would be able to be measured
by other means.Good, Mr. Mathman, this is the end of our
analogy.MR. MATHMAN: And?HC: What?MR. MATHMAN: Are there other
criteria to find out whether R is countable or not?HC: Well, in
our analogy A cannot be measured with linear units, but if we
arrange orthogonally three of them, then we can. Perhaps the
reals have an undiscovered property which may be used as
criterion to count them. Think about it Mr. Mathman, think
about it.MR. MATHMAN: Ill do
it.***************************Proof of Point 2INITIAL
ASSERTIONS1) We can represent any real number using a given
positional system of numeration. For ease we will use the
decimal notation.2) Cantor uses a transfinite method (the
diagonal argument) to prove that the premise N <-> R is false,
and we will use another transfinite constructive method to
prove that the transfinite construction N <-> R is impossible,
and therefore we cannot assume the bijection N <->
R.Proposition: The transfinite construction N <-> R is not
possibleProof: Every real number within the interval (0, 1)
has as first decimal digit one of the ten digits of the
decimal system of numeration, i. e. it must be of the form
0.0, 0.1, 0.2, , 0.9. As you can see for this first digit
there are 10 ^ 1 possibilities. For the second digit we have
0.00, 0.01, 0.02,, 0.99. Consequently, there are 10 ^2
possible combinations for the two first digits, 10 ^3 for the
first three digits, and so on. When the number of digits is
infinite we get R.Now, when we have only a digit of the
decimal expansion of the reals, we make the one-to-one
correspondence 0 <-> 0.0, 1 <-> 0.1, 2 <-> 0.2, , 9 <-> 0.9.
Next, with two digits we begin the 1-1 correspondence 0 <->
0.00, 1 <-> 0.01,, 10 <-> 0.10, 11 <-> 0.11, , 99 <-> 0.99.
With three digits, we continue doing the same, and so on.And
now is when the impossible transfinite construction appears.
While we keep doing the 1-1 correspondence within the finite
decimal expansion, everything will go fine. But, what natural
number would correspond to the decimal expansion of the number
e? By induction it is obvious that it should be 71828182, i.
e., a number with infinite nonzero digits, which doesnt belong
to N (Point 1). Conclusion:The transfinite construction N <->R
is not possible if we do not admit naturals with an infinite
number of nonzero digits. If some day we admit that, then the
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>I think we have
to center the tread, because at this moment it looks >>like
the chats in the auditorium before the concert.>POINTS>>Point
1:>>The surjection f: N -> R is not possible because the
definition of >>naturals and reals do not allow it.So what is
it about the definitions of the naturals and the reals
that>allows you to confidently state that such a surjection is
not possible?This question has been answered in my latest
version.>Obviously we cannot use the definitions of the
naturals and the rationals >to prove that there is no
surjection f : N -> Q, since we know that there>is such a
surjection. This means that you cannot use the argument
thatr>N is a proper subset of R (N is a proper subset of Q and
there exists a >surjection f : N -> Q).>There is a thread
dedicated to this stuff and, as you can see, it is not so
clear that Q is countable. I also have some arguments against
this proof, but if you don't mind I will leave untreated the
theme at moment.>>Point 2:>>Cantor clearly stated that we can
say that an infinite set K is >>countable if and only if it is
possible to establish a one-to-one >>correspondence between K
and N, i.e. f: N <-> K.>Point 3:>>As the one-to-one
correspondence N <-> R is not possible (Point 1), Your claim
is to the effect that it is trivial to demonstrate that>such a
one-to-one correspondence does not exist. Let's see
your>trivial proof.>As probably you have already seen, it is a
really trivial proof. You can find a less annoying version of
the whole proof in the thread *A mathematical proof of the
Cantors goof?* There you dont need to play the role of Mr.
Na.95ve if you want to refute it.>>the criterion settled by
Cantor (Point 2) is not valid to resolve >>whether R is
countable or not.Point 1 means that R is not countable, so it
does provide an answer,>except that you have not deemed it
necessary to provide the trivial>proof that makes the
non-existence of a surjection trivially obvious >to you, but
not to others. Point 2 is valid. It is the DEFINITION>of
countability of a set. If you can show that R does not satisfy
>Cantor's criterion for countability as espoused in Point 2,
then R>is not countable. What is so difficult about this
point?>Perhaps the analogy in the latest version it will help
you to grasp the idea I want to transmit.>> In other words, we
cannot use this >>criterion with R. (See note below)Point 1
implies that R is uncountable. What more do you want?>Once
again, the analogy may help you to see the difference.>>Point
4: >>If a proof (any) comes to the conclusion that R is not
countable >>because it is not possible to accomplish the
criterion established in >>Point 2,In other words, we conclude
that R is uncountable because it fails >the criterion for
countability.>YES>> then the proof is useless because that
criterion is not >>valid for that purpose (Point 3).Yes, it
is. It is the DEFINITION of countability of a set. It
lays>down the one and only criterion for countability. R fails
that >criterion, so R does not satisfy the definition for
countability,>so R is not countable.NO. If for you *the money
only can be counted if and only is it is counted by hand* is a
definition, for me is a CRITERION that obviously fails in some
cases.> Note that the proof could be >>correct and the
conclusion too.>DEFINITION>>Useless proof: A correct proof
that proves nothing.You have not demonstrated why we should
accept that there is a >trivial proof of the nonexistence of a
surjection f : N -> R.>Now I did>>QUESTION>>Does somebody know
any other criteria (different from the one >>established in
Point 2) to resolve whether R is countable or not?It is
Cantor's definition for countability. This means that
any>infinite set which satisfies the criterion is countable,
and any>infinite set which does not satisfy the criterion is
uncountable.>NO. It is Cantors criterion. If there are not
other criteria, then this is another problem.>> If >>you know
one of them, I would like to know it please (but first be
>>sure that it doesn't involve implicitly the criterion of
Point 2).>NOTE: >>Someone could interpret that if N <-> R is
not possible, then R is >>uncountable. This conclusion is
true, but in this case, it will imply >>that R is uncountable
because we haven't got a suitable tool to count >>its
elements, since the properties of N and R are incompatible. It
>>will never imply that R has more elements than N.We know that
R has more elements than N since there exists an injection>f :
N -> R, but there is no surjection f : N -> R. That is ALL
that>you need.>Probably you are already thinking that this not
as obvious as it looks like.Nicolas de la Foz>David McAnally
Despite anything you may have heard to the contrary,> the rain
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>In
Foz) said:>Point 3:>>As the one-to-one correspondence N <-> R
is not possible (Point 1), >>the criterion settled by Cantor
(Point 2) is not valid to resolve >>whether R is countable or
not.No. If you stipulate that R is uncountable, then it
immediately>follows that R is uncountable. If you don't
stipulate it, a reductio>ad absurdum is perfectly reasonable,
although not necessary. A>reductio ad absurdum does not assume
a false statement: it proves that>the statement must be false
by proving that it leads to a>contradiction.>It is not the
same to be INTRINSICALY uncountable, that to be uncountable
because we havent the suitable means to do it. Please, read
carefully my latest version of the proof, and if necessary
have a look to the analogy just to grasp the idea.(you can
find this poof in the tread *A mathematical proof of the
Cantors goof?*)>>Point 4: >>If a proof (any) comes to the
conclusion that R is not countable >>because it is not
possible to accomplish the criterion established>>in Point 2,
then the proof is uselessIncorrect.>because that criterion is
not valid for that purpose (Point 3). No, it is your Point 3
that is not valid.>DEFINITION>>Useless proof: A correct proof
that proves nothing.What do you mean by proves nothing?>An
useless job.>>QUESTION>>Does somebody know any other criteria
(different from the one >>established in Point 2) to resolve
whether R is countable or not?Do you know a definition of
frequency other than the number of cycles>per unit of time?
The definition is the definition, and any proof>ultimately
comes down to showing that it satisfies the definition.>Why do
you always take out the things out of context? As G.9adel and
Cohen established, there are undecidable statements.
Concluding whether the reals are countable or not it probably
is one of them.By the way, someone asked what the heck G.9adel
has to do with this. Here you can find the answer. I hope you
don't have to regret it>>NOTE: >>Someone could interpret that
if N <-> R is not possible, then R is >>uncountable. This
conclusion is true, but in this case, it will>>imply that R is
uncountable because we havent got a suitable tool>>to count its
elements,No. It will imply that no such tool exists.>Sorry but,
is it not the same? If such a tool doesnt exist, then we havent
got a suitable one to count R.>>since the properties of N and R
are incompatible.You still haven't defined what you mean by
that statement in>Mathematical terms.>Statement: sentence or
affirmation that can be true, false or undecidable (within the
current mathematics).Nicolas de la Foz>-- > Shmuel (Seymour J.)
Metz, SysProg and JOAT>not reply to
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> I think we have to center the tread,
because at this moment it looks >> like the chats in the
auditorium before the concert.>>>> POINTS>> Point 1:>> The
surjection f: N -> R is not possible because the definition of
>> naturals and reals do not allow it.This statement reqires
proof, as it is not self evident.See latest version.>>>
Point 2:>> Cantor clearly stated that we can say that an
infinite set K is >> countable if and only if it is possible
to establish a one-to-one >> correspondence between K and N,
i.e. f: N <-> K.More precisely, a set K is countable if there
is a surjective map from N >to K and countably infinite (or
some say denumerable) if there is a >bijective map.Good
mathematician!>>> Point 3:>> As the one-to-one
correspondence N <-> R is not possible (Point 1), >> the
criterion settled by Cantor (Point 2) is not valid to resolve
>> whether R is countable or not. In other words, we cannot
use this >> criterion with R. (See note below)You assert it to
be impossible, but you provide no proof. In >mathematics, no
assertion need be accepted without proof. Since you >provide
no proof, we may assume that you do not know whether the reals
>are countable.See latest version.>>> Point 4: >> If a proof
(any) comes to the conclusion that R is not countable >>
because it is not possible to accomplish the criterion
established in >> Point 2, then the proof is useless because
that criterion is not >> valid for that purpose (Point 3).
Note that the proof could be >> correct and the conclusion
too.>>There were, at last count, well over 100 proofs of the
Pythagorean >theorem. Which ones are invalid? If all of them
are correct, all of them are valid. Why?>> DEFINITION>>
Useless proof: A correct proof that proves nothing.>Obviously
you never have done an useless work. Otherwise you would know
what a useless proof is. >This is an empty definition, since
it can never be instantiated. A >correct proof, by its own
definition, proves something. You are defining >the memebers
of the empty set.This Futz is nearly as wacky as JSH.This
statement requires proof, as it is not self evident.Nicolas de
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we have to center the tread, because at this moment it looks
>>like the chats in the auditorium before the
concert.>>POINTS>>Point 1:>>The surjection f: N -> R is not
possible because the definition of >>naturals and reals do not
allow it.This needs proof. Although it can easily be proved, so
in some>sense it's true. In exactly the same way it is true to
say that>sqrt(2) is irrational because the defintion of the
rationals>do not allow it to be rational. So if Point 1 makes
the standard>proof of the uncountability of R invalid it also
makes the standard>proof of the irrationality of sqrt(2)
invalid. (And it makes _any_>proof by contradiction
invalid.)>If sqrt(2) cannot be rational by definition, what
else it can be? Irrational. Good. And nowIf N <-> R cannot be
by definition, what else it can be? NOTHING. Go with an
optionless premise to a proof by contradiction and youll get
NOTHING (or rubbish).>>Point 2:>>Cantor clearly stated that we
can say that an infinite set K is >>countable if and only if it
is possible to establish a one-to-one >>correspondence between
K and N, i.e. f: N <-> K.>>Point 3:>>As the one-to-one
correspondence N <-> R is not possible (Point 1), >>the
criterion settled by Cantor (Point 2) is not valid to resolve
>>whether R is countable or not. In other words, we cannot use
this >>criterion with R. (See note below)Huh? This makes just
as much sense as saying because sqrt(2)>rational is not
possible, it is not valid to resolve whether sqrt(2)>is
rational by using the definition of the word 'rational'.See
the analogy in the latest version to grasp the concept.>You
really are making no sense - these rules you state
about>what's valid are simply not correct, they're just things
you made up.>Please, prove it in the latest version.Nicolas de
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>So what you're
saying is the sequence .9^2, .9^3, .9^4 ....9^2 = .81>.9^3 =
.729>.9^4 = .6561>etc>the sequence .99^2, .99^3, .99^4
....99^2 = .9801>.99^3 = .970299>.99^4 = .96059601>etcthe
sequence .999^2, .999^3, .999^4 ... .999^2 = .998001>.999^3 =
.997002999>.999^4 = .996005996001>etc>tend to 1 making the
sequence .999...^2, .999...^3, .999...^4 tend to 1.Is that
what you are saying?Garry Denke, Geologist>Denoco Inc. of
Texas No, she is not saying anything like that. She is saying,
as is clear, that you do not know basic arithmetic! In the
first place, >.999^4 = .996005996001 is not a SQUARE as you
keep asserting. Your own example show that the SQUARES do tend
to 1:.9^2= .81, 0.99^2= 0.9801, 0.999^2= 0.998001, 0.9999^2=
0.99980001, etc. Any school boy with a calculator should know
that, as long as a< 1, different powers of a, a^2, a^3, a^4,
... tend to 0. But that tells you NOTHING about what happens
as a itself goes to 1. If a-> 1 (for example, .9, .99, .999,
.9999, etc.) the a^n for any FIXED n goes to 1. This has been
pointed out to you repeatedly but you don't seem to understand
simple arithmetic. Are you really a geologist? Don't geologist
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>I would like to know if this is a
mathematical proof of the Cantors >goof, or conversely it is
just another mathematical goof about the >Cantors proof. A
MATHEMATICAL PROOF IN THREE ACTSPROLEGOMENON[The naturals]A
direct outcome of the definition of the natural numbers is
that * Point 1: There are not naturals with infinite nonzero
digits.>* Point 2. As a result of Point 1, it is not possible
to assume the >bijection N <-> R. >- Proof: (see below). #
FIRST ACT[Cantor and Human Curiosity]>HC: Will it be possible
to count the elements of R?>C: Perhaps.>HC: But how?>[Cantor
thinks for a while, and states a criterion]>C: Well, I think
it is easy. Without getting into extreme >formalities, we
would be able to say that an infinite set K would be
>countable, if and only if we would be able to establish a
bijection >between K and N, i.e. N <-> K.>HC: Thats great! It
sounds elegant, logical and intuitive. However, >I think we
will have a little problem using it with R.>C: Why?>HC:
Because Point 2 tells us that we cannot propose that bijection
>with R. So, your criterion doesnt work in this case. We will
have to >look for another criterion. Its a pity, because it
was nice.>C: Yes, thats true. # SECOND ACT>[Mrs. Perfect
Proof1, Mrs. Perfect Proof2, Mrs. Perfect Proof_n and >Human
Curiosity]>answer to your curiosity is that the reals are
uncountable, because >it is no possible to achieve the
bijecci.97n N <-> R.>PP2: Yes I have reached the same
conclusion.>HC: Why?>PP2: For the same reason that PP1.>HC:
And you?>PP_n: Exactly the same that my colleagues.>HC: Well
to conclude that R is not countable from that criterion,
>because it cannot be used with R.>PP1: Oh, we are very sorry.
We did our best. # FINAL ACT>[Mr. Mathman and Human
Curiosity]>MR. MATHMAN: Do you know, Mr. HC? If we admit
Cantors criterion to >count R, as we positively know that N
<-> R is not possible (it >doesnt matter why), it is clear
that R will be necessarily >uncountable. What do you think
about it?>HC: Well Mr. Mathman, we can do so, but it will not
convince anybody. >The problem is that we know that Point 2
invalidates Cantors >criterion. Therefore, R would be
certainly uncountable, but no >because R be intrinsically
uncountable, but because we dont have a >valid criterion to
count it.>MR. MATHMAN: Well then, perhaps there are other
different criteria to >count the reals.>HC: I dont know it,
but if we put a simple analogy perhaps we could >clarify our
thoughts. Lets try it. >We may take the following equivalence
table N <==> M = A set of linear units of measure > R <==> A =
The air in a room Being our goal to find out whether A is
measurable or not, we have >the following analogy:POINT 1: A
cannot be measured with linear unities, so we cannot >assume M
-> A.>CRITERION: A set D will be lineally measurable if and
only if it is >possible M -> D.>POINT 2: We cant use CRITERION
to find out whether A is measurable >or not, because POINT 1
tells us that A cant be measured in this way.>PROOFS: We have
found out that A is not measurable, because CRITERION
>case.>PROPOSAL: We can admit CRITERION to measure A, and
consequently A is >not measurable. Well, no one will accept
this, because POINT 1 tells >us that we cannot apply CRITERION
to measure A, and also because A >would be able to be measured
by other means.>Good, Mr. Mathman, this is the end of our
analogy.>MR. MATHMAN: And?>HC: What?>MR. MATHMAN: Are there
other criteria to find out whether R is >countable or not?>HC:
Well, in our analogy A cannot be measured with linear units,
but >if we arrange orthogonally three of them, then we can.
Perhaps the >reals have an undiscovered property which may be
used as criterion to >count them. Think about it Mr. Mathman,
think about it.>MR. MATHMAN: Ill do
it.***************************Proof of Point 2>INITIAL
ASSERTIONS>1) We can represent any real number using a given
positional system >of numeration. For ease we will use the
decimal notation.>2) Cantor uses a transfinite method (the
diagonal argument) to prove >that the premise N <-> R is
false, and we will use another >transfinite constructive
method to prove that the transfinite >construction N <-> R is
impossible, and therefore we cannot assume >the bijection N
<-> R.Proposition: The transfinite construction N <-> R is not
possibleProof: >Every real number within the interval (0, 1)
has as first decimal >digit one of the ten digits of the
decimal system of numeration, i. >e. it must be of the form
0.0, 0.1, 0.2, , 0.9. As you can see for >this first digit
there are 10 ^ 1 possibilities. For the second digit >we have
0.00, 0.01, 0.02,, 0.99. Consequently, there are 10 ^2
>possible combinations for the two first digits, 10 ^3 for the
first >three digits, and so on. When the number of digits is
infinite we get >R.>Now, when we have only a digit of the
decimal expansion of the reals, >we make the one-to-one
correspondence 0 <-> 0.0, 1 <-> 0.1, 2 <->0.2, , 9 <-> 0.9.
Next, with two digits we begin the 1-1 >correspondence 0 <->
0.00, 1 <-> 0.01,, 10 <-> 0.10, 11 <-> 0.11, , >99 <-> 0.99.
With three digits, we continue doing the same, and so on.>And
now is when the impossible transfinite construction appears.
>While we keep doing the 1-1 correspondence within the finite
decimal >expansion, everything will go fine. But, what natural
number would >correspond to the decimal expansion of the number
e? By induction it >is obvious that it should be 71828182, i.
e., a number with infinite >nonzero digits, which doesnt
belong to N (Point 1). Conclusion:>The transfinite
construction N <->R is not possible if we do not >admit
naturals with an infinite number of nonzero digits. If some
day >we admit that, then the reals would be countable.>Nicolas
de la Foz If you want us to comment on a proof, then you will
have to show us a proof. I saw a lot of words, a few terms
that might have been intended as mathematics but were not
defined (for example measurable by linear units). There are
quite a lot of non-sequiturs (for example, you define
countable set as meaning that there exist a one to one
correspondence between the set and N (which is correct) and
then assert that that criterion doesnt work for the real
numbers since it is said in Point 2 that no such
correspondence. That criterion works quite nicely, thank you:
it says that the set of real numbers is NOT countable. What
did you think a criterion was for?) While Point 2, that there
is no one to one correspondence between the set of real
numbers and the set of positive integers is true, your
purported proof, that (to summarize) the set of real numbers
contains number with an infinite number of nozero digits while
no integer does, therefore, there can be no such
correspondence, is completely invalid. The set of all rational
numbers also includes numbers with an infinite number of
non-zero digits (1/3 for example) but there IS such a
correspondence- the set of rational numbers IS countable.
Finally, I can't see what this has to do with Cantor's proof.
You don't actually say anything about Cantor's proof. By the
way, you end with Conclusion:>The transfinite construction N
<->R is not possible if we do not >admit naturals with an
infinite number of nonzero digits. If some day >we admit that,
then the reals would be countable. Yes, and if we define
countable to mean simply is infinite then the reals would be
countable- and if we defined blue to mean is infinite then the
reals would be blue- but we wouldn't have learned anything
important in either case. It's not a matter of admitting
naturals with an infinite number of nonzero digits. The
naturals, as defined in our basic arithmetic, do not have
such. If you want to try to change elementary school
arithmetic just to make the reals countable (and so make
mathematics trivial and unable to do most of the things we
rely on it for) I don't think you will be very sucessful.