mm-14 Hal Puthoff has held very high national security clearances and his main interest for 30 + years now is figuring out the advanced military WMD implications of the alien ?ing saucers. Hal and I are competitors in this exotic weaponry field of legendary black ops . PS Addressing some of Hal? technical specifics below: coherence. You?e missed the whole point! Hal JS: I do not see how your idea of vacuum coherence is same as what I am talking about? HP: It is simply that we are talking about different aspects of the vacuum when we use the word stiffness. You use it to characterize how difficult it is for the stress-energy tensor to bend spacetime. JS: True. I mean G/c^4 = (String Tension)^-1 Space-Time gets more stiff as the String Tension increases. Guv(Space-Time Curvature) = (String Tension)^-1Tuv(Mass-Energy Sources) for non-exotic vacuum with / zpf = 0. It is only this definition of stiffness that is relevant to the practical achievement of metric engineering. That? my key point here that IMHO you and Eric Davis et-al do not yet get. HP: I? using it to characterize how difficult it is to polarize virtual vacuum electron-positron pairs with an electric field. JS: This is not a good use of the term since Feynman? QED shows you can never make a large enough vacuum polarization for practical metric engineering using external EM fields. If you have a theoretical argument to the contrary I would like to see it. I am confident that the saucers you and I are both keenly interested in, if The Truth be known ;-), ? by the approach you take here. I doubt you have any experimental evidence either. Please refute me if you can? HP: When the vacuum dielectric constant K rises near a mass, it means that it is easier (less stiffness in my sense) for an electric field to separate charge. JS: Prove that. I mean you have K = e^2GM/c^2 I forget your formula off-hand for K in presence of an electric field. This requires that your M source in K also have a net charge Q I presume. Is that correct? Oh yes you have that in your Mitre paper but the numbers do not work and support what I say above. Also since when do saucers carry large amounts of net electric charge? Also, what is your relation of K to the physical speed of light? Looking only at radial null geodesic ds^2 = K^-1(cdt)^2 - Kdr^2 ds^2 = 0 Therefore dr/dt = c/K And you mistakenly interpret K as an index of refraction n. In fact physically, as any standard text on GR shows dT = goo^1/2dt dR = grr^1/2dr Therefore, when ds^2 = 0 in radial direction in SSS case Physically measured radial speed of light = dR/dT = (grr/goo)^1/2(dr/dt) In your model goo = K^-1 grr = K Therefore, grr/goo = K^2 Therefore dR/dT = c so that n = 1 ALWAYS in this situation! So what are you talking about? You then go into a vague over-intepretation, seemingly motivated by Yilmaz? bi-metric confusion as seen in your Tables I & II, to try to give a physical meaning to dr/dt = c/K, which is completely confused IMHO and which has no relevance at all to real metric engineering. Your argument here is Medieval like counting Angels on a head of a pin to my mind at least. HP: It is that sense that vacuum coherence effects in the charge distribution are induced. JS: You have never given an explicit definition of vacuum coherence the way I have. My definition is in accord with mainstream physics. What is your definition and how does it appear in your mathematical model. I do not think that Feynman? QED has vacuum coherence . That was later added in Green? function formalisms for BCS superconductors by Gorkov et-al in Moscow in 60? or 70? based on Oliver Penrose? ODLRO for reduced quantum density matrices of fermions. Where in your papers do you even write vacuum coherence ? HP: There is nothing wrong with what you say. You just missed the boat when you thought that there was something wrong with what I say, since you do not think of the vacuum virtual charge pairs as having this special electrical relationship with spacetime effects. JS: False. My vacuum coherence is explicitly, using standard second-quantization notation and the Gorkov formalism e.g. P.W. Anderson in A Career in Theoretical Physics World Scientific Vacuum Coherence = <0|e+(x)e-(x)|0> where e+(x) destroys a virtual off mass shell positron at x e-(x) destroys a virtual off mass shell electron at x |0> is the physical vacuum with ODLRO that admits CURVED space-time whose LNIF local metric is the strain tensor guv(Curved Space-Time) = Global Flat Minkowski Metric + du(x),v + dv(x),u , means partial derivative and du(x) = Lp^2(arg <0|e+(x)e-(x)|0>),u Lp^2 = hG/c^3 String Tension = Vacuum Stiffness = hc/Lp^2 Now that? what I mean mathematically by vacuum coherence and the vacuum virtual charge pairs as having this special ?lectrical? relationship with spacetime effects . What, in contrast, do you mean mathematically by your words in terms of your model? I am just trying to understand the real meaning behind your smooth honeyed words. ;-) Whether either of us are right or wrong is another issue. Right now I do not understand what your words in ordinary language really mean until I see some kind of a math model to go with them. HP: It is this latter interpretation that distinguishes the PV approach from the standard. BTW, to set another record straight, I do not explore PV as an absolute replacement for Einsteinian GR as you often attribute to me. JS: That? good. Some of your PR people have given a very different impression. With friends like them you do not need enemies. :-) HP: (For example, in PV analysis of rotating dumbbells, they generate only 2/3rds of the radiation power calculated from standard GR and inferred from observation of PSR 1913 + 16.) My interest is PV has to do with exploring certain geometries and consequences from an engineering standpoint. A corollary of this is that I also do not approach metric engineering from the standpoint of the brute force stress-energy tensor approach that you also attribute to me. JS; That? all your papers show brute force like your Mitre paper. Are HP: Rather, I am interesting in exploring the underlying EM aspects of spacetime that undergird the PV (but not usual GR) approach. = A am using an Exponential Moving Average with time-series data. I am using smoothing constants (SC) of 0.70 and 0.98 I remember vaguely that there is a simple formula that roughly equates the values of SC to simple moving average equivalent. Something like saying Using SC=0.70 is equivalent to a 5-day moving average. Can someone confirm this - send me the formula or tell me is a rough sense, what would SCs of 0.7 and 0.98 equate to in simple moving average terms. Jay = A N day EMA uses the constant 2/(1+N). I know that is not quite the answer to your question, but the equivalence would depend on the data so it may be quite a good average answer. > A am using an Exponential Moving Average with time-series data. I am using > smoothing constants (SC) of 0.70 and 0.98 I remember vaguely that there is a simple formula that roughly equates the > values of SC to simple moving average equivalent. Something like saying > Using SC=0.70 is equivalent to a 5-day moving average. Can someone confirm > this - send me the formula or tell me is a rough sense, what would SCs of > 0.7 and 0.98 equate to in simple moving average terms. Jay = > A rat is in a chamber in a maze. There are 3 doors to the chamber. > Door_1 will make the rat return to the chamber after 3 minutes. > Door_2 will make the rat find its way out after 2 minutes. > Door_3 will make the rat return to the chamber after 5 minutes. > The probabilities that the rat takes the doors 1, 2 and 3 are > resp. 1/2, 1/6 and 1/3. > Find the expected value and the variance of the total time spent > in the maze. [...] Then we *tried* the following to calculate E[X^2]: > E[X^2|T=1] = (3+E[X])^2 > E[X^2|T=2] = 4 > E[X^2|T=3] = (5+E[X])^2 > > There is your problem: E[X^2 | T=1] = E[(3 + X)^2], etc. BTW, you could also use the conditional variance formula: Var(X) = E Var(X|T) + Var(E[X|T]). -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu = A rat is in a chamber in a maze. There are 3 doors to the chamber. > Door_1 will make the rat return to the chamber after 3 minutes. > Door_2 will make the rat find its way out after 2 minutes. > Door_3 will make the rat return to the chamber after 5 minutes. > The probabilities that the rat takes the doors 1, 2 and 3 are > resp. 1/2, 1/6 and 1/3. > Find the expected value and the variance of the total time spent > in the maze. > > >[...] > Then we *tried* the following to calculate E[X^2]: > E[X^2|T=1] = (3+E[X])^2 > E[X^2|T=2] = 4 > E[X^2|T=3] = (5+E[X])^2 > > > There is your problem: E[X^2 | T=1] = E[(3 + X)^2], etc. Ha yes of course. Had I not skipped a logical step by immediately writing E[X|T=1] = 3+E[X] , but rather first E[X|T=1] = E[3+X] , resulting in the former expression, then I would also have written the correct expression E[X^2|T=1] = E[(3+X)^2] Blessings and dangers of the automatic pilot of analogy ;-) I checked the result and it makes perfect sense now. BTW, you could also use the conditional variance formula: Var(X) = E Var(X|T) + Var(E[X|T]). Indeed. I had found it in http://www.ma.utexas.edu/users/mks/384G03/condmv3.pdf but decided not to try to use it since it was not covered in the course notes. Dirk (and Bert) Vdm = > How do you produce the material conditional symbol that looks kind of like > supset but is longer and thinner? Use the horseshoe from the Postscript Symbol font. If you?e using LaTeX, try usepackage{pifont} def horseshoe{ mathrel{ hbox{ Pisymbol{psy}{201}}}} begin{document} [A horseshoe B } end{document} and see if you?e happy. -- rar = >No, that is wrong. The Indian philosophical thought - Sanatana >dharma, or the way of life beyond the scope of time - is completely >different from the modern and dominant Jewish thinking [...] > > This frame of mind, of course, serves to lend additional credence > to the otherwise unbelievable notion that the Swastika actually > originated in India. The sign of the swastika relates to good health and well being, from the Indian perspective. = >This frame of mind, of course, serves to lend additional credence >to the otherwise unbelievable notion that the Swastika actually >originated in India. > > The sign of the swastika relates to good health and well being, from > the Indian perspective. Several Native American peoples also used it, and they?e been here for at least 12,000 years. Maybe it? something instinctive , hard- wired or built-in about right angles or something. If the Indian use is Vedic wouldn? that mean it? ultimately Persian? D. -- It does not matter how many innocent babies die. .............................................................. ..... = >> contempt black skin. >And thats why most important menifestation of god in hinduism like >vishnu, >shiva,ram,krishna have black skin.Stupid propoganda machine learn the >facts right before spewing venom against hinduism. That may be so. But, one is disadvantage if one? skin color is kala in > India. Why? Dan DNA testing shows that India was founded by white men who bred with the darker women there. They established the caste system based on color or varna (related to varnish). When India was once again conquered by the British, they noticed the idea of the Aryan or Indo-European language group. For years, the Nazis were ridiculed for digging for Aryans in places like Tibet. But they were recently vindicated when tall, fair Caucasian mummies were found even further west, in China. These 4,000 year old people were incredibly advanced. They had horses, cattle, the wheel, modern textiles and practiced surgery. Even their clothes looked modern. One mummy looks like a handsome white man asleep in a jogging suit. It now seems that the myth of a mass migration of tall, fair people across Eurasia speaking a proto-Aryan language is perfectly true. The swastika really was their symbol too. So much for the Jewish bull we?e been fed for all these years. See: www.white-history.com = > contempt black skin. >> And thats why most important menifestation of god in hinduism like >> vishnu, >> shiva,ram,krishna have black skin.Stupid propoganda machine learn the >> facts right before spewing venom against hinduism. > >That may be so. But, one is disadvantage if one? skin color is kala in >India. Why? Dan DNA testing shows that India was founded by white men who bred with the > darker women there. They established the caste system based on color or > varna (related to varnish). When India was once again conquered by the British, they noticed the > idea of the Aryan or Indo-European language group. For years, the Nazis were ridiculed for digging for Aryans in places like > Tibet. But they were recently vindicated when tall, fair Caucasian mummies > were found even further west, in China. These 4,000 year old people were > incredibly advanced. They had horses, cattle, the wheel, modern textiles and > practiced surgery. Even their clothes looked modern. One mummy looks like a > handsome white man asleep in a jogging suit. I saw an episode in discovery channel where they where showing a 6000 year old mummy riding a minivan. I know its hard to believe, but fact is stranger than fiction as one white man put it. Dan It now seems that the myth of a mass migration of tall, fair people across > Eurasia speaking a proto-Aryan language is perfectly true. The swastika > really was their symbol too. So much for the Jewish bull we?e been fed for all these years. See: www.white-history.com = >> >> > >contempt black skin. >> > And thats why most important menifestation of god in hinduism like >> > vishnu, >> > shiva,ram,krishna have black skin.Stupid propoganda machine learn the >> > facts right before spewing venom against hinduism. >> >> That may be so. But, one is disadvantage if one? skin color is kala in >> India. Why? >> >> Dan > >DNA testing shows that India was founded by white men who bred with the >darker women there. They established the caste system based on color or > varna (related to varnish). > >When India was once again conquered by the British, they noticed the the >idea of the Aryan or Indo-European language group. > >For years, the Nazis were ridiculed for digging for Aryans in places like >Tibet. But they were recently vindicated when tall, fair Caucasian > mummies >were found even further west, in China. These 4,000 year old people were >incredibly advanced. They had horses, cattle, the wheel, modern textiles > and >practiced surgery. Even their clothes looked modern. One mummy looks like > a >handsome white man asleep in a jogging suit. I saw an episode in discovery channel where they where showing a 6000 year > old mummy riding a minivan. I know its hard to believe, but fact is stranger > than fiction as one white man put it. Have a look, if you dare, sr saliva. They sure looked better than spics. http://library.?wlesslogic.com/china.htm http://www.pbs.org/wgbh/nova/chinamum/taklamakan.html Dan >It now seems that the myth of a mass migration of tall, fair people > across >Eurasia speaking a proto-Aryan language is perfectly true. The swastika >really was their symbol too. So much for the Jewish bull we?e been fed for all these years. > >See: www.white-history.com > > = >> > contempt black skin. >> And thats why most important menifestation of god in hinduism like >> vishnu, >> shiva,ram,krishna have black skin.Stupid propoganda machine learn the >> facts right before spewing venom against hinduism. That may be so. But, one is disadvantage if one? skin color is kala in >India. Why? Dan > yes because Indians are racist pigs, just see chamar kooly and his contempt for kala people. = >: >> I am trying to find a primitive function for y = (1 - x^2)^2 but have >> >>Typo. It? supposed to be y = (1 - x^2)^(1/2). Mathematica gives (x * sqrt(1-x^2) + arcsin(x)) / 2. Looks correct. >Maybe a clever trigonometric substitution like x = sin t: Int[ (1 - x^2)^(1/2) dx ] >= Int[ (cos t)^2 dt ] Shouldn? it be Int[ (cos^2(t))^(1/2) dt ] = Int[ cos t dt ]? I? missing something. >= Int[ 1/2 * ( 1 + cos 2x) ] >= 1/2 * [ t + 1/2 * sin 2x ] >= t/2 + sin(2x)/4 >= t/2 + (sin t cos t) / 2 Substituting back gives: = arcsin(x) / 2 + (x cos(arcsin x)) / 2 >= arcsin(x) / 2 + x * Sqrt(1-x^2) / 2 Presto! -- /Torbj .9arn Svensson Diaz Please visist this site. http://www.againsttcpa.com/ = > Shouldn? it be Int[ (cos^2(t))^(1/2) dt ] = Int[ cos t dt ]? I? > missing something. The differential (?-something? has to be substituted as well: INT (1 - x^2)^(1/2) dx = INT (1 - (sint(t)^2)^(1/2) d(sin(t)) = INT (cos(t)^2)^(1/2) * d(sin(t))/dt dt = INT cos(t) * cos(t) * dt = INT cos(t)^2 dt etc... You can take whatever is behind the ?? to the front by differentiating, as can already be seen from the notation: dy = = >>Typo. It? supposed to be y = (1 - x^2)^(1/2). >>Mathematica gives (x * sqrt(1-x^2) + arcsin(x)) / 2. Looks correct. >>Maybe a clever trigonometric substitution like x = sin t: >> >> Int[ (1 - x^2)^(1/2) dx ] >>= Int[ (cos t)^2 dt ] Shouldn? it be Int[ (cos^2(t))^(1/2) dt ] = Int[ cos t dt ]? I? >missing something. We must also substitute the differential accordingly: x = sin t => dx = cos t dt which gives the extra cos t. If this were a definite integral, the integration limits would also change. = >Let X be compact Hausdorff, let {A_i} be a collection of closed > >connected subsets of X that is simply ordered by proper inclusion. Prove Y = {intersection of A_i over all i} is non-empty and connected. >How can I prove this? Exercise: > compact S, closed sets { Ki }, / { Ki } subset open U > ==> some K1,..Kn with K1 / ../ Kn subset U Now assume some open disjoint U,V with / C subset U /V > and use the exercise. I don? think Hausdorff is needed. > / C ???? What is C? I don? think I understand your proof. > / = cap = intersection ---- = > Let X be compact Hausdorff, let {A_i} be a collection of closed > >connected subsets of X that is simply ordered by proper inclusion. > >Prove Y = {intersection of A_i over all i} is non-empty and > >connected. > >Exercise: >compact S, closed sets { Ki }, / { Ki } subset open U >==> some K1,..Kn with K1 / ../ Kn subset U > >Now assume some open disjoint U,V with / C subset U /V >and use the exercise. I don? think Hausdorff is needed. / C ???? What is C? I don? think I understand your proof. > Whoops, C = { A_i } >/ = cap = intersection = > >> >Let X be compact Hausdorff, let {A_i} be a collection of closed >> >connected subsets of X that is simply ordered by proper inclusion. >> >Prove Y = {intersection of A_i over all i} is non-empty and >connected. >> >> >How can I prove this? >> >> Exercise: >> compact S, closed sets { Ki }, / { Ki } subset open U >> ==> some K1,..Kn with K1 / ../ Kn subset U >> >> Now assume some open disjoint U,V with / C subset U /V >> and use the exercise. I don? think Hausdorff is needed. >> / C ???? What is C? C is the family of closed connected subsets A_i; so / C would be equal to your Y. >I don? think I understand your proof. Assume you have proven the exercise: so you have proven that if an intersection of closed subsets of a Haussdorf compact set is contained in an open set U, then there is a finite subcollection whose intersection is already contained in U (this is in fact simply the finite intersection property for the compact set S-U, and the collection of closed subsets K_i-U; recall that a set is compact if and only if any collection of closed subsets whose intersection is empty must have a finite subcollection whose intersection is already empty). Now, assume you have two open disjoint subsets U and V, such that Y (the intersection of all the A_i) is contained in U union V. Since U union V, by the exercise, there must be a finite collection, A_1,...,A_n, such that the intersection of A_1, A_2,...,A_n is already contained in U union V. Since the A_i are linearly ordered, it follows that the intersection of A_1,...,A_n is equal to one of them, say A_n; so A_n is contained in the union of the open disjoint subsets U and V. You know that A_n is connected. So what does this mean? And you know that the intersection of all A_i is contained in A_n. So what does this mean? IN conclusion, can you prove that: For all disjoint open subsets U and V, if Y is contained in U union V, then either Y is contained in U or Y is contained in V ? And if so, have you managed to prove the original statement? (Well, you still need to prove nonempty, but that again follows from the finite intersection property for X). -- = It? not denial. I? just very selective about what I accept as reality. --- Calvin ( Calvin and Hobbes ) = Arturo Magidin magidin@math.berkeley.edu > > > >> >Let X be compact Hausdorff, let {A_i} be a collection of closed > >> >connected subsets of X that is simply ordered by proper inclusion. > >> >Prove Y = {intersection of A_i over all i} is non-empty and > >> >connected. > >> > >> >How can I prove this? > >> > >> Exercise: > >> compact S, closed sets { Ki }, / { Ki } subset open U > >> ==> some K1,..Kn with K1 / ../ Kn subset U > >> > >> Now assume some open disjoint U,V with / C subset U /V > >> and use the exercise. I don? think Hausdorff is needed. > >> > > >/ C ???? What is C? C is the family of closed connected subsets A_i; so / C would be > equal to your Y. >I don? think I understand your proof. Assume you have proven the exercise: so you have proven that if an > intersection of closed subsets of a Haussdorf compact set is contained > in an open set U, then there is a finite subcollection whose > intersection is already contained in U (this is in fact simply the > finite intersection property for the compact set S-U, and the > collection of closed subsets K_i-U; recall that a set is compact if > and only if any collection of closed subsets whose intersection is > empty must have a finite subcollection whose intersection is already > empty). Now, assume you have two open disjoint subsets U and V, such that Y > (the intersection of all the A_i) is contained in U union V. Since U > union V, by the exercise, there must be a finite collection, > A_1,...,A_n, such that the intersection of A_1, A_2,...,A_n is already > contained in U union V. Since the A_i are linearly ordered, it follows > that the intersection of A_1,...,A_n is equal to one of them, say A_n; > so A_n is contained in the union of the open disjoint subsets U and V. You know that A_n is connected. So what does this mean? And you know that the intersection of all A_i is contained in A_n. So > what does this mean? IN conclusion, can you prove that: For all disjoint open subsets U and V, if Y is contained in U union V, > then either Y is contained in U or Y is contained in V > You mean for disjoint open subets U and V those union is all of X, AND Y is connected, then Y is contained in U or Y is contained in V? Or is this also true for any disjoint open subsets U and V in X where Y is connected and contained in U union V? > ? And if so, have you managed to prove the original statement? (Well, > you still need to prove nonempty, but that again follows from the > finite intersection property for X). > -- > = > It? not denial. I? just very selective about > what I accept as reality. > --- Calvin ( Calvin and Hobbes ) > Arturo Magidin > magidin@math.berkeley.edu > Consider the following subspaces of R^2 : E = (Z x R) / (R x Z) (the infinite grid) and B = {(x,y) in R^2 : ||(x,y) - (-1,0)|| = 1 or ||(x,y) - (1,0)|| = 1} (the figure eight). How can I describe E being a covering space for B? What is an explicit formula for the covering map? And, how can I show that E is not simply-connected? Michael = >Consider the following subspaces of R^2 : E = (Z x R) / (R x Z) (the >infinite grid) and B = {(x,y) in R^2 : ||(x,y) - (-1,0)|| = 1 or ||(x,y) - >(1,0)|| = 1} (the figure eight). How can I describe E being a covering space for B? What is an explicit >formula for the covering map? And, how can I show that E is not >simply-connected? Michael > W. Dale Hall already told you about the covering map. To use this map to show that B is not simply-connected, you can apply the covering homotopy theorem, which should appear early in your algebraic topology text. For example, consider the loop L in B that goes once around the right circle (starting and ending at the base point (0, 0)). Draw the lift L~ of that loop in E. How imagine that L can be continuously deformed to a point in B, keeping its end points fixed. That homotopy would lift to a continuous deformation of L~ to a point in E, keeping its end points fixed. It should be easy to see why that? impossible (if necessary, review the proof that the circle is not simply-connected). This shows that B contains a loop that is not homotopically trivial; that is, B is not simply-connected. The argument is quite general. John = > Consider the following subspaces of R^2 : E = (Z x R) / (R x Z) (the > infinite grid) and B = {(x,y) in R^2 : ||(x,y) - (-1,0)|| = 1 or ||(x,y) - > (1,0)|| = 1} (the figure eight). > > How can I describe E being a covering space for B? What is an explicit > formula for the covering map? And, how can I show that E is not > simply-connected? > > Michael Let the left circle be labeled L; the right one labeled R. Now, on the big grid, assign labels L and R to each of the grid edges, subject to the constraint that, at each lattice point, two edges are labeled L, and two are labeled R. Next, orient the two circles, and then orient the edges so that at each lattice point, one L edge is oriented towards, and the other away from that point, and likewise one R edge is directed towards, the other away from that point. There? a covering map for you. Well, that labeling is not the covering map, but it tells you where to put all the edges in constructing a map, and how to orient them. BTW, all the lattice points get mapped to the unique 0-simplex of B. As far as showing E is not simply-connected, note that it is a locally-finite 1-dimensional simplicial complex that contains a non simply-connected subcomplex (a copy of the circle). Use the fact that the homology vanishes in dimensions >= 2 together with the long exact sequence of the pair (E,C) where C is the image of one such copy of that circle. Dale. = >Consider the following subspaces of R^2 : E = (Z x R) / (R x Z) (the >infinite grid) and B = {(x,y) in R^2 : ||(x,y) - (-1,0)|| = 1 or ||(x,y) - >(1,0)|| = 1} (the figure eight). > >How can I describe E being a covering space for B? What is an explicit >formula for the covering map? And, how can I show that E is not >simply-connected? > >Michael > > Let the left circle be labeled L; the right one labeled R. Now, on the big grid, assign labels L and R to each of the > grid edges, subject to the constraint that, at each lattice > point, two edges are labeled L, and two are labeled R. Next, > orient the two circles, and then orient the edges so that at > each lattice point, one L edge is oriented towards, and the > other away from that point, and likewise one R edge is > directed towards, the other away from that point. There? a covering map for you. Well, that labeling is not > the covering map, but it tells you where to put all the edges > in constructing a map, and how to orient them. BTW, all the > lattice points get mapped to the unique 0-simplex of > B. As far as showing E is not simply-connected, note that it > is a locally-finite 1-dimensional simplicial complex that > contains a non simply-connected subcomplex (a copy of the > circle). Use the fact that the homology vanishes in > dimensions >= 2 together with the long exact sequence of > the pair (E,C) where C is the image of one such copy of > that circle. > I have only taken a first semester topology course (i.e. first 7 chapters of algebraic topology in munkres)...hence I don? know what a simplicial complex or homology or exact sequence means...is there any other way to prove E is not simply connected? > Dale. > = ... stuff deleted ... >> >>As far as showing E is not simply-connected, note that it >>is a locally-finite 1-dimensional simplicial complex that >>contains a non simply-connected subcomplex (a copy of the >>circle). Use the fact that the homology vanishes in >>dimensions >= 2 together with the long exact sequence of >>the pair (E,C) where C is the image of one such copy of >>that circle. >> > I have only taken a first semester topology course (i.e. first 7 chapters of > algebraic topology in munkres)...hence > I don? know what a simplicial complex or homology or exact sequence > means...is there any other way to prove E is > not simply connected? >Dale. >> > I? pretty sure of that. Let? start this way: What spaces have you seen that fail to be simply-connected, and how have you seen that fact proven? Dale. = ... stuff deleted ... >> > >>As far as showing E is not simply-connected, note that it > >>is a locally-finite 1-dimensional simplicial complex that > >>contains a non simply-connected subcomplex (a copy of the > >>circle). Use the fact that the homology vanishes in > >>dimensions >= 2 together with the long exact sequence of > >>the pair (E,C) where C is the image of one such copy of > >>that circle. > > > >I have only taken a first semester topology course (i.e. first 7 chapters of >algebraic topology in munkres)...hence >I don? know what a simplicial complex or homology or exact sequence >means...is there any other way to prove E is >not simply connected? > > > > >>Dale. > >> > > > I? pretty sure of that. Let? start this way: What spaces have you seen that fail to be > simply-connected, and how have you seen that > fact proven? > Maybe I can prove this by drawing pictures and using a reasoning argument? > Dale. > = ... stuff deleted ... >> > >>As far as showing E is not simply-connected, note that it > >>is a locally-finite 1-dimensional simplicial complex that > >>contains a non simply-connected subcomplex (a copy of the > >>circle). Use the fact that the homology vanishes in > >>dimensions >= 2 together with the long exact sequence of > >>the pair (E,C) where C is the image of one such copy of > >>that circle. > >> > > >I have only taken a first semester topology course (i.e. first 7 chapters of >algebraic topology in munkres)...hence >I don? know what a simplicial complex or homology or exact sequence >means...is there any other way to prove E is >not simply connected? > > >>Dale. > >> > > > I? pretty sure of that. Let? start this way: What spaces have you seen that fail to be > simply-connected, and how have you seen that > fact proven? > I forgot to mention how I saw the facts proven : For the circle, the standard way using the real line. For SO(3), using the fact that the ball of radius pi modded out by an equivalence relation where v ~ w iff ||v|| = ||w|| = pi and v = -w and another method using quaternions. RP^n I believe can be derived from knowing fundamental group of SO(3) > Dale. > = > >> >> >> >>... stuff deleted ... >> >> >As far as showing E is not simply-connected, note that it >is a locally-finite 1-dimensional simplicial complex that >contains a non simply-connected subcomplex (a copy of the >circle). Use the fact that the homology vanishes in >dimensions >= 2 together with the long exact sequence of >the pair (E,C) where C is the image of one such copy of >that circle. > I have only taken a first semester topology course (i.e. first 7 >> > chapters of >>algebraic topology in munkres)...hence I don? know what a simplicial complex or homology or exact sequence means...is there any other way to prove E is not simply connected? >Dale. > >>I? pretty sure of that. Let? start this way: >> >>What spaces have you seen that fail to be >>simply-connected, and how have you seen that >>fact proven? >> > I forgot to mention how I saw the facts proven : > > For the circle, the standard way using the real line. > For SO(3), using the fact that the ball of radius pi modded out by an > equivalence relation where v ~ w iff ||v|| = ||w|| = pi and v = -w > and another method using quaternions. > RP^n I believe can be derived from knowing fundamental group of SO(3) >>Dale. >> OK, so it appears that you know that if there is a covering map Y ---> X for which Y is connected, and for which the preimage of any element of X has more than a single point, then X cannot be simply-connected. That is, if X were simply-connected, and assuming Y connected, then any covering map as given above would need to be a homeomorphism (i.e., a 1-fold covering). Here? a double covering of your big grid space. First, I?l construct the space E#E (just a made-up notation, nothing official). If you?e seen the map z |--> z^2 of the punctured complex plane to itself, it won? seem like such a weird construction. Take two copies of your grid, call them E_1 and E_2; let? use the notation (i,j)_1 to denote the vertex (i,j) in E_1, and similarly (i,j)_2 to denote the vertex (i,j) in E_2. Next, for each positive integer n, delete the edge joining (n,0)_1 to (n,1)_1, and the edge joining (n,0)_2 to (n,1)_2. This amounts to slitting each plane along the horizontal ray y=1/2, x>=1/2. Finally, attach E_1 to E_2 by inserting edges (n,0)_1 to (n,1)_2 and (n,0)_2 to (n,1)_1, for each positive integer n. The resulting space is what I?l call E#E. The map from this new space to E takes each (i,j)_1 and each (i,j)_2 to the point (i,j) in E. If you do this construction, it? not too difficult to see that the mapping I?e described E#E ---> E is a covering map. It? simple to see that each vertex of E is hit by precisely two elements of E#E. Finally, E#E is easily seen to be connected. Thus, E cannot be simply-connected. > Dale = > > >> > >> > >> >... stuff deleted ... > >> > >> > >As far as showing E is not simply-connected, note that it > >is a locally-finite 1-dimensional simplicial complex that > >contains a non simply-connected subcomplex (a copy of the > >circle). Use the fact that the homology vanishes in > >dimensions >= 2 together with the long exact sequence of > >the pair (E,C) where C is the image of one such copy of > >that circle. > > > I have only taken a first semester topology course (i.e. first 7 > >> >chapters of > >>algebraic topology in munkres)...hence > I don? know what a simplicial complex or homology or exact sequence >>means...is there any other way to prove E is > not simply connected? > > > > > >Dale. > > >> > > > >>I? pretty sure of that. Let? start this way: > >> > >>What spaces have you seen that fail to be >simply-connected, and how have you seen that > >>fact proven? > >> > > >I forgot to mention how I saw the facts proven : > >For the circle, the standard way using the real line. >For SO(3), using the fact that the ball of radius pi modded out by an >equivalence relation where v ~ w iff ||v|| = ||w|| = pi and v = -w >and another method using quaternions. >RP^n I believe can be derived from knowing fundamental group of SO(3) > > > >>Dale. > >> > > OK, so it appears that you know that if there is a covering map Y ---> X for which Y is connected, and for which the preimage of any element > of X has more than a single point, then X cannot be simply-connected. That is, if X were simply-connected, and assuming Y connected, then > any covering map as given above would need to be a homeomorphism (i.e., > a 1-fold covering). Here? a double covering of your big grid space. First, I?l > construct the space E#E (just a made-up notation, nothing official). > If you?e seen the map z |--> z^2 of the punctured complex plane > to itself, it won? seem like such a weird construction. Take two copies of your grid, call them E_1 and E_2; let? use the > notation (i,j)_1 to denote the vertex (i,j) in E_1, and similarly > (i,j)_2 to denote the vertex (i,j) in E_2. Next, for each positive integer n, delete the edge joining > (n,0)_1 to (n,1)_1, and the edge joining (n,0)_2 to (n,1)_2. > This amounts to slitting each plane along the horizontal ray > y=1/2, x>=1/2. Finally, attach E_1 to E_2 by inserting edges > (n,0)_1 to (n,1)_2 and (n,0)_2 to (n,1)_1, for each positive > integer n. The resulting space is what I?l call E#E. The map from this new space to E takes each (i,j)_1 and each (i,j)_2 > to the point (i,j) in E. If you do this construction, it? not too difficult to see that > the mapping I?e described E#E ---> E is a covering map. It? simple to see that each vertex of E is > hit by precisely two elements of E#E. Finally, E#E is easily > seen to be connected. Thus, E cannot be simply-connected. Dale > Michael = ... stuff deleted ... >> > >>As far as showing E is not simply-connected, note that it > >>is a locally-finite 1-dimensional simplicial complex that > >>contains a non simply-connected subcomplex (a copy of the > >>circle). Use the fact that the homology vanishes in > >>dimensions >= 2 together with the long exact sequence of > >>the pair (E,C) where C is the image of one such copy of > >>that circle. > > > >I have only taken a first semester topology course (i.e. first 7 chapters of >algebraic topology in munkres)...hence >I don? know what a simplicial complex or homology or exact sequence >means...is there any other way to prove E is >not simply connected? > > > > >>Dale. > >> > > > I? pretty sure of that. Let? start this way: What spaces have you seen that fail to be > simply-connected, and how have you seen that > fact proven? > SO(3), RP^n and the circle. But the proof for SO(3) was difficult to follow. > Dale. > = >What spaces have you seen that fail to be >simply-connected, and how have you seen that >fact proven? > SO(3), RP^n and the circle. But the proof for SO(3) was difficult to > follow. > Please, what are those spaces? = >> SO(3), RP^n and the circle. But the proof for SO(3) was difficult to >> follow. > Please, what are those spaces? SO(3) - special orthogonal group of 3x3 matrices RP^n - real projective plane in n dimensions -- Ren .8e Meyer Student of Physics & Mathematics Zhejiang University, Hangzhou, China = > >> SO(3), RP^n and the circle. But the proof for SO(3) was difficult to > >> follow. >Please, what are those spaces? SO(3) - special orthogonal group of 3x3 matrices 3x3 matrices with determinate = 1 or is that O(3) ? > RP^n - real projective plane in n dimensions > Ouch, I did ask. = >> RP^n - real projective plane in n dimensions >> >Ouch, I did ask. It consists of the equivalence classes of elements of R^{n+1} other than (0,0,...,0) under the equivalence relation that (a1,...,a_{n+1}) ~ (b1,...,b_{n+1}) if there is some c<>0 such that c*a1=b1, c*a2=b2, ..., c*a_{n+1}=b_{n+1}. Each element with the last (n+1-st) coordinate <>0 is equivalent to a unique (a1,...,an, 1). These correspond one-to-one with the points of an ordinary n dimensional space R^n. Then for purposes of projective geometry, we add the points at infinity which have to be expressed as (a1,...,an, 0). It? treated as being a point at infinity on the lines parallel to the vector (a1,...,an). Multiplying (a1,...,an) by a nonzero constant c leaves it parallel to the same lines. Projective geometry is a surprisingly useful idea. Keith Ramsay = > >> >> SO(3), RP^n and the circle. But the proof for SO(3) was difficult to >> >> follow. >> > Please, what are those spaces? >> >> SO(3) - special orthogonal group of 3x3 matrices > 3x3 matrices with determinate = 1 or is that O(3) ? that? be SL_3 O(3) is matrices {X | (Xu,Xv)=(u,v)} for the usual inner product. Or equivalently the ones satisfying XX^t = I They do have determinant 1 SO(3) is the subgroup with determinant 1. > >> RP^n - real projective plane in n dimensions >> > Ouch, I did ask. = >> SO(3) - special orthogonal group of 3x3 matrices > 3x3 matrices with determinate = 1 or is that O(3) ? Yes. >> RP^n - real projective plane in n dimensions > Ouch, I did ask. ;-) Everybody has a bad day. Rene. -- Ren .8e Meyer Student of Physics & Mathematics Zhejiang University, Hangzhou, China = One of the hard things in this day and age is to find something new, especially in mathematics. What I?e done is use Usenet to test out some discoveries of mine, and see if anyone could successfully challenge them either on correctness or uniqueness. No one could. Obviously there are people willing to argue that point, but in the big scheme of things they don? matter because it really matters what I can get published. I have help beyond what you see talked about on Usenet. What I did was talk to enough people to see if anyone could find a at this time I feel very confident that I can move forward, get *help* on writing a couple of math papers in standard format, and get published. That process will be invisible to you, as you will only see the end result if I succeed. Here? what I did stepped out: 1. I brainstormed a lot of ideas and put them forward for scrutiny. Most failed to be useful. 2. The few ideas that were useful often turned out to be difficult to understand, so I talked about them until I? worked out how everything worked. 3. By talking them out and challenging people I opened the door for someone to point out errors or point out if they were not new. My assessment of Usenet is that it is a talkers?forum versus a doers? That meant that I could put out ideas without much fear of anyone stealing them. Talk them out thoroughly, and get a good feel for them, and then move on to other venues. I feel that I?e used Usenet quite effectively. Mostly getting what I needed involved challenging people in various ways, as well as finding ways to be entertaining while doing it. There were more than enough posters willing to step up and be used in this process. I *did* hope that possibly Usenet might also offer a short-cut, but the information I?e provided has not moved beyond Usenet into mainstream math society from what I?e determined. Mathematicians don? seem to read in the places I?e posted, so the interaction was, you could say, adiabatic. So I?l be looking now to publish my find of the error with the ring of algebraic integers, having determined that I need to re-write the paper currently under review at the Southwest Journal of Pure and Applied Math, so I?l contact them in a bit. I?l also be writing up a paper on my prime counting function. I think I should have both papers completed and published within two years. In the meantime, who knows what else I might talk about, as my math research continues. James Harris = > One of the hard things in this day and age is to find something new, > especially in mathematics. What I?e done is use Usenet to test out > some discoveries of mine, and see if anyone could successfully > challenge them either on correctness or uniqueness. > > No one could. > Not so. Example: you said that if you have a polynomial of the form P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3, where v = -1 + m*f^2, and m, u, and f are integers, with f prime and m coprime to f, then P(x)/f^2 can be factored in the form P(x)/f^2 = (b1*x + u)*(b2*x + u)*(b3*x + u*f) [1] where b1, b2, and b3 are algebraic integers. I said not. Let m = 1, f = 5, and u = 1. Then v = 24, and v^3 + 1 = 13825 = 25*553. It is easily verified that P(x)/f^2 = 553*x^3 - 72*x + 5. If this is factored in the form [1] as you claimed, then -u/b1 = -1/b1 is a root of Q(x) = P(x)/f^2. That is, Q(-1/b1) = 553*(-1/b1)^3 - 72*(-1/b1) + 5 = 0. Multiply through by b1^3: 5*b1^3 + 72*b1^2 - 553 = 0. The expression on the left is a *non-monic* polynomial in b1 with integer coefficients, and it is *irreducible* over the rationals. Therefore b1 cannot be an algebraic integer. Therefore your claim is contradicted and must be false. This argument was posted repeatedly. You never refuted it and you never retracted your claim. > Obviously there are people willing to argue that point, but in the big > scheme of things they don? matter because it really matters what I > can get published. > I thought you had already published some of what you did: with the Mega Society, and then of course at various times on various websites and your math for profit site: lots of exposure. > I have help beyond what you see talked about on Usenet. > > What I did was talk to enough people to see if anyone could find a > at this time I feel very confident that I can move forward, get *help* > on writing a couple of math papers in standard format, and get > published. > > That process will be invisible to you, as you will only see the end > result if I succeed. > > Here? what I did stepped out: 1. I brainstormed a lot of ideas and put them forward for scrutiny. > Most failed to be useful. > > 2. The few ideas that were useful often turned out to be difficult to > understand, so I talked about them until I? worked out how everything > worked. > That? your view. The opposite view is that those ideas were wrong. Counterexamples were given, as above. > 3. By talking them out and challenging people I opened the door for > someone to point out errors or point out if they were not new. > > My assessment of Usenet is that it is a talkers?forum versus a > doers? > Above I gave a rigorous argument that refutes one of your central methods and claims. It and similar arguments have been offered to you many times. You have both ignored them and denied their relevance. You did not refute them or find errors in them. In such cases you were the talker and others were the doers. > That meant that I could put out ideas without much fear of anyone > stealing them. I do not know any instance of anyone here EVER stealing your ideas. Given your view that mathematicians are lying, deceitful, dishonest, etc., this should tell you something: namely, if your ideas had merit, some of us sneaky snakes-in-the-grass would have stolen them, written them up according to our secret Style Manual, and reaped the rewards (publication, tenure, salary, etc). This has not happened. You must be wondering why. How do you rationalize this fact away? > Talk them out thoroughly, and get a good feel for > them, and then move on to other venues. > > I feel that I?e used Usenet quite effectively. Mostly getting what I > needed involved challenging people in various ways, as well as finding > ways to be entertaining while doing it. There were more than enough > posters willing to step up and be used in this process. > > I *did* hope that possibly Usenet might also offer a short-cut, but > the information I?e provided has not moved beyond Usenet into > mainstream math society from what I?e determined. Mathematicians > don? seem to read in the places I?e posted, so the interaction was, > you could say, adiabatic. > > So I?l be looking now to publish my find of the error with the ring > of algebraic integers, having determined that I need to re-write the > paper currently under review at the Southwest Journal of Pure and > Applied Math, so I?l contact them in a bit. > Why do you think it needs re-writing? If you think now it needs to be re-written, it is unethical and inconsiderate to let the referees and editors continue to work on it. It should be withdrawn until you have amended it. Nora B. > I?l also be writing up a paper on my prime counting function. > > I think I should have both papers completed and published within two > years. > > In the meantime, who knows what else I might talk about, as my math > research continues. > James Harris = > One of the hard things in this day and age is to find something new, > especially in mathematics. What I?e done is use Usenet to test out > some discoveries of mine, and see if anyone could successfully > challenge them either on correctness or uniqueness. No one could. And thus the cycle repeats. = >One of the hard things in this day and age is to find something new, >especially in mathematics. What I?e done is use Usenet to test out >some discoveries of mine, and see if anyone could successfully >challenge them either on correctness or uniqueness. No one could. _Except_ for the people who found it very easy to point out errors in yours proofs of FLT and that supposed problem with the algebraic integers (where as often happens your reasoning is not even wrong until you explain why the things you say should be algebraic integers should be), and the people who?e pointed out that there? nothing new about The Prime Counting Function, this is an excellent point. >Obviously there are people willing to argue that point, but in the big >scheme of things they don? matter because it really matters what I >can get published. Really? Then you don? matter. I? surprised to hear you say this, after having so many journal submissions rejected. >I have help beyond what you see talked about on Usenet. What I did was talk to enough people to see if anyone could find a >at this time I feel very confident that I can move forward, get *help* >on writing a couple of math papers in standard format, and get >published. (Unless we?e considering the notion of correct proof as part of the format ...) >That process will be invisible to you, as you will only see the end >result if I succeed. Here? what I did stepped out: 1. I brainstormed a lot of ideas and put them forward for scrutiny. >Most failed to be useful. 2. The few ideas that were useful often turned out to be difficult to >understand, so I talked about them until I? worked out how everything >worked. 3. By talking them out and challenging people I opened the door for >someone to point out errors or point out if they were not new. My assessment of Usenet is that it is a talkers?forum versus a >doers? That meant that I could put out ideas without much fear of anyone >stealing them. Um,... >Talk them out thoroughly, and get a good feel for >them, and then move on to other venues. I feel that I?e used Usenet quite effectively. Mostly getting what I >needed involved challenging people in various ways, as well as finding >ways to be entertaining while doing it. There were more than enough >posters willing to step up and be used in this process. I *did* hope that possibly Usenet might also offer a short-cut, but >the information I?e provided has not moved beyond Usenet into >mainstream math society from what I?e determined. Mathematicians >don? seem to read in the places I?e posted, so the interaction was, >you could say, adiabatic. So I?l be looking now to publish my find of the error with the ring >of algebraic integers, having determined that I need to re-write the >paper currently under review at the Southwest Journal of Pure and >Applied Math, so I?l contact them in a bit. Really? How did you determine that the paper needs to be rewritten? Just curious. >I?l also be writing up a paper on my prime counting function. I think I should have both papers completed and published within two >years. In the meantime, who knows what else I might talk about, as my math >research continues. >James Harris ************************ David C. Ullrich = > One of the hard things in this day and age is to find something new, > especially in mathematics. What I?e done is use Usenet to test out > some discoveries of mine, and see if anyone could successfully > challenge them either on correctness or uniqueness. It is hard but lots of other people manage it. Have you seen how many papers Luzstig has written, for example? > > No one could. > > Please could you answer all the questions I?e raised (I?l append them at the bottom) or point out where you?e refuted them. > Here? what I did stepped out: > > 1. I brainstormed a lot of ideas and put them forward for scrutiny. > Most failed to be useful. Is that litotes? > gain cut and pasted off James? own site > Besides finding an over one hundred year old definition >error 1) Oh boy, don? even start on that one >I have a partial difference equation whose sum over > a certain range can be used to count prime numbers. 2) Presumably you are going to justify why that? true at some point >Here is the partial difference equation and instructions >for summing. >dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], > 3) and we are to infer then that this dS(x,y) is S(x,y) - S(x,y-1)? Or is it S(x,y) - S(x-1,y) or S(x,y) - S(x/y,y-1) >S(x,1) = 0, p(x, y) = ?or(x) - S(x, y) - 1, and S(x,y) is the sum of dS from dS(x,2) to dS(x,y). > 4) Ok, but what is S(x,y)? >Note that it? a *discrete* function, so for you >programmers that means you need to use int? or long? >or some discrete variable type. So you only want the integer square root value, like for >sqrt(10), you want 3. For programmers you get S as the sum of dS from dS(x,2) to dS(x,y) >means you sum up to and *including* dS(x,y). Note: p(x,sqrt(x)) here gives the same value as the >traditional pi(x). > 5) Why? what does p do? is this the thing that counts composites? Why haven? you said what it is? Is this the thing you claimed counted primes and then decided it counted composites. >For faster calculations you can use dS(x,y) = [ p(x/y, sqrt(x/y)) - p(y-1, sqrt(y-1)][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))] when sqrt(x/y) < y-1. > 6) You are keen on this formula, what does it do? Still waiting for an explanation of it. I think we can safely infer, that it is supposed to count something in the range 0 to x, but what does y have to do with it? Is there a ?rst y primes?bit, or is it ?rimes less than y? >There are other speed-ups, and it can be fun finding them. For those of you who want an actual program, I? including >a quick and straight-forward Java implementation. It has >the one speed-up of checking to see if [ p(y, sqrt(y)) - p(y-1, sqrt(y-1))] is non-zero before doing the other calculation, and it >also prints out some prime numbers. Feel free to modify it. >Have fun! 8) Well, I? still none the wiser. How are you justifying S counts primes? 9) Why does p count composites, or is it by definition a composite counter, and you?e proved somewhere, not on your web page, that it satisfies the recurrence relation you keep chucking in? I? love to examine your maths, but there isn? any there! You?e said how to get S as a sum of dS? ok. But I still need to know how to work out what p does don? I? p(x,y) = ?or(x) - S(x,y) + 1 was it? So if I were to put that together I might be able to get something out. I can then find/use the recurrence relation to work out what I presume to be p(x,y) - p(x,y-1) I can? argue with you on whether or not that recurrence relation is valid because you?e not said what p is, or perhaps more clearly, what meaning the y has. Please tell me what it means, then we can decide if it satisfies that relation. 10) And if it does all that, let? then decide if it? a rewriting of Legendre? formula. which, does the same thing as yours as it counts composites. As you?e not explained how you count composites I don? if it? the same way. Painstakingly typeset from your latest web page. >In a previous post I gave a partial difference equation >whose sum can be used to count primes numbers--a first > in recorded human history--and now I? like to talk a >little bit about why it? so important. But it?l take >a bit of calculus. Just a little bit so don? panic if > your calculus is rusty. In calculus for a single variable equation you can get >a differential equation by considering f?x) = [f(x) - f(x-dx)]/dx as dx --> 0. Now if dx = 1, as your first approximation, then you have df(x) = f(x) - f(x-1), which is a difference equation. Now, you can go back from a difference equation to a >differential equation, or from a partial difference >equation to a partial differential equation, which is >what I?l demonstrate with my partial difference equation. Remember from before I have dS(x,y) = [p(x/y, y-1) - p(y-1,sqrt(y-1))][ p(y, sqrt(y)) - p(y-1, sqrt(y-1))], where p(x, y) = ?or(x) - S(x, y) - 1, so it follows that dS(x,y) = [p(x/y, y-dy) - p(y-dy, sqrt(y-dy))][ p(y, sqrt(y)) - p(y-dy,sqrt(y-dy))], > 1) Why haven? all your y? got dy? with them? >where p(x, y) = x - S(x, y) + C, where C comes in from integral calculus. 2) What calculus? You?e not done any yet have you? What?e you integrated? Now I can divide both sides by dy to get dS(x,y)/dy = [p(x/y, y-dy) - p(y-dy, sqrt(y-dy))][ p(y, sqrt(y)) -p(y-dy, sqrt(y-dy))]/dy, where hopefully you can see similarities with what >I had before, though now notice that with more than > one variable, things are a little more complicated, > and in fact I? approaching what? called a partial > differential equation. Now letting dy approach 0, in the limit I have S?(x,y) = [p(x/y, y) - p(y, sqrt(y))] p?(y, sqrt(y)), where some may be looking for p?y, sqrt(y)) on the >end as a directional derivative instead of a partial, > but I have my reasons for seeing it as a partial >which are beyond the scope of this more general post. 3) What? p?(y,sqrt(y))? I mean a partial derivative wrt x of a function of y is zero (independent variables). So, does it mean partial d by du of p(u,v) evaluated at u=y v= sqrt(y)? I mean, the quantity ?n the end?is in the limit going to be d by dy of p(y,sqrt(y) no partials at all. Let? show that isn? equal to what i infer. P(u,v) = v^2 Exercise for reader to show as 1 isn? equal to 0 there must be some other interpretation for what you mean. In the original post I seem to remember you saying you weren? sure about this step. Now you are saying it? beyond the scope of the post. So what is going on? Now then, taking the partial derivative of >p(x, y) = x - S(x, y) + C, with respect to y, I have p?(x, y) = - S?(x, y), and making the substitution >I finally have p?(x,y) = -[p(x/y, y) - p(y, sqrt(y))] p?(y, sqrt(y)). Where I?e connected my partial difference equation, >to a partial differential equation, and now the number >of primes numbers is seen to be a first approximation to > an integration for a continuous function. 4) And? are the solutions of the difference equation and differential equation related necessarily? Example from off topic: the positive rationals as a group under multiplication are isomorphic to integer coefficient polynomials in one variable under addition. A poly has a degree but it? meaning less to talk of a degree of a rational number as there are many isomorphisms all yielding different degrees for the same rational 5) Might do the prime counter in this afternoon? coffee break. 6) These are just my questions, I? not claiming there must be mistakes, merely parts I don? agree with. = ... >Here? what I did stepped out: > >1. I brainstormed a lot of ideas and put them forward for scrutiny. >Most failed to be useful. > > Is that litotes? ... An understatement, of course, but not litotes in my estimation. The following rephrasings exhibit litotes: Not a few failed to be useful. Not a one but failed to be useful. Most were not a little useless. I find it difficult to put a litotic spin on an already-negative verb like failed; does anyone know how? -jiw = > One of the hard things in this day and age is to find something new, > especially in mathematics. What I?e done is use Usenet to test out > some discoveries of mine, and see if anyone could successfully > challenge them either on correctness or uniqueness. > > No one could. Nonsense. Each so called discovery of yours has been either repeatedly shown to be complete nonsense, except one that has repeatedly shown to be a repeat of something that is 200 years old and is completely irrelevant nowadays. > My assessment of Usenet is that it is a talkers?forum versus a > doers? www.cbau.freeserve.co.uk Source code for a prime counting function that beats the crap out of everything you have ever done. ONE THOUSAND TIMES faster. = Now, I always love it when you explain your past behavior with fanciful surveys and the like. I am thrilled to read an account which says (paraphrased), See, you people only *thought* I was an asshole and an idiot. In fact, I *had* to act this way, for I was on a secret mission... Now some might suggest that these posts lack credibility, since each secret mission seems to have little to do with the previous secret mission, but this criticism is easily dispelled. All previous claims about secret missions were simply a ruse necessary to accomplish the secret mission which we now discuss. Any, I believe that this post is especially important, since it gives a blueprint that anyone could follow, so I just want to be sure I understand it. [...] > What I did was talk to enough people to see if anyone could find a > at this time I feel very confident that I can move forward, get *help* > on writing a couple of math papers in standard format, and get > published. That process will be invisible to you, as you will only see the end > result if I succeed. Here? what I did stepped out: 1. I brainstormed a lot of ideas and put them forward for scrutiny. > Most failed to be useful. 2. The few ideas that were useful often turned out to be difficult to > understand, so I talked about them until I? worked out how everything > worked. 3. By talking them out and challenging people I opened the door for > someone to point out errors or point out if they were not new. So, by following these steps, you have succeeded beyond anyone? wildest dreams. In eight short years, this plan will have produced efficiency in action! And all it took was very regular participation in Usenet, including years of dishing out argument, abuse and insult and receiving much of same, together with some similarly brilliant interactions in other fora. My assessment of Usenet is that it is a talkers?forum versus a > doers? That meant that I could put out ideas without much fear of anyone > stealing them. With this, I agree, but it has less to do with Usenet as a talkers?forum than with the brilliant success of other bits of your plans. > Talk them out thoroughly, and get a good feel for > them, and then move on to other venues. I feel that I?e used Usenet quite effectively. Mostly getting what I > needed involved challenging people in various ways, as well as finding > ways to be entertaining while doing it. Others may disagree here, but I certainly agree. You?e been very entertaining. > There were more than enough posters willing to step up and be used > in this process. I think so, too. Myself included. Say, I wonder: do my .sigs help advance your plan, too? I? just fishing for acknowledgment in your [...] > So I?l be looking now to publish my find of the error with the ring > of algebraic integers, having determined that I need to re-write the > paper currently under review at the Southwest Journal of Pure and > Applied Math, so I?l contact them in a bit. Joshing aside, I hope that you contact the journal sooner rather than later. If you plan on a rewrite, then don? waste the referee? time. -- Jesse F. Hughes My baby don? allow me in the kitchen and I?e come to love her decision. -- Bad Livers = > One of the hard things in this day and age is to find something new, > especially in mathematics. What I?e done is use Usenet to test out > some discoveries of mine, and see if anyone could successfully > challenge them either on correctness or uniqueness. Congratulations on conducting and completing your ?bjective?tests. > No one could. Hold it, Wacky! That statement is simply FALSE. Here are four (4) challenges you have so far *failed* to meet: 1) You previously stated that given ?bc=p? where ?? ??and ??are algebraic integers and ??is an integer (or a prime integer, or some such subset of the algebraic integers), it is possible for ?/a?or ?/b?or ?/c?to be outside the ring of algebraic integers. This claim is false. It has been repeatedly demonstrated to be false, yet you continued to make the claim. You never retracted your claim, or posted any supporting example. 2) You repeatedly claimed that there are numbers which *should* be algebraic integers, but which are not. You were then challenged to produce some (or even just one) of these numbers. You *never* met that challenge, yet you continued to make the claim. 3) Recently, you claimed that your algorithm for solving the prime counting problem could do things which no other method could do. When challenged to simple *name* any of these things, you failed to do so. To date, you have produced no details on the capabilities of your algorithm which are beyond those of other, similar algorithms, and have therefore *failed* to meet the challenge. Nevertheless, you still make the claim. 4) You claim that your difference equation *leads* to a so-called ?artial differential equation?which provides a continuous solution to the prime counting problem. By your posted method, *every* difference equation leads to a differential equation, by simply replacing the finite difference of unity with a fractional difference, which is then treated as a differential. You have been advised that such an exercise produces an equation which does *not* solve the original problem, unless unity is stuck back in for the misapplied differential. You have not produced any evidence, no proof and no data to support your claim that this result has any connection with the prime counting problem, whatsoever. You have *failed* the clearly stated challenge, which has been on the table for some time. Hence, your claim that no one could successfully challenge your discoveries is not only false, it is ludicrous. Your above statement in this post that no one could challenge your claims, which you have posted in the presence of a public record containing voluminous counter examples of which you are well aware, is worse than simply wrong, it is patently dishonest. You are a liar. Wacky, isn? it? But, hey, it? just basic math. Yup, yup, yup. Put up, or SHUT UP. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com = Eat a deep-dish dog penis pizza pie and die screaming with many sharp objects in your bottom. = > One of the hard things in this day and age is to find something new, > especially in mathematics. What I?e done is use Usenet to test out > some discoveries of mine, and see if anyone could successfully > challenge them either on correctness or uniqueness. No one could. Obviously there are people willing to argue that point, but in the big > scheme of things they don? matter because it really matters what I > can get published. I have help beyond what you see talked about on Usenet. What I did was talk to enough people to see if anyone could find a > at this time I feel very confident that I can move forward, get *help* > on writing a couple of math papers in standard format, and get > published. That process will be invisible to you, as you will only see the end > result if I succeed. Here? what I did stepped out: 1. I brainstormed a lot of ideas and put them forward for scrutiny. > Most failed to be useful. 2. The few ideas that were useful often turned out to be difficult to > understand, so I talked about them until I? worked out how everything > worked. 3. By talking them out and challenging people I opened the door for > someone to point out errors or point out if they were not new. My assessment of Usenet is that it is a talkers?forum versus a > doers? That meant that I could put out ideas without much fear of anyone > stealing them. Talk them out thoroughly, and get a good feel for > them, and then move on to other venues. I feel that I?e used Usenet quite effectively. Mostly getting what I > needed involved challenging people in various ways, as well as finding > ways to be entertaining while doing it. There were more than enough > posters willing to step up and be used in this process. I *did* hope that possibly Usenet might also offer a short-cut, but > the information I?e provided has not moved beyond Usenet into > mainstream math society from what I?e determined. Mathematicians > don? seem to read in the places I?e posted, so the interaction was, > you could say, adiabatic. So I?l be looking now to publish my find of the error with the ring > of algebraic integers, having determined that I need to re-write the > paper currently under review at the Southwest Journal of Pure and > Applied Math, so I?l contact them in a bit. I?l also be writing up a paper on my prime counting function. I think I should have both papers completed and published within two > years. In the meantime, who knows what else I might talk about, as my math > research continues. > James Harris And what if those papers aren? accepted? You going to continue your nonsense? David Moran = > > One of the hard things in this day and age is to find something new, > especially in mathematics. What I?e done is use Usenet to test out > some discoveries of mine, and see if anyone could successfully > challenge them either on paper explicitly empirically demonstrated symbolically and numerically that you are an ineducable psychotic trolling idiot. http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav http://w0rli.home.att.net/youare.swf http://www.mazepath.com/uncleal/sunshine.jpg http://www.you-moron.com/ Hey stooopid loud troll James Harris, put up or shut up, http://www.rsasecurity.com/rsalabs/challenges/factoring/ faq.html http://www.rsasecurity.com/rsalabs/challenges/factoring/ numbers.html http://www.crank.net/harris.html It? not every braying jackass that gets a whole page at crank.net Is a $10,000 prize no questions asked too small to justify your submission of two little prime numbers? d5959fd%40posting.g oogle.com> James Harris Somebody said it couldn? be done, But James with a chuckle replied That maybe it couldn?, but he would be one Who wouldn? say so till he? tried. So James buckled right in with the trace of a grin On his face. If he worried he hid it. James started to sing and he tackled the thing And James never ing could do it. Somebody scoffed: Oh, you?l never do that; At least no one has ever done it ; But James took off his coat and he took off his hat, And the first thing we knew he? begun it. With a lift of his chin and a bit of a grin, Without any doubting or quiddit, James started to sing and he tackled the thing And James never ing could do it. There are thousands to tell James it cannot be done, There are thousands to prophesy failure; There are thousands to point out to James, one by one, How hopeless that task set before you. But just buckle in with a bit of a grin, James take off your coat and go to it; Just start to sing as you tackle the thing And James, you?l never ing do it. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! = Hey stooopid loud troll James Harris, put up or shut up, http://www.rsasecurity.com/rsalabs/challenges/factoring/ faq.html >http://www.rsasecurity.com/rsalabs/challenges/factoring/ numbers.html >http://www.crank.net/harris.html > It? not every braying jackass that gets a whole page at crank.net Is a $10,000 prize no questions asked too small to justify your >submission of two little prime numbers? > The $10,000 prize has already been claimed. rich = >What I did was talk to enough people to see if anyone could find a >at this time I feel very confident that I can move forward, get *help* >on writing a couple of math papers in standard format, and get >published. The best wishes for success. I hope you recognize that most people will withold judgement until they see the actual evidence of publication. Some will be downright also offer a short-cut, but >the information I?e provided has not moved beyond Usenet into >mainstream math society from what I?e determined. There are no shortcuts to good research. > Mathematicians >don? seem to read in the places I?e posted, so the interaction was, >you could say, adiabatic. There are many very good mathematicians who read sci.math . Undoubtedly there are many more who never read it. There are relatively few mathematicians who would read sci.cognitive. >So I?l be looking now to publish my find of the error with the ring >of algebraic integers, having determined that I need to re-write the >paper currently under review at the Southwest Journal of Pure and >Applied Math, so I?l contact them in a bit. When it? published I may take a look. Until that time, I remain skeptical as to whether there could be the kind of error you claim to have found. = > > >What I did was talk to enough people to see if anyone could find a > >at this time I feel very confident that I can move forward, get *help* > >on writing a couple of math papers in standard format, and get published. > > The best wishes for success. > > I hope you recognize that most people will withold judgement > until they see the actual evidence of publication. Some will > be possibly Usenet might also offer a short-cut, but > >the information I?e provided has not moved beyond Usenet into mainstream math society from what I?e determined. > > There are no shortcuts to good research. Usenet has been a short-cut in important ways though it hasn? been a short-cut to acceptance which is what I was talking about with that comment. Remember I? not a mathematician. Let that soak in for a moment. I? NOT a mathematician. However, I? talking about two papers in rather fundamental areas of mathematics. I brainstormed on Usenet and used Usenet posters to critique ideas. It was a fairly efficient process in that respect. Oddly, no one provided positive input in terms of providing important ideas that I incorporated into my work. That was a blessing and a surprise. In any event, there? more fun to posting on Usenet than just research. Like, I like to argue. And here I seem to find lots of people willing to argue as well, though too many cheat by depending primarily on insults and other cheapshots. But hey, I?l take the weak along with the occasional good debater, as that? just more practice! James Harris important ways though it hasn? been a > short-cut to acceptance which is what I was talking about with that > comment. Remember I? not a mathematician. Let that soak in for a moment. I? NOT a mathematician. However, I? talking about two papers in rather fundamental areas of > mathematics. I brainstormed on Usenet and used Usenet posters to critique ideas. > It was a fairly efficient process in that respect. Oddly, no one > provided positive input in terms of providing important ideas that I > incorporated into my work. That was a blessing and a surprise. In any event, there? more fun to posting on Usenet than just > research. Like, I like to argue. And here I seem to find lots of > people willing to argue as well, though too many cheat by depending > primarily on insults and other cheapshots. But hey, I?l take the weak along with the occasional good debater, as > that? just more practice! James Harris OK, James. Now be good boy and go back to your playpen. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com = I was trying to figure out how the double angle formula is derived from the double angle formula as indicated in trig books. sin(a)cos(b)+cos(a)sin(b) - > i found in one book that they just switch the b? to a?; now, at first I found in another trig book about switching one angle symbol with another, and I was at a total loss at what they were talking about. What were these angle symbols? Finally, I picked up another trig book, and it indicated that the angle symbols were a and b. O.K. so sin(a)cos(b)+cos(a)sin(b) gets turned into sin(a)cos(a)+cos(a)sin(a). Now, I? thinking factor out the sins right? So, we get sin(cos + cos). No, it must be factor out the cos? So, we have cos(sin + sin)? and then 2sincos? The book just does not explain! More: I? looking at this one other trig book, and in it is one introductory paragraph about where the identities come from; if I recall correctly, it said they come from expressing the sides and angles of geometric problems in terms of trigonometric expressions. Where is this in my modern day for the last who knows editions of trig books? What else am I missing here that gives some idea of what I? trying to solve? I mean that when I first took a trig class, the teacher started talking not about solving right triangles as I?e come to see that most of high school intro? to trig in pre-algebra and watered down geometry classes show you, but all this mapping of 45,-45,90, and 30-60-90 triangles as the only definite exact solutions of trig functions, and then the identities and trig equations. At which point, I? asking how does all that relate to solving right triangles and other naive questions. I mean this book was pretty old. What is the definitive trig book that shows everything and explains everything? And yes, I suspect I have the same questions for all the rest of mathematics education. What quality of education am I really getting? Yes, I still want to see my questions about the double angle derivation from the addition formulas answered please. = .... I found in one book that they just switch the b? to a? .... > > I? looking at this one other trig book, and in it is one introductory > paragraph about where the identities come from; if I recall correctly, > it said they come from expressing the sides and angles of geometric > problems in terms of trigonometric expressions. Where is this in my > modern day for the last who knows editions of trig books?.... If you? like an intelligent modern introduction to trigonometry, I think it? still worth while to hunt down a copy of Clayton W. Dodge, The Circular Functions, Prentice-Hall, 1966. Ken Pledger. = >I was trying to figure out how the double angle formula is derived >from the double angle formula as indicated in trig books. You mean from the addition formula. >sin(a)cos(b)+cos(a)sin(b) - > i found in one book that they just >switch the b? to a?; now, at first I found in another trig book >about switching one angle symbol with another, and I was at a total >loss at what they were talking about. What were these angle symbols? >Finally, I picked up another trig book, and it indicated that the >angle symbols were a and b. O.K. so sin(a)cos(b)+cos(a)sin(b) gets >turned into sin(a)cos(a)+cos(a)sin(a). Now, I? thinking factor out >the sins right? So, we get sin(cos + cos). Hell no. You get sin(a)*(cos(a)+cos(a)). But why bother? You have the same thing twice: sin(a)*cos(a). And if you take a number x, and you add it to itself, you always get 2x: x+x = 2x. So here you have sin(a)cos(a) + sin(a)cos(a) = 2sin(a)cos(a). (Remember: cos(a) and sin(a) are just numbers, and since multiplication is commutative, cos(a)*sin(a) = sin(a)*cos(a)). >the cos? So, we have cos(sin + sin)? and then 2sincos? The book just >does not explain! Basic algebra. If you do not get that, you should be learning basic algebra, not trigonometry. >I? looking at this one other trig book, and in it is one introductory >paragraph about where the identities come from; if I recall correctly, >it said they come from expressing the sides and angles of geometric >problems in terms of trigonometric expressions. Where is this in my >modern day for the last who knows editions of trig books? I have no idea what you mean with the last statement. >What else >am I missing here that gives some idea of what I? trying to solve? A clue? -- = It? not denial. I? just very selective about what I accept as reality. --- Calvin ( Calvin and Hobbes ) = Arturo Magidin magidin@math.berkeley.edu > Is there an arithmetic compression formula or algorithm that can be used to store > a numeric 64bit or numeric 32bit quantity in a compressed format into a numeric > 16bit quantity? You should try the comp.compression newsgroup, as it is a great forum for these question. Simply stated, it is impossible to losslessly compress an arbitrary bit sequence, as shown by the Counting Theorem. However, most files are bit sequences contain a large amount of redundant data, and if you know the formats in advance, you can factor those redundancies out. However, the unfortunate side effect is that it will necessarily expand other. As a brief look at your example, lets take the 32-bit quantities first. There are 2^32 possible bit sequences. Not being very adventuresome, lets say you merely wish to compress down to 31 bits for each sequence. You therefore have 2^31 possible compressed sequences to map to. Obviously (by what? known as the Pigeon Hole Principle), only half of your original set can be losslessly compressed down to them. The other half might be ?gged (say with a leading 1 bit) as being uncompressed, increasing them to 33 bits. And so it goes. On average, for every n bits you save losslessly, only 1/2^n of them get to be compressed. This is why random data can? be compressed; only sequences you target can be compressed. Now you also understand why text compression algorithms do very badly on images, as imaage compression algorithm don? work on text. Jonathan Hoyle Gene Codes Corporation = > Question: > - Is there a computable system as described above, for which all the > axioms concering numbers, as layed down by the Ancient Greeks (I have > to look this up), are valid for the set C? What are those axioms? There is no such computable system which has the least upper bound and density properties. Thomas <87d6avq30p.fsf@becket.becket.net> <3fd8e65a$0$249$19deed1b@news.inter.NL.net> <87llpieu4b.fsf@becket.becket.net> = > Question: >> - Is there a computable system as described above, for which all the >> axioms concering numbers, as layed down by the Ancient Greeks (I have >> to look this up), are valid for the set C? What are those axioms? There is no such computable system which has the least upper bound and > density properties. Can you brie? sketch why? Also, what density property do you have in mind? -- ?very man who has ever lived holds tight to the out by love[...] Therefore, you will marry Guinevere. You do not want advice --- only agreement.?Merlin sighed... -- John Steinbeck = > > > >> Question: > >> - Is there a computable system as described above, for which all the > >> axioms concering numbers, as layed down by the Ancient Greeks (I have > >> to look this up), are valid for the set C? > >What are those axioms? > >There is no such computable system which has the least upper bound and >density properties. > > Can you brie? sketch why? > > Also, what density property do you have in mind? Density is the property that for any A and B with AP?is an isomorphism, then define f*:C-C?as f*(x) = LUB{f(p):p in P and p <= x}. It is easy to check that f* is an order isomorphism. 3) Look this one up. 4) Cantor. Thomas = with these kind of things you have to be very carefull with the things you are proving. I think the people that don? like Cantor have the greatest problem with the fact that whatever system you use, R contains numbers which can not be expressed. At least I can say about it, it is a fair concern. If I have time, I will check the proof you sketched. Of course, it is correct. However, if you don? allow that you refer to objects/numbers that are not computable, the proof maybe does not hold. I have to check this once. You should realize, that if R as uncountable set is disbanded, certain proof constructions in general are disbanded. You can? take one thing out. If there is an alternative for the Cantor system, them at least some things should still work. If the computable numbers are taken, then at a certain moment there is a wish to switch to the axioms of real numbers. This of course, must be possible and those axioms must apply to the computable numbers. If not, then mathematics is extremely limited and Cantor is much better. I am not sure why such alternative would not work. For instance, if you take a limit based on computable numbers, you get a computable number again. This counts for all operation you can do on real numbers. And you can? do the diagolization within the computable system, because of the Turing Halting problem. Interesting research.... Lucas >> Question: - Is there a computable system as described above, for which all the axioms concering numbers, as layed down by the Ancient Greeks (I have to look this up), are valid for the set C? >> >> What are those axioms? >> >> There is no such computable system which has the least upper bound and >> density properties. > >Can you brie? sketch why? > >Also, what density property do you have in mind? Density is the property that for any A and B with A which is strictly greater than A and strictly less than B. The proof proceeds in the following three steps: 1) Any two countable unbounded dense linearly ordered sets are > isomorphic. > 2) Any two sets with that have LUB property and have a countable > unbounded dense subset, are isomorphic. > 3) R, defined by Dedekind cuts, has the LUB property and a countable > unbounded dense subset. > 4) R is uncountable. > Sketches of proofs: 1) Go from set A to set B, back and forth, each time you pull an > element out of B, you find an element of A that corresponds in an > isomporphism that you build up one element at a time. Because the > sets are dense, you can always find an element that is order > isomorphic to the one you just picked from the other set. Since > the sets are countable, and you alternate which set you pick an > element from, you exhaust both sets and have a complete > isomorphism. 2) Take the isomorphism between the two countable dense subsets. You > can construct an isomorphism between the sets themselves by using > the LUB property. That is, if the sets are C and C? and the > countable dense subsets are P and P? and f:P->P?is an > isomorphism, then define > f*:C-C?as f*(x) = LUB{f(p):p in P and p <= x}. > It is easy to check that f* is an order isomorphism. 3) Look this one up. 4) Cantor. Thomas > <87d6avq30p.fsf@becket.becket.net> <3fd8e65a$0$249$19deed1b@news.inter.NL.net> <87llpieu4b.fsf@becket.becket.net> <87n09ykzvv.fsf@phiwumbda.org> <874qw5q5lt.fsf@becket.becket.net> = >> >> Can you brie? sketch why? >> >> Also, what density property do you have in mind? Density is the property that for any A and B with A which is strictly greater than A and strictly less than B. The proof proceeds in the following three steps: 1) Any two countable unbounded dense linearly ordered sets are > isomorphic. > 2) Any two sets with that have LUB property and have a countable > unbounded dense subset, are isomorphic. > 3) R, defined by Dedekind cuts, has the LUB property and a countable > unbounded dense subset. > 4) R is uncountable. -- Jesse Hughes Well, you know as soon as you have a new number I will be happy to add it to the list. Don? try those childish tit-for-tat games with me. -- Ross Finlayson on Cantor? theorem. = > The proof proceeds in the following three steps: > > 1) Any two countable unbounded dense linearly ordered sets are > isomorphic. > 2) Any two sets with that have LUB property and have a countable > unbounded dense subset, are isomorphic. > 3) R, defined by Dedekind cuts, has the LUB property and a countable > unbounded dense subset. > 4) R is uncountable. Shades of Monty Python: Four steps sir -- yes, four! . I think the rest of the post was correct, even if I have trouble counting to four. Thomas = -------------------------------------------------------------- -------------- ------ > ... >I don? know what a random number is. But the existence of non >computable numbers is perfectly well defined. > May I introduce a new question. Suppose we have a computable system where we can define some computable > irrational numbers. Call the set of numbers that can be defined with this > computable system, the set C. Question: > - Is there a computable system as described above, for which all the > axioms concering numbers, as layed down by the Ancient Greeks (I have > to look this up), are valid for the set C? If the answer is NO, then Cantor? system is rather unavoidable, but if > the answer is YES... > I hope someone addresses your question... My point can be summarised here by what is a non computable irrational? . This is the only space left on the number line after UTM(Z). sqrt(2), pi, e, all line up in the same UTM(Z) listing that captures all rationals. NCIs are impossible to encapsulate anyhow, non repeating, non terminating, impossible to represent their digital representation, they are, as I here define it *random* digit sequences. Reals - Computables = infinite random digit sequences There is no way to represent them, my deciding factor they aren? considered real numbers is the fact that at *some* level numbers take on a physical characterisic, their transfer, their role as a parameter, a message, identifier, and Shannon? theory on information transfer is quite clear that a *sole* infinite sequence will corrupt. This is a platonic viewpoint more than mathematical I know. What I would like is to see how far number theory gets WITHOUT Cantors proof, because there is so much mutually supporting high level theory once it is accepted there are cyclic arguments coming down that claim to refute countability of numbers. Herc = > This is the only space left on the number line after UTM(Z). > sqrt(2), pi, e, all line up in the same UTM(Z) listing that captures > all rationals. So the real line doesn? have the least upper bound property according to you? That? the question here: does it, or doesn? it. Any dense set with the least upper bound property is uncountable. Most people think that the least upper bound property is a key part of their intuitions about the number line. Thomas = So the real line doesn? have the least upper bound property according >to you? That? the question here: does it, or doesn? it. > It? just an axiom, right? So there is no real answer to your question. I? just as happy with it as without it. Come to think of it, that a *lot* of math goes away without it, means there is less math I don? know. I suppose that? good, so maybe I don? like it too much after all. Time for dinner, this is giving me a headache. rich = > It? just an axiom, right? So there is no real answer to your > question. Sure there is: we can consider sets with the least upper bound property, and sets without. It turns out that sets with the LUB property have lots of wonderful properties that come along: we can do analysis, for example. Nothing prevents someone from working in the universe of constructible numbers. It doesn? follow that the universe of real numbers is not also interesting. Thomas = > Almost 4 years ago, there was a very long thread on this very topic. > See the thread starting at news.globalcenter.ne t > > As you mention, there are three ways to approach this problem; each > boils down to a question of convergence. If we have a sequence {a_n}, > we might ask what happens to that sequence as n tends to infinity, and > thus, pass to the limit to define what happens at infinity. However, > to define a limit, we must first define a topology. > But the question can be answered without appealing to any sort of topology. > > ------- > There are standard topologies on both the reals and the integers, so > that if the a_n are reals or integers, we use those topologies. > For example, let a_n be the number of balls in the bucket after the n^th > placement/removal. That sequence converges to infinity as n tends to > infinity. > But that has nothing to do with the question that was asked. > >> The question asked was, how many balls? Counting how many balls are >> in the bucket at any given time would seem to have something to do with >> the question. Just because we don? know which balls are in the bucket >> does not mean there are no balls in the bucket. Counting the balls at some time before noon has nothing to do with the >bucket? contents at noon, if there are still balls to be moved in and >out. Counting the balls at noon is what matters. Your approach reminds me of the old joke that when you have a hammer, >every problem begins to resemble a nail. Here you have three hammers, >slightly different in design, but they are all basically the same tool. >Yet nowhere do you address the question of whether your tool is the >appropriate one for the job. Not only that, your tools give con?cting >answers, which is evidence that at least some of them are being >incorrectly applied. > Suppose the problem were asked in a slightly different way. Suppose >> that instead of ball #n being removed at step n, we remove ball #10n, >> that is, the last ball added. Here I don? think you will disagree that >> there are 9 balls added to the bucket at each step that stay there for >> all time. Thus, at noon, there are infinitely many balls. Right answer, wrong reason. The reason there are infinitely many balls >at noon is that a_n(t_0) = 1 for each n that is not a multiple of 10, and >there are infinitely many such n. Therefore a(t_0) = sum_{n=0}^oo >a_n(t_0) = +oo. > Suppose one ball is removed at each step, but we are not told which ball >> that is. What happens at noon. All we know is that at step n, there >> are 9n balls in the bucket. How many balls are in the bucket at noon? >> Does our knowledge of which balls are in the bucket alter how many balls >> are in the bucket? If the ball to be removed is selected at random at each step, then I can >conclude that with probability 1, there are no balls left at noon. ------- > Let us define a_n to be the function, after the n^th placement/removal, > which maps the set of balls to {0,1}, where 1 means that ball is in the > bucket and 0 means it is not. If we use the topology of pointwise > convergence on these functions, the sequence {a_n} converges to the > function which maps all the balls to 0, since at some point, each ball > is removed from the bucket, never to be put back. > If we view a_n to be a function defined in the time domain rather than in the domain of natural numbers, such that a_n(t) = 1 if ball n is in the bucket at time t and a_n(t) = 0 otherwise, then we find that a_n(t_0) = 0 for each n, where t_0 = noon. Thus a(t_0) = sum_n=1^oo a_n(t_0) = 0. > Although this reasoning leads to the same conclusion as your argument, there is an essential difference. Your argument uses limits as t->t_0, but mine does not. The only limit that appears in my argument is the one that says the sum of an infinite collection of zeros is zero. Your argument depends on justifying the step of introducing a pointwise topology and using it to answer the original question, while my argument depends on no such artifice. > >> You are summing an infinite number of zeroes. Just because they are >> zeroes does not excuse you from having to take a limit, trivial as it >> may be. Taking a limit involves a topology. Didn? I already say that? Actually, I do see a way to formulate the >argument that does not involve a limit at all, but that was not the point >I was trying to make in that paragraph, where I specifically said I was >taking a limit and I identified exactly where it was. The point is not >whether a topology is involved, but *which* topology is involved, and how >it relates to the stated problem. ------- > let us use the discrete topology on this sequence. With this topology, > the functions converge to a function f if after some point, all the a_n > are equal to f. Thus, using this topology, the {a_n} do not converge. >>Again, I don? see that this argument has anything to do with the question that was asked. > >> It does in the sense that although we know that past step n there are >> more that 9n balls in the bucket, the set of balls keeps changing. >> Thus, the set of balls in the bucket never settles to any limit. It >> doesn? settle to the empty set since there is an increasing number of >> balls in the bucket, yet no particular ball is in the bucket forever. If no particular ball remains in the bucket until noon, and no ball is >returned to the bucket having once been removed, then no ball is in the >bucket at noon. Thus, each of the replies is correct depending on how you define the > convergence. Since the question asks, how many balls , it seems to me > I would say an infinite number. If the question were which balls , > balls. > And this is strong evidence that there is something wrong in your reasoning. If you know which balls , then you automatically know how many balls . Every set has a cardinality. If your reasoning leads to inconsistent answers, then your reasoning is wrong. > >> If you know which balls, then you know how many balls; but if you know >> how many balls, you don? necessarily know which balls. Oh, come, now. You know better than that. If I demonstrate if A then >B , it is useless to retort that I have not proved the converse, since I >have not even stated the converse or relied on it in any way. If the answer to which balls is none , then the answer to how many >balls cannot possibly be infinitely many . This demonstrates that >something is wrong with the application of your chosen tools. >Looking only at >> the cardinality of the balls in the bucket, we get that there are an >> infinite number of balls at noon. No, we do not get that. Not unless you add an unwarranted assumption >concerning continuity at noon. The function is obviously discontinuous >at a great many points (each time balls are added or removed), so why >should we expect continuity at noon? After all, noon (t=t_0) is a limit >point of the set of discontinuities of the function a = sum a_n, and >therefore it should not be surprising that there is also a discontinuity >at t_0. >However, looking at which balls are >> in the bucket, we cannot say as that information keeps changing. Each of the changes takes place before noon, and the state of each ball >is unchanging once that ball is removed. > Let us pose a different problem. Suppose we have two balls, #1 and #2. >> At each step above we swap which ball is in the bucket. At each step >> there is one ball in the bucket. How many balls are in the bucket at >> noon? Which ball is in the bucket at noon? Although we can answer the >> first question pretty easily, we can? answer the second. If we use b_1(t) and b_2(t) as the characteristic functions of the two >balls, then we find that in this case (unlike the original problem) >b_1(t_0) and b_2(t_0) cannot be determined from the stated information, >and therefore the approach that I described fails. But so what? This >problem obviously is not well-posed, but the original problem does not >have that defect. In that case we can determine a_n(t_0) for each n. If we don? assume any topology or convergence, the conditions of the problem do not say anything whatsoever about what happens at noon. The problem specifies what happens at any time before noon, but without continuity, I can specify that all the balls that have been thrown out are back in, or that no balls are there, or that every other ball is in the bucket with no contradiction. The only way to deduce from what has gone on before what happens at noon is if there is some sort of continuity and along with that comes a topology that allows you to evaluate the limit. It is like saying that because f(x) = x for all x > 0 implies f(0) = 0 without saying that f is continuous at 0. All I am trying to demonstrate is that the sequence of balls in the bucket can reach different limits depending on the topology you use. The way the problem is stated, without a topology, there is no limit and hence we can say nothing about what happens at noon. Rob Johnson take out the trash before replying = > If we don? assume any topology or convergence, the conditions of the > problem do not say anything whatsoever about what happens at noon. By that standard, there is almost no such thing as a word problem that can be answered without assuming some unstated facts. It is an unstated fact that if a ball is removed from the bucket before noon and is not subsequently returned, then that ball is not in the bucket at noon. This is so without regard to what may happen to other balls in the meantime. > The problem specifies what happens at any time before noon, but without > continuity, I can specify that all the balls that have been thrown out > are back in, or that no balls are there, or that every other ball is > in the bucket with no contradiction. The only way to deduce from what > has gone on before what happens at noon is if there is some sort of > continuity and along with that comes a topology that allows you to > evaluate the limit. It is like saying that because f(x) = x for all > x > 0 implies f(0) = 0 without saying that f is continuous at 0. It? like saying that if you have 5 apples and I take 2, then you may have 7 left because someone else may have given you 4 more. That answer cannot be excluded by what was stated in the original question, but a reasonable reading of the problem is to assume that there are no apples (or balls, or transitions, or whatever) other than those that are explicitly mentioned. > All I am trying to demonstrate is that the sequence of balls in the > bucket can reach different limits depending on the topology you use. > The way the problem is stated, without a topology, there is no limit > and hence we can say nothing about what happens at noon. I have never denied that applying different topologies will produce different results. In fact, that is an essential part of my argument. The fact that you can produce contradictory answers by using different topologies is evidence that at least some of those answers must have resulted from incorrect reasoning. -- Dave Seaman Judge Yohn? mistakes revealed in Mumia Abu-Jamal ruling. = > If we don? assume any topology or convergence, the conditions of the >> problem do not say anything whatsoever about what happens at noon. By that standard, there is almost no such thing as a word problem that >can be answered without assuming some unstated facts. It is an unstated fact that if a ball is removed from the bucket before >noon and is not subsequently returned, then that ball is not in the bucket >at noon. This is so without regard to what may happen to other balls in >the meantime. That is precisely the assumption for pointwise convergence. If you take that as an unstated fact, then yes, you do get the limit you claim. It is the fact that that fact is not given that makes the problem ambiguous to start with. >> The problem specifies what happens at any time before noon, but without >> continuity, I can specify that all the balls that have been thrown out >> are back in, or that no balls are there, or that every other ball is >> in the bucket with no contradiction. The only way to deduce from what >> has gone on before what happens at noon is if there is some sort of >> continuity and along with that comes a topology that allows you to >> evaluate the limit. It is like saying that because f(x) = x for all >> x > 0 implies f(0) = 0 without saying that f is continuous at 0. It? like saying that if you have 5 apples and I take 2, then you may >have 7 left because someone else may have given you 4 more. That answer >cannot be excluded by what was stated in the original question, but a >reasonable reading of the problem is to assume that there are no apples >(or balls, or transitions, or whatever) other than those that are >explicitly mentioned. This is just not so. Without the assumption of pointwise convergence, that you take as an unstated fact, nothing is implied at noon. Without that unstated fact, or another similar unstated fact, the rules telling what to do at any time before noon do not tell you how to pass to the limit and say anything about the contents at noon. >> All I am trying to demonstrate is that the sequence of balls in the >> bucket can reach different limits depending on the topology you use. >> The way the problem is stated, without a topology, there is no limit >> and hence we can say nothing about what happens at noon. I have never denied that applying different topologies will produce >different results. In fact, that is an essential part of my argument. >The fact that you can produce contradictory answers by using different >topologies is evidence that at least some of those answers must have >resulted from incorrect reasoning. You have denied that topology has anything to do with the result; that without resorting to topology, you can get your result. The fact that different results can be obtained by assumptions that do not contradict the other conditions only means that the problem was ambiguous to start with. Given the premises of the problem as stated, it is neither provable nor contradictory that the bucket be empty at noon. Rob Johnson take out the trash before replying = >>It is an unstated fact that if a ball is removed from the bucket before >>noon and is not subsequently returned, then that ball is not in the bucket >>at noon. This is so without regard to what may happen to other balls in >>the meantime. > That is precisely the assumption for pointwise convergence. If you take > that as an unstated fact, then yes, you do get the limit you claim. It > is the fact that that fact is not given that makes the problem ambiguous > to start with. The same assumption operates if I ask, You have five apples. I take away two. How many are left? Are you claiming that every first-grade arithmetic problem is an exercise in topology? The balls-in-the-bucket problem is no more ambiguous than the apples problem. There may be more balls than there are apples, but if you focus your attention on just one ball or one apple at a time, exactly the same principle applies. The problem specifies what happens at any time before noon, but without continuity, I can specify that all the balls that have been thrown out are back in, or that no balls are there, or that every other ball is in the bucket with no contradiction. The only way to deduce from what has gone on before what happens at noon is if there is some sort of continuity and along with that comes a topology that allows you to evaluate the limit. It is like saying that because f(x) = x for all x > 0 implies f(0) = 0 without saying that f is continuous at 0. >>It? like saying that if you have 5 apples and I take 2, then you may >>have 7 left because someone else may have given you 4 more. That answer >>cannot be excluded by what was stated in the original question, but a >>reasonable reading of the problem is to assume that there are no apples >>(or balls, or transitions, or whatever) other than those that are >>explicitly mentioned. > This is just not so. Without the assumption of pointwise convergence, > that you take as an unstated fact, nothing is implied at noon. For which ball is it not so? Does ball #1 have this property? Do you agree that ball #1 is removed before noon? Does the problem say that ball #1 is returned to the bucket after it is removed? I can conclude that ball #1 is not in the bucket, just as surely as I can conclude that you do not have the apple I just took. Topology has nothing to do with it. One of the principles of ZF is the axiom schema of separation (sometimes stated as a separate axiom schema, sometimes derived from the more is a set and phi(x) is a predicate, then there is a set Y = { x in X : phi(x) }. For X substitute N, the set of natural numbers, and for phi(x) substitute ball x is removed from the bucket before noon and is not subsequently replaced. Then Y = N = the set of n such that ball n is not in the bucket at noon. No topology is involved. >Without > that unstated fact, or another similar unstated fact, the rules telling > what to do at any time before noon do not tell you how to pass to the > limit and say anything about the contents at noon. There is no limit to pass to. The rules, when applied to any single ball, are complete and unambiguous. If we know the location of each single ball at noon, then we know the location of the entire collection. All I am trying to demonstrate is that the sequence of balls in the bucket can reach different limits depending on the topology you use. The way the problem is stated, without a topology, there is no limit and hence we can say nothing about what happens at noon. >>I have never denied that applying different topologies will produce >>different results. In fact, that is an essential part of my argument. >>The fact that you can produce contradictory answers by using different >>topologies is evidence that at least some of those answers must have >>resulted from incorrect reasoning. > You have denied that topology has anything to do with the result; that > without resorting to topology, you can get your result. I deny that your chosen tool is applicable to the problem at hand, but I do not deny that the tool exists. Since the problem does not mention limits or continuity, it is a mistake to assume anything about them. The ZF approach avoids topology altogether. >The fact that > different results can be obtained by assumptions that do not contradict > the other conditions only means that the problem was ambiguous to start > with. > Given the premises of the problem as stated, it is neither provable nor > contradictory that the bucket be empty at noon. Which step in the ZF proof do you not accept? What is the smallest n such that ball n can be outside the bucket before noon and not subsequently moved, and yet be inside the bucket at noon? -- Dave Seaman Judge Yohn? mistakes revealed in Mumia Abu-Jamal ruling. = It is an unstated fact that if a ball is removed from the bucket before noon and is not subsequently returned, then that ball is not in the bucket at noon. This is so without regard to what may happen to other balls in the meantime. > That is precisely the assumption for pointwise convergence. If you take >> that as an unstated fact, then yes, you do get the limit you claim. It >> is the fact that that fact is not given that makes the problem ambiguous >> to start with. The same assumption operates if I ask, You have five apples. I take >away two. How many are left? Are you claiming that every first-grade >arithmetic problem is an exercise in topology? No, the apples problem is a two step problem. There is no infinite set of steps which requires a topology to say what happens in the limit. >The balls-in-the-bucket problem is no more ambiguous than the apples >problem. There may be more balls than there are apples, but if you focus >your attention on just one ball or one apple at a time, exactly the same >principle applies. The balls-in-the-bucket problem defines what happens at every finite step, but does not specify how to pass to the limit. Therein lies the ambiguity. Would you agree that a ball is in the bucket at noon if and only if that ball was put in at some time before noon and then never taken out after that; and that a ball is not in the bucket if it is either never put in the bucket or was taken out some time before noon and never put back in? What I have just described is precisely the assumption of pointwise convergence of the indicator functions of whether the balls are in the bucket or not. Perhaps this is the only thing that makes sense to you. In any case, under that assumption, I agree that there will be no balls in the bucket at noon (and I have said so before). The problem is that without this (or another) assumption, the state at noon is not derivable from the states before noon. Perhaps you think that this is part of the statement of the problem. If so, then I can see why you are so convinced that this is the only way to look at the problem. This seems to be our difference, that you assume this as given and I do not. If so, I have no disagreement with you over the outcome of the process; I just don? see this as necessarily an assumption for the problem. Since replacing that assumption yields differing results, I wanted to point out that the cause for the ambiguity noted by the OP is the choice of these assumptions, and that each assumption relates to its own topology. If you still think I am full of crap, then so be it. I don? think that there is much point to us arguing about this further. Are we agreed on this at least? Rob Johnson take out the trash before replying = >The problem really only has one answer, despite the fact that it >appears as a paradox. If we merely redescribe the same problem using >only mathematical and set terms, we get the answer right away. >Consider the following redescription: Let the set S0 = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 } {1}. Now >recursively define the set Sn as follows: Sn = Sn+1 U { 10n+1, 10n+2, ..., 10n+10 } {10n+1}. Let S = lim n->inf (Sn). > And this makes sense because the set limit exists, i.e., lim inf S_n = lim sup S_n = {}. (For readers unfamiliar with limits of sets, lim inf A_n = U{I{A_m: m >= n}: n = 1 to infty}, where I indicates intersection. For lim sup, interchange union and intersection.) get the listener to think about possible interpretations of abstract notions involving non-finite sets. To make matters more interesting, compare the original problem (with numbered balls) to one where the balls are unlabeled an indistinguishable. More precisely: Problem 1. At times 12 - 2^(1-n) (in hours) for n = 1, 2 ,..., you add ten identical balls to the basket, then remove one of balls in the basket. Problem 2 (LMW? original post). At times 12 - 2^(1-n) for n = 1, 2 ,..., you add ten balls, labeled 10n-9, 10n-8, ..., 10n, to the basket, then remove the ball with the least label. In each case, how many balls are in the basket at noon (= 12)? I first encountered this problem presented in these two cases. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu = > > Problem 1. At times 12 - 2^(1-n) (in hours) for n = 1, 2 ,..., you > add ten balls to the basket, then remove one of balls in the > basket. > Let? be more specific. When we remove a ball, we do so at random, with uniform distribution on the balls present, independent of past or future choices of which ball to remove. What is the probability that ball 1 is eventually removed? Ball 2? All of them? I?l get you started: probability that ball 1 is never removed is an infinite product, (9/10)(18/19)(27/28)..., evaluate this. So, as pointed out, we cannot say for sure how many balls are left. But we can provide probabilistic answers like: the expected number of balls left, the probability that ball 72 is left, etc. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ = > > Problem 1. At times 12 - 2^(1-n) (in hours) for n = 1, 2 ,..., you > add ten identical balls to the basket, then remove one of balls in the > basket. Which one to remove at each step will wildly affect the outcome of the result. You leave the problem undefine. Not saying which ones to remove in Problem #1 is like saying: Start with the set N of natural numbers. Now, remove an infinite number of them, I don? care which ones. How many do you have left? It certainly matters which ones you take out, as Infinity - Infinity is indeterminate. If I take out all of them and be left with an empty set; I could take out just the even ones and be left with an infinite set, or I can say take out all n > n0 (for some n0) and be left with a finite number. For Problem #1, if you are always removing the last ball, then you are left with an infinite number, as the set S = { 1, 2 ... 8, 11, 12 .. 18, ... }. If on the other hand, you remove the last ball for a finite number of steps, but the first one for an infinite number of steps, you end up with S being a finite set. Always removing the first one is essentially the original problem. Jonathan Hoyle Gene Codes Corporation = >>Problem 1. At times 12 - 2^(1-n) (in hours) for n = 1, 2 ,..., you >>add ten identical balls to the basket, then remove one of balls in the >>basket. >> >> Which one to remove at each step will wildly affect the outcome of the >result. You leave the problem undefine. Not saying which ones to remove in Problem #1 is like saying: Start >with the set N of natural numbers. Now, remove an infinite number of >them, I don? care which ones. How many do you have left? It >certainly matters which ones you take out, as Infinity - Infinity is >indeterminate. If I take out all of them and be left with an empty >set; I could take out just the even ones and be left with an infinite >set, or I can say take out all n > n0 (for some n0) and be left with a >finite number. For Problem #1, if you are always removing the last ball, then you are >left with an infinite number, as the set S = { 1, 2 ... 8, 11, 12 .. >18, ... }. If on the other hand, you remove the last ball for a >finite number of steps, but the first one for an infinite number of >steps, you end up with S being a finite set. Always removing the >first one is essentially the original problem. > No, you have a basket filled with some number of indistinguishable balls. At each epoch, you reach in and remove one. You can even shake them up first, if you like. I think would most say that, after the nth epoch, there are 9n indstigushable balls in the basket. At noon, there are infinitely many indistinguishable balls in the basket. Some other paradoxes (Unlike JH, I do not deny their exsitence.): Problem 3. Same set up as #2, but remove ball labeled 10n (rather than n) on nth step (the same as JH? last ball ). Most, as Jonathan, will say that there are infinitely balls at noon Why should the label of the specific ball removed make a difference? Problem 4. Labeled balls (as in #2 and #3). At step n, remove both balls labeled n and 10n. Switch the labels these balls, then return the ball newly labeled 10n back in the basket. Discard the ball newly labeled n. I think JH will insist there are no balls left in the basket at noon. But how does this differ from Problem 3? Problem 5. Labeled balls. At the nth epoch, you remove one of the remaining balls at random (uniformly amongst the remaining balls). In this case, my probability calculations say that, with probability 1, the basket is empty at noon. (My calculations may be erroneous - if you get a different answer, ask and I will share. Basically, the probability that any specific ball remains is 0. I don? *think* I have made an unjustifiable reordering of limits anywhere.) Yet how does this differ from the #1 (unlabeled balls)? As I have pointed out before, the very interesting subtlety here is dealing with the concept of infinity. I do not agree with JH? assessment that the answer to Problem 2 is so clear cut. Implicitly, one is dealing with modeling - i.e., the careful translation of a word problem into a well-defined mathematical structure. Or perhaps it is a matter of set theory: One can speak of a set of numbers, but not of a set of balls. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu = > No, you have a basket filled with some number of indistinguishable > balls. At each epoch, you reach in and remove one. You can even shake > them up first, if you like. I think would most say that, after the nth > epoch, there are 9n indstigushable balls in the basket. At noon, there > are infinitely many indistinguishable balls in the basket. If we agree to map this to Set Theory, you can? say these are indistinguishable elements. In Set Theory, each element is distinct and separate, and it absolutely matters which ones you remove. And just because the result is the same at any finite step, it absolutely matters which ones you remove at lim Sn. Calling them indistinguishable doesn? change this fact. Constantly removing the ones that first went in (labelled or not) at each step will leave you with an empty set; constantly removing the last one in at each step leaves you with an infinite number. A much more simplified example of your paradox is to start with the set of natural numbers and remove one at each step. Well, it DEPENDS on which ones you remove! Removing n at step n leaves you with the empty set. Removing element 2n at step n leaves you with an infinite sized set. If you do not specify which ones you take out, you cannot determine the answer. That doesn? make your story a paradox; it merely makes it indeterminant. > Problem 3. Same set up as #2, but remove ball labeled 10n (rather than > n) on nth step (the same as JH? last ball ). Most, as Jonathan, will > say that there are infinitely balls at noon Why should the label of the > specific ball removed make a difference? The label is not the issue. The label is merely a convenient way of distinguishing them. I suppose they could be distinguished by color, weight, set of molecules that make them, whatever. But if they are not the same ball, they are distinct. And the solution to the problem (very simply!) lies in knowing: for each ball (labelled or not), if there exists a step n in which it is removed. > Problem 4. Labeled balls (as in #2 and #3). At step n, remove both > balls labeled n and 10n. Switch the labels these balls, then return > the ball newly labeled 10n back in the basket. Discard the ball newly > labeled n. I think JH will insist there are no balls left in the basket > at noon. But how does this differ from Problem 3? In problem #4, a ball initially labelled n will eventually be removed, relabelled 10n and put back. However, every integer, no matter how many times you increase it tenfold, remains finite. There are an infinite number of times each ball is removed and an infinite number of times it is put back (albeit with a new label)...a classic infinity minus infinity problem. The solution, as before, is noting that any non-empty collection of natural numbers has a least element. If a ball exists in set S = lim Sn, it must have a label, and that label must have a positive finite number. But Label n was removed at step n, and this is true with every n. Therefore, the set of integers on the remaining labels has no smallest value. So there are no labels, and thus no balls. The set is empty. > Problem 5. Labeled balls. At the nth epoch, you remove one of the > remaining balls at random (uniformly amongst the remaining balls). In > this case, my probability calculations say that, with probability 1, the > basket is empty at noon. (My calculations may be erroneous - if you get > a different answer, ask and I will share. Basically, the probability > that any specific ball remains is 0. I don? *think* I have made an > unjustifiable reordering of limits anywhere.) Yet how does this differ > from the #1 (unlabeled balls)? I have not really examined this case, but I would agree. For example, the probability that Ball #1 would be removed in Step 1 is 1/10, in Step 2 is 1/19, in Step n is 1/(9n+1), the sum of these independent trials appears to be 1. It differs from the unlabelled balls example because you have specified a rule to determine which ball to remove. Given the rule for each step n, we can determine a solution (even if it? only a probability distribution of answers). > As I have pointed out before, the very interesting subtlety here is > dealing with the concept of infinity. I do not agree with JH? > assessment that the answer to Problem 2 is so clear cut. Implicitly, > one is dealing with modeling - i.e., the careful translation of a word > problem into a well-defined mathematical structure. Or perhaps it is a > matter of set theory: One can speak of a set of numbers, but not of a > set of balls. The subtley here I think is not infinity at all, but your concept of indistinguishable elements. This concept does not map onto Set Theory, and therefore this is where I see our disjunction of understanding comes from. If you do not map this problem to Set Theory, I would be at a loss as to how you can model it. Physics will not allow acts which are performed faster than light, so it? not even possible to describe. It seems to me that this problem has a solution only within the field of mathematics, where infinity (particular within sets) is very well defined. Jonathan Hoyle Gene Codes Corporation = > Some other paradoxes (Unlike JH, I do not deny their exsitence.): If you move the same object infinitely many times, its final state may not be well defined. That is not a problem with the original version, where each ball is moved exactly twice. -- Dave Seaman Judge Yohn? mistakes revealed in Mumia Abu-Jamal ruling. = > Let a_n = n*Pi^2/6 - sum(1/(k*C(n+k-1,n)),k=1..infinity), > > where C(n+k-1,n) is binomial coefficient n+k-1 take n . > > Then {a_n} is a sequence of rational numbers which starts > out {0,2,15/4,49/9,1025/144,5269/600,...}. > > Is there a formula, perhaps involving Bernoulli or Stirling > numbers for a_n? > Jim Buddenhagen Let? note H_m(n)= sum_{k=1}^n 1/k^m then a_n= n*H_2(n-1) for n>0 Hoping this helped, Raymond = > >Let a_n = n*Pi^2/6 - sum(1/(k*C(n+k-1,n)),k=1..infinity), where C(n+k-1,n) is binomial coefficient n+k-1 take n . Then {a_n} is a sequence of rational numbers which starts >out {0,2,15/4,49/9,1025/144,5269/600,...}. Is there a formula, perhaps involving Bernoulli or Stirling >numbers for a_n? >Jim Buddenhagen > Let? note H_m(n)= sum_{k=1}^n 1/k^m then a_n= n*H_2(n-1) for n>0 Hoping this helped, > Raymond It helped a lot! I think you said about all one can about this. Jim Buddenhagen -- To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE = > I know guys like Abel and Galois did their best work early and then died, > but can people give me some examples of mathematicians who did their best > work in their early 20s (say), then continued to work for a long time but Speaking stricly for myself, I did my most original work at age 30 and the second most at age 36, 30 years ago. I have remained (and even remain) active, but nothing like those two works. What I am doing now is best described as being of marginal interest. = > Speaking stricly for myself, I did my most original work at age 30 and > the second most at age 36, 30 years ago. I have remained (and even > remain) active, but nothing like those two works. What I am doing now > is best described as being of marginal interest. Sort of like Fermat. -- = I noticed this pattern or series and implemented in TP7: {$N+} {KBH code} Var mp, p: extended; Begin WriteLn; p:= 2; While (mp < 99999999999999999999999999999999999999.0) Do Begin mp:= Exp(Ln(2) * p) - 1; WriteLn(mp:30:0); p:= mp; End; ReadLn; End. The first four outputs are Mersenne Primes... The obvious question is whether or not the fifth output would be prime...That? (2^((2^127) - 1)) - 1 . = > The obvious question is whether or not the fifth output would be > prime...That? (2^((2^127) - 1)) - 1 . A well known question. See http://primes.utm.edu/mersenne/index.html near the bottom. --- J K Haugland http://www.neutreeko.com = > I noticed this pattern or series and implemented in TP7: > {$N+} > {KBH code} > Var > mp, p: extended; > Begin > WriteLn; > p:= 2; > While (mp < 1.7e+38) Do > Begin > mp:= Exp(Ln(2) * p) - 1; > WriteLn(mp:30:0); > p:= mp; > End; > ReadLn; > End. The first four outputs are Mersenne Primes... > The obvious question is whether or not the fifth output would be > prime...That? (2^((2^127) - 1)) - 1 . > Oh, I see at a popular web reference source the subject of double-Mersenne . It lists MM2, MM3, MM5, and MM7 all of the form of 2^((2^n) - 1)) - 1 where n is the numerical part of MMn. But that? kinda difficult because it? not clear what is being paired to what. I would do it like this: 2^2 - 1 = 3 2^3 - 1 = 7 2^3 - 1 = 7 2^7 - 1 = 127 2^5 - 1 = 31 2^31 - 1 = 2147483647 2^7 - 1 = 127 2^127 - 1 = Mp And the reference source says that 2^((2^31) - 1) - 1 is composite. So basically I suggested that in the case of a double-Mersenne to look for a triple or in the case of a triple to look for a quad...Or in the case of a quad to look on further. = > I noticed this pattern or series and implemented in TP7: {$N+} {KBH code} Var > mp, p: extended; Begin > WriteLn; > p:= 2; > While (mp < 1.7e+38) Do {Numerical limit not required in theory} > Begin > mp:= Exp(Ln(2) * p) - 1; > WriteLn(mp:30:0); > p:= mp; > End; > ReadLn; > End. The first four outputs are Mersenne Primes... The obvious question is whether or not the fifth output would be > prime...That? 2^((2^127) - 1) - 1 . > Oh, the subject title in relating to the subject should say Four of the known Mersenne Primes instead of Four known Mersenne Primes . Suppose, this series pattern were to start at p = 5 instead of p = 2. (2^5) - 1 = 31 which is a Mersenne Prime and (2^31) - 1 = 2147483647 which is a Mersenne Prime. So here the question is whether or not (2^2147483647) - 1 is prime or not. That? an exponent about 100 times larger than the exponent of the largest known Mersenne Prime and I don? know what the computational demand would be for that... = > I mean the second possibility. This space is compact with Hausdorff > metric. > I consruct a finite dimensional measure such that mu(A) = mu( cup > A_i) with A = cup A_i. Then apply the extension theorem. Does it work > ? There are lots of probability measures on the space of open or closed subsets of [0,1]. I don? think there? any particularly distinguished one that deserves to be called uniform . Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 >> Is it true that there exists a (uniform) probability measure on the >> space of open or closed subsets of [0,1] ? At least two possibilities come to mind: 1. Is there a uniform probability measure on [0,1] that assigns a > measure to all open sets? Or one that does so to all closed sets? 2. Is there a measure on the space of open subsets of [0,1] that is > uniform ? Dale = CLUE posted immediately below original post. > Let c(1,k) = 1, for all positive integers k. > > For all m >= 2, > > Let c(m,k) = --- > - ln(k) > ------- > c(j,k) > ln(m) / > --- > j|m > j > (which, in linear-mode, is...) > c(m,k) = > > -(ln(k)/ln(m)) sum{j|m, j (the sum is over the positive PROPER divisors, j, of m.) > > So, let, for every real x > 1, f(x) = oo > --- --- > > > ( > c(k,m/k) ) /m^x > / / > --- --- > m=1 k|m > (which in linear mode is...) > > f(x) = > > sum{m=1 to oo} (sum{k|m} c(k,m/k) ) /m^x. > (Inner-sum is over all positive divisors, k, of m) > > > Puzzle: > > What is a closed-form (in terms of known functions) for f(x)?? > function, but in a very unexpected way (in my opinion). Leroy Quet = > ... >of solutions for n=3...7; some solution times in seconds; >and some solution-sizes, ie, number of rectangles in solutions: >n #sols time Solution-sizes > ... >8 >2330 >5400 14 16-27 >(n=8 has run about 1.5 hours so far, on my 450MHz Athlon) > > n=8 finished this morning, after 48.5 cpu hours (1000 times longer > than at n=7), with 3434 solutions. I don? plan to devote the years > of computer time that exhaustive searches for n>8 would take; however, > I ran n=9 to 12 for a few seconds each with following results: > n Smallest-solution-size-seen > Run-time and sol-count in that time > 9 9 (10s, 91) > 10 14 (10s, 34) > 11 19 (10s, 12) > 12 16 (70s, 46) > I also tested 13 to 20 at about 15 minutes each but got > no solutions. > > Here? the reported 9-cover at n=9 -- > Cells Prime Corner Form > 18 743 0 6 2x9 > 18 739 0 4 2x9 > 14 577 1 2 2x7 > 10 223 0 0 2x5 > 9 373 0 8 1x9 > 6 401 6 0 2x3 > 2 151 8 2 2x1 > 2 107 5 0 2x1 > 2 7 0 2 2x1 > (Corner = top left corner? row and column numbers) > -jiw Oh, wow... Perhaps, if anyone is interested in persuing my OB puzzle (reposted below), an n of > 9 or 10 would make things much more interesting therefore!... > OB puzzle (that I have not solved myself): > > For the n= 5 case, place the integers 1 through 25 > into the grid AS YOU WISH so as to: > 1) maximize the number of prime-rectangles you can > subdivide the square into. > 2) minimize the number of prime-rectangles you can > subdivide the square into. > > If there is no obvious way to PROVE the number of primes is > minimal/maximal, then we might as well make a contest out > of this for sci.math/rec.puzzles Leroy Quet = So, if a bubble of vacuum decay (aka: decay of the false vacuum ) is coming at the Earth from outer space at light speed, we would not know this until too late, of course. As a matter of fact, the entire universe could be a Swiss cheese of giant expanding VDk voids, and we would be...oblivious. For we could not observe any of them until one hit us, because of the fact they are expanding at light speed. (I guess) So, if vacuum decay is actually existent, then, from the theories of such, how common are the vacuum-decay bubbles? What I am really asking is... what is the likelihood (as conjecured) that the Earth will be hit by a bubble in, say, the next year? (I am asking for the likelihood based upon the number of conjectured VDk bubbles {and their expected sizes as of now} which would exist in our area of the universe, if any Quet = I? going to go out on a limb and claim that the chance is zero. However, please inform me if one does hit Earth. I will eat my own words. ~ Chris = > What I am really asking is... what is the likelihood (as conjecured) > that the Earth will be hit by a bubble in, say, the next year? Presumably less than 1/ten billion, since it hasn? happened yet, and we?e half way through such an interval. xanthian. -- = > I am reposting this from March: threadm=b4be2fdf. reposting because, given that my Mathematica is busted, I was > wondering if someone could calculate numerically 1 of the 2 constants > I describe below. > (If the sums converge, which they do if the world is not insane...) > > Are either of the constants recognized by the Plouffe Inverter? > > I am also wondering again if the sequence mod 2 is in the EIS as what Let A(1) = 1, a sequence with only a single element. > > >Let A(m+1) = > >{A(m), a(m)+1, A(m)}, > > >a concatenation, where a(m) is the m_th term of the sequence A(m). > > >( A(m) has 2^m -1 elements. So a(m) is not the last element of A(m) for m >= 2, it should be emphasized.) > > >The sequence begins: > 1,2,1,3,1,2,1,2,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,2,1,2,1,3,1,2,1 ,... > > > >Noteworthy: > > >The average of the terms approaches: > > >1 + sum{k=1 to oo} a(k)/2^(k+1) > > >Again, a(k) is the k_th term of the original sequence, the sequence whose terms we are calculating the average of the terms of... > > >A closed-form for this sequence is, I think: > > >If e(m) is a nonnegative integer such that: > >2^e(m) is the highest power of 2 dividing m; > > >then: > > >a(m) is the number of e()? where: > > >e(e(e(...e(m)...))) = 0. > > >-- > >It might be more natural, for some purposes, to define A(1) = 0. Then we will have the same sequence minus 1. > > >(In this case, a(m) would be the number of e()? needed to achieve an odd integer, instead of 0.) > > >And, in this case, the average of the terms will by 1 less, but so > >will the a()?. So the average, with a?k) = a(k)-1, a(k) = original > >a(k): > 1/2 + sum{k=1 to oo} a?k)/2^(k+1) > > >-- > > >I THINK that the a(k)-sequence, taken mod 2, might be likely to be > the absolute values of the first differences of the Thue-Morris > sequence > >according to the OEIS. (paraphrased title??) > > PS: the sums are: > > 1 + sum{k=1 to oo} a(k)/2^(k+1) > > and/or > > 1/2 + sum{k=1 to oo} a?k)/2^(k+1), > which are the differing by exactly 1. I have some more to say about this sequence which is not necessarily constant-related. (But perhaps this thread? name is still apt, given how the constant can be derived from the sum including the terms themselves which are to be averaged.) A Dirichlet/power-series sum relation: {a(k)} is as defined above. Where the sums both converge: (sum{m=1 to oo} a(m) /m^x ) / zeta(x) = 1 + (1 - Quet = > Awhile back I posted this linier equation --- > > e^2 - (1/(n-2))/(1/n) = 0 > > I had no closed form for n at the time. > > The closed form for (n) in this case I just discovered --- > > n = (1/((1/((1/(e^2 - 7))*2))+3))+2 where n satisfies this equation > --- > > e^2 - (1/(n-2))/(1/n) = 0 where n = 2.3130352854993313036361612469... > > Where the cf of n = [2:3,5,7,9,11,13,15,17,19,21,23,25,27,...] > > This cf contains all the odd integers ---->oo after even integer 2. As you will note, the last two terms on the right in the closed form > for (n) are the first two terms in its cf in reverse order. > > This cf. is a simple ordered progression of an irrational (n) even > more significant than e? cf. because of its simplicity. > > Where the cf. of e = [2:1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1...] > > Strange how (e^2) has no order in its cf but the (n) in my equation > has. > > This constant (n) probably disserves a spot in OEIS because of its > well ordered continued fraction derived from e^2. > > Dan Experimenting further with pi using the same linier equation --- pi^2 - (1/(n-2))/(1/n) = 0 The closed form for (n) in the case for pi^2 --- n = (1/((1/((1/(pi^2 - 9))*2))+4))+2 n = 2.225489199919036005327825502278560437281694... Cf for n = [2:4,2,3,2,1,95,1,3,1,3,1,5,17,1,21,...] Again note, the last two terms on the right in the closed form for (n) reversed, are the first two terms in its cf. Pi^2 (n) cf is understandable, like pi has mixed integers in its cf. with no apparent order. For the golden ratio (phi)--- phi^2 - (1/(n-2))/(1/n) = 0 Closed form for phi^2 (n)--- n = (1/((1/((1/(phi^2 -2))*2))+.5))+2 n = sqrt(5)+1 cf for n = [3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4...] which becomes the fourth metal mean where the first metal mean is phi! The metal means along with the golden mean have the simpliest order of all in there infinite cf?. Dan = >Awhile back I posted this linier equation --- e^2 - (1/(n-2))/(1/n) = 0 Why the obfuscation of (1/(n-2))/(1/n) rather than n/(n-2) ? Phil -- Unpatched IE vulnerability: Web Archive buffer over?w Description: Possible automated code execution. Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0303/107.html = > >> Awhile back I posted this linier equation --- >> >> e^2 - (1/(n-2))/(1/n) = 0 > Why the obfuscation of (1/(n-2))/(1/n) rather than n/(n-2) ? > > Phil Phil, You are right! It is confusing even though it is the same thing but redundant. Dan = Whoops! typo JS: I am confident that the saucers you and I are both keenly interested in, if The Truth be known ;-), ? by the approach you take here. Should be: JS: I am confident that the saucers you and I are both keenly interested in, if The Truth be known ;-), DO NOT ? by the approach you take here. = Here is another very elegant solution to B6, due to JOEL ZINN, from Texas A&M University: Use the identity: |z|=(constant) int_0^ infty Re(1-e^(itz))/t^2 dt (the constant equals pi/2 I think but here it doesn? matter anyway) Replace z by f(x)+f(y) and f(x) respectively, integrate, use Fubini and compare the integrands. Enjoy !!! Ciprian Pop = > Here is another very elegant solution to B6, due to JOEL ZINN, from > Texas A&M University: > Use the identity: > > |z|=(constant) int_0^ infty Re(1-e^(itz))/t^2 dt > > (the constant equals pi/2 I think but here it doesn? matter anyway) > Replace z by f(x)+f(y) and f(x) respectively, integrate, use Fubini and > compare the integrands. Could you elaborate on that last step ( compare the integrands )? -- A. = Here is another very elegant solution to B6, due to JOEL ZINN, from >Texas A&M University: >Use the identity: |z|=(constant) int_0^ infty Re(1-e^(itz))/t^2 dt (the constant equals pi/2 I think but here it doesn? matter anyway) >Replace z by f(x)+f(y) and f(x) respectively, integrate, use Fubini and >compare the integrands. Well, the last time I didn? follow something you suggested I was applying Fubini to the wrong double integral. So maybe I? just misinterpreting this one as well. But: It _seems_ as though the idea is to show that (*) int_0^1 int_0^1 (1-cos((f(x)+f(y))t)) dx dy >= int_0^1 (1-cos(f(x)t)) dx. That would prove the result, but (*) is not true. (For example, say f(x) = 1 for all x and t = pi.) I just have it all wrong again, (*) is not supposed to have anything to do with it, right? >Enjoy !!! Ciprian Pop ************************ David C. Ullrich = So, if a bubble of vacuum decay (aka: decay of the false vacuum ) is >coming at the Earth from outer space at light speed, we would not know >this until too late, of course. As a matter of fact, the entire universe could be a Swiss cheese of >giant expanding VDk voids, and we would be...oblivious. For we could >not observe any of them until one hit us, because of the fact they are >expanding at light speed. [snip - more crap] This isn? a relevant subject for either sci.math or rec.puzzles... adam = > > >So, if a bubble of vacuum decay (aka: decay of the false vacuum ) is coming at the Earth from outer space at light speed, we would not know > >this until too late, of course. > > >As a matter of fact, the entire universe could be a Swiss cheese of > >giant expanding VDk voids, and we would be...oblivious. For we could > >not observe any of them until one hit us, because of the fact they are > >expanding at light speed. > [snip - more crap] > > This isn? a relevant subject for either sci.math or rec.puzzles... Whether or not the hypothesis is preposterous, the question asked is perfectly serious. Given the distributions of radii, velocities and rates of expansion, he asks a perfectly valid mathematical question. So get off your high horse. You?e a SELF-APPOINTED censor, who has contributed nothing to the newsgroup, so far as I can see, as opposed to LeRoy, who is a regular and frequently asks valid (if usually less quixotic questions. --Ron Bruck = Its messy, but You solve the curve to get y1=f(x),y2 =f(x) and the limits of x Then bung them into the double integral integrating over y from y1 to y2 >then over x. > That is not the easy way if you have had polar coordinates. Put x-1=rcos(t), y+2=rsin(t) and r dr dt for dydx. You get int_0^2Pi int_0^1 7(1+rcos(t))r dr dt which is easy. --Lynn = > > >Perhaps this is a counter example? There are no counterexamples - any monotonic function is > in fact differentiable except on a set of measure zero. > (See any book on real analysis, meaning a book that > starts with measure theory...) >Define: > f(0)=0 > >f(x+1)=(f(x)+1)/2 > >f(1/x)=1-f(x) > > >This defines f(x) for rational x>=0. The irrationals can be > >filled in using limits to define a continuous monotonic function > >which has derivative equal to zero at all rationals as well as any other point at which the function is differentiable. I don? follow your definition of f. But there are in fact > non-constant strictly increasing functions that have > derivative equal to 0 almost everywhere; how does that > give a counterexample? ************************ David C. Ullrich Sorry about the hasty post. I can see what you mean about this function. It has slope equal to zero almost everywhere. Its inverse has slope equal to zero almost everywhere also. It seems as if it is made of infinitesimal stairsteps. How about stretching it out by defining g(x+f(x))=f(x). It seems to me that g(x) would have slopes of 0 and 1 densely mixed together. Over a given interval the average slope can be computed, but when decreasing the interval to find the derivative, which of the slopes, 0 or 1, would in general never be known for sure. This doesn? seem like genuine convergence of the limit towards the derivative. = >Perhaps this is a counter example? >> >> There are no counterexamples - any monotonic function is >> in fact differentiable except on a set of measure zero. >> (See any book on real analysis, meaning a book that >> starts with measure theory...) >> >> >Define: >>f(0)=0 >> >f(x+1)=(f(x)+1)/2 >> >f(1/x)=1-f(x) >This defines f(x) for rational x>=0. The irrationals can be >filled in using limits to define a continuous monotonic function >> >which has derivative equal to zero at all rationals as well as any >> >other point at which the function is differentiable. >> >> I don? follow your definition of f. But there are in fact >> non-constant strictly increasing functions that have >> derivative equal to 0 almost everywhere; how does that >> give a counterexample? >> >> ************************ >> >> David C. Ullrich >Sorry about the hasty post. I can see what you mean about this function. It has slope equal to >zero almost everywhere. Its inverse has slope equal to zero almost >everywhere also. It seems as if it is made of infinitesimal >stairsteps. How about stretching it out by defining g(x+f(x))=f(x). It seems >to me that g(x) would have slopes of 0 and 1 densely mixed together. >Over a given interval the average slope can be computed, but when >decreasing the interval to find the derivative, which of the slopes, >0 or 1, would in general never be known for sure. This doesn? seem >like genuine convergence of the limit towards the derivative. I? not sure exactly what you?e trying to construct here. But if you?e trying to construct a monotone function which is not differentiable almost everywhere you can stop now - that? impossible. This is a well-known theorem. Honest - see any book on real analysis. (As I think I mentioned before, there are lots of books with the words real analysis in the title that are too elementary to include this result. See a book on real analysis that develops the theory of the Lebesgue integral, aka measure theory . Look for a thing called the Lebesgue Differentiation Theorem.) ************************ David C. Ullrich = > > I want to find the variable P such that P will satisfy the following > two conditions: > 1.) P = x*u > 2.) P = y*g + 1. Given that x and y are known. There are a lot of values for P for > which the following will hold. I believe the problem can be solved using the Chinese Remainder Theorem. Type Chinese Remainder Theorem in Google search and you will have an extensive list of URLs that describe this elegant method. I have the code in Matlab, and if you like, I can send it to you. = > I believe the problem can be solved using the Chinese Remainder > Theorem. Type Chinese Remainder Theorem in Google search and you > will have an extensive list of URLs that describe this elegant method. > I have the code in Matlab, and if you like, I can send it to you. To actually solve the problem using the CRT, you must first get the inverses 1/x mod y and 1/y mod, which pretty much solves this problem directly. The CRT does show that P is unique mod xy, though, as well as how to get all the other solutions, and is generally useful in a great many situations that arise when you?e doing this sort of work. = > Cute. When the results are presented in the form box, they end with some raw html code. Oh yes, make the lines longer, it?l be easier to read. = > > > Cute. When the results are presented in the form box, they end with some > raw html code. Oh yes, make the lines longer, it?l be easier to read. Well if you read above the box it should print the results in long-line format. It? in the box so that you can copy the results to your webpage and have it look nice. The one above the box doesn? have any HTML though. Did you get L^infinity[0,1] by any chance? That one has the best description. = > forgot what the little numbers at the top and botom of the integral sign are > called. > > One of my problems has a triangle with verticies at (0,0), (1,1), and (0,1) > and I had to do something to it to end up with this: [int from 0 to 1] {[integral from 0 to y] x dx} dy > > My book showed it to me this way and I figured why couldn? I switch the > order of integration: [int from 0 to 1] {[int from x to 1] x dy} dx > > I solved both and the answers matched (I hope I integrated correctly). This > was easy because the shape I was integrating over was simple and had simple > numbers. My question is do you have any hints/tips/sites that show how to > determine the correct numbers/variables in the integrand. Maybe this example will help a little: The region will be the triangle formed by y = x, y = 2 - x and the y-axis. The shadow (projection) of the region on the x-axis is the interval [0, 1]. For each x in that interval, the vertical line through x enters the region at y = x and exits it at y = x - 2. In other words, the segment inside the region goes from y = x to y = x - 2. So the innermost integral will be int(f(x, y),y=x..2 -x). The possibilities for x are 0 <= x <= 1 so the integral is int(int(f(x, y),y=x..2 - x),x = 0..1). On the other hand, the projection of the region on the y-axis is the interval [0, 2] For any y in that interval the horizontal line through y enters the region at x = 0 and exits it at ... hmmm. For 0 <= y <= 1 it leaves at x = y. For 1 <= y <= 2 it exits at x = 2 - y. So, our integral is int(int(f(x, y),x = 0..y),y = 0..1) + int(int(f(x, y),x = 0..2 - y),y=1..2). -- http://www.crbond.com = > http://www.crbond.com |Pretty much universally, those people that cranks like to cite as people who |had ideas that some thought were crackpot ideas and turned out to be |revolutionaries were experts in the fields that they revolutionized. I don? know whether cranks often cite him, but Wegener and continental drift is a good example of someone who was widely regarded as wrong. The book _Great Feuds in Science_ has chapters titled things like Pope Urban VIII versus Galileo , Newton versus Leibniz and so on, but the chapter on this episode is titled Wegener versus Everybody . :-) And Wegener won! But apparently it was generally agreed that the evidence he cited was actual evidence. The idea that continents moved just wasn? believed to be the correct explanation until some kind of plausible mechanism was found. On the face of it, continents pushing through rock just sounds implausible. Keith Ramsay = KRamsay says... >I don? know whether cranks often cite him, but Wegener and continental >drift is a good example of someone who was widely regarded as wrong. >The book _Great Feuds in Science_ has chapters titled things like > Pope Urban VIII versus Galileo , Newton versus Leibniz and so on, >but the chapter on this episode is titled Wegener versus Everybody . :-) >And Wegener won! But apparently it was generally agreed that the evidence he cited was >actual evidence. The idea that continents moved just wasn? believed >to be the correct explanation until some kind of plausible mechanism >was found. On the face of it, continents pushing through rock just >sounds implausible. Another idea that has moved from crackpot territory to acceptance is the idea that the dinosaurs were killed by a planet-wide disaster (collision of the earth with a comet, asteroid, or large meteor). In general, paleontologists have been reluctant to consider catastrophes as the explanation for extinctions---they prefer gradual, natural-selection type explanations. -- Daryl McCullough Ithaca, NY = |Another idea that has moved from crackpot territory to acceptance |is the idea that the dinosaurs were killed by a planet-wide disaster |(collision of the earth with a comet, asteroid, or large meteor). |In general, paleontologists have been reluctant to consider catastrophes |as the explanation for extinctions---they prefer gradual, natural-selection |type explanations. Was it ever crackpot and not just speculative? Keith Ramsay = |> |> >Basically my crime as far as math people are concerned is that I |started posting several years back ideas that I thought might solve |> >Fermat? Last Theorem. A couple of times I became very excited when I |> >thought I? actually solved it. A couple of times ? Trying to minimize this kind of thing later on just looks silly. If you really don? remember, go look up your old postings on Google (probably [tm]). |> Maybe the problem is that people tried to help you, putting in time |> them. | |No maybe about it, that? not what happened. | |I had an idea, maybe wacky, that I could find a simple and short proof |of Fermat? Last Theorem, and when I started posting, the insults |started ?ing! the discussion. I was a bit surprised at how little hostility there was. Compare Dik Winter? remarks last year in . On April 12 of that first year, 1996, roughly a month and a half after starting, you apologized and said you would go away. Then on July 21 you apologized again and said you would go away: |I realized just now just how seriously I abused the helpfulness and |kindness of people from this newsgroup and at colleges and universities various times for being so helpful. In these two apologies in particular, you thanked not just a few individuals, but all of sci.math as a group. How do you square this with your recollection of insults ?ing when you started posting ? It? true Matthew Wiener called you a retard when you asked for him not to respond to you, but that seems to have been an isolated incident. He? also especially prone to write like that. [...] |In all that noise there were occasionally a few posters who would |sometimes post nicely, but that was always temporary. No, it went on for a long time to begin with, and even during some of the harsher discussions there have been some people (e.g. Kevin Brown) who were consistently nice , for months at a time, until they no longer wanted to spare the time to continue the discussion (whereupon they disappeared nicely). In the first year, it appears you complained much more about not getting enough responses, rather than about getting unpleasant ones. Keith Ramsay = Can anyone please recommend a good 3d graphing software. Steven = You can see my site done with this software. http://perso.wanadoo.fr/roger.assouly/ roger.assouly@wanadoo.fr lodCb.6430$Nq2.186@news02.roc.ny... > Can anyone please recommend a good 3d graphing software. > Steven = > Can anyone please recommend a good 3d graphing software. > Steven GnuPLot. There? a tutorial here: http://www.cs.uni.edu/Help/gnuplot/ = >> The Prophet Nobody known to the wise as toolshed37@yahoo.com, opened >> the Book of Words, and read unto the people: >> >Why are people so resistant to answering questions about GRE scores? >> >Of all the times I?e asked this question, I?e never once gotten a >> >number back as an answer. Even my advisor was like Don? worry so >> >much about GRE scores. Just worry about doing your best . Well if >> >> People don? answer it because, frankly, it? an ill-formed question. >> GRE scores _don?_ get you into good universities. Good >> recommendations/research history gets you into good universities. You >> need at least a good GRE to get into a mediocre university, but even a >> great GRE won? get you into a great university. The subject GRE covers >> competence, not excellence; a GRE score above a certain level >> guarantees you know a good survey of undergrad math -- knowing more >> above that simply means you have a good memory. The general GRE may >> honestly be of more value to math programs, since the main thing is not >> so much what you know as how you think, and the general is honestly a >> better predictor of that. >> >> But, y?now, don? let GRE scores bother you. If you think they?e >> important, study hard. Apply everywhere you? apply anyways. I don? >> see how knowing the relationship between GREs and acceptance (one which >> is actually tenuous, as I say above) helps you in any way. >> >know GRE scores aren? the final word on everything, but I hope that >> >for once someone can finally give me a straight answer, and hopefully >> >respond with 3 numbers/intervals corresponding to the above 3 ranges >> >of schools. >> >> The numbers/intervals you?e asking for don? exist. If it would make >> you happy, I could say ?00-1000?for all three (I? guessing at the >> lower bound at which ?ompetence?is assessed). > > So with a GRE score of 600 you? get into MIT? Wrong. What about with > a GRE score of 700? Wrong. What do you mean by you? get into MIT? If you mean every student with a 600 would get into MIT, then no. The same holds for 990. If you mean its possible to get into MIT with a 600, then yes. Having a too low GRE wont exclude you because there is no such thing as too low . Being low in one area can be made up for in another. I? willing to bet that there is at least 1 student in recent history at MIT with a 600 GRE. > People seem to confuse my question with what scores will get you into a > good university? However, nowhere did I ask that question. I said > what scores are good if you want to go to such and such university? > Hence a score which would cause the reviewer to throw away your > application would not be considered good. No such score. > So the correct response > should be an interval which cause the reviewer to probably not throw > away your application. These intervals do exist, as a few people have > pointed out that. For example, if you want to go to a top 10 university > your GRE score should be in the top 95 percentile or so. Yes, I > understand that doesn? mean anything other than getting in the top 95 > percentile will cause them to not eliminate you. But after all, that? > what I was asking about. Dont be so hostile if your questions have no answers. Graduate admissions isnt a scientific process. Its mostly holistic. The entire picture is considered, rather than any obsession over a low GRE. Shaun = this isn? an answer to your question, but related. when exactly should one take the gre? 3rd year? 4th year? and are you allowed to take it multiple times and choose which scores are sent to universities? if there is an faq about this kind of stuff you can just point me there instead of good is fairly arbitrary, but if we > consider the three cases: > > 1) Want to get into a top 10 school > 2) Want to get into a top 20 school > 3) Want to get into an upper Tier 2 school > Then for each of the above, what would be the subject test > score/percentile that one would recommend? = |I was wondering about the following equation: |a^b == c (mod n) {for 0 <= a <= n-1} |For a given value of _b_, is there a heuristic that can give the |max/min/average of how many unique values of _c_ there are in terms of |n? (I? most interested in the average, but max and min might be nice |too) Part A. A general property of f. -------------------------------- Given b, let f(n) be the number of c mod n that are b-th powers mod n. f(n) is what is known as a multiplicative function. A function f(n) is called multiplicative if for any relatively prime m and n, we have f(mn)=f(m)f(n). (If f(mn)=f(m)f(n) for all integers m and n, then f is called totally multiplicative.) Each number n can be written uniquely as a product of powers of distinct primes (greater than 1), and the value of f(n) is the product of the value of f at those prime powers. Part B. The value of f at primes. --------------------------------- The value at primes is the first step toward figuring out the value at the prime powers: |I know that if _n_ is prime and _b_ is 2 then the max&min&avg is about |1/2. [that is, ~(1/2)*n] However, I also want to consider composites. |And when b=2, it looks like the avg ~ 1/4. [ie, ~(1/4)*n] However, I |didn? look at very many values of n so this may go up or down. When b=2 and p is an odd prime, then f(p)=(p+1)/2. If p and q are distinct odd primes, then f(pq)=f(p)f(q)=(p+1)(q+1)/4 which as you observe is on the order of pq/4. There? some elementary number theory which you might find interesting here. If p is a prime, then the congruences classes mod p besides 0 form a group under multiplication. It? always a cyclic group with p-1 elements. Now by some basic group theory, the number of elements of a cyclic group of order e that are b-th powers is e/d where d is the greatest common divisor of e and b. So the number of b-th powers is (p-1)/d where d is the greatest common divisor of b with p-1, plus 1 for the congruence class 0 mod p. If you haven? seen anything like this before, this is not supposed to be entirely obvious! I don? know how much number theory or algebra you know already, but it? all the kind of thing you can learn from books on elementary number theory and algebra. Try some examples and it should help you see how it goes. Let b=3. Modulo 11 the nonzero classes are powers of 2: 1 2 4 8 5 10 9 7 3 6 (One says that 2 is a primitive root mod 11.) If we take the cubes of each of these, they are the powers of 8, which go around the cycle in steps of three: 1 8 9 6 4 10 3 2 5 7 But since 3 is relatively prime to 10, this hits everything. On the other hand, modulo 13 the nonzero classes are powers of 2 again: 1 2 4 8 3 6 12 11 9 5 10 7 but because 13-1 is divisible by 3, the cubes hit only every third one before returning to 1: 1 8 12 5 which together with 0 gives f(13)=(13-1)/3+1=5. Part C. The value of f at powers of primes. ------------------------------------------- Determining the value of f for powers of primes is a bit trickier. The values c not divisible by p are a cyclic group again when p is not 2. There are (p-1)*p^{k-1} such congruence classes, so the number which are b-th powers are (p-1)*p^{k-1}/d where d is the greatest common divisor of (p-1)*p^{k-1} with b. When p does not divide b (and only a few primes p divide b), the common factor between b and (p-1)*p^{k-1} is just the same as the common factor between b and p-1. When p=2 and k>1, the congruence classes mod 2^k of c? not divisible by 2 are a product of a group of order 2 with a cyclic group of order 2^(k-2). (If I remember correctly, each c can be written as (-1)^i * 5^j where i=0,1 and j=0,...,2^{k-1}-1.) If b is odd, all of them are b-th powers. If b is even, taking a b-th power eliminates the factor of (-1)^i, and the number of c? that are b-th powers is 2^{k-2}/d where d is the greatest common divisor of b with 2^{k-2}. Now to deal with all the c?, consider that either c==0 mod p^k, or c can be written as p^l * c?where 0<=l= (p-1)p^(k-1)/2 which for large p is on the order of n/2. The small values are for n which have many prime factors p where p-1 has a common divisor with b. The question of the average value of f is a little tricky. Partly this is because the value of f is obtained by *multiplying* together a bunch of factors. For an analogy, suppose we took 1024 people and assigned them the binary numbers with 10 bits in them. Then we give each of them 2^k dollars, where k is the number of 1 bits in their number. The most common amount that they get is $32, where half of their bits are 1? and half 0?. That? also the median amount. The average amount, however, is $(3/2)^10 or approximately $57.66. The amounts given out correspond to the terms of (1+2)^10 expanded using the distributive law into 2^10 terms: 1*1*1*...*1, 1*1*1*...*2, etc. (The actual income distribution is allegedly something like log-normal over its middle range, as if it were the result of a lot of little independent factors multiplied together.) So if we succeed in computing the average, it? liable to be above the typical or most common value, being dragged upward by the minority of n for which n has few prime divisors with p-1 having a common factor with b. It? been awhile since I tried to work out an estimate of the average value of a multiplicative arithmetic function... probably I?e only read it done. I have LeVeque? _Topics in Number Theory_ open to the section titled Average order of magnitude . Hmm. Doesn? seem so helpful in this case. (Some of the earlier sections explain some of the stuff I have written above, but much more polished of course, so you might want to look at it.) I? pretty sure that the ratio of the average value of f from 1 to n to n decreases gradually toward 0. My main question is how slowly. Let me try to guess based on heuristics. Suppose for example b=3. Then in a certain sense half of the primes are == 1 (mod 3), and half are == 2 (mod 3). Each prime == 1 mod 3 cuts the value of f(p) down by approximately a factor of 3. So it has some vague resemblance to the scenario above. If I remember correctly, the typical number of prime factors of a large number n grows like log log n. If the number n ?pped a coin log log n times, and for each time it came up heads, divided itself by 3, then the median value it got would be n/3^(log log n/2), and the average value would be n/(3/2)^(log log n). Ignore the fact that log log n isn? an integer for the moment; this is just a crude heuristic argument.... We can presumably get a lower bound using the fact that the number of primes p == 2 (mod 3) that are <= n grows like n/(2log n). For each of them, f(p)=p. So (5+11+17+...)/n gives us a lower bound on the average. That lower bound >= Cn/log n for a suitable constant C>0. Note that (3/2)^(log log n) = (log n)^(log (3/2)), and log (3/2) < 1. This is good, because it shows the crude heuristic doesn? immediately contradict the lower bound we get from looking just at primes! Since it looks like taking this further would require more work than I want to do later this evening, I?l just stop with a guess. I? not so sure that the handwaving above gives the right exponent, but I would expect that there is some exponent r, 0r. The value of r would depend on b, of course. It occurs to me to wonder what happens to this exponent for highly divisible b. I would also guess that this problem has been solved at some time, and it just happens not to be easy to find on the web. Keith Ramsay