mm-52 === >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't 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't 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 ?ite subcollection whose> intersection is already contained in U (this is in fact simply the> ?ite 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 ?ite 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 ?ite 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, NO! That would imply that X is disconnected, and there was nothing inthe statement of the problem that implies this. No: a SUBSET C of X is connected if and only if, for any two opendisjoint subsets U and V such that C is contained in U union V, C iseither contained in U or is contained in V. No assumption need be madethat U union V is all of X, unless C is equal to X (in which case, Xcontained in U union V implies that U union V is all of X).>AND Y is>connected, then Y is contained in U or Y is contained in V?YOU ARE TRYING TO PROVE THAT Y IS CONNECTED. Why would you ->assume<-that Y is connected?You seem to have confused yourself beyond all reckoning.Here is the problem again:>Let X be compact Hausdorff, let {A_i} be a collection of closedconnected subsets of X that is simply ordered by proper inclusion.>Prove Y = {intersection of A_i over all i} is non-empty and>connected.You need to prove two things: (1) Y is non-empty; and (2) Y is connected.To prove Y is connected, you must prove that: If U and V are disjoint open subsets of X, such that Y is contained in the union of U and V, then Y is contained in U or Y is contained in V.Now, the hint you were given was to prove that:> Exercise: compact S, closed sets { Ki }, /{ Ki } subset open U ==> some K1,..Kn with K1 /../ Kn subset UThat is: prove the following, true for ANY compact set S:(*)If S is a compact set, and {K_i} is a family of closed subsets of S, then for every open set W contained in S: If the intersection of all the K_i is contained in W then there is a ?ite subset K_1,....,K_n of {K_i} such that the intersection of K_1,...,K_n is already contained in W.So, ?st: assume you have managed to prove this, and recall that Y isthe intersection of {A_i}, where each A_i is closed and connected.We want to prove that Y is connected. So let U and V be disjoint opensubsets of X, such that Y is contained in the union of U and V. Sincethe union of U and V is open, you can apply the exercise (*) byletting S=X, and {K_i}={A_i}, W=U union V. That means that, since theintersection of all A_i is contained in W, there is a ?ite subset,A_1,...,A_n, whose intersection is already contained in W.Since the A_i are assumed to be linearly ordered, the intersection ofany ?ite subset is just the smallest element of the subset. So say,without loss of generality, that A_1 is contained in A_2, contained inA_3,...., contained in A_n. Thus, the intersection of A_1,...,A_n isjust A_1.(*) tells you that A_1 is contained in U union V. You also know thatA_1 is connected (by assumption), and if a connected subset iscontained in the union of two open disjoint sets, then it is containedin one or the other. So A_1 is contained in U or A_1 is contained inV.Now, Y is the intersection of ALL A_i, so Y is contained in eachA_i. In particular, Y is contained in A_1. So Y is contained in A_1,and A_1 is contained in either U or in V. So Y is contained in U or Yis contained in V.In summary, you have proven that if Y is contained in the union of twodisjoint open subsets of X, then Y is contained in one of them. Thisimplies that Y is connected.To ?ish the problem, assuming (*), you just need to prove that Y isnonempty. This follows from the ?ite intersection property forcompact sets: X is compact if and only if for every family {C_i} of closed subsets of X, if the intersection of all C_i is empty, then there is a ?ite subset C_1,...,C_n, such that the intersection of C_1,...,C_n is empty.(If you don't know this, here's how to prove it: let U_i = X-C_i, andshow that the assumption that the intersection is empty means that{U_i} is an open cover of X; then ?d a ?ite subcover, and look atthe intersection of the corresponding C_i).Since {A_i} is a family of closed subsets of X, and the hypothesis onthe A_i imply that the intersection of any ?ite collection of A_i'sis nonempty, this implies (by contrapositive) that the intersection ofall of them is not empty either; so Y is non empty.The above explains how to prove what you want to prove, provided youcan prove (*).So now you need to prove (*). Prove it.-- === ) === Arturo Magidinmagidin@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 andconnected.>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't 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't 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 ?ite subcollection whose> intersection is already contained in U (this is in fact simply the> ?ite 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 ?ite 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 ?ite 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 unionV,> then either Y is contained in U or Y is contained in VYou mean for disjoint open subets U and V those union is all of X,> NO! That would imply that X is disconnected, and there was nothing in> the statement of the problem that implies this.> No: a SUBSET C of X is connected if and only if, for any two open> disjoint subsets U and V such that C is contained in U union V, C is> either contained in U or is contained in V. No assumption need be made> that U union V is all of X, unless C is equal to X (in which case, X> contained in U union V implies that U union V is all of X).>above theorem...I knewthat if U,V formed a separation of X and Y is a connected subspace of X,then Y is contained in U or V. I am ok now I think....THANK YOU for yourlengthy explanation after this!Steven>AND Y is>connected, then Y is contained in U or Y is contained in V?> YOU ARE TRYING TO PROVE THAT Y IS CONNECTED. Why would you ->assume<-> that Y is connected?> You seem to have confused yourself beyond all reckoning.> Here is the problem again:>Let X be compact Hausdorff, let {A_i} be a collection of closedconnected subsets of X that is simply ordered by proper inclusion.>Prove Y = {intersection of A_i over all i} is non-empty and>connected.> You need to prove two things:> (1) Y is non-empty; and> (2) Y is connected.> To prove Y is connected, you must prove that:> If U and V are disjoint open subsets of X, such that Y is> contained in the union of U and V, then Y is contained in U or Y> is contained in V.> Now, the hint you were given was to prove that:> Exercise:> compact S, closed sets { Ki }, /{ Ki } subset open U> ==> some K1,..Kn with K1 /../ Kn subset U> That is: prove the following, true for ANY compact set S:> (*)If S is a compact set, and {K_i} is a family of closed subsets of> S, then for every open set W contained in S:> If the intersection of all the K_i is contained in W> then there is a ?ite subset K_1,....,K_n of {K_i} such that> the intersection of K_1,...,K_n is already contained in W.> So, ?st: assume you have managed to prove this, and recall that Y is> the intersection of {A_i}, where each A_i is closed and connected.> We want to prove that Y is connected. So let U and V be disjoint open> subsets of X, such that Y is contained in the union of U and V. Since> the union of U and V is open, you can apply the exercise (*) by> letting S=X, and {K_i}={A_i}, W=U union V. That means that, since the> intersection of all A_i is contained in W, there is a ?ite subset,> A_1,...,A_n, whose intersection is already contained in W.> Since the A_i are assumed to be linearly ordered, the intersection of> any ?ite subset is just the smallest element of the subset. So say,> without loss of generality, that A_1 is contained in A_2, contained in> A_3,...., contained in A_n. Thus, the intersection of A_1,...,A_n is> just A_1.> (*) tells you that A_1 is contained in U union V. You also know that> A_1 is connected (by assumption), and if a connected subset is> contained in the union of two open disjoint sets, then it is contained> in one or the other. So A_1 is contained in U or A_1 is contained in> V.> Now, Y is the intersection of ALL A_i, so Y is contained in each> A_i. In particular, Y is contained in A_1. So Y is contained in A_1,> and A_1 is contained in either U or in V. So Y is contained in U or Y> is contained in V.> In summary, you have proven that if Y is contained in the union of two> disjoint open subsets of X, then Y is contained in one of them. This> implies that Y is connected.> To ?ish the problem, assuming (*), you just need to prove that Y is> nonempty. This follows from the ?ite intersection property for> compact sets:> X is compact if and only if for every family {C_i} of closed> subsets of X, if the intersection of all C_i is empty, then there> is a ?ite subset C_1,...,C_n, such that the intersection of> C_1,...,C_n is empty.> (If you don't know this, here's how to prove it: let U_i = X-C_i, and> show that the assumption that the intersection is empty means that> {U_i} is an open cover of X; then ?d a ?ite subcover, and look at> the intersection of the corresponding C_i).> Since {A_i} is a family of closed subsets of X, and the hypothesis on> the A_i imply that the intersection of any ?ite collection of A_i's> is nonempty, this implies (by contrapositive) that the intersection of> all of them is not empty either; so Y is non empty.> The above explains how to prove what you want to prove, provided you> can prove (*).> So now you need to prove (*). Prove it.> -- > === === === > === ===> Arturo Magidin> magidin@math.berkeley.edu>===| at 08:30 AM, kramsay@aol.com (KRamsay) said:|>Conesetter's remark was in reply to David Ullrich, who said that the|>analytic continuation construction produced a unique Riemann|>surface.||Which it does.Of course.|It's not clear to me what the referent of what you get is, but the|Riemann surface as a whole is well de?ed, and does not depend on a|choice of path, not even independent of homotopy.Shmuel Metz:|?whether he meant (a) (correct but not an objection to what Ullrich wastelling him) that the _function element_ you get when you continue a functionelement along a path depends on the path, or (b) (incorrect, but would atleast contradict what Ullrich said) that the _Riemann surface_ you get inthis construction depends on a choice of path you choose to take to eachpoint. Or maybe he had something else in mind.My puzzling remark here is just pointing out the (b) is incorrect. TheRiemann surface you get doesn't depend on a choice of path at all. Onedoesn't choose a path at the outset. I mean this in the most trivial andobvious sense. The Riemann surface of log z might get called R_{log z},not R_{log z, C} for some path C. Obviously it doesn't depend on a homotopyclass of a path either.|>Note that if something (like the function element reached by|>continuing along a path which depends on the homotopy class of the|>path), it follows logically that also it depends on the path.||uniqueness. What you get depends on the path. What makes the|construction unique is precisely the fact that it does not depend on|the path within the homotopy class. You have to read his depends in|context.I could be wrong, but I don't see much sign that he meant varies withina homotopy class when he said it depends on the path. If he meant thefunction element you arrive at when you continue along a path, and we takedepends on the path more literally, then that statement is correct, atleast by itself.Varying within a homotopy class as well would not make his objection make anymore sense. If we had some functional F on a family of paths C which startsat a basepoint p, we could consider {(C,F(C))}, and it would be a uniqueconstruction, whether F varies within a homotopy class or not. I would saythat it's the fact that one is taking the collection of all paths that makesthe construction independent of a choice of any one given path. It's likemaking a manifold not depend on your choice of coordinate system by de?ingits atlas to be all possible compatible coordinate charts.The fact that a homotopy of paths (that includes only paths along which thefunction can be continued) lifts to a homotopy on the constructed spaceimplies nice geometric properties of the mapping from the constructed spaceback to the original base space (complex plane or Riemann sphere).Incidentally, I like the construction in Narasimhan. He de?es a topologyon the set of all function elements at all points, and then considers theconnected components of this space. The connected component of a functionelement is essentially the Riemann surface we're considering here.|>Shmuel Metz seems to enjoy saying no to people. :-)||He also enjoys saying yes ;-)Balance.Keith Ramsay === >message>What with the war on terror, and Canada's puny military and corrupt>police force, the only way to insure peace and safety for that>country, is for it to be annexed by America.>It is the only way foward for Canada aka Canuckistan.>---------------------------------> Well, another solution is the liberation of U.$.A by Canadians and the>return of democracy to the Americans peoples.> I am sure Canadians will be welcome with (french) kisses and ?>when>they will topple the Little Dictator and his gang of corporate crooks.> Patriot act will be scraped; free press will be restaured, Health Care>will be available to all citizens,>Yeah, like it is in Canada. If you wait. And wait. And wait. And... oops,>died already.What crap! I had to have surgery. I found out the week before Christmas last year. I had the surgery by Feb. 15th. Three weeks ago, we got a call to go to the local hospital. My four year old grandson was taken by ambulance to Sick Children's Hospital in Toronto. This was done within hours of his arrival at the local hospital. They did the biopsy the NEXT DAY, which was a Saturday. They had the diagnosis of cancer by Monday night, and he's into his second week of treatment.And you know what? None of it cost anything but what the taxes pay for?We aren't great and there are some waits, but I'll take ours over American medical care anyday.My grandson is lucky, this type of cancer is usually caught late, but this was found early and hadn't spread. It's also very fast developing, because it's a lymphoma. But, they caught it and caught it fast.D> Or you can join an American HMO where they deny you care because they> don't want to pay out the money. ooopps one dead wife...mine!> Capitalist medicine is an abomonation and a human rights violation.> THOM>But at least the bigtime politicians and bureaucrats get their healthcare!>(By paying cash in Yankee hospitals...)>-- >zimriel sbc dot> at global net>.>http://pages.sbcglobal.net/zimriel/blog/zimblog.htmlbecause everyone else is doing it> === What with the war on terror, and Canada's puny military and corrupt> police force, the only way to insure peace and safety for that> country, is for it to be annexed by America.> It is the only way foward for Canada aka Canuckistan.>You seem to forget that the Canadians kicked our ass in two>con?...> Really? Which two?1775http://www.civilization.ca/cwm/chrono/1000invasion_e.html you apparently didn't read it very well. PLUS Canada didn't exist> then, it was just another British puppet and the forces that worked> their way north went after the British, not the so called Canadians.>By that logic, those that were repelled weren't american, either.I agree with that. Until we won the war of independence we wereEnglishman, English subjects and English citizens.> The area then was commonly called New France and the name Kanata> is an indian word for cluster of villages or houses.>And...?The reference to Canada in the 1500's was probably pronounced andspelled Kanata> Canada was formed in 1867 with the Constitution Act of 1867, and in> Canada's original constitution the name of the new dominion would be .> . . Canada.>And...?1812http://www.civilization.ca/cwm/chrono/1774invasion_repelled_ e.html> that was a war with the British. Canada did not exist then.> THOM>Wow, you really are a nation in denial...Notice I'm writing from Australia?THOM> === > Can you please show where the mathematical reasoning is wrong?> JSBeen there, done that. === > Can you please show where the mathematical reasoning is wrong?> JS> Been there, done that.I'm afraid not. The Schoenfeld Theorem remains astoundingly true. === > The Schoenfeld Theorem:> An in?ite bounded sequence of random numbers contains all ?ite> bounded sequences of numbers.> [snip]> Trivially disproven by example in Hofstadter's Godel, Escher, Bach. > It's only 777 pages long. If you look at each page for one second you> can ?d the table within 12 minutes.> What are you babbling on about now?> An in?ite random sequence of integers bound by [n,m] contains all> ?ite sequences of integers within [n,m] (not necessarily bound by n,m though).> This is proven. It's called the SCHOENFELD THEOREM.> JS> Your proof is equivalent to a reductio ad absurdum and, in general,> this kind proofs are weak. The weakness of your proof is due to your> missing step of proving the existence of such in?ite sequence, which> you intuitively assume. Your intuition may be fooling you.> Speci?ally:I'm afraid you have it all wrong. R = sequence of random integers bound by [n,m].Any in?ite sequence bound by [n,m] with the following properties canbe an instance of R.1. The probability of drawing x from R where n <= x <= m is1/[(m-n)+1].2. The probability of drawing x after a ?ite sequence S is1/[(m-n)+1]^(|S| + 1) where S : some ?ite sequence where all elements are within [n,m](sequence is not necessarily bound by [n,m]) |S|: cardinality of SSo Property 1 says that any number has an equally probable chance ofbeing drawn from R. But this is not a suf?ient condition for R to berandom as many other sequences have this property.e.g. [1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3]The probability of drawing 1, 2 or 3 are all the same, yet we can makethe obvious inference that it is not random since the probability ofdrawing a the sequence [1,3,2] is not the same as drawing [1,2,3]).So wihout cluttering usenet with so much detail, property 2 ?es thisproblem and enforces that R has totaly entropy.So, all in?ite sequence of integers bound by [n,m] with properties 1and 2 are valid instances of R.Now you are telling me that no such sequences exist? I'm afraid not.> An event E with probability zero to never occur during an experiment> is equivalent to the certain event S, with P(S) = 1, if and only if,> the event E is proven to be part of the experiment. In other words,> your theorem is not really a theorem, but just a corollary of a> theorem that would prove the existence of your in?ite sequence. You> could probably use the Axiom of Choice to de?e such sequence but> that's about it, all you can have is an axiomatic declaration and the> proof is not constructive.> Therefore, show proof that your in?ite sequence exists in N. (It> does not)I do not contend that in?ite sequences exist in anything, I contendthat FINITE sequences occur in?itely in INFINITE RANDOM SEQUENCES(see above for de?ition of a RANDOM sequence).JS === > PS: You might not know it yet but a real is really just a 2-tuple of> an integer and an in?ite sequence of integers.However, as this clearly does not apply to Dedekind reals or Cauchy reals, it must only apply to Schoen?ld reals, which no one but Schoen?ld ever uses. === A misconception. When I draw a real number out of the reals in the> range [0,1], the probability of drawing exactly 0.5 is 0, as is the> probability of drawing exactly whatever other number in that range.> Nevertheless a number is drawn in that range, and it even can be 0.5.> Only when you draw from a *?ite* list, the probability 1 means> certain and the probability 0 means can not occur.> Your argument does not apply here.> Why not? You claim that probability ?1' and ?0' do mean that something> is certain, resp. can not occur. I show you a case where that is simply> false. In the same way it is false with your theorem.It does not apply because the case you imply talks about theprobability of drawing an in?ite sequence of integers where my casetalks about the probability of drawing a ?ite sequence of integers.When you draw a real, you are really drawing a 2-tuple (we'll callthis a Schoenfeld Real from now on).The Schoenfeld Real is a real de?ed as a 2-tuple written (I,[M])where I is the integer whole part and M is the mantissa de?ed as anin?ite sequence of digits.Schoenfeld Real = (I, [M])The purpose of the Schoenfeld Real is to reduce a real into aninteger-based representation.e.g. the real 100.0123450_ can be written as a Schoenfeld Real asfollows,100.123450_ = (100, [0,1,2,3,4,5,0,0,.....]) or100.0_ = (100, [0,0,.....])However, I know you probably think this is all farce since youprobably only understand a number by the quantity it denotes, not it'srepresentation. So to evaluate a Schoenfeld Real into a real you mayuse the Schoenfeld Real Evaluator: = I + sum(i=0, +inf) M_i / r^iwhere I : integer whole part of the Schoenfeld Real M : manitissa digit sequence of the Schoenfeld Real M_i: i'th element of M r : radix of the desired number system You might also care to know that a digit is an integer element fromthe set [0, r-1] where r is the radix.Anyway, the point is that drawing Schoenfeld Real's and drawingintegers are very different.So if you want to disprove the Schoenfeld Theorem, you mustdemonstrate that a probability of drawing a FINITE SEQUENCE OFINTEGERS FROM AN INFINITE SEQUENCE OF INTEGERS is 0 or 1 and yet theycan or never get drawn respectively.Until you do this, I'm afraid the Schoenfeld Theorem remainsastoundingly true.JS === > When you draw a real, you are really drawing a 2-tuple (we'll call> this a Schoenfeld Real from now on).You are *soooo* cute!-- What I've learned is that [mathematicians are] the gatekeepers, andseem to have almost absolute power when it comes to mathematics. -- James Harris, on All I Really Ever Needed to Know I Learned in /Ghostbusters/. === [snip]> When you draw a real, you are really drawing a 2-tuple (we'll call> this a Schoenfeld Real from now on).What additional evidence might be needed to clinch the ststement that thisman Schoenfeld is a twittule?[snip]Franz Heymann === Franz Heymann: >[snip] > When you draw a real, you are really drawing a 2-tuple (we'll call > this a Schoenfeld Real from now on). >What additional evidence might be needed to clinch the ststement that this >man Schoenfeld is a twittule? None, really, but if you look back over the posts he's made tosci.physics, you'll see a similar pattern and other things he'snamed after himself in case you have any doubts. === > Franz Heymann: >[snip]> When you draw a real, you are really drawing a 2-tuple (we'll call> this a Schoenfeld Real from now on).>What additional evidence might be needed to clinch the ststement that this>man Schoenfeld is a twittule?> None, really, but if you look back over the posts he's made to> sci.physics, you'll see a similar pattern and other things he's> named after himself in case you have any doubts.Is he or anyone else keeping a list? Otherwise, it's ever so slightlypossible that we'll have some naming con?somewhere down theroad.-- Jesse HughesI often told you of the dangers of hubris, and most importantly ofall, I TOLD you that I wanted to change the institution of mathematicsworldwide. -- James Harris, on the evils of pride <87y8tga0zo.fsf@phiwumbda.org> === > Franz Heymann: [snip]> When you draw a real, you are really drawing a 2-tuple (we'll call> this a Schoenfeld Real from now on).>What additional evidence might be needed to clinch the ststement that this>man Schoenfeld is a twittule?> None, really, but if you look back over the posts he's made to> sci.physics, you'll see a similar pattern and other things he's> named after himself in case you have any doubts.> Is he or anyone else keeping a list? Otherwise, it's ever so slightly> possible that we'll have some naming con?somewhere down the> road.Actually, you know, that sounds like a great idea, Jesse. I'll bewhen he names something new. I don't read sci.physics and I don'treally expect to read Schoenfeld threads too often.-- Evariste Galois was clearly a passionate man. He tried to kill theking of France for instance... Remember that the French Revoloution*did* happen, so he wasn't really out of his times. -- JSH on Galois(1811 - 1832) pre?uring the French Revolution (1789) === > [snip]> When you draw a real, you are really drawing a 2-tuple (we'll call> this a Schoenfeld Real from now on).> What additional evidence might be needed to clinch the ststement that this> man Schoenfeld is a twittule?I don't know about the others, but I'll need a de?ition of twittulebefore I assent to the claim.-- Sure, [my Usenet presence is] like Shaq playing against you in yourbackyard, but that has its perks, as I ?d ways to have my fun *and*I can send messages to certain people in the United States Governmentwithout concern that the rest of you understand them. -- James Harris === > A misconception. When I draw a real number out of the reals in the> range [0,1], the probability of drawing exactly 0.5 is 0, as is the> probability of drawing exactly whatever other number in that range.> Nevertheless a number is drawn in that range, and it even can be 0.5.> Only when you draw from a *?ite* list, the probability 1 means> certain and the probability 0 means can not occur.> Your argument does not apply here.> Why not? You claim that probability ?1' and ?0' do mean that something> is certain, resp. can not occur. I show you a case where that is simply> false. In the same way it is false with your theorem.>It does not apply because the case you imply talks about the>probability of drawing an in?ite sequence of integers where my case>talks about the probability of drawing a ?ite sequence of integers.It does apply because you talk about drawing that ?ite sequencefrom a given in?ite sequence of random integers. Any statement you make about a sequence of random integers needs to bequali?d with something about the probability that that statementis true - it's _possible_ (but only with probability 0) for yoursequence of random integers to be the sequence 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...,which does not satisfy the conclusion of your theorem.>When you draw a real, you are really drawing a 2-tuple (we'll call>this a Schoenfeld Real from now on).>The Schoenfeld Real is a real de?ed as a 2-tuple written (I,[M])>where I is the integer whole part and M is the mantissa de?ed as an>in?ite sequence of digits.>Schoenfeld Real = (I, [M])>The purpose of the Schoenfeld Real is to reduce a real into an>integer-based representation.>e.g. the real 100.0123450_ can be written as a Schoenfeld Real as>follows,>100.123450_ = (100, [0,1,2,3,4,5,0,0,.....]) or>100.0_ = (100, [0,0,.....])>However, I know you probably think this is all farce since you>probably only understand a number by the quantity it denotes, not it's>representation. So to evaluate a Schoenfeld Real into a real you may>use the Schoenfeld Real Evaluator:> = I + sum(i=0, +inf) M_i / r^i>where > I : integer whole part of the Schoenfeld Real> M : manitissa digit sequence of the Schoenfeld Real> M_i: i'th element of M> r : radix of the desired number system >You might also care to know that a digit is an integer element from>the set [0, r-1] where r is the radix.>Anyway, the point is that drawing Schoenfeld Real's and drawing>integers are very different.>So if you want to disprove the Schoenfeld Theorem, you must>demonstrate that a probability of drawing a FINITE SEQUENCE OF>INTEGERS FROM AN INFINITE SEQUENCE OF INTEGERS is 0 or 1 and yet they>can or never get drawn respectively.>Until you do this, I'm afraid the Schoenfeld Theorem remains>astoundingly true.>JS****David C. Ullrich === >It does apply because you talk about drawing that ?ite sequence>from a given in?ite sequence of random integers. Any statement >you make about a sequence of random integers needs to be>quali?d with something about the probability that that statement>is true - it's _possible_ (but only with probability 0) for your>sequence of random integers to be the sequence> 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...,>which does not satisfy the conclusion of your theorem.Maybe 1 isn't a Schoenfeld Real? I think Harris is getting a worthycompetitor.>So if you want to disprove the Schoenfeld Theorem, you must>demonstrate that a probability of drawing a FINITE SEQUENCE OF>INTEGERS FROM AN INFINITE SEQUENCE OF INTEGERS is 0 or 1 and yet they>can or never get drawn respectively.Huh? Assume the in?ite sequence of natural numbers. The probabilitythat a given number n is found at position k is exactly lim 1/N asN->oo = 0 but we are still guaranteed to ?d the number at position nevery time.>Until you do this, I'm afraid the Schoenfeld Theorem remains>astoundingly true.Astounding it might be and in some sense true, but SchoenfeldTheorem it's not. === > (If you were discussing Mandarin literature in the same way you> discuss mathematics, I'd know you were cranking on that subject,> too, even though I know *nothing* about Mandarin literature.)> The key to being a crank is that you can't know *nothing*...you have> to know *something*. For example, you know that Mandarin literature> comes from China, you might well know that the Mandarin's were the> bureaucratic class, and you might know that Confucius was highly> revered especially by the Mandarins.> I think that small bit of knowledge (which just about exhausts mine, I> think) is plenty to be a crank about Mandarin literature. ;)> ThomasOne of the ubiquitous clues identifying JSH as a crank is that herepeatedly posts math arguments which he calls ?irrefutable' proofs. Whenother posters identify errors, he launches into tirades and haranguesaccusing them of being liars, cheaters or simply ?dog'. He oftenaccuses them of attacking math itself or basic algebra. When he ?allyrealizes that there was indeed an error, he sometimes posts an OOPS!followup. Then the cycle repeats. A Google search on James Harris andOOPS! reveals over 80 hits. Note that it does not require much more thancasual reading of these threads, certainly not any in-depth mathematicalbackground, to identify his behavior as ?crankish' or ?cranklike', or tosimply dismiss James as a ?crank'.--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 === > One of the ubiquitous clues identifying JSH as a crank is that he> repeatedly posts math arguments which he calls ?irrefutable'> proofs. Quite right! I had this discussion once with someone whose Latin isway weaker than mine. He was insisting that such-and-such was theonly way to construe a particular sentence. I showed him the correctparse, and why that made so much more sense in context. To be fair,it was a hard case out of Virgil, so it's not like he was an idiot.But he resisted, and resisted.I showed him three separate translations, which all agreed with myreading, and still to no avail. One day something clicked, and the OOPS happened. But this person was not a crank. Ever after, when he had a hardsentence to construe, he would ask me what I thought, to the pointwhere it was even a mild annoyance. Will James do that?Thomas === > One of the ubiquitous clues identifying JSH as a crank is that he> repeatedly posts math arguments which he calls ?irrefutable' proofs. > Quite right! Really? Do you believe that Wiles's work is an *irrefutable* proof ora refutable one? And please answer if you believe there is such athing as a refutable proof!As a sidenote, I've not said that my proofs are irrefutable proofs, asI've said they are proofs. At other times I've said that I haveproofs, and that since they are proofs they are irrefutable.>I had this discussion once with someone whose Latin is> way weaker than mine. He was insisting that such-and-such was the> only way to construe a particular sentence. I showed him the correct> parse, and why that made so much more sense in context. To be fair,> it was a hard case out of Virgil, so it's not like he was an idiot.> But he resisted, and resisted.I've had my arguments independently checked.James Harris <3c65f87.0312111543.8baadfd@posting.google.com> <87wu91g7md.fsf@becket.becket.net> <3FDA710F.874B75CD@ix.netcom.com> <87y8the8rf.fsf@becket.becket.net> <3c65f87.0312130615.7d5d979@posting.google.com> === > As a sidenote, I've not said that my proofs are irrefutable proofs, as> I've said they are proofs. At other times I've said that I have> proofs, and that since they are proofs they are irrefutable.And on other occasions, you've admitted that these things which youformerly called proofs and hence formerly alleged were irrefutablewere... wrong.So what does that mean?-- A set having three members is a single thing wholly constituted byits members but distinct from them. After this, the theologicaldoctrine of the Trinity as ?three in one' should be child's play. --Max Black, _Caveats and Critiques_ === >[...]>I've had my arguments independently checked.Well yes of course, for example plenty of people here havechecked them and found them to be incorrect.But probably you mean to imply that people have checkedthem and determined that they're correct. If you wantpeople to react to this assertion with anything other thangales of laughter you need to tell us who did the checking...>James Harris****David C. Ullrich === [snip]> Really? Do you believe that Wiles's work is an *irrefutable* proof or> a refutable one? And please answer if you believe there is such a> thing as a refutable proof!There is *certainly* such a thing as a refutable argument. And you *always* present your arguments, nomatter how ill-conceived or simply false, as *irrefutable proofs*. This is the mathematical equivalent ofcounterfeiting. But, do a Google search on James Harris and OOPS!. You will ?d over 80 hits! Many ofthose follow your post of your latest ?proof'.> As a sidenote, I've not said that my proofs are irrefutable proofs, as> I've said they are proofs. At other times I've said that I have> proofs, and that since they are proofs they are irrefutable.It isn't your assurance that makes them proofs. It is correctness. Your track record at mislabeling faultyarguments as irrefutable proofs (or simply ?proofs', if you prefer), is legion. Whether the arguments youpost actually represent proofs has nothing to do with your insistence that they contain no errors, but onlyin whether *in fact* they contain no errors. What you have asserted amounts to the claim thay you areinfallible. Hahahah....--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 === >prove if x is an irrational number then sqroot of x is irrational???The contrapositive of your statement is that the square of a rational is rational, which is both equivalent and trivial! === Dear Sir/Madam,Given an undirected graph, every edge in it will be in a speci?color. For example, there are 4 lines in the graph. Two lines are red.One line is yellow. Another line is blue. Then there are totally 3kinds of colors in this graph. I further assume there are n cut setsin the graph. In one cut set, the lines are either red or yellow. Sothere are 2 kinds of color in this cut set.With the de?ition stated above, my objective is to ?d a cut setthat contains the least kinds of colors. I am wondering if thisproblem is an NP complete problem.HelenX-Face: #@BO:7A_K1T_2!j{yJ {i,P&OwQ Dear Sir/Madam,> Given an undirected graph, every edge in it will be in a speci?> color. For example, there are 4 lines in the graph. Two lines are red.> One line is yellow. Another line is blue. Then there are totally 3> kinds of colors in this graph. I further assume there are n cut sets> in the graph. In one cut set, the lines are either red or yellow. So> there are 2 kinds of color in this cut set.> With the de?ition stated above, my objective is to ?d a cut set> that contains the least kinds of colors. I am wondering if this> problem is an NP complete problem.> Helen>begin 666 member.gifM1TE&.#=A0`+`( ``````/___RP`````0`+```#XR/J< OI1I[4!Y5Y4<#%0`[`endbegin 666 wedge.gifM1TE&.#EA!P`(`( ``````/___R'Y! $```$`+ `````'``@```(,C&&)J,'>(X)'J'>L*` #L``endbegin 666 neq.gifM1TE&.#EA@`*`/ ``/___P```'Y! $`````+ `````*``H```(2A ^!&+KF.7CNMV@A?9M For a peasant like me, this is unde?ed. All I can imagine an undirected> graph to be is a piece of squared paper with two axes and a line ?> throught the air. URL:http://mathworld.wolfram.com/UndirectedGraph.html URL:http://mathworld.wolfram.com/CutSet.html URL:http://mathworld.wolfram.com/EdgeColoring.html--WARNING: I cannot be held responsible for the above They're coming tobecause my cats have apparently learned to type. take me away, ha ha! === Let i = sqrt(-1).i^i is kind of an interesting number. Note that i = cos(pi/2) + i*sin(pi/2) = exp(i*pi/2).Therefore i^i should be [exp(i*pi/2)]^i = exp(i*i*pi/2) = exp(-pi/2) = .2078 On the other hand, (i^i)^4 = i^{i*4} = (i^4)^i = [(-1)*(-1)]^i = 1^i = 1,which implies that i^i is a fourth root of unity: that is i^i = 1, i, -1, or -i.Say for example i^i = 1. Then (i^i)^2 = (-1)^i = 1. Therefore (-1) = 1^{1/i} = 1, a contradiction.So instead say i^i = -1. Obtain the same contradiction.So instead try i^i = i. This implies that [i^i]^2 = (-1)^i = -1.Noting that -1 = cos(pi) + i*sin(pi) = exp(i*pi), one concludes that (-1)^i = [exp(i*pi)]^i = exp(-pi) = -1, whichis clearly false.So ?ally, say i^i = -i. This leads to the same contradictionas the previous case.Thus all the fourth roots of 1 are eliminated, even thoughi^i satis?s the equation x^4 = 1.Note, incidentally, that exp(-pi/2) is not a fourth rootof 1 either!So what is i^i, really ?Nora B. === In sci.math, Nora Baron<36024859.0312121757.613e264f@posting.google.com>:> Let i = sqrt(-1).> i^i is kind of an interesting number. > Note that i = cos(pi/2) + i*sin(pi/2) = exp(i*pi/2).> Therefore i^i should be [exp(i*pi/2)]^i = exp(i*i*pi/2) = exp(-pi/2) = .2078 > On the other hand, > (i^i)^4 = i^{i*4} = (i^4)^i = [(-1)*(-1)]^i = 1^i = 1,> which implies that i^i is a fourth root of unity: that is> i^i = 1, i, -1, or -i.> Say for example i^i = 1. Then (i^i)^2 = (-1)^i = 1. Therefore> (-1) = 1^{1/i} = 1, a contradiction.> So instead say i^i = -1. Obtain the same contradiction.> So instead try i^i = i. This implies that [i^i]^2 = (-1)^i = -1.> Noting that -1 = cos(pi) + i*sin(pi) = exp(i*pi), one > concludes that (-1)^i = [exp(i*pi)]^i = exp(-pi) = -1, which> is clearly false.> So ?ally, say i^i = -i. This leads to the same contradiction> as the previous case.> Thus all the fourth roots of 1 are eliminated, even though> i^i satis?s the equation > x^4 = 1.> Note, incidentally, that exp(-pi/2) is not a fourth root> of 1 either!> So what is i^i, really ?> Nora B.i^i = exp(i * ln(i)) = exp(i * (i * pi * (4n+1)/2))=(sort of) exp(pi * (4n+3)/2), for any integer n.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Mathcad agrees with a single answer which implies its is true for i= -SQR(-1) as well-- Bruce Harveybruce@bearsoft.co.ukThe Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft> Let i = sqrt(-1).> i^i is kind of an interesting number.> Note that i = cos(pi/2) + i*sin(pi/2) = exp(i*pi/2).> Therefore i^i should be [exp(i*pi/2)]^i = exp(i*i*pi/2) = exp(-pi/2) =.2078> On the other hand,> (i^i)^4 = i^{i*4} = (i^4)^i = [(-1)*(-1)]^i = 1^i = 1,> which implies that i^i is a fourth root of unity: that is> i^i = 1, i, -1, or -i.> Say for example i^i = 1. Then (i^i)^2 = (-1)^i = 1. Therefore> (-1) = 1^{1/i} = 1, a contradiction.> So instead say i^i = -1. Obtain the same contradiction.> So instead try i^i = i. This implies that [i^i]^2 = (-1)^i = -1.> Noting that -1 = cos(pi) + i*sin(pi) = exp(i*pi), one> concludes that (-1)^i = [exp(i*pi)]^i = exp(-pi) = -1, which> is clearly false.> So ?ally, say i^i = -i. This leads to the same contradiction> as the previous case.> Thus all the fourth roots of 1 are eliminated, even though> i^i satis?s the equation> x^4 = 1.> Note, incidentally, that exp(-pi/2) is not a fourth root> of 1 either!> So what is i^i, really ?> Nora B. === > (i^i)^4 = i^{i*4} = (i^4)^i = [(-1)*(-1)]^i = 1^i = 1,more than one invalid step in this line! === How painful is a branch to the eye? === > (i^i)^4 = i^{i*4} = (i^4)^i = [(-1)*(-1)]^i = 1^i = 1,This is not quite correct. (i^i)^4 = i^{i*4} is correct (assuming principalvalue) but i^{i*4} does NOT equal (i^4)^i === > So what is i^i, really ?It's one of the values e^(-Pi/2 + 2nPi), n an integer, depending on the branch of the logarithm chosen. If it's the principal branch, then it's e^(-Pi/2).The laws of exponents for positive numbers raised to a real powers do not extend to complex numbers raised to complex powers. Given the ambiguity in the de?ition of z^w, this is not too surprising. === The problem is that I^(4I) != (I^4)^I> Let i = sqrt(-1).> i^i is kind of an interesting number.> Note that i = cos(pi/2) + i*sin(pi/2) = exp(i*pi/2).> Therefore i^i should be [exp(i*pi/2)]^i = exp(i*i*pi/2) = exp(-pi/2) =.2078> On the other hand,> (i^i)^4 = i^{i*4} = (i^4)^i = [(-1)*(-1)]^i = 1^i = 1,> which implies that i^i is a fourth root of unity: that is> i^i = 1, i, -1, or -i.> Say for example i^i = 1. Then (i^i)^2 = (-1)^i = 1. Therefore> (-1) = 1^{1/i} = 1, a contradiction.> So instead say i^i = -1. Obtain the same contradiction.> So instead try i^i = i. This implies that [i^i]^2 = (-1)^i = -1.> Noting that -1 = cos(pi) + i*sin(pi) = exp(i*pi), one> concludes that (-1)^i = [exp(i*pi)]^i = exp(-pi) = -1, which> is clearly false.> So ?ally, say i^i = -i. This leads to the same contradiction> as the previous case.> Thus all the fourth roots of 1 are eliminated, even though> i^i satis?s the equation> x^4 = 1.> Note, incidentally, that exp(-pi/2) is not a fourth root> of 1 either!> So what is i^i, really ?> Nora B. === >Whereas James makes valuable contributions to sci.math and math world>the detractors and their negativity add nothing, you might try not>posting for a month and instead really read and try to understand>James's posts.I've been reading James' posts for well over ?e years now. He's aninstigator. Doug === >Whereas James makes valuable contributions to sci.math and math world>the detractors and their negativity add nothing, you might try not>posting for a month and instead really read and try to understand>James's posts.>I've been reading James' posts for well over ?e years now. He's an>instigator. >Don't feed the troll.It's a sad little turdlet who has nothing better to do than -badly- fake otherpeople's posts. === > He's no founder of Lawsonomy, but his work is not without a certain > kind of merit. I think that his name ranks up there as a pioneer of > sorts (but not one without peers, even on Usenet -- still, I've always > found him more engaging than the competition).Eh, no. The pioneer is (in my opinion) Archimedes Plutonium. I alwaysfound Archimedes also quite entertaining. And he has never gone to thepoint of abusing his opponents (as far as I know).-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > He's no founder of Lawsonomy, but his work is not without a certain> kind of merit. I think that his name ranks up there as a pioneer of sorts (but not one without peers, even on Usenet -- still, I've always> found him more engaging than the competition).>Eh, no. The pioneer is (in my opinion) Archimedes Plutonium. I always>found Archimedes also quite entertaining. And he has never gone to the>point of abusing his opponents (as far as I know).Sure he has. Someone else has pointed out that he's tried to get people ?ed, etc, without mentioning the tactics: At one time (Idon't know whether this is still true) he would totally misquotepeople, claiming that so and so said he hated Jews, whatever,using a direct quote, when the person involved simply neversaid any such thing. (Which makes him unique, or did until recently - John Correydid that at least once, inserting a word into a _quote_ thattotally changed the meaning.)****David C. Ullrich === > He's no founder of Lawsonomy, but his work is not without a certain> kind of merit. I think that his name ranks up there as a pioneer of> sorts (but not one without peers, even on Usenet -- still, I've always> found him more engaging than the competition).> Eh, no. The pioneer is (in my opinion) Archimedes Plutonium. I always> found Archimedes also quite entertaining. And he has never gone to the> point of abusing his opponents (as far as I know).Archimedes takes me back. De?itely a pioneer, at least in mathematical Crankdom.-- Will Twentyman <87iskmardu.fsf@phiwumbda.org> === > He's no founder of Lawsonomy, but his work is not without a certain> kind of merit. I think that his name ranks up there as a pioneer of> sorts (but not one without peers, even on Usenet -- still, I've always found him more engaging than the competition).> Eh, no. The pioneer is (in my opinion) Archimedes Plutonium. I> always found Archimedes also quite entertaining. And he has never> gone to the point of abusing his opponents (as far as I know).I think Archimedes Plutonium is in many ways the archetype Usenetcrank, but James has pioneered a distinct style. For whatever reason, I've never found AP very entertaining. I'm notsure why. Some cranks are just fun to read and others (like AP) arefun to read about, without reading the original sources.James is just funny.-- Jesse HughesShe moaned, in pain and pleasure, as, in a confused whirlwind, sheglimpsed an image of Saint Sebastian riddled with arrows, cruci?dand impaled. --Mario Vargas Llosa on category theory === > He's no founder of Lawsonomy, but his work is not without a certain> kind of merit. I think that his name ranks up there as a pioneer of> sorts (but not one without peers, even on Usenet -- still, I've always> found him more engaging than the competition).> Eh, no. The pioneer is (in my opinion) Archimedes Plutonium. I always> found Archimedes also quite entertaining. And he has never gone to the> point of abusing his opponents (as far as I know).AP has attempted to have people ?ed from their jobs, kicked out of school,and has harassed the faculty and employers of various posters. === > He's no founder of Lawsonomy, but his work is not without a certain> kind of merit. I think that his name ranks up there as a pioneer of sorts (but not one without peers, even on Usenet -- still, I've always> found him more engaging than the competition).> Eh, no. The pioneer is (in my opinion) Archimedes Plutonium. I always> found Archimedes also quite entertaining. And he has never gone to the> point of abusing his opponents (as far as I know).>Well, unless you count his Go to hell (Styx, that is), which, IIRC,appeared several years ago. AP was quite a bit feistier in those days.Nowadays, I agree with you that he's (a) quite entertaining and(b) inoffensive.Rick === The attempt to draw parallels between the advncement of human thought(mathematics) and negemtropy in biology and society is interesting but doesnot compare like with like. I suppose that seeming ?useless' evolutionarydevelopments (peacocks tail, ? etc) are a bit like the super?ofhuman scienrti? and artiastic endeavour.signi?ant effect on human actions. One cannot escape the fact that humanbeliefs do have a transforming effect on the environment; witness religiousbeliefs and modern day capitalist ideology.That there is no ?ite limit to theoretical complexity seems to imply thatthere is a limit on human comprehension, however much aided by arti?ialmeans. Ashby's Law captures this idea: Only variety overcomes variety. Aquick appraisal of modern society should soon convince the observer that allis lost as far as the human individual is concerned. The evolution of socialsystems has passed the point of controllability by human agencies(governments). This might lead to a breakdown in coordination on a worldscale or it might lead to a totalitarian system which limits further growthin complexity at the expense of individual freedom. This is a far cry fromthe ivory towers of Goedel and his ilk.Comments belowTony Thomas> The brain and nervous system evolved so an organism could control its> actions and predict their consequences.'Organisms developed brains and nervous systems and so were able to bettercontrol and predict their actions.' The did not evolve in order to doanything, it was an accidental variation which survived.. Evolutionary theory implies that the> organism will attempt to choose actions most bene?ial to itself, its> offspring and other closely related individuals.Natural selection in conjunction with environmental forces leads to thesurvival of some individuals rather than others. There is no voluntaryaction to this effect by organisms which bene?s themselves or theiroffspring.> In the 1930's, Goedel proved that no consistent nontrivial mathematical> system could be complete.This applies to logical, mathematical or other mental constructs but not tothe real world.It is superstition to suppose that ?logic' determines or limits ?reality' inany way. One consequence is that, even in an ideal> deterministic universe where all the laws and initial conditions areknown,> it is not possible to always predict the consequences of ones choices.Chaos theory illustrates why non-linear systems are not predictable.In a system as large as a universe (whatever that might mean) it seemslikely that any subsystem of that universe would be unable to compute thepossible future states of the whole system. Predictions, therefore, dependon limiting the variables affecting the future of the subsystem.To take a particular example. An undetected cosmic event may suddenly affectthe solar system and therefore interfere with predictions in the region ofthis system. Hypotheses that a knowing being could isolate all the necessaryvariables, ?d an invariant relation between them and make omniscientpredictions is just plain daft. For> example there is no general way to determine if a computer program thathas> stopped responding will ultimately start responding again. This is trueeven> though all the computer's actions are determined and predictable.This is a contradiction, since the halting was not predicted. Anyway, ahalted programis no longer a functioning system and so has no future state that could bepredicted. For any> given computer program there is some axiom of mathematics that coulddecide> the issue but no ?ite collection of axioms can solve the problem forevery> computer program. The only way to explore this full hierarchy of> mathematical truth is with a process like biological evolution that> considers an ever increasing number of alternatives without limitingitself> to any particular path or set of paths. In evolution a path is a sequenceof> species each descendant from the previous. In mathematics a path is a> sequence of axioms of increasing complexity.Evolution too will halt when life is no longer sustainable, for whateverreason.Consider the ?ite differentiation of a very long string of binary digits.The process will either oscillate and repeat itself or halt when the stringis reduced to zeros.> Does Goedel's result establish a direction for evolution albeit one that> requires ever expanding diversity?Diversity is limited by the possibility of combining the elements in astable way. For example, the atomic table does not continue inde?itely,presumably because of limits on the size of stable atoms. Molecularcombinations are much more diverse, so one could assume that biologicalentities are even more diverse still. Purely geometrical forms arein?itely various as can be seen from topology. However, there are only?e regular solids, because of the constraints placed on their de?ition. The axioms that decide more complex> problems involve more complex and subtle levels of indirection and self> re?n. A de?ing characteristic of the human mind is its ability to> think indirectly and self re?ely in very complex ways. This allowsit> to recognize and use complex and subtle patterns in the environment.Science> is a major achievement of this process. Humanity has created andenormously> effective body of knowledge often based on extremely subtle and indirect> patterns. The human mind, capable of such an achievement, could have only> evolved on a planet that supported an enormous diversity of life.>The brain system limits what can be comprehended if only because of shortterm memory.Long term memory and the ability to recall and organise data is a furtherconstraint. Very complex computer programs are not usually held in theprogrammer's memory but large chunks often are.> Mathematics itself provides perhaps the strongest argument against this> position. Mathematicians have been aware of the limitations of any single> path approach to expanding mathematics for decades, but this has not ledto> attempts to expand mathematics by making it more diverse.This is surely wrong. Is there anyone living today who has any clear idea ofthe scope and content of all the mathematical disciplines? I should havethought that the body of mathematics had long exceeded the ability of thehuman individual to know it all. On the contrary> the ideal of absolute truth that motivated the invention of mathematics> still holds sway.Mathematics was not invented it evolved like other human memes. The axioms of Zermelo Frankel set theory plus the axiom of> choice (ZFC) seem more than adequate for virtually all problems of> mathematics that have practical implications for the physical world.>I should have thought that this discipline above all was the most uselessand impractical.The fact that it grew out of metamathematics demonstrates this. The aim ofset theory was to unify and explain mathematics under a single system . Theresult has been less than satisfactory.> The answer to this lies in the way evolution drives itself. Biological and> cultural evolution have created the environment in which we compete today.> The subtlety of thought that an organism needs is mostly determined by the> subtlety of thought of its fellow creatures. ZFC does seem to havecaptured> much of the subtlety that human reasoning is capable of. That does notmake> it an end point of useful mathematics even if we have not evolvedculturally> to the point where we need more subtle levels of reasoning and pattern> recognition.ZFC in the context of culture as a whole is supremely irrelevant.> The draft of my book that explores and expands on these ideas and related> ones is at http://www.WhatWillBe.com> --> Paul Budnik> Mountain Math Software> http://www.mtnmath.com> 408 353 3824> === > I think the people that don't 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.behavior I cannot predict; there is the secret thought of my closestfriend; there are the thoughts within me which are too deep forwords--all of these are hidden to me. They exist anyhow. The worlddoes not revolve around what I happen to be able to express.> If I have time, I will check the proof you sketched. Of course,> it is correct. However, if you don't 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't take one thing out.You can play constructivist mathematics if you want to. Nothing wrongwith that. But insisting that my game must be worthless because itisn't the same as your game is a rather petulant exercise, isn't it?> 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 is not necessarily so. The computable numbers do not have theLUB property.Thomas === > I tried to solve another task on this issue today and failed. {/-(> First of all, I wanted to know if there is a general approach to solve> tasks of the form: You have given a fraction of the form P(x)/Q(x).> There P(x) and Q(x) are polynoms. You have to express this fraction> in the form sum(a_k*(m*x+n)^k, k, 0, in?ity). ?> [m, n are constants and a_k represents k constants.]> If somebody knows a general approach or a website (understandable> for a beginner) there this is explained then please tell me. (I somehow> must understand this matter.)> Anyway, this was my task today:> ###> task:> Express 14 / ( 8*(x^2-1) ) as an in?ite sequence of the form:> sum(a_k*(2-x)^k, k, 0, in?ity).> ____________> |my attempt(s):> 14 / ( 8*(x^2-1) ) = (7/4) * ( 1 / ( (x-1)(x+1) ) )> partial fraction decomposition (Do I name it right?): 1 = A*(x+1) + B(x-1)> if x = 1> => A = 1/2> if x = -1> => B = -1/2> ==> (7/4) * ( 1 / ( (x-1)(x+1) ) )> = (7/4) * ( 1/( 2*(x-1) ) - 1/( 2*(x+1) ) )> = (7/8) * ( 1/(x-1) - 1/(x+1) ) ***> Attempt N 1:> *** = (7/8) * ( -1/(1-x) - 1 / ( 1-(-x) ) )> = (7/8) * ( -sum( x^k, k, 0, in?ity) - sum( (-x)^k, k, 0,> in?ity) )> = (7/8) * ( -sum( x^k - (-1)^k * x^k, k, 0, in?ity) )> = (7/8) * ( -sum( ( 1 - (-1)^k ) * x^k, k, 0, in?ity) ) = ?> I didn't know how to transform this to the demanded form.> Attempt N 2:> *** = (7/8) * ( 1/(1-2+x) + 1/(1-2-x) )> = (7/8) * ( 1 / ( 1 - (2-x) ) + 1 / ( 1-(2+x) ) )> = (7/8) * ( sum( (2-x)^k, k, 0, in?ity ) +> sum( (2+x)^k, k, 0, in?ity ) ) = ?> I didn't know how to simplify (2-x)^k + (2+x)^k, so> that I get (..)*(2-x)^k {the demanded form}.> ###> That's it. It would be very nice if you told me a general> approach for these tasks.As long as the denominator splits into a product of distinct linearfactors, your method will work. That is, you'll get a linearcombination of terms of the form c/(k-x), which you can expand into ageometric series by writing it as = (c/k) * 1/(1-x/k).But out of curiosity, what would you do if you encountered a repeatedlinear factor, say 1/(1-x/k)^2? Or an irreducible quadratic, like1/(1-x+x^2)? You're going to have to handle these to solve youragenda. And you'll have to know about the digamma function to do thelast. To do THAT you're going to have to learn a lot more aboutin?ite series.--Ron Bruck === > Epilog> The Interesting Question has already been satisfactorily answered now.> But I thought that some people might enjoy a nice, pertinent passage> (which I only just now discovered) in the excellent _Concrete Mathematics_> by Graham, Knuth, and Patashnik.> On page 481 (2nd ed.), where m is the number of terms of Stirling's> asymptotic expansion of ln(n!) to be used, they say> ... if n is ?ed and m increases,> the error bound |B_(2m+2)|/((2m+2)(2m+1)n^(2m+1)) decreases to a certain> point and then begins to increase. Therefore the approximation reaches a> point beyond which a sort of uncertainty principle limits the amount by> which n! can be approximated.> David>You can ?d an asymptotic series for n!>rather than that for ln(n!) in>A new derivation of Stirling's Approximation to n!,> G. and J. Marsaglia, Am. Math. Monthly, 97,No. 9, 1990.>Mixing TeX and * for multiplication:>n! = n^n e{-n}sqrt{2pi n}*> [1 + 3a_3/n + 3*5a_3/n^2 + 3*5*7a_5/n^3 + 3*5*7*9a_9/n^4+...]>where the a's satisfy the recursion,> (n+1)a_n = a_{n-1} - 2a_2a_{n-1} - 3a_3a_n-2- ... - (n-1)a_{n-1}a_2>initialized with a_1=1, a_2=1/3.I don't have easy access to the AMM from 1990. Is your new derivationsimilar to the one I mentioned earlier in this thread at http://www.whim.org/nebula/math/stirling.htmlIf not, would you give a brief description of your derivation or a webRob Johnson take out the trash before replying === >Epilog>The Interesting Question has already been satisfactorily answered now.>But I thought that some people might enjoy a nice, pertinent passage>(which I only just now discovered) in the excellent _Concrete Mathematics_>by Graham, Knuth, and Patashnik.>On page 481 (2nd ed.), where m is the number of terms of Stirling's>asymptotic expansion of ln(n!) to be used, they say>... if n is ?ed and m increases,>the error bound |B_(2m+2)|/((2m+2)(2m+1)n^(2m+1)) decreases to a certain>point and then begins to increase. Therefore the approximation reaches a>point beyond which a sort of uncertainty principle limits the amount by>which n! can be approximated.Yes, but you knew this before. All that GKP is saying here is that theasymptotic series doesn't converge for any n. They are saying nothingabout the minimum error that can be attained by choosing m as a functionof n.Rob Johnson take out the trash before replying === A small comment in this month's issue of _Mathematics Magazine_>brings to mind an interesting question about Stirling's asymptotic>expansion of the gamma function which has been brooding in my mind>for several months now. On page 370, merely to give an example of>the use of Bernoulli numbers, Peter Knopf says... an extension of Stirling's formula allows us to compute n! to>arbitrarily good precision:>n! = (n/e)^n Sqrt(2 pi n) exp( sum(k=1, oo, B_(2k)/((2k-1) 2k n^(2k-1))) )>The questionable and interesting part is the claim of arbitrarily>good precision. After all, the expansion does not converge, being>merely asymptotic.>[snip]> The error, using a number of terms approximately pi n, de?itely seem> to be small enough to make the error less than 1/2.>Always, or just until n surpasses some rather large value? That is the>unanswered question.> On , [17] says that> asymptotically,> n-1 n 2n> B ~ (-1) 4 sqrt(pi n) ( ---- )> 2n pi e> Plugging this into the formula you quote above, I get that the minimum> term in the expansion is on the order of sqrt(2/pi)(n/e^{2pi+1})^n.> n (n/e^{2pi+1})^n> 1 7*10^{-4}> 10 2*10^{-22}> 100 5*10^{-117}> 1000 9*10^{-164}> which is in pretty close agreement with your values of>n error>1 -3*10^(-4)10 -9*10^(-23)>100 2*10^(-117)>1000 -4*10^(-164)> However, when n >= 1456 (e^{2pi+1}), it looks as if things might start> to blow up.>I hadn't commented on this before for a couple of reasons. (1) As I noted>previously in this thread, |error| starts to increase at n = 536. So I>thought n = 1456 was simply wrong. I would have said (and still do say)>that things _start_ to blow up, slowly, beginning at n = 536. But I see now>that I must have misinterpreted what you meant by things might start to>blow up. Rather, I now suppose that you meant that error has, by then,>gotten suf?iently big that we can no longer get n! precisely after>rounding. I now believe that is correct; see below. (2) My intuition had>told me that error probably wouldn't get suf?iently big to preclude>getting n! precisely after rounding until n was _much_ bigger than 1000. I>was thinking of n = 10000, or more perhaps. Well, my intuition was>apparently wrong in this case.the formula I give above indicates the minimum error occurs when n isabout e^{2 pi}, which is about 535.5, very close to 536.>I decided to try to discover, approximately, the smallest n such that the>asymptotic series cannot be used to give n! precisely after rounding. It>wasn't dif?ult (and I'm now suitably embarrassed that I hadn't done it>before!) Surprise: I got n = 1456. (Well, not a surprise to you readers.>But remember that I had thought n would need to be bigger than that, and so>it surprised me.) And it's good now to have independent con?mation of the>value which Rob obtained.>BTW, I'm not claiming (and I don't suppose that Rob is either) that the>smallest such n is precisely 1456. The smallest such n might be a little>more or less than 1456.I haven't done the calculation; is 1456 the smallest integer for whichthe asymptotic series fails to have an error less than 1/2?Rob Johnson take out the trash before replying === n! = (n/e)^n Sqrt(2 pi n) exp( sum(k=1, oo, B_(2k)/((2k-1) 2k n^(2k-1))) )[snip]>I decided to try to discover, approximately, the smallest n such that>the asymptotic series cannot be used to give n! precisely afterrounding. It wasn't dif?ult (and I'm now suitably embarrassed that I>hadn't done it before!) Surprise: I got n = 1456. (Well, not a surprise>to you readers. But remember that I had thought n would need to be>bigger than that, and so it surprised me.) And it's good now to haveindependent con?mation of the value which Rob obtained.BTW, I'm not claiming (and I don't suppose that Rob is either) that the>smallest such n is precisely 1456. The smallest such n might be a little>more or less than 1456.> I haven't done the calculation; is 1456 the smallest integer for which> the asymptotic series fails to have an error less than 1/2?I haven't done the calculation either. It would surely be rather tedious.But our independent estimates agree that the smallest such n should beclose to 1456. That's good enough to satisfy me.and separately>Epilog>The Interesting Question has already been satisfactorily answered now.>But I thought that some people might enjoy a nice, pertinent passage>(which I only just now discovered) in the excellent _Concrete>Mathematics_ by Graham, Knuth, and Patashnik.>On page 481 (2nd ed.), where m is the number of terms of Stirling's>asymptotic expansion of ln(n!) to be used, they say>... if n is ?ed and m increases,>the error bound |B_(2m+2)|/((2m+2)(2m+1)n^(2m+1)) decreases to a certainpoint and then begins to increase. Therefore the approximation reaches a>point beyond which a sort of uncertainty principle limits the amount by>which n! can be approximated.> Yes, but you knew this before. All that GKP is saying here is that the> asymptotic series doesn't converge for any n. They are saying nothing> about the minimum error that can be attained by choosing m as a function> of n.Certainly. I didn't mean to mislead anyone into thinking that they hadsomehow answered the basic question I had posed. I just enjoyed their turnof phrase (...a sort of uncertainty principle limits...) and thought thatsome other people might enjoy it too.David === n! = (n/e)^n Sqrt(2 pi n) exp( sum(k=1, oo, B_(2k)/((2k-1) 2k n^(2k-1))) )[snip]>I decided to try to discover, approximately, the smallest n such that>the asymptotic series cannot be used to give n! precisely afterrounding. It wasn't dif?ult (and I'm now suitably embarrassed that I>hadn't done it before!) Surprise: I got n = 1456. (Well, not a>surprise to you readers. But remember that I had thought n would need>to be bigger than that, and so it surprised me.) And it's good now tohave independent con?mation of the value which Rob obtained.>BTW, I'm not claiming (and I don't suppose that Rob is either) that>the smallest such n is precisely 1456. The smallest such n might be a>little more or less than 1456.> I haven't done the calculation; is 1456 the smallest integer for which> the asymptotic series fails to have an error less than 1/2?> I haven't done the calculation either. It would surely be rather tedious.> But our independent estimates agree that the smallest such n should be> close to 1456. That's good enough to satisfy me.See below.> and separately>Epilog>The Interesting Question has already been satisfactorily answered>now. But I thought that some people might enjoy a nice, pertinentpassage (which I only just now discovered) in the excellent _Concrete>Mathematics_ by Graham, Knuth, and Patashnik.On page 481 (2nd ed.), where m is the number of terms of Stirling's>asymptotic expansion of ln(n!) to be used, they say>... if n is ?ed and m increases,>the error bound |B_(2m+2)|/((2m+2)(2m+1)n^(2m+1)) decreases to acertain point and then begins to increase. Therefore the approximation>reaches a point beyond which a sort of uncertainty principle limits>the amount by which n! can be approximated.> Yes, but you knew this before. All that GKP is saying here is that the> asymptotic series doesn't converge for any n. They are saying nothing> about the minimum error that can be attained by choosing m as a> function of n.> Certainly. I didn't mean to mislead anyone into thinking that they had> somehow answered the basic question I had posed. I just enjoyed their> turn of phrase (...a sort of uncertainty principle limits...) and> thought that some other people might enjoy it too.Oy! Well, GKP didn't answer the question in the passage I'd quoted. But, asfate would have it, I just now stumbled across their answer to my veryquestion (of course, while looking for something unrelated). If I'd beenaware of the following, I would never have started this thread.p. 491, exercise 9.26...If we now round to the nearest integer, knowing that 10! is an integer,we get an exact result.Is it always possible to calculate n! in a similar way, if enough terms ofStirling's approximation are computed?...p. 594, answer to above...The absolute error in ln n! is therefore too large to determine n! exactlyby rounding to an integer, when n is greater than about e^(2 pi + 1).Hmm. It pays to look at their exercises!-------------------------------------------------- What does a drowning analytic number theorist say?log log log log ...--------------------------------------------------David === =I've recently been looking at little and big endian memory storage but haveencountered a little confusion when it comes to transferring or copying databetween architectures of different endianness and would be grateful for someclari?ation.For instance if we had a byte addressable memory with a 32 bit word size ona little endian architecture containing, say, the following multibyte dataitems (e.g. address 1 to 2 stores one 16 bit item, 3 to 6 a 32 bit itemetc):Address 1 = aAddress 2 = bAddress 3 = cAddress 4 = d...Address n = zif we were to copy this data to a big endian architecture of the same sizewhat would we store at each address? Would it actually be the same layout,given that we read out from address 1 to n from the little memory then thebig memory considers what was at address 1 as the most signi?ant byte andso the ordering remains unchanged? Would the ordering be different for abyte by byte transfer compared to a word by word transfer--I can't see anyreason why? And ?ally, if we also had one byte data items at certainaddresses would their order be any different from the rest of the itemscopied over (I am aware that if we were to compare one byte items on alittle and big endian architecture their order would be the same but I waswondering if there is a special case for copying one byte items or do wetreat them the same as all other bytes which I think must be the case sincewe're copying by byte or word and the mechanism doesn't care whether acertain byte belongs to a larger multi-byte value or not and hence treatsthem all the same)? === In sci.math, Perception:> I've recently been looking at little and big endian memory storage but have> encountered a little confusion when it comes to transferring or copying data> between architectures of different endianness and would be grateful for some> clari?ation.> For instance if we had a byte addressable memory with a 32 bit word size on> a little endian architecture containing, say, the following multibyte data> items (e.g. address 1 to 2 stores one 16 bit item, 3 to 6 a 32 bit item> etc):> Address 1 = a> Address 2 = b> Address 3 = c> Address 4 = d> ...> Address n = z> if we were to copy this data to a big endian architecture of the same size> what would we store at each address? Would it actually be the same layout,> given that we read out from address 1 to n from the little memory then the> big memory considers what was at address 1 as the most signi?ant byte and> so the ordering remains unchanged? Would the ordering be different for a> byte by byte transfer compared to a word by word transfer--I can't see any> reason why? And ?ally, if we also had one byte data items at certain> addresses would their order be any different from the rest of the items> copied over (I am aware that if we were to compare one byte items on a> little and big endian architecture their order would be the same but I was> wondering if there is a special case for copying one byte items or do we> treat them the same as all other bytes which I think must be the case since> we're copying by byte or word and the mechanism doesn't care whether a> certain byte belongs to a larger multi-byte value or not and hence treats> them all the same)?>Your speci?ation might be better rendered using C-style or C++-stylestructures. For example, your spec above seems to be the structurestruct Examp { short item16; int item32;};with the proviso that int on these hypotheticalarchitectures is 4 bytes -- which isn't always a givennowadays, because of 64-bit architectures. (I should alsopoint out that historically int was 2 bytes, way back inthe old PDP 11/70 16-bit architecture days.)So now we declare struct Examp a = {1,2} (the typical Cinitialization syntax, which stores 1 into a.item16 and2 into a.item32). What are we storing?If we de?e struct Examp * aPtr = &a (the ?&' operator isthe location of a variable, in this case a; aPtr is a pointerto a), and we hypothesize that aPtr has the value 0x112234(pointers being traditionally written in hex), we can thenstipulate that in little-endian memory we will store thefollowing bytes.0x112234: 10x112235: 00x112236: 20x112237: 00x112238: 00x112239: 0In big-endian the ordering is quite different:0x112234: 00x112235: 10x112236: 00x112237: 00x112238: 00x112239: 2(I'm assuming here that there's no padding between item16 and item32.Most modern compilers place a hidden 16-bit item between them,to attempt to make accesses to item32 more ef?ient. Thereason has to do with how memory is organized; the microprocessormight fetch 32-bit items as a single 32-bit fetch or as a pair of32-bit fetches, lopping off the bits it doesn't like. Or perhapsit does 2 16-bit fetches. The single 32-bit fetch is faster,but requires a 32-bit aligned address. This hidden item willmost likely be initialized to random values.)One can verify what one's machine is actually doing in a numberof slightly hackish ways; for example, if one's using C or C++,one can write the program snippet unsigned char * bPtr = ((unsigned char *) aPtr); for(int i = 0; i < sizeof(a); i++) { printf(%p: %02xn, &bPtr[i], bPtr[i]); }which should work in all cases; however the usual caveats apply.As you've no doubt already noticed a naive transfer of data(e.g., fread()/fwrite() over a socket) from a little-endianto a big-endian machine will screw up things very quickly.One project I worked on simply speci?d at the top ofthe ?e what endianity was used during writing; if themachine reading the ?e was not the same endianity itwas expected to ?e individual components of eachstructure as it was reading. However, this won't workif transferring from a 32-bit to a 64-bit architecturewithout very careful data declarations.More modern code either uses one endianity during writingor uses less ef?ient but more portable means such asASCII or XML to store data on ?es or send it over thenetwork.Java also has a number of software classes to circumvent this problem.HTH-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === and where they come from?bccIf Yilmaz et al. are right that static weak ?ld n-body Einstein GR solutionsdon't exist, then this may turn out to be a real problem after all.I think the exponential metric in the isotropic radial coordinate has serious problems independent of the precise metric model, Yilmaz or Einstein because of topological reasons. I mean there are an in?ity of isotropic coordinate patches outside the turning point boundary at GM/c^2 for a single curvature coordinate. In Einstein's GR this ratio is only 2:1, i.e. 2 coordinate patches outside the event horizon at 2GM/c^2 in the Penrose-Kruskal diagram with 4 coordinate patches covering the entire vacuum manifold.That is, in any SSS metric theory look at the relation between isotropic and curvature radial coordinates. You get a polynomial of degree N with the curvature radial coordinate as a control parameter. The N roots for isotropic r, at least when real, each de?e a coordinate patch. In the exponential case that Hal uses in his PV model, N --> in?ity. In Einstein's GR N = 2.On Gravity LensingThe general SSS metric, for simplicity, independent of the action and/or local metrical ?ld equation, is in the curvature radial coordinateds^2 = gtt(r)(cdt)^2 - grr(r)dr^2 - r^2(dtheta^2 + sin^2thetadphi^2)A light ray obeys,ds^2 = 0Using Einstein's theorydT = gtt(r)^1/2dtdT is what a clock at rest relative to the center of symmetry in the LNIF measures between neighboring events P and P + dP.dR = grr(r)^1/2drdR is what a radar or a measuring rod measures.where t, r are local coordinates in the rest LNIF that is a point on a timelike non-geodesic.LNIF's can only exist if there are non-gravity forces.Suppose the light ray also has a component along a tangent to a longitudinal great circle at r so that, dphi = 00 = gtt(r)(cdt)^2 - grr(r)dr^2 - r^2dtheta^2c^2dT^2 = dR^2 + r^2dtheta^2c^2 = (dR/dT)^2 + r^2(dtheta/dT)^2Therefore, the speed of light is always c.However, we have a right triangle above;1 = (vr/c)^2 + c^-2 r^2w^2vr/c = cosChiw = physical angular speed of light pulse = gtt^-1/2w'w' = dtheta/dt1 = (grr/gtt)c^-2(dr/dt)^2 + c^-2r^2gtt^-1(dtheta/dt)^2 = cos^2Chi + sin^2ChitanChi = sinChi/cosChi = grr^1/2(dr/dt)/(dtheta/dt) = grr^1/2(dr/dtheta)Let's take Einstein's theory as a de?ite example.gtt = 1 - 2GM/c^2rgrr = (1 - 2GM/c^2r)^-1outside event horizon at r* where 1 - 2GM/c^2/r* = 0only when r > 2GM/c^2 outside the throat of radius r* of this non-traversable wormhole that pinches off crushing you if you try to ?ough it.as an asymptotically ?herically symmetric static wormhole Einstein-Bridge solution of Ruv = 0i.e. Tuv(Non- Exotic Vacuum) = 0The gravity lens effect has one obvious measuretanChi = (1 - 2GM/c^2r)^-1/2(dr/dtheta)Note that tanChi = in?ity at the event horizon, i.e. the light ray is trapped in a circular orbit con?ed to the surface of the event horizon. There is no radial component.Now this is pretty. Much prettier than Hal's model IMHO. Einstein is a genius in terms of pure aesthetics. Beauty is Truth.Bottom line, gravity lensing does give information about r the curvature radial coordinate where the concentric surface area is 4pir^2 with an entropy of 4pir^2/Lp^2 c-bits. One curvature radial coordinate has more than one isotropic radial coordinate corresponding to it. The map is not 1-1.I'll have to give you a copy of Brian Tupper's 1974 Nuovo Cimento paperwhich deals with a parametrized class of geometrodynamic theories that areconsistent with the four classical tests of GR. Also, a 1999 paper by Alley,Yilmaz et al. that restates the entire n-body argument in the context of morerecent empirical data....That is what I call semantic incoherence.JS: This is to be expected from Noether's theorem! It is trivially sothat the bigger theory always violates Sacred Cows of the smallerearlier theory e.g. Newton's Absolute Time as a slaughtered sacredcow. But Einstein, The Great Rabbi, did it Kosher!PZ: But then how do you explain the existence of these conservedenergy-momentum integrals in Newtonian theory --arguably the most successful physical theory of all time?JS: Trivial - action at a distance in globally ?ce time. No problemthe translation group symmetry works.PZ: : Are you saying that you don't need to recover the classical conservationprinciples in the Newtonian limit?JS: No I said just the opposite. I told you why it works. Curvature meansviolation of the translational symmetry from which conservation ofmomenergy comes. Of course one hopes that the compensating gauge ?ld,in this case the geometrodynamic guv ?ld restores the broken symmetry.It does do that in the sense of locally conservation of stress-energy densitycurrents when all dynamical degrees of freedom are included.Tuv(Non Exotic Vacuum) + Tuv(Exotic Vacuum) + Tuv(Matter) = 0The total covariant 4-divergence vanishes, but not the individual termsseparately in the practical metric engineering regime whose technologywe see in the saucers from the brane worlds next door is my educatedguess. Brian Greene thought he was joking on NOVA, the joke may be onhim - let us hope. :-)More anon. === SeeReplacing Einstein's SR and GRA Uni?d Classical Theory of the Electric, Magnetic and GravitationalForcesBruce Harveyhttp://users.powernet.co.uk/bearsoft/Paper7.html-- Bruce Harveybruce@bearsoft.co.ukThe Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft> and where they come from?> bcc> If Yilmaz et al. are right that static weak ?ld n-body Einstein GR> solutions> don't exist, then this may turn out to be a real problem after all.> I think the exponential metric in the isotropic radial coordinate has> serious problems independent of the precise metric model, Yilmaz or> Einstein because of topological reasons. I mean there are an in?ity of> isotropic coordinate patches outside the turning point boundary at> GM/c^2 for a single curvature coordinate. In Einstein's GR this ratio is> only 2:1, i.e. 2 coordinate patches outside the event horizon at 2GM/c^2> in the Penrose-Kruskal diagram with 4 coordinate patches covering the> entire vacuum manifold.> That is, in any SSS metric theory look at the relation between isotropic> and curvature radial coordinates. You get a polynomial of degree N with> the curvature radial coordinate as a control parameter. The N roots for> isotropic r, at least when real, each de?e a coordinate patch. In the> exponential case that Hal uses in his PV model, N --> in?ity. In> Einstein's GR N = 2.> On Gravity Lensing> The general SSS metric, for simplicity, independent of the action and/or> local metrical ?ld equation, is in the curvature radial coordinate> ds^2 = gtt(r)(cdt)^2 - grr(r)dr^2 - r^2(dtheta^2 + sin^2thetadphi^2)> A light ray obeys,> ds^2 = 0> Using Einstein's theory> dT = gtt(r)^1/2dt> dT is what a clock at rest relative to the center of symmetry in the> LNIF measures between neighboring events P and P + dP.> dR = grr(r)^1/2dr> dR is what a radar or a measuring rod measures.> where t, r are local coordinates in the rest LNIF that is a point on a> timelike non-geodesic.> LNIF's can only exist if there are non-gravity forces.> Suppose the light ray also has a component along a tangent to a> longitudinal great circle at r so that, dphi = 0> 0 = gtt(r)(cdt)^2 - grr(r)dr^2 - r^2dtheta^2> c^2dT^2 = dR^2 + r^2dtheta^2> c^2 = (dR/dT)^2 + r^2(dtheta/dT)^2> Therefore, the speed of light is always c.> However, we have a right triangle above;> 1 = (vr/c)^2 + c^-2 r^2w^2> vr/c = cosChi> w = physical angular speed of light pulse = gtt^-1/2w'> w' = dtheta/dt> 1 = (grr/gtt)c^-2(dr/dt)^2 + c^-2r^2gtt^-1(dtheta/dt)^2 = cos^2Chi +> sin^2Chi> tanChi = sinChi/cosChi = grr^1/2(dr/dt)/(dtheta/dt) = grr^1/2(dr/dtheta)> Let's take Einstein's theory as a de?ite example.> gtt = 1 - 2GM/c^2r> grr = (1 - 2GM/c^2r)^-1> outside event horizon at r* where 1 - 2GM/c^2/r* = 0> only when r > 2GM/c^2 outside the throat of radius r* of this> non-traversable wormhole that pinches off crushing you if you try to ?rough it.> as an asymptotically ?herically symmetric static wormhole> Einstein-Bridge solution of Ruv = 0> i.e. Tuv(Non- Exotic Vacuum) = 0> The gravity lens effect has one obvious measure> tanChi = (1 - 2GM/c^2r)^-1/2(dr/dtheta)> Note that tanChi = in?ity at the event horizon, i.e. the light ray is> trapped in a circular orbit con?ed to the surface of the event> horizon. There is no radial component.> Now this is pretty. Much prettier than Hal's model IMHO. Einstein is a> genius in terms of pure aesthetics. Beauty is Truth.> Bottom line, gravity lensing does give information about r the> curvature radial coordinate where the concentric surface area is 4pir^2> with an entropy of 4pir^2/Lp^2 c-bits. One curvature radial coordinate> has more than one isotropic radial coordinate corresponding to it. The> map is not 1-1.> I'll have to give you a copy of Brian Tupper's 1974 Nuovo Cimento paper> which deals with a parametrized class of geometrodynamic theories that are> consistent with the four classical tests of GR. Also, a 1999 paper byAlley,> Yilmaz et al. that restates the entire n-body argument in the context of> more> recent empirical data.> ...> That is what I call semantic incoherence.> JS: This is to be expected from Noether's theorem! It is trivially so> that the bigger theory always violates Sacred Cows of the smaller> earlier theory e.g. Newton's Absolute Time as a slaughtered sacred> cow. But Einstein, The Great Rabbi, did it Kosher!> PZ: But then how do you explain the existence of these conserved> energy-momentum integrals in Newtonian theory --> arguably the most successful physical theory of all time?> JS: Trivial - action at a distance in globally ?acetime. No problem> the translation group symmetry works.> PZ: : Are you saying that you don't need to recover the classical> conservation> principles in the Newtonian limit?> JS: No I said just the opposite. I told you why it works. Curvature means> violation of the translational symmetry from which conservation of> momenergy comes. Of course one hopes that the compensating gauge ?ld,> in this case the geometrodynamic guv ?ld restores the broken symmetry.> It does do that in the sense of locally conservation of stress-energy> density> currents when all dynamical degrees of freedom are included.> Tuv(Non Exotic Vacuum) + Tuv(Exotic Vacuum) + Tuv(Matter) = 0> The total covariant 4-divergence vanishes, but not the individual terms> separately in the practical metric engineering regime whose technology> we see in the saucers from the brane worlds next door is my educated> guess. Brian Greene thought he was joking on NOVA, the joke may be on> him - let us hope. :-)> More anon.> === > The obvious question is whether or not the ?th output would be> prime...That's (2^((2^127) - 1)) - 1 .> A well known question. See> http://primes.utm.edu/mersenne/index.html> near the bottom.>There's no mention of one pair ? into another pair...(but that'slikely just a consequence of the Mersenne Primes being close together whenthe values are small). They adhere to pairs and there is a project sieving2^((2^61) - 1) - 1 . === > The obvious question is whether or not the ?th output would be> prime...That's (2^((2^127) - 1)) - 1 .> A well known question. See> http://primes.utm.edu/mersenne/index.html> near the bottom.> There's no mention of one pair ? into another pair...(but that's> likely just a consequence of the Mersenne Primes being close together when> the values are small). They adhere to pairs and there is a project sieving> 2^((2^61) - 1) - 1 .>Oh, there it is...it's the Catalan sequence. === > I haven't tried to prove it, but it occurred to me after posting that if one> does the same construction with the closed interval [a,b], but leaves out> (a,-1) and (b,+1), then one ought to get the maximal ideal space of the> algebra of functions on [a,b] with right and left hand limits at every point,> which ought to be a Banach algebra in the sup norm.> Of course, even if that is correct, it isn't a name...>Why not {+-oo} / R with order -oo < r < oo, r in Rand r < s when r,s in R and r less than s,a homeomorph of [0,1] ? === > I think that> 2^n = sum (nCk) (k ranges from 0 to n)> (by nCk I mean n!/(k!(n-k)!))> but can't prove it. does anyone have any hints? Induction, perhaps?-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes /039 53 36 N / 086 11 55 W === > I think that> 2^n = sum (nCk) (k ranges from 0 to n)> (by nCk I mean n!/(k!(n-k)!))> but can't prove it. does anyone have any hints? 2^n = (x+1)^n, when x = 1. === Instead of looking for solutions to nonlinear differential equations using solutions as series of polynomials, why not try solutions as series of rational powers, ie. y=Sum(x^(u/v)), where the sum would be over both ?u' and ?v', with u != n*v to eliminate redundancy (!= is ?not equal'). Instead of going to in?ity, it would be over a fairly small set of integers, say 0<=u<=5, and 1<=v<=5. Has anybody ever tried this? === > Instead of looking for solutions to nonlinear differential equations using> solutions as series of polynomials, why not try solutions as series of> rational powers, ie. y=Sum(x^(u/v)), where the sum would be over both ?u'> and ?v', with u != n*v to eliminate redundancy (!= is ?not equal').> Instead of going to in?ity, it would be over a fairly small set of> integers, say 0<=u<=5, and 1<=v<=5. Has anybody ever tried this?>The orginal idea expressed is good.Perhaps you should investigate theta series or psi series. === > Instead of looking for solutions to nonlinear differential equations using > solutions as series of polynomials, why not try solutions as series of > rational powers, ie. y=Sum(x^(u/v)), where the sum would be over both ?u' > and ?v', with u != n*v to eliminate redundancy (!= is ?not equal'). > Instead of going to in?ity, it would be over a fairly small set of > integers, say 0<=u<=5, and 1<=v<=5. Has anybody ever tried this?Replacing x by t^v in the original DE, and expressing it in terms ofdy/dt etc, i.e. eliminating x entirely, in principle you can ?d theusual series solutions in the form y = sum(t^u).---------------------------------------------------- -----------------------John R Ramsden (jr@adslate.com)--------------------------------------------- ------------------------------Eternity is a long time, especially towards the end. Woody Allen === > Instead of looking for solutions to nonlinear differential equations using > solutions as series of polynomials, why not try solutions as series of > rational powers, ie. y=Sum(x^(u/v)), where the sum would be over both ?u' > and ?v', with u != n*v to eliminate redundancy (!= is ?not equal'). > Instead of going to in?ity, it would be over a fairly small set of > integers, say 0<=u<=5, and 1<=v<=5. Has anybody ever tried this?It obviously does not work. What solutions of this type has thedifferential equation f' = f, in your opinion?Jose Carlos SantosA question Say I have a numberit is the summation of other numbersgiven the following set of rules how can I calculate which sets of numbers sum to the original number.e.g. d = a + b + c;and d = 10;what are the possible values of a,b and c.rulesall numbers are unique and > 0 and < d (10)i.e a < b < c a,b and c > 0a,b and c < d (10 in example)so all possible values are 1 + 2 + 7 = 101 + 4 + 5 = 102 + 3 + 5 = 102 + 2 + 6 = 10 *Illegal* breaks a < b ruleso there is 3 possible combinations my actual problem involves a much larger dataset and iterating over the possible values runs into the billions of iterationsI wish to short circuit that and ?d all possible combinations in the least possible timeand thanks all for previous help. its what has got me to this point :)ChrisP.S. I hope my explanation has not made a possibly simple problem seem complex. === This is computing.For a = 0 to d FOR b = a To d FOR c = b TO d IF a+b+c = d THEN PRINT a,b,c next nextnext-- Bruce Harveybruce@bearsoft.co.ukThe Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft> A question> Say I have a number> it is the summation of other numbers> given the following set of rules how can I calculate which sets of> numbers sum to the original number.> e.g. d = a + b + c;> and d = 10;> what are the possible values of a,b and c.> rules> all numbers are unique and > 0 and < d (10)> i.e> a < b < c> a,b and c > 0> a,b and c < d (10 in example)> so all possible values are> 1 + 2 + 7 = 10> 1 + 4 + 5 = 10> 2 + 3 + 5 = 10> 2 + 2 + 6 = 10 *Illegal* breaks a < b rule> so there is 3 possible combinations> my actual problem involves a much larger dataset and iterating over the> possible values runs into the billions of iterations> I wish to short circuit that and ?d all possible combinations in the> least possible time> and thanks all for previous help. its what has got me to this point :)> Chris> P.S. I hope my explanation has not made a possibly simple problem seem> complex. === > This is computing.> For a = 0 to d> FOR b = a To d> FOR c = b TO d> IF a+b+c = d THEN PRINT a,b,c> next> next> next> You program excerpt cycles through a = 0 and a >= d/3 -2 and b >= d/3 + 1 and c = d, all of which are prohibited by the rules. Consider:For a = 1 to d/3 -1 FOR b = a + 1 TO (d - a - 1)/2 c = d-a-b PRINT a,b,c nextnextThis runs much faster, and covers all the allowable cases.> -- > Bruce Harvey> bruce@bearsoft.co.uk> The Alternative Physics Site> http://users.powernet.co.uk/bearsoft> A question> Say I have a number> it is the summation of other numbers> given the following set of rules how can I calculate which sets of> numbers sum to the original number.> e.g. d = a + b + c;> and d = 10;> what are the possible values of a,b and c.> rules> all numbers are unique and > 0 and < d (10)> i.e> a < b < c> a,b and c > 0> a,b and c < d (10 in example)> so all possible values are> 1 + 2 + 7 = 10> 1 + 4 + 5 = 10> 2 + 3 + 5 = 10> 2 + 2 + 6 = 10 *Illegal* breaks a < b rule> so there is 3 possible combinations> my actual problem involves a much larger dataset and iterating over the> possible values runs into the billions of iterations> I wish to short circuit that and ?d all possible combinations in the> least possible time> and thanks all for previous help. its what has got me to this point :)> Chris> P.S. I hope my explanation has not made a possibly simple problem seem> complex.> === > This is computing.> For a = 0 to d> FOR b = a To d> FOR c = b TO d> IF a+b+c = d THEN PRINT a,b,c> next> next> next>That will generate a few inadmissible sets, e.g. 0,0,10 and has anunnecessary loop and if statement. I think the following should do the job(but please check this thoroughly OP).for i = 1 to ?n-3)/3) for j = i+1 to ?n-i-1)/2) print i,j,n-i-jor in Pythondef f(n): for i in range(1, (n-3)/3+1): for j in range(i+1, (n-i-1)/2+1): print i,j,n-i-j> f(13)1 2 101 3 91 4 81 5 72 3 82 4 72 5 63 4 6>Duncan === Your de?ition is for in?ite subsets of the reals to be calledrecognizable. Then, you ask which particular numbers arerecognizable. However, you gave no de?ition for a real number to berecognizable. Perhaps you mean to ask which numbers are inrecognizable sets.If you are actually asking about which sets are recognizable, then thecountable subsets that include 0 (and only the countable subsets thatinclude 0) of the reals are recognizable. Take any countable set S ofthe reals. Since it is countable, there is a bijection between S andthe integers, Z. Let g be this bijection. Then let f(x) = g^{-1}( 1+ g(x)). f essentially increases the index of a point in the set S. Then for any point x in S, g(s) is an integer, and f^{g(0)-g(x)}(x) =0.If a set does not include 0 and f is a bijection on that set, thenobviously, for every x in the set and every k an integer, f^k(x) isnot 0.Suppose that S is uncountable, 0 is in S, and f is any bijection on S. Then the set of all f^{-k}(0) for k an integer is at most countable. Thus, the cardinality of all x such that f^k(x) = 0 is at mostcountable. Thus, there is some y in S so that f^k(y) is never 0 forany integer k.~ Chris > Call an in?ite subset X of the real numbers recognizable if and only if there> is a bijection f:R->R such that for all x in X there is an integer k (possibly> dependent on x) with f^k(x)=0. [f^2(x)=f(f(x)), ect.]> Say a real number x is recognizable iff x in X for some recognizable X.> What numbers are recognizable?> rich === >Your de?ition is for in?ite subsets of the reals to be called>recognizable. Then, you ask which particular numbers are>recognizable. However, you gave no de?ition for a real number to be>recognizable. Perhaps you mean to ask which numbers are in>recognizable sets.>A real number x is recognizable iff x in X for some recognizable X. As youvery nicely show this means all reals are recognizable. So my de?ition waskind of a dumb one. What I want is the recognizable numbers to be, well,recognizable. They should include the rationals, the algebraics, and perhapssome subset of the familar transcendentals, but *not* all the reals.There is a fair amount of wiggle room in the de?ition. For example, if werequire X to be a dense subset of [0,1] and f:[0,1]->[0,1], that complicatesthings a little. Is it enough to meet my goals? Probably not. Perhaps somestructure on X or restriction on f is needed to avoid being able to append anarbitrary r to X. I really don't know how to get at what I want. Intuitively,the de?ition seems to be a good one, but perhaps it's just another dumbidea...happens all the time during the full moon. >If you are actually asking about which sets are recognizable, then the>countable subsets that include 0 (and only the countable subsets that>include 0) of the reals are recognizable. Take any countable set S of>the reals. Since it is countable, there is a bijection between S and>the integers, Z. Let g be this bijection. Then let f(x) = g^{-1}( 1>+ g(x)). f essentially increases the index of a point in the set S.> Then for any point x in S, g(s) is an integer, and f^{g(0)-g(x)}(x) =>0.>If a set does not include 0 and f is a bijection on that set, then>obviously, for every x in the set and every k an integer, f^k(x) is>not 0.>Suppose that S is uncountable, 0 is in S, and f is any bijection on S.> Then the set of all f^{-k}(0) for k an integer is at most countable. >Thus, the cardinality of all x such that f^k(x) = 0 is at most>countable. Thus, there is some y in S so that f^k(y) is never 0 for>any integer k.> Call an in?ite subset X of the real numbers recognizable if and only if>there> is a bijection f:R->R such that for all x in X there is an integer k>(possibly> dependent on x) with f^k(x)=0. [f^2(x)=f(f(x)), ect.]> Say a real number x is recognizable iff x in X for some recognizable X.> What numbers are recognizable?> rich === >Proof. Given an arbitrary real number r, de?e g_r:R->R to be the>function g_r(x) = x+r, and let X_r be:>X_r = {kr: k an integer}.g_r is a bijection; and (g_r)^k(x) = x+kr for every integer k. In>particular, for every x in X_r, x=kr for some k, we have >(g_r)^{-k}(x)= x -kr = 0.>Since r is in X_r, it follows that any real number is in a>recognizable set, and therefore, that every real number is>recognizable. QED> Are there any simple restrictions on k and/or f that will make the> recognizable numbers a proper subset of R that includes the> algebraic numbers?It seems unlikely -- the functions g_r are about as nice as youcould imagine, and with them, every real is recognizable with krestricted to 1.You might be able to get interesting limitations on what *sets* X arerecognizable.Dale === Fish, or cut bait.Solve problems.Borel says almost all sequences are normal and have equal zero and onedensities, combinatorics say almost no sequences do, I think half do.How are reconciled Borel and asymptotic combinatorics?Sets of numbers contain only sets. Ross === > Fish, or cut bait.> Solve problems.> Borel says almost all sequences are normal and have equal zero and one> densities, combinatorics say almost no sequences do, I think half do.I presume Borel is Emile Borel (1871-1956).Who is combinatorics?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === I have an idea what a proof is. In my mind there's a concept of whata theory is, and a theorem, and what an axiom is. I have an inklingof what a conjecture is.What is it when two mathematical statements, each true, contradicteach other?Generally it's a sign that one or the other is false.What is it when the combinatoric analysis of binary sequencescontradicts with accepted statements about the prevalence of normalnumbers?It's not a paradox. There are no paradoxes. Thus, I suspect an error in my reasoning. I've corrected some errors. Others presumably remain to correct. Yet, on this point, in over amonth sci.math collected has not presented an error to explain awaythis discrepancy. This is where errors presented on sci.math areoften readily corrected.There are no paradoxes. It's interesting to read the phrase prime at in?ity. I conjecturethat there is a composite at in?ity. Then again, I think theuniverse is in?ite, in?ite in 3-D space.I haven't made any progress in determing the rational functions torepresent the Stirling number s(n+1, n+1-x) except as already stated. Once past 3-cubes it just isn't so simple. The literature describesclosed forms for integer values of n and x, just not rationalfunctions, except for those described. There probably are others, yetthere may not be: summing the products of k-subsets of an n-set.A method to enumerate k-subsets of an n-set was discussed, we candiscuss it.An approximation for n!, one of many, was discussed, able to beaccurate to several digits. Some description of the factors of s(n+1,n+1-x) was described.I wish I knew more about complex analysis, that is where trying tointerpolate among the functions describing s(n+1, n+1-x) for positivereal n and x less than n goes.I think about the digit summation congruence factorizationprepreprocessor, I've been reviewing some of the processor literatureand its requirements are low in cycle count. It's only good for smallMersenne factors, binary rep-units. Are any Mersenne numbers integerpowers?Consider the reformulated Gamma function, if you will. Picture this: scalar in?ities.Ross--Ross A. Finlaysonhttp://www.tiki-lounge.com/~raf/It's always one more. === I haven't the foggiest...On the off chance you don't know, the argument of (1-3i)^4 is fourtimes the argument of 1-3i and the modulus of (1-3i)^4 is the fourthpower of the modulus of 1-3i.> Does anyone know why for example my calculate can produce the argument and> modulus for 1 - 3i no problems at all, but if I enter (1-3i)^4 I get a Math> Error? === If you need to work it out by hand, here's what you'd do.r=sqrt(1^2+3^2)=sqrt(10)theta=arctan(-3)=-1.25sqrt(10)^4( cos(4*-1.25)+isin(4*-1.25))100(.283+.96i)28.3+96i-- David MoranChief MeteorologistOklahoma Storm Team> Does anyone know why for example my calculate can produce the argument and> modulus for 1 - 3i no problems at all, but if I enter (1-3i)^4 I get aMath> Error?> === > Does anyone know why for example my calculate can produce the argument and> modulus for 1 - 3i no problems at all, but if I enter (1-3i)^4 I get a Math> Error? === > Does anyone know why for example my calculate can produce the argument and> modulus for 1 - 3i no problems at all, but if I enter (1-3i)^4 I get a Math> Error?> What make and model of calculator do you have? === In message <2a0cceff.0312101407.2225db91@posting.google.com>, Edward|This stirs some vague recollection: there are some privileged|irrational numbers which can be speci?d with a ?ite amount of|information. Obviously _all_ irrational numbers cannot be so|speci?d, or they in fact would be countable. So most irrational|numbers are poor lost souls which not only have non-repeating decimal|representations, but can't even be named in any meaningful way --they|are unknowable.The concept of speci?ation has some pitfalls here. Ordinarily, onedeals only with speci? means of speci?ation. If we ? a languageL, each number r, 0<=r<1, which can be speci?d (using a ?iteamount of information) *using L* ordinarily corresponds to a formulaP(n) in L having one free natural number variable, which is true whenthe n-th bit of r is 1, and is false when the n-th bit of r is 0.Given a number we can de?e, r, we can also de?e P(n) to be then-th bit of r is 1, and given P(n), we can de?e r to be the numbergiven by the binary expansion whose n-th bit is 1 for the n satisfyingP(n) and 0 otherwise.This depends on the language, though. There is no ultimate language,however. Once you have such a language L, you can always extend it toa bigger language in which more reals can be de?ed.That being said, it's still true that in a loose, informal sense, oneusually (i.e., in mainstream mathematics) thinks of the typical realnumber as having a binary expansion which is so totally arbitrary thatthere won't be any way at all to de?e it.One would like to make this precise, but as far as I know there's noespecially precise meaning to it. For instance, you can't say thatthere exists a real number r which can't be speci?d in any languageL. For each r, we can construct a language L_r based on r, which hasas one of its basic terms P(n) de?ed to be true when the n-th bitof r is 1.The concept of speci?ation is in much the same predicament as theconcept of proof. We have an informal understanding of each, butno mathematical de?ition of either one as an absolute concept.Instead logicians consider the concept of proof within a given system,and speci?ation within a given system.|(This was the subject of some (IBM?) news release|within the past decade, possibly one of those over-hyped newsreleases|which appear regularly, repeating essentially known results as ifwere|fresh revelation).||Can somebody remind me what this concept is called? Do these|unknowable irrational numbers correspond to non-computable functions?I would bet that you saw an IBM press release about some work byGregory Chaitin. The last I heard he was working for them. He's donesome work on _uncomputable_ numbers. Uncomputable numbers correspondto uncomputable functions giving their binary or decimal expansions.most famous example is called Chaitin's Omega. Omega is uncomputable,but it's the upper bound of an increasing, computable sequence ofrational numbers. Such a sequence converges but not constructively.The sequence lies between 0 and 1, so for each epsilon>0, we can saythat the sequence will not move upward by epsilon more than 1/epsilontimes. Nonconstructively, that implies that eventually the sequencewill cease to move by more than epsilon. Since this holds for everyepsilon>0, one says that the sequence converges.The reason this is not constructive is that you have no method fordetermining when the sequence is ?ally done moving by more thanepsilon. It is sort of tricky. It can hang around a very long timeapparently done with moving by increments of more than 1/2^1000000,only to trick you by making such a jump. That's because it makes thesesteps based on whether Turing machines stop. Chaitin de?ed it interms of LISP programs. If the n-th LISP program is found to stop,then 1/2^n is added to the sum (or something like that).In short, Chaitin studies uncomputable irrational numbers, but oneswhich can be de?ed.Now I see there was some discussion following your question whichreferred to some particular ways of de?ing numbers. Giving analgorithm which computes it is certainly a way to de?e a number,just not the only way. Algebraic numbers are numbers which are rootsof nonzero polynomials with integer coef?ients, and they are allnot only computable, but relatively ef?iently computable. They'renot the only computable numbers by a long shot. It turns out ordinaryfamiliar transcendental numbers like pi and e are also not so hardto compute either. One could cook up a number which was very hard tocompute, but these numbers don't seem to appear individually inmathematics very often.|Yes, it is perfectly consistent to believe that the only real numbers|that exist are computable reals. However, Cantor's theorem is *still*|valid: Under this interpretation, it says that there is no computable|listing of all computable reals.Note that it's consistent only if one drops the law of excludedcan prove that Chaitin's Omega exists and is uncomputable.Constructive mathematics drops these two assumptions because they'renonconstructive. Cantor's theorem is still valid (although one useshis ?st proof, rather than the diagonal argument, because one isinterested in real numbers given by convergent sequences, rather thanby decimal expansion. Given a convergent sequence it's not alwayspossible to compute its decimal expansion.)The computable real numbers can be encoded by giving a correspondingTuring machine and then turned into natural numbers. The reason thisdoesn't constructively give an enumeration of the computable realsis that not every natural number is an encoding of a real number, andthere's no method for determining which does and which does not inevery case.If we assume the law of excluded middle (as is usually done), we canprove that there is an enumeration of the computable real numbersobtained by de?ing the real number corresponding to a natural numberwhose corresponding Turing machine doesn't de?e one just to be 0(just to keep it from being unde?ed). The diagonal number is anuncomputable real. But the sequence isn't computable, since there'sno way in general to tell whether a Turing machine de?es a realnumber.Keith Ramsay === --www.StealthHostiing.com You rule Truman. http://tinyurl.com/iky4 Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! >Is the truman living in Townsville? I've been hearing stuff, yeah.Webmasters help the TRUEman by joining www.theBanner.net Current:1 Goal:1000--------------------------------------------------- -------------------------------> It seems odd Information Theory is wiped out of existence from a single> ?proof'> that is self referential, can anyone give me an actual number that isn't> computable?> Herc> You would think that it was impossible to specify a number that wasn't> computable - the act of specifying the number would seem to require> specifying the algorithmic process which generates it.> Well, you would be wrong. Here is exactly what you asked for:> http://mathworld.wolfram.com/ChaitinsConstant.html> Is this completely fair? If there is a Turing Halting problem that> is undecidable within our own system of reasoning? Is it then still> a proper de?ed number?Halt does not exist. There is only an oracle that gives its values.The Halting proof begins with Halt having 2 values : 0 or 1.The proof then shows it doesn't exist.In reality it has 2 known values plus an undetermined_as_yet.I'm unsure if a speci? algorithm can be proven to have no resolution at all on its halting value,that would be a 4th value.We could construct a similar number.x1 = 1 iff God is real, zero otherwise. (non computable!)x2 = 1 iff there are 2 gods of equal powerx3 = 1 iff there are 3 gods of equal power...I added the ...of equal power so more than one proposition can hold.Now we can construct the non computable irrational!longlastingduels = E(xn) 2^-nhttp://tinyurl.com/yo9wA large discussion on making numbers from unknowns.Herc === In sci.math, |-|erc --> www.StealthHostiing.com You rule Truman. http://tinyurl.com/iky4> Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky!>Is the truman living in Townsville? I've been hearing stuff, yeah.> Webmasters help the TRUEman by joining www.theBanner.net Current:1 Goal:1000> ------------------------------------------------------------ ----------------------> It seems odd Information Theory is wiped out of existence> from a single ?proof' that is self referential, can anyone> give me an actual number that isn't computable?> Herc> You would think that it was impossible to specify a number that wasn't> computable - the act of specifying the number would seem to require> specifying the algorithmic process which generates it. Well, you would be wrong. Here is exactly what you asked for:> http://mathworld.wolfram.com/ChaitinsConstant.html> Is this completely fair? If there is a Turing Halting problem that> is undecidable within our own system of reasoning? Is it then still> a proper de?ed number?> Halt does not exist. There is only an oracle that gives its values.> The Halting proof begins with Halt having 2 values : 0 or 1.> The proof then shows it doesn't exist.> In reality it has 2 known values plus an undetermined_as_yet.> I'm unsure if a speci? algorithm can be proven to have no resolution at all on its halting value,> that would be a 4th value.> We could construct a similar number.> x1 = 1 iff God is real, zero otherwise. (non computable!)> x2 = 1 iff there are 2 gods of equal power> x3 = 1 iff there are 3 gods of equal power> ...> I added the ...of equal power so more than one proposition can hold.> Now we can construct the non computable irrational!> longlastingduels = E(xn) 2^-nIrrational is right. :-) If one wanted to structure thisone would probably want a more logical organization, suchas the cardinality of the set of all gods. If we de?ethis set as G, then card(G) would be 1 for monotheism,>1 for any polytheistic religion (Wicca, for instance,stipulates a cardinality of 2: one male, one female;classical Greek mythology stipulates quite a number,as does Norse), and 0 for some variants of atheism.card(G) is of course highly disputable in light of the veryunde?ed notion of what a deity is and which religion isthe One True Truth(tm), but from it one might computecard(G) * (card(G)-1) / 2the number of possible god-pairings. In the case ofcertain religions the numbercard(G) * card(P)might be relevant as well, where P is the set of all people.Zeus in particular loved to hang out with the ladies. :-)Demigods would probably be put in a separate set D, as opposedto contributing 1/2 to card(G) -- assuming any of this makeslogical sense at all.As for the halting problem: if one stipulates that one cancode it in a series of addresses P_k (with input beinga pointer to the ?st instruction of the algorithm),returning 1 in R0 if the program halts, and 0 if it doesn't,I can code the following, somewhere else. [*] Q_1: PUSHAB Q_1 Q_2: CALLS #1, P_1 Q_3: TESTL R0 Q_4: JNZ Q_2 Q_5: HALT(PUSHAB = Push Address of Byte, CALLS = Call using Stack, TESTL = Test Longword JNZ = Jump if NonZero)Does Q_1 halt? If it does, then R0 is 1 at Q_3, andthe program loops. If it doesn't, then R0 is 0 at Q_3,and the program halts.Therefore R0 is unde?ed at Q_3, and the original assumptionis not tenable.> http://tinyurl.com/yo9w> A large discussion on making numbers from unknowns.> Herc>[*] This is a corrupted variant of VAX assembly code, if anyone really cares. :-) VAX also de?es a CALLG, which uses a static argument list, for example. There are also PUSHAW, PUSHAL, and PUSHAQ.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === > Is this completely fair? If there is a Turing Halting problem that> is undecidable within our own system of reasoning? Is it then still> a proper de?ed number?What do you mean by properly de?ed? === > What do you mean by properly de?ed? What do you mean by predicative proof step? And so on. === > In message , Daryl McCullough ... > _I_ was referring to Ed's earlier words in the same paragraph: >This stirs some vague recollection: there are some privileged >irrational numbers which can be speci?d with a ?ite amount of >information. > where I took the ?ite amount of information to mean coef?ients of > a [?ite] polynomial. If you take the name of a number to be the set > of polynomial coef?ients which de?es it, then clearly the > transcendentals, even pi and e, don't have a name in that sense.But that is an extremely limited view on ?ite amount of information. >Then it follows that the algebraic/transcendental distinction >is not what Ed was getting at. Algebraic numbers are no >less unknowable than computable transcendental numbers. > But they are less unnameable.In what way? By not being roots of some speci? polynomials? I wouldsay that Euler's gamma is better nameable than any root of x^5 + 3x + 2.But although Euler's gamma is computable, it is not known to which classof computable numbers it belongs (rational? algebraic? transcedental?).-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === ---------------------------------- ------------------------------------------------> That only leaves two types of numbers left that qualify for irrational,> non computable and random numbers.> I was silent the ?st time, but this claim is obviously complete> nonsense: the de?ition of irrational is ... well, you know what it> is ... not non computable, and a random number is a meaningless> idea in this context. The number line contains numbers, not random> numbers. But the OP is still to be thanked for giving an occasion to> display foils of learnedness.>Perhaps this is taken out of context, note the *left*.I started with reals, subtracted rationals, subtracted computable irrationals...that leaves non computable (irrational) numbers. The reason I use the term *random* is Ihave 2 seperate arguments,1/ NC numbers don't exist2/ If Cantors proof still holds then the uncountable numbers are just white noise, they are unable to be reproduced, non terminating, hence random.Herc === > In message <2a0cceff.0312101407.2225db91@posting.google.com>, Edward > I think you're thinking of the distinction between algebraic (roots of a > polynomial) and transcendental (the rest) ? > I wonder if there are any other possibilities - read on... Wrong.... > _What's_ wrong?... >Well, this is incorrect. > _What_ is incorrect?I think this has been clari?d. I commented on your I think you're thinkingof. See the word named in a meaningful way? > Yes, that's in Ed Green's posting, not mine. Take it up with him, don't > yell wrong at me.You should have seen the word wrong in a wider context. >?ite space you can calculate an arbitrary amount of digits. > That's not very precise. I think you mean that it is said to be > computable if your algorithm can calculate a rational which approximates > to it with an arbitrarily small error bound. Note that the thing you > _calculate_ is rational.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === _What_ is incorrect?> I think this has been clari?d. I commented on your I think> you're thinking of.Well, *that's* confusing. Did you claim that Richard was wrong when hesaid that Edward Green was thinking of, let's say, X? That is, Edwardwas *not* thinking of X? Or did you allege that Richard was wrongwhen he reported that he *thinks* Edward was thinking of X -- in fact,Richard was *not* thinking that Edward was thinking of X, but wasthinking something else, and he only mistakenly reported that histhoughts were about Edward and his thinking of X?I think I need a good lie-down.-- If you see math knowledge as a tool--as a hammer--with whichyou can attack other people then ... you defeat rational discourse.I get to call my proof the Hammer. It's more powerful than *any*physical object. It is overwhelming force. -- Two JSH quotes === > In message , Daryl McCullough >Richard Herring says...>So most irrational numbers are poor lost souls which not onlyhave non-repeating decimal representations, but can't even be>named in any meaningful way -- they are unknowable...>Can somebody remind me what this concept is called?>Richard Herring responds:> I think you're thinking of the distinction between algebraic (roots of a polynomial) and transcendental (the rest) ?> I wonder if there are any other possibilities - read on...>Dik Winter said Wrong because Ed Green is clearly talking about>numbers that are unknowable that can't even be named in any>meaningful way. Those descriptions are not true of the numbers>pi and e, for example. pi and e are just as knowable assquare-root(2). So the algebraic/transcendental distinctiondoesn't capture what Ed was talking about.> clari?ation. _I_ was referring to Ed's earlier words in the same paragraph:>This stirs some vague recollection: there are some privileged>irrational numbers which can be speci?d with a ?ite amount of>information.> where I took the ?ite amount of information to mean coef?ients of > a [?ite] polynomial. Ah. There's the rub.> If you take the name of a number to be the set > of polynomial coef?ients which de?es it, then clearly the > transcendentals, even pi and e, don't have a name in that sense.Right. But this is not a useful sense of name. ;)A name would be a ?ite string which determines the number. Determines in this sense could probably be taken to mean that, witha ?ite amount of effort, we could determine the n-th digit. Whatelse could it mean?I see however that e and pi are not proper names in this sense,so maybe we should say the name labels an algorithm(s) whichdetermines the n-th digit.[I've had this joke idea that I would claim to have a computablenumber in mind, named Fred, and that my proof that I wasn't lying,was that whenever anybody asked for any ?ite digit of Fred at all, Iwould consult my algorithm, and provide the requested digits! Mydigits of Fred, which I made up randomly as needed. ;-) Nobody couldever prove I was cheating.This may have some serious moral, I'm not sure. :-)] I'm still trying to grok Dik Winter's comment about the list ofcomputable numbers ... the Cantor argument proves that the realsaren't countable by ?st assuming a list is given, then writing a newone not on the list. Ok. But either the computable numbers arecountable or not -- we seem to be proving they are not? There issomething very paradoxical going on here, typically!Maybe ... offhand ... the computable numbers are indeed countable, andwe could produce an unambiguous list giving the number of all _ ?itealgorithms_ which might produce one. But the proposed algorithm ofthe Cantor argument is not a ?ite algorithm, since it requires us todo an in?ite amount of work a priori to make the list!But I'm not sure if this is a fatal objection to the paradox, though... we could in principle just evaluate each element of the list outto the digit we proposed to change by the Cantor argument. But is thealgorithm ?ite even now? We may run into the halting problem,since while we can list and number all ?ite strings of symbols which_might_ be algorithms, we may not be able to guarantee we can ID thosewhich are _really_ algorithms, and not nonsense, in ?ite time.All arguments in this general area seem to degenerate into the sameargument -- you wind up with some annoying Cantor-like argumentproducing an example of an item not on your list, which should havebeen. === ... > Maybe ... offhand ... the computable numbers are indeed countable, and > we could produce an unambiguous list giving the number of all _ ?ite > algorithms_ which might produce one. But the proposed algorithm of > the Cantor argument is not a ?ite algorithm, since it requires us to > do an in?ite amount of work a priori to make the list!That just about is it. The complete list requires us to do an in?iteamount of work just to determine the n-th element of the list when n islarge enough. Suppose we de?e an algorithm as something that has asinput an integer and as output (when it terminates) a decimal digit. Thecomputable numbers are de?ed by all such algorithms that terminatefor each ?ite input value. We can enumerate the ?ite algorithms forwhich it is obvious that they de?e a computable number. But that isnot suf?ient, we also have to enumerate the ?ite algorithms for whichit is *not* obvious that they de?e a computable number.For instance de?e the following (Collatz' hailstorm): T(n): c := 1; while n != 1 do c +:= 1; if(n % 2 = 0) n /:= 2 else n := n * 3 + 1 ?od; return c % 10.This is a ?ite algorithm. Its output is always a decimal digit. Itsinput are the integers. Does it de?e a computable number? We do notknow, because we do not know whether it terminates for all n. Can wedetermine in ?ite time whether it must be on the list? Perhaps, ourmathematical knowledge is not good enough to determine this. And thereare many more such ?ite algorithms.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === --www.StealthHostiing.com You rule Truman. http://tinyurl.com/iky4 Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! >Is the truman living in Townsville? I've been hearing stuff, yeah.Webmasters help the TRUEman by joining www.theBanner.net Current:1 Goal:1000--------------------------------------------------- -------------------------------> ...> Maybe ... offhand ... the computable numbers are indeed countable, and> we could produce an unambiguous list giving the number of all _ ?ite> algorithms_ which might produce one. But the proposed algorithm of> the Cantor argument is not a ?ite algorithm, since it requires us to> do an in?ite amount of work a priori to make the list!> That just about is it. The complete list requires us to do an in?ite> amount of work just to determine the n-th element of the list when n is> large enough. Suppose we de?e an algorithm as something that has as> input an integer and as output (when it terminates) a decimal digit. The> computable numbers are de?ed by all such algorithms that terminate> for each ?ite input value. We can enumerate the ?ite algorithms for> which it is obvious that they de?e a computable number. But that is> not suf?ient, we also have to enumerate the ?ite algorithms for which> it is *not* obvious that they de?e a computable number.> For instance de?e the following (Collatz' hailstorm):> T(n):> c := 1;> while n != 1 do> c +:= 1;> if(n % 2 = 0) n /:= 2> else n := n * 3 + 1> ? od;> return c % 10.> This is a ?ite algorithm. Its output is always a decimal digit. Its> input are the integers. Does it de?e a computable number? We do not> know, because we do not know whether it terminates for all n. Can we> determine in ?ite time whether it must be on the list? Perhaps, our> mathematical knowledge is not good enough to determine this. And there> are many more such ?ite algorithms.The quality of the list is not important. What's important is that the list coversall the reals, if 99% of programs are garbage we still have a mapping from N to R.We use a multiple capable UTM that parses encoded TMs in parallel. It does a fetch cycleon UTM(1), then it does a fetch cycle on UTM(2), then UTM(1), then UTM(2), thenUTM(3), then UTM(1). Each well behaved TM index will get properly evaluated.I don't see Halting coming into it. We can de?e computation as non terminating sinceit ?s with non terminating numbers, or we can index the nth digit of each number, ie usen as a parameter on classically de?ed terminating programs.Herc === > We use a multiple capable UTM that parses encoded TMs in parallel.> It does a fetch cycle on UTM(1), then it does a fetch cycle on> UTM(2), then UTM(1), then UTM(2), then UTM(3), then UTM(1). Each> well behaved TM index will get properly evaluated.Right, but when we come to ?ure out the Nth digit of the Nthmachine's number, we might never get that digit, because the relevantmachine, on the input N, simply doesn't terminate.The result is that we can't compute the Nth digit of the diagonalizednumber. === -------------------------------------- --------------------------------------------> We use a multiple capable UTM that parses encoded TMs in parallel.> It does a fetch cycle on UTM(1), then it does a fetch cycle on> UTM(2), then UTM(1), then UTM(2), then UTM(3), then UTM(1). Each> well behaved TM index will get properly evaluated.> Right, but when we come to ?ure out the Nth digit of the Nth> machine's number, we might never get that digit, because the relevant> machine, on the input N, simply doesn't terminate.> The result is that we can't compute the Nth digit of the diagonalized> number.Basically the list is dynamic, but Cantor's proof can be applied at any timeusing -1 for no value yet. Though I like the idea of the write once sectionfor the TMs.Herc === > where I took the ?ite amount of information to mean coef?ients of > a [?ite] polynomial. >Ah. There's the rub.> If you take the name of a number to be the set > of polynomial coef?ients which de?es it, then clearly the > transcendentals, even pi and e, don't have a name in that sense.>Right. But this is not a useful sense of name. ;)>A name would be a ?ite string which determines the number. Just out of curiousity, what is the name of, say, 2^(1/19)+3^(1/23)+5^(1/29)? rich === I just had a linear algebra ?al, and one of the true/false questions was:If matrix A and matrix B are diagonalizable, is (A+B) diagonalizable?I couldn't ?d anything in the textbook that gave an answer. Searching onthe net, I found an entry atonline.kitp.ucsb.edu/online/gravity03/ reula/pdf/Reula.pdfThat said The sum of two diagonalizable matrices is not necessarilydiagonalizable, but gives no proof.Anyone have a proof? === > I just had a linear algebra ?al, and one of the true/false questions was:> If matrix A and matrix B are diagonalizable, is (A+B) diagonalizable? I couldn't ?d anything in the textbook that gave an answer. Searching on> the net, I found an entry at> online.kitp.ucsb.edu/online/gravity03/ reula/pdf/Reula.pdf> That said The sum of two diagonalizable matrices is not necessarily> diagonalizable, but gives no proof.> Anyone have a proof?Take A equal to1 00 0and B equal to0 10 1Then A and B are diagonalisable, but A + B is not.Jose Carlos Santos === > I just had a linear algebra ?al, and one of the true/false questions was:> If matrix A and matrix B are diagonalizable, is (A+B) diagonalizable?>Take A equal to>1 0>0 0>and B equal to>0 1>0 1>Then A and B are diagonalisable, but A + B is not.This, of course, applies only to n x n matrices where n > 1 :) === perhaps does anyone have a written exposition of the groups of order8? it seems that my messy work does not catch the quaternion group.> im trying to ?d all groups of order eight. i already know what these> groups are, but i am trying to derive it, mainly from the sylow> theorems. will this approach work? for example, I was able to work> with groups of order 21=3*7, because I was able to ?d a semidirect> product representation of the group, but 8=2^3 doesnt really lead me> anywhere. is there another method?> thanks === >perhaps does anyone have a written exposition of the groups of order>8? it seems that my messy work does not catch the quaternion group.Here is summary of the argument I would use. First de?e D8 and Q8, justto prove that they exist.Case 1. G has an element of order 8. G is cyclic (C8).Case 2. All elements have order 1 or 2. G = C2 x C2 x C2.Case 3. G has an element x of order 4. |G:| = 2, so is normal in G. Let y in G - . So y^-1 x y in . Since y^-1 x y has same order as x, y^-1 x y = x or x^-1. Also y^2 in . If y^2 = x or x^-1, then y has order 8, and we are back in Case 1. So can assume y^2 = 1 or x^2. Each element of G can be written uniquely as y^i x^j with 0<=i<=1, 0<=j<=4, and it is is easy to show that a knowledge of y^2 and y^-1 x y enables us to multiply two of these elements and write it in the same form - hence the isomorphism class of the group is determined by y^2 and y^-1 x y. We have four possibilities: (i) y^-1 x y = x, y^2 = 1. Then G = C4 x C2. (ii) y^-1 x y = x, y^2 = x^2. Since (yx)^2 = 1, this also gives G = C4 x C2. (iii) y^-1 x y = x^-1, y^2 = 1, G = D8. (iv) y^-1 x y = x^-1, y^2 = x^2, G = Q8.Derek Holt. === > Induction hypothesis - If the k graphs intersect at exactly (n+1)> vertices, there is a k-coloring of G.> We note that : m = k(k+1)/2> Induction step - To prove that if the k graphs intersect at n vertices> (n < m), there is a k-coloring of G.> Consider an intersection point (i,j,...,l) [There is atleast one> intersecting point (i,j,...,l) because n < m]. We construct G' from G> such that the intersection point (i,j,...,l) in G is split as> (i,j,...) and (i,l) in G'.Doesn't that mean K_i will have k+1 points in G'?> By induction hypothesis, G' has a> k-coloring. A k-coloring of G' is also a k-coloring of G by assigning> the color of (i,j,...) in G' to (i,j,...,l) in G.How do you know that will be a *proper* k-coloring of G? === > 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've been here> for at least 12,000 years. Maybe it's something instinctive, hard-> wired or built-in about right angles or something.> If the Indian use is Vedic wouldn't that mean it's ultimately Persian?No,> D.> --> It does not matter how many innocent babies die. ............................................................. ......> === Arindam Banerjee posted: > 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've been here> for at least 12,000 years. Maybe it's something instinctive, hard-> wired or built-in about right angles or something.> If the Indian use is Vedic wouldn't that mean it's ultimately Persian?> D.> It does not matter how many innocent babies die. ............................................................. ......> No,I posted the following in 1994: The Holy Swastik(a) in Bharat (India) is clockwise as well as counter-clockwise. Please read the following adaptation of a past OM OMOMOMOMOM good luck insignia. Boy Scouts wore it engraved on OM OM their belt-buckles, the British used it on a World War OM OM OM OM I pin and private U.S. clubs were proud to stamp their OM OM membership tokens with it. But the origins of the OMOMOMOMOMOMOMOMOM Swastik: go as far back as man's history itself, and OM OM the diagram has been used in many cultures for OM OM OM OM millennia. OM OM Protected from a parching Sun, 3,500 years ago OMOMOMOMOM OM Babylonian stone carvers squatted inside an open-airthatched hut. A goddess was emerging from a large stone. Onto her decoratedgirdle the sculptors neatly cut a bent-spoked symbol, a Swastik:. On theother side of the planet, American Indians have scribed the spoked sign ofgood luck into salmon colored seashells, healing sticks, woven garments andpottery, As long as we can remember, they say. Two thousand miles to thesouth, the Mayans of the Yucatan chiseled it into temple diagrams. Once moored to the ancient highland cultures of Asia Minor, thebent-ribbed wheel later voyaged around the Mediterranean, through Egypt andGreece, northward into Saxon lands, Scandinavia and west to Scotland andIreland. In Bharat (India), the Swastik: also has been always present --sometimes turned clockwise, sometimes counter -- with a variety of Tantricleft-handed and right-handed interpretations for each. For the Vedicauspiciousness. There are numerous other meanings. Clearly, the word Swastik: itself comes from the Ved(as): Swasti --well being from the Sanskrit Su,well, and As,to be. When Buddhismfomented its birth out of Bharat's spiritual wellspring, it inherited theright-angled emblem. On the backs of monks, the little double-jointed designBuddha's feet -- and splayed into a spectrum of decorative meanderingSwastik:(s), the classical, geometric oriental motif. Nineteenth century Americans picked up the symbol from the AmericanIndians. Boy scouts tightened their britches with brassy Swastik: bucklesand a U.S. World's Fair minted ?Swastik: commemorative coins. Deformedexcept in Bharat where it retains its auspicious identity and aura ofblessings. Jai Maharaj http://www.mantra.com/jai Om Shanti http://www.mantra.com/holocausthttp://www.hindu.orghttp:// www.hindunet.orgThe truth about Islam and Muslimshttp://www.?m/~jai/satyamevajayate o Not for commercial use. Solely to be fairly used for theeducational purposes of research and open discussion. The contents ofthis post may not have been authored by, and do not necessarily representthe opinion of the poster. The contents are protected by copyright lawand the exemption for fair use of copyrighted works.considered or answered if it does not contain your full legal name,are not necessarily those of the poster. === Does anyone know a simple proof of the fact that (at least over a?ld) the rank of the symmetric matrix A^T A is the same as the rankof A?It is easy to see that the null space of A^T A contains the null spaceof A, so it is enough to see the reverse inclusion (which I haven'tbeen able to do). === > Does anyone know a simple proof of the fact that (at least over a> ?ld) the rank of the symmetric matrix A^T A is the same as the rank> of A?> It is easy to see that the null space of A^T A contains the null space> of A, so it is enough to see the reverse inclusion (which I haven't> been able to do).If A^t A x = 0, then 0 = = = ||Ax||^2, thereforeAx = 0.--Ron Bruck === That's a nice proof if the coef?ients of A are real. And ifA-transpose is replaced by A-transpose-conjugate, then the proofclearly works for complex coef?ients. But the OP appears to haveasked for the case that A has coef?ients in an arbitrary ?ld.Consider A = [ 0 b ] [ 0 d ]Then A^T = [ 0 0 ] [ b d ]so A^T A = [ 0 0 ] [ 0 b^2+d^2 ]So over a ?ld of characteristic p, if we take b and d to be nonzeromod p, but so that p divides b^2+d^2, then A and A^T have rank 1, butA^T A has rank 0. Thus rank(A^T A) = rank(A) is not true incharacteristic p.And for the OP's question, even over the complex numbers, if we takeb=1 and d=i, then A^T A is the zero matrix.> Does anyone know a simple proof of the fact that (at least over a> ?ld) the rank of the symmetric matrix A^T A is the same as the rank> of A?> It is easy to see that the null space of A^T A contains the null space> of A, so it is enough to see the reverse inclusion (which I haven't> been able to do).> If A^t A x = 0, then 0 = = = ||Ax||^2, therefore> Ax = 0.> --Ron Bruck === > If A^t A x = 0, then 0 = = = ||Ax||^2, therefore> Ax = 0.them as linear transformations with respect to the standard basis)--sothis works at least over R. I think this fact is true over any ?ld,though, is it not?-Chad === > If A^t A x = 0, then 0 = = = ||Ax||^2, therefore> Ax = 0.>them as linear transformations with respect to the standard basis)--so>this works at least over R. I think this fact is true over any ?ld,>though, is it not?Many people are fond of ?lds which contain a square root of -1, call it i. (You might also, of course, call it 1 in case the characteristic is 2.) Over such a ?ld, the matrix A = [1, i; 0, 0] may make you unhappy.Lee Rudolph === > Many people are fond of ?lds which contain a square root of > -1, call it i. (You might also, of course, call it 1 in case > the characteristic is 2.) Over such a ?ld, the matrix > A = [1, i; 0, 0] may make you unhappy.That answers the question then. It's always harder to prove something that isn't true.(Actually in this case it is A A^T that is zero, but [1, 0;i, 0] works for the original question.)-Chad === > Michigan State University grad student Michael Shafer has succeeded in> identifying the largest known prime number to date, using a distributed> computer network of more than 200,000 computers located around the world.> The new number is 6,320,430 digits long and is only the 40th Mersenne prime> to have ever been discovered (Mersenne primes are an especially rare breed> that take the form of 2-to-the-power-of-P, where P is also a prime number).Slight error. A power of two from which one is subtracted yielding a prime. The entire project indicates clearly that these people have way too much time on their hands.Bob Kolker> === > When I read it the ?st time I found the most surprising not that he> had discovered a large power of 2 but that he had discovered one that> is prime. There must be something wrong with the core of mathematics.Calling James ... === > When I read it the ?st time I found the most surprising not that he> had discovered a large power of 2 but that he had discovered one that> is prime. There must be something wrong with the core of mathematics.> Calling James ...James === > norm topology, it's called a C*-algebra. The spectrum of a=20 C*-algebra A is the set of all homomorphisms=20> x: A -> C,=20> where C is the complex numbers. Though it's not immediately obvious,> ...> the idea. There's an easy way to make the spectrum of a commutative> C*-algebra A into a topological space: we say x_i -> x precisely when> x_i(a) -> x(a)> for all elements a of A. With this topology any element a of A gives=20> a continuous complex function on the spectrum, de?ed by this clever=20> formula:> I don't understand what you mean by> x_i -> x> x_i(a) -> x(a)> What has the index i to do with this? x_i and x are the above> mentioned homomorphisms, right? Could you explain a little more> details?x_i -> x means that x_i is a sequence of homomorphisms indexed by i thatconverges to the homomorphism x, while x_i(a) -> x(a) means that x_i(a) is asequence of complex numbers that converges to the complex number x(a).Without additional structure, the set of all homorphisms from A to C doesn'tallow for a notion of convergence. The set of complex numbers C is used togive the set of all homomorphisms this extra structure.A topological space is a set X together with a collection w of subsets of Xthat satis?s:i) both X and the empty set are in the collection w;ii) the intersection of any 2 members of the collection w is itself in thecollection w;iii) the union of all the members of an arbitrary subcollection is a memberof the collection w.The members of the collection w are the open subsets of X. This is key toextending the notion of continuity to spaces more general than R^n.If, like the set of complex numbers C with the standard distance function, atopological space is also a metric space, than knowing the topology isequivalent to knowing all the conergent sequences. For more generaltopological spaces, convergent sequences have to be generalized toconvergent nets.In the above, x and x_i (for every natural number i) are homomorphisms,while x(a) and x_i(a) are complex numbers. This allows convergence in C to beused to *de?e* convergence in the set of all homomorphisms for A to C.One possible source of confusion is that physics course often gloss over thedifference between f and f(x) for a mapping f. Supposef : X -> Y .Then f is the abstract concept of the mapping, and f(x), i.e., f evaluatedany element of the set X, is an element of the set Y, so there is bigdifference between f and f(x). Physics books often write f(x) when they meanf.In the above, John makes this distiction between the mapping (andhomomorphism) x and the complex number x(a).George === > Green's functions often have logarithmic terms in them, so they are> often irrational functions.> So irrational( function)s can be Green(?s fuctions).RR === However, I always questioned if saying that |N| < |R|> has any more meaning than saying that irrational numbers> are green.> LucasSurely it should be easy to tell the colour of the irrationals. Drawthe real line and then stand back a bit. Since the irrationalsdominate the line (the rationals are a tiny subset), the colour of theline as a whole is presumably the colour of the irrationals. Usuallythey look blue to me but sometimes black. I remember in university,they would sometimes look white (against the background of ablackboard anyway). They seem sort of anti-chameleon and try to standout from their background.Now ?ding the colours of the rationals and other countable subsetssuch as the algebraic or computable numbers is a challenge. How canyou pick them out from that mass of non-computable numbers? I triedjust drawing a few rationals by themselves but however ?e a pen Iused, I got a mass of non-computable numbers with them.J === > [...]> Now somewhat seriously:> Besides being a mathematician, I am also a musician. As such, I am aware> that some people experience synesthesia. The composer Scriabin was famous> for this, experiencing certain color sensations when he heard certain> pitches. I would not be greatly surprised if there were some people who> experience certain color sensations when contemplating certain numbers.> Does anyone know of any instances of such?: Hearing Colors, Tasting Shapes Feature Article psychophysics, : synaesthesia, synesthesia, neuroscience In the extraordinary : world of synesthesia, senses mingle together--revealing some of : the brain's mysteries People with synesthesia--whose senses : blend together--are providing valuable clues to understanding the : organization and functions of the human brain 15-APR-03 When : Matthew Blakeslee shapes hamburger patties with his hands, he: experiences a vivid bitter taste in his mouth. ... There were some perceptual tests done that showed that thesynesthetes were not fooling themselves. One test used a pattern made inscattered 6's and 9's, similar to the patterns of light and darkreds and greens that are used to test for color-blindness.They were able to immediately pick out the shape made of 6'sthe way a person with normal perceptions could not.Jim Burns === If you want to answer any homwork in future contact with me on my === let I=k[x_1,x_2,..,x_n] is polinomial ring over ?ld of char=0 andI_n - subspase gomogenius polinomial of power n. Let sl_2 - 3 -dimesional simple lie algebra wich act at I_n in usual way. How ?d airreducible components of decomposition of this representation? Need?d something like as formulae of (Klebsh-Gordon)for tensor product. === > let I=k[x_1,x_2,..,x_n] is polinomial ring over ?ld of char=0 and> I_n - subspase gomogenius polinomial of power n. Let sl_2 - 3 -> dimesional simple lie algebra wich act at I_n in usual way. How ?d a> irreducible components of decomposition of this representation? Need> ?d something like as formulae of (Klebsh-Gordon)for tensor product.Try ?ding the highest weight elements, counting dimensions (?d itscharacter) and semi-simplicity. === > Is there a mathematical term for the opposite point of contact of a> circle's diameter?> Or, to put it another way, any point on a circle's circumference 180> degrees opposite to another point on the same circle's circumference.>These phrases are perhaps even more used in common language, than in>mathematics. The two points are said to be> diametrically opposite (or opposed)> in diametric opposition> etc.I usually call the point referenced the antipode. See http://mathworld.wolfram.com/AntipodalPoints.htmlRob Johnson take out the trash before replying === Is there a mathematical term for the opposite point of contact of a> circle's diameter?> Or, to put it another way, any point on a circle's circumference 180> degrees opposite to another point on the same circle's circumference.>These phrases are perhaps even more used in common language, than in>mathematics. The two points are said to be> diametrically opposite (or opposed)> in diametric opposition> etc.>be a different matter. Aren't there those who are trying to legislate thatmarriage must be between exactly two persons of diametrically oppositesexes? So, if that were the law, then marrying yourself would clearly be> but I usually call the point referenced the antipode. See> http://mathworld.wolfram.com/AntipodalPoints.htmlI have no antipathy for antipode. I would have mentioned it in myoriginal response, but just didn't think to do so.David === Would somebody please explain the notation___lim f(x)x->clim f(x)~~~x->cas used Theorem II, equation (2,2) inA. E. Taylor, L'Hospital's RuleAmer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24.an interesting short paper giving various proofs of lim(x->c) f'(x)/g'(x) = lim(x->c) f(x)/g(x)when lim(x->c) f(x) = lim(x->c) g(x) = 0or lim(x->c) |g(x)| = oo---- === >Would somebody please explain the notation>___>lim f(x)>x->c>lim f(x)>~~~>x->c>as used Theorem II, equation (2,2) in>A. E. Taylor, L'Hospital's Rule>Amer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24.>an interesting short paper giving various proofs of> lim(x->c) f'(x)/g'(x) = lim(x->c) f(x)/g(x)>when lim(x->c) f(x) = lim(x->c) g(x) = 0>or lim(x->c) |g(x)| = ooI would assume that the notation means lim sup and lim inf. Take alook at http://mathworld.wolfram.com/SupremumLimit.html, and the linkthere to In?um Limit.Rob Johnson take out the trash before replying === >Would somebody please explain the notation>___>lim f(x)>x->c>lim f(x)>~~~>x->cThese are presumably the lim sup and the lim inf.The most intuitive explanation for what the lim sup is isprobably this:You know that lim_{x->c} f(x) = L if and only if f(x_n) -> Lfor every sequence x_n such that x_n <> c but x_n -> c.The lim sup of f(x) as x -> c is the _largest_ lim f(x_n)where x_n -> c (and x_n <> c), while the lim inf is thesmallest such lim f(x_n). (So lim_{x->c} exists ifand only if the lim sup equals the lim inf; if and onlyif there is only one such lim f(x_n).)Alternately, you can de?e lim sup_{x->c} f(x) = lim_{d ->0+} sup{f(t) : 0 < |t - c| < d}.The reason these things are useful is that _any_ functionhas a lim sup and a lim inf at _every_ point (if we allowplus and minus in?ity as a value), while not every functionhas a limit at every point.>as used Theorem II, equation (2,2) in>A. E. Taylor, L'Hospital's Rule>Amer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24.>an interesting short paper giving various proofs of> lim(x->c) f'(x)/g'(x) = lim(x->c) f(x)/g(x)>when lim(x->c) f(x) = lim(x->c) g(x) = 0>or lim(x->c) |g(x)| = oo>----****David C. Ullrich === > [...]> But check this out:> http://www.bearnol.pwp.blueyonder.co.uk/Math/Wow, James has been busy. I had no idea he'd, um, ?tackled' so many problems!In fairness though, his inclusion of the books of Diophantus on his pagemakes it more useful than it otherwise would be, assuming they're completeand not augmented by too many (or _any_ ;-) proofs by James! It could dowith a bit of decent formatting though, such as a separate frame forfootnotes by Sir Thomas Heath and others.(A couple of the missing books (including IV I think) were discovered afew years ago and published by Springer, and it looks like James hasn'tAlso I'm not convinced he's a crank in the usual sense, more of a clown(in his maths antics at least). He lacks that dogmatism of always beingright, or at least publicly declaring it.If anything it's the the reverse, when the proofs he occasionally postson sci.math are shot to bits and he acknowledges the criticisms and goesaway to lick his wounds for a month or two!--------------------------------------------------------- ------------------John R Ramsden (jr@adslate.com)--------------------------------------------- ------------------------------Eternity is a long time, especially towards the end. Woody Allen === > [...] But check this out:> http://www.bearnol.pwp.blueyonder.co.uk/Math/> Wow, James has been busy. I had no idea he'd, um, ?tackled' so many problems!For those who don't know Ramsden is an alias of David Rusin, but Idon't see his point here in posting that link.Anyone have any idea what Rusin is up to here?James Harris === > [...]> But check this out:> http://www.bearnol.pwp.blueyonder.co.uk/Math/> Wow, James has been busy. I had no idea he'd, um, ?tackled' so many problems!>For those who don't know Ramsden is an alias of David Rusin, but I>don't see his point here in posting that link.>Anyone have any idea what Rusin is up to here?Or any idea how James got the idea that Ramsden and Rusin werethe same person?>James Harris****David C. Ullrich <3c65f87.0312131113.663a3a6c@posting.google.com> === > [...]> But check this out:> http://www.bearnol.pwp.blueyonder.co.uk/Math/> Wow, James has been busy. I had no idea he'd, um, ?tackled' so many problems!>For those who don't know Ramsden is an alias of David Rusin, but I>don't see his point here in posting that link.>Anyone have any idea what Rusin is up to here?> Or any idea how James got the idea that Ramsden and Rusin were> the same person?>It ?st came up, without explanation, in the thread, JSH: Morecleanup, Ramsden = Rusin? (You might remember this, since you postedin that thread.) Google has no record of James mentioning the namesRamsden and Rusin in the same post prior to that.I'm bumfuzzled as to how it happened. It's really one of the morecurious features of James's diatribes. As far as I know, he has neverhinted as to how he stumbled on this stunning fact or what evidencemade it all come clear to him.Here's the entirety of that ?st post, sans sig.,----| Hey, am I the last to know that John R Ramsden| (jr@redmink.demon.co.uk) was (or is) David Rusin postin under an| alias?`----Dave contributed on rare occasions to JSH threads both before andafter this Ramsden guy appeared, so it's unclear what Dave's gamereally is. Actually, that's not true. I *do* know what he's up to,but since Jesse Hughes is also a pseudonym of Dave Rusin, I'm nottelling.-- I'm talking about mathematics--hard, brutal, extreme ... pushing yourmind beyond the limits to understand what no one else can becausethey're afraid to risk it all, to lose their freaking worthless mindsin the push to know. --James Harris, for the Nike Derivator === > |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?Indeed. And currently there are scientists that doubt it again. Butthe same holds for continental drift. That was at the start also justspeculative. And similar for relativity. Explaining natural phenomenais just that, speculative. Theories in those sciences are valid whenthey best explain what is happening or what has happened, and they arestill better when they are able to predict things that can be veri?d.Nevertheless, they may be (and in many instances are) superseded by newertheories that give a still better explanation.Crackpots in mathematics are of a different class. They just statesomething is true which has been proven false by mathematical theorems,(or that something is false which has been proven true).However, they refuse to indicate which part of the original proof isincorrect, nor are their proofs either complete or logical. The mosteasy ones in mathematics are those that claim the pi is rational, thathave the trisection of an angle, and what you want. More subtle arethose that show an easy proof of FLT, a disproof of Cantor, or evena disproof of FLT (who was it in this newsgroup that has shown that FLTwas false, some years ago? He was from the Philippines I think).Amusing was TLeko, who has shown that z = z is not an analytical function(he even has a paper about it in some proceedings).-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > Crackpots in mathematics are of a different class. They just state> something is true which has been proven false by mathematical> theorems, (or that something is false which has been proven true).> However, they refuse to indicate which part of the original proof is> incorrect, nor are their proofs either complete or logical. The> most easy ones in mathematics are those that claim the pi is> rational, that have the trisection of an angle, and what you want.> More subtle are those that show an easy proof of FLT, a disproof of> Cantor, or even a disproof of FLT (who was it in this newsgroup that> has shown that FLT was false, some years ago? He was from the> Philippines I think). Amusing was TLeko, who has shown that z = z> is not an analytical function (he even has a paper about it in some> proceedings).This is a common trend, I think, and it's not just true of math.The crackpot sees that some academics are very famous for havingbucked the trend, and shown that what everyone else took to be true,is actually wrong. They want, they *need* that glory, and so theybuck every trend they can ?d.It is no accident that you describe math cranks as being focused oneither claiming something true which has been proved false, or theopposite. Cranks generally do not attack open problems (FLT was theexcept for a long time), but rather want the glory of having shown thecommon wisdom to be wrong.In math this is particularly pathetic, because the way math isstructured makes the odds of such an event extraordinarily low. Butinside every crank is that little voice that says maybe I'll get thefame Betrand Russell got for his proof that the old set theory wasinconsistent. The crackpot does not know that Russell would have been just as famouseither way. That Russell's greatest achievements were constructiveenterprises, extending knowledge, rather than tearing down mistakes ofthe past.Scholarship is nearly always a matter of extending knowledge. In myown ?ld (12 century philosophy) I am amazed at how much has yet tobe uncovered, discussed, analyzed, probed, considered, and related.It can be fun to show that someone else's dating of some importantdocument is wrong by ten years, but that's a minor footnote. Fame andimportance come by hard work extending the borders.Remember Newton, who saw further by standing on the shoulders ofgiants? That is a laudible attitude. The crackpot wants to tear downthe giants and stand in their place.Einstein may have showed that Newtonian physics was wrong, but he didnot tear it down (as evidenced by the curriculum in every freshmanphysics class in the country). A recent issue in my own area is the status of the letters of Abelardand Heloise. Always taken to be genuine, about a hundred years agosome folks bucked the trend, and said these are a forgery, writtenlong after by other people. The book is now basically closed though,and those people were wrong: better linguistic tools and moreknowledge of history and the writings of Abelard have proven quiteconclusively that the letters are genuine. The crackpot thinks wow, I would love to have turned around knowledgethat way. But both times, the ?st ?of these aren'tgenuine, to the second reversal back to the traditional opinion, were*advances*, and built heavily on what went before. The people who didthe work proving the letters were genuine did not proclaim theirbrilliance and the foolishness of those who thought they werespurious--quite the opposite--they owed their success to the attentionand techniques of the people who had formed the earlier, incorrect,conclusion.Thomas === > Crackpots in mathematics are of a different class. They just state> something is true which has been proven false by mathematical> theorems, (or that something is false which has been proven true).> However, they refuse to indicate which part of the original proof is> incorrect, nor are their proofs either complete or logical. The> most easy ones in mathematics are those that claim the pi is> rational, that have the trisection of an angle, and what you want.> More subtle are those that show an easy proof of FLT, a disproof of> Cantor, or even a disproof of FLT (who was it in this newsgroup that> has shown that FLT was false, some years ago? He was from the> Philippines I think). Amusing was TLeko, who has shown that z = z> is not an analytical function (he even has a paper about it in some> proceedings).> This is a common trend, I think, and it's not just true of math.What evidence do you have that it's common?That is, of the numerous contacts made to mathematicians throughvarious forums, what actual evidence do you have to support yourclaims?I'm curious because I've seen a strange need in math society to ?dsome kind of scapegoat for hostility.Mathematicians dump on cranks with extreme vitriol. > The crackpot sees that some academics are very famous for having> bucked the trend, and shown that what everyone else took to be true,> is actually wrong. They want, they *need* that glory, and so they> buck every trend they can ?d.Really? Again, I ask, what is your evidence to support that bit ofmindreading?It seems to me that mathematicians have a lot of hatred and ?d atarget.Certainly there are people who present various ideas which aredebatable, but that happens in just about any ?ld.In English literature, there are battles over who was WilliamShakespeare.> It is no accident that you describe math cranks as being focused on> either claiming something true which has been proved false, or the> opposite. Cranks generally do not attack open problems (FLT was the> except for a long time), but rather want the glory of having shown the> common wisdom to be wrong.And again, based on what evidence?Now then critical thinking involves dealing with *reality* and notjust what you think is true.Can you give some factual support for your claims?Or are you relying on beliefs you feel certain are shared with others?James Harris <87smjpe8c5.fsf@becket.becket.net> <3c65f87.0312130556.6f723689@posting.google.com> === > In English literature, there are battles over who was William> Shakespeare.I think you'll ?d it was William Shakespeare.-- Well *supposedly* a correct and profound math paper can get publishedin a ?reputable journal' which means that the journals I've faced sofar may lose a lot of their luster once the full story comes out.--- James Harris, on the quality of math journals rejecting his paper <87smjpe8c5.fsf@becket.becket.net> <3c65f87.0312130556.6f723689@posting.google.com> <87iskkbqrw.fsf@phiwumbda.org> === > In English literature, there are battles over who was William> Shakespeare.> I think you'll ?d it was William Shakespeare.This is not true.... Erm, wait a bit. Here we go:THIS IS NOT TRUE!!I can prove that William Shakespeare was actuallyANOTHER MAN OF THE SAME NAME!!! ! ! Yours,Jim Burns === [snip]> Can you give some factual support for your claims?> Or are you relying on beliefs you feel certain are shared with others?> James HarrisYou have no standing to expect factual support for a claim from other posters. You have forfeit the rightto demand evidence or proof by your own consistent failure to provide the same for your own claims.Where, for example, is the evidence (proof, factual support) for your claim that *your* so-called ?partialdifferential equation' solves the prime counting problem? Where is the data? Where are your results? Putup, 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 === This (below) ws a remarquably lucid description of the crackpot attitude.> Crackpots in mathematics are of a different class. They just state> something is true which has been proven false by mathematical> theorems, (or that something is false which has been proven true).> However, they refuse to indicate which part of the original proof is> incorrect, nor are their proofs either complete or logical. The> most easy ones in mathematics are those that claim the pi is> rational, that have the trisection of an angle, and what you want.> More subtle are those that show an easy proof of FLT, a disproof of> Cantor, or even a disproof of FLT (who was it in this newsgroup that> has shown that FLT was false, some years ago? He was from the> Philippines I think). Amusing was TLeko, who has shown that z = z> is not an analytical function (he even has a paper about it in some> proceedings).> This is a common trend, I think, and it's not just true of math.> The crackpot sees that some academics are very famous for having> bucked the trend, and shown that what everyone else took to be true,> is actually wrong. They want, they *need* that glory, and so they> buck every trend they can ?d.> It is no accident that you describe math cranks as being focused on> either claiming something true which has been proved false, or the> opposite. Cranks generally do not attack open problems (FLT was the> except for a long time), but rather want the glory of having shown the> common wisdom to be wrong.> In math this is particularly pathetic, because the way math is> structured makes the odds of such an event extraordinarily low. But> inside every crank is that little voice that says maybe I'll get the> fame Betrand Russell got for his proof that the old set theory was> inconsistent.> The crackpot does not know that Russell would have been just as famous> either way. That Russell's greatest achievements were constructive> enterprises, extending knowledge, rather than tearing down mistakes of> the past.> Scholarship is nearly always a matter of extending knowledge. In my> own ?ld (12 century philosophy) I am amazed at how much has yet to> be uncovered, discussed, analyzed, probed, considered, and related.> It can be fun to show that someone else's dating of some important> document is wrong by ten years, but that's a minor footnote. Fame and> importance come by hard work extending the borders.> Remember Newton, who saw further by standing on the shoulders of> giants? That is a laudible attitude. The crackpot wants to tear down> the giants and stand in their place.> Einstein may have showed that Newtonian physics was wrong, but he did> not tear it down (as evidenced by the curriculum in every freshman> physics class in the country).> A recent issue in my own area is the status of the letters of Abelard> and Heloise. Always taken to be genuine, about a hundred years ago> some folks bucked the trend, and said these are a forgery, written> long after by other people. The book is now basically closed though,> and those people were wrong: better linguistic tools and more> knowledge of history and the writings of Abelard have proven quite> conclusively that the letters are genuine.> The crackpot thinks wow, I would love to have turned around knowledge> that way. But both times, the ?st ?of these aren't> genuine, to the second reversal back to the traditional opinion, were> *advances*, and built heavily on what went before. The people who did> the work proving the letters were genuine did not proclaim their> brilliance and the foolishness of those who thought they were> spurious--quite the opposite--they owed their success to the attention> and techniques of the people who had formed the earlier, incorrect,> conclusion.> Thomas === I have (-12-5i)^-3 and I have to ?d the argument and modulus.I correctly got 13^-3 = 4.552E-4 as the modulus.However I got the primary argument as -2.747 * -3 = 8.24, but my answerstell me that I should have got 5.8968. What have I done wrong? === argument and modulus.> I correctly got 13^-3 = 4.552E-4 as the modulus.> However I got the primary argument as -2.747 * -3 = 8.24, but my answers> tell me that I should have got 5.8968. What have I done wrong?I would do it like this: (5,12,13) is aPythagorean triple, i.e. 5^2 + 12^2 = 13^2.Let z=-12-5i, with conjugate z_=-12+5i.The modulus squared of z is z*z_=5^2+12^2=13^2.So the modulus of z is 13 and the modulus of1/z is 1/13. Thus the modulus of your number,1/z^3 is 1/13^3. (?multiply the moduli').Taking arguments of the equation z*(1/z)=1gives arg(z)+arg(1/z)=0 (?add the arguments'),i.e. arg(1/z)=-arg(z).But arg(z)=arctan(5/12)+Pi. Thus:arg(1/z) = -arctan(5/12)-Pi,and so argument of 1/z^3 (your number) isthis number added to itself 3 times:arg(1/z^3) = -3*arctan(5/12)-3*Pi.Finally, since arguments are determined onlyup to a multiple of 2*Pi, we can add 4*pi tothis and get:arg(1/z^3) = -3*arctan(5/12)+Pi, or approximately1.9572.Jim Buddenhagen-- To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE === >I have (-12-5i)^-3 and I have to ?d the argument and modulus.>I correctly got 13^-3 = 4.552E-4 as the modulus.>However I got the primary argument as -2.747 * -3 = 8.24, but my answers>tell me that I should have got 5.8968. What have I done wrong?What answers tell you you should get 5.8968? Using the half-tangentformula, we get tan(arg(z)/2) = Im(z)/(|z|+Re(z))so that the argument of -12-5i is 2*atan(-5/(13-12)) = -2.7468 radians.Thus, you seem to be correct in saying that the argument of (-12-5i)^-3is -3*-2.7468 = 8.2404. The only thing further I would suggest is toreduce the argument mod 2 pi, giving 1.9572 as the argument.Rob Johnson take out the trash before replying === >I have (-12-5i)^-3 and I have to ?d the argument and modulus.>I correctly got 13^-3 = 4.552E-4 as the modulus.>However I got the primary argument as -2.747 * -3 = 8.24, but my answers>tell me that I should have got 5.8968. What have I done wrong?> What answers tell you you should get 5.8968?Sorry I should have been clearer. The answer given to me by my lecturer forthe problem.> Using the half-tangent> formula, we get> tan(arg(z)/2) = Im(z)/(|z|+Re(z))> so that the argument of -12-5i is 2*atan(-5/(13-12)) = -2.7468 radians.> Thus, you seem to be correct in saying that the argument of (-12-5i)^-3> is -3*-2.7468 = 8.2404. The only thing further I would suggest is to> reduce the argument mod 2 pi, giving 1.9572 as the argument.Phew. So I was correct then... I freaked out for a minute there. === >I have (-12-5i)^-3 and I have to ?d the argument and modulus.>I correctly got 13^-3 = 4.552E-4 as the modulus.However I got the primary argument as -2.747 * -3 = 8.24, but my answers>tell me that I should have got 5.8968. What have I done wrong?> What answers tell you you should get 5.8968?> Sorry I should have been clearer. The answer given to me by my lecturer for> the problem.Lecturers make mistakes, sometimes! === > I have (-12-5i)^-3 and I have to ?d the argument and modulus.> I correctly got 13^-3 = 4.552E-4 as the modulus.> However I got the primary argument as -2.747 * -3 = 8.24, but> my answers tell me that I should have got 5.8968. What have I> done wrong?The argument is found by scaling the original expression to unitlength, i.e. dividing by the modulus, giving: (-12/13 - i.5/13)^-3Assuming the expression in brackets is cos(t) + i.sin(t), andnoting that cos(t + PI) = - cos(t) and sin(t + PI) = - sin(t),the smallest negative solution is t = arcsin(5/13) - PI.The ?al argument is this multiplied by -3 and then, to obtain theprinciple angle, the largest multiple of 2.PI subtracted that leavesthe result non-negative.P.S. As -12-5i = (2-3i)^2 it is marginally easier sign-wise to workout the argument as - arcsin(3/13) * -6, although that's a bit ofa ?trick' that exploits the particular numerical values in thisproblem but wouldn't work nearly so easily for every problem ofthe same type.-------------------------------------------------------- -------------------John R Ramsden (jr@adslate.com)--------------------------------------------- ------------------------------Eternity is a long time, especially towards the end. Woody Allen === > P.S. As -12-5i = (2-3i)^2 it is marginally easier sign-wise to work> out the argument as - arcsin(3/13) * -6DOH! That should read -12-5i = (1-5i)^2 / 2, so that the argument is - arcsin(5/(2.SQRT(13))) * -6.Or you could argue that -12-5i = i.(2 + 3i)^2, and the argument of2 + 3i is arcsin(3/SQRT(13)), and multiplying by i adds PI/2 to this,etc.---------------------------------------------------- -----------------------John R Ramsden (jr@adslate.com)--------------------------------------------- ------------------------------Eternity is a long time, especially towards the end. Woody Allen === >Let a_n = n*Pi^2/6 - sum(1/(k*C(n+k-1,n)),k=1..in?ity),>where C(n+k-1,n) is binomial coef?ient 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's 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 Well I didn't provide a proof of your nice result so... In fact I found it very interesting and investigated this generalization : f^m_n= 1/(n+1)*sum_{k=1}^oo 1/(k^m*C(n+k,n+1)) (n->n+1 and I divided by n+1 for simpli?ation) I'll use a more common notation for harmonic numbers of mth order H^(m)_n= sum_{k=1}^n 1/k^m and H_n= sum_{k=1}^n 1/k I will further need this recursive function : G^m_n(r)= H^(m)_n if r=0 = sum_{k=1}^n G^m_k(r-1)/k if r>0 The previous result was : f^1_n= zeta(2) -H^(2)_n or = zeta(2) -G^2_n(0) I got too : f^2_n= zeta(3) -H_n*zeta(2) +G^2_n(1) f^3_n= zeta(4) -H_n*zeta(3) +G^1_n(1)*zeta(2) -G^2_n(2) f^4_n= zeta(5) -H_n*zeta(4) +G^1_n(1)*zeta(3) -G^1_n(2)*zeta(2) +G^2_n(3) (Note that G^1_n(1)=sum_{k=1}^n H_k/k = (H^(2)_n+(H_n)^2)/2 and so on...) More generally : f^m_n= zeta(m+1) -sum_{i=0}^{m-2} (-1)^i*G^1_n(i)*zeta(m-i) +(-1)^(m)*G^2_n(m-1) Hoping you'll like this, Raymond === > The point is that the characteristic class of the ses in Ext(S_1,B) comes> from one in Ext(S_1,S_2). This comes from an ses looking like> 0 -> S_1 (+) S_2 -> S_1 (+) N -> S_1 -> 0> where> 0 -> S_2 -> N -> S_1 -> 0> is an ses.I think you have lost me. I spoke to my advisor yesterday. It looks likethat I can get through without the above considerations in the specielabout Jordan-Holder and a bit about getting information from Ext groups.-- Michael Knudsen === In short: Gravity is the sheer force that weighs us down; called weight. http://newsone.net/ -- Free reading and anonymous posting to 60,000+ groups NewsOne.Net prohibits users from posting spam. If this or other posts === > In short: Gravity is the sheer force that weighs us down; called weight.Go away, Head. === > In short: Gravity is the sheer force that weighs us down; called weight.Did someone say something different? === ABSTRACT. The synchronization of clocks in SR is speculative chatter, and this speculative chatter about Synchronizations of clocks in SR has not ACTUAL TECHNICAL EMBODYING In CONCRETE TECHNICAL DEVICES.http://groups.google.com/groups?selm= e16a4a22.0311290740.179ebe19%40posting.google.com === == === > [Sarfatti]: Van Flandern . [made] false statements that GPS does not> require GR.> [Undeniable]: Van Flandern is right regarding the corrections due to> GPS clocks. Those corrections are only done once before the clocks are> placed in orbit.> [Pyriform]: Every GPS receiver ever made contains algorithms to apply> relativistic corrections required by the eccentricity of the satellite> orbits. Von Flandern is talking out of his arse.> Given that the small eccentricity corrections and clock> imperfections are tens of nanoseconds instead of tens of thousands and> are periodic rather than cumulative, they are irrelevant to the point.> My actual remarks appear in What the Global Positioning System tells us> about relativity, in Open Questions in Relativistic Physics, F.> Selleri, ed., Apeiron, Montreal, pp. 81-90 (1998); also available at> http://metaresearch.org/cosmology/gps-relativity.asp; and in What the> Global Positioning System tells us about the twins paradox, Episteme #6> pt. II, http://www.dipmat.unipg.it/~bartocci/ep6/ep6-van? (2002/12/21); MRB 11, 39-46 (2002). My actual statement (in the second> of these references) is:> Yet the Global Positioning System is designed in such a way that, after> the individual clock rates are adjusted once pre-launch for the> predicted relativity effects, all satellite clocks in all orbits remain> in synchronization with one another and with all ground clocks without> need for further consideration of relativity corrections, with the> exception of one small correction needed for the slight non-circularity> of the orbits.The trajectory to ?al rout of a method of synchronization of clocks in the theory of relativity uncloses a new method of synchronization of atomic clocks leaning on technology of VLBI interferometry.http://groups.google.com/groups?selm= e16a4a22.0311280708.73e98d8a%40posting.google.comhttp:// groups.google.com/groups?selm=77vg46%24mai%241%40nnrp1. dejanews.com Principles of work of VLBI. The microwave interferometr with superlong basis consists of two radio telescopes were on a very large distance from each other. Before experiment or after him, the atomic clocks are synchronized. accepted by an antenna. Simultaneously with a signal, the time scores received from the atomic clock, are written on a videotape. After ending experiment we have two videotapes with signal dataand scores of time. The interference picture is received after dataprocessing of these videotapes on the computer.Fixed font !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! There are two graphic schemes illustrating the description: The microwave interferometer with superlong basis. Part 1. Block scheme.-> radio-telescope 1-> hydrogenous-> atomic-> parabolic antenna 1 tape 1 clock 1-> -> -> [ microwave ] -> [ receiver + ] [videotape] [hydrogen ]-> ) )-P->[analog-to-digital]--->[recorder ]<---[frequency]-> /^ [ converter ] ^ ^ [standard ]-> / | ^ | | |-> / polarizer 1 | radio-signals time-marks |-> microwave / | |-> radiation |__________________________________|-> ^-> reference frequency from hydrogenous atomic clocks->-> -> for synchronization of atomic clock-> [transportable caesium]-> [ frequency standard ] ^ |[snip] === Length of basis ~ the Earth diameter === === | V-> radio-telescope 2->-> hydrogenous-> atomic-> parabolic antenna 2 tape 2 clock 2-> -> -> [ microwave ]-> [ receiver + ] [videotape] [hydrogen ]-> ) )-P->[analog-to-digital]--->[recorder ]<---[frequency]-> /^ [ converter ] ^ ^ [standard ]-> / | ^ | |-> / polarizer 2 | radio-signals time-marks |-> microwave / | |-> radiation |__________________________________|-> ^-> reference frequency from hydrogenous atomic clock->-> ->->->. The microwave interferometer with superlong basis. Part 2.. ----------------------------------------------------------.. Interference fringes. ^. |. [videotape 1] ------> [ COMPUTER ] <--------- [videotape 2]. ^ ^. | |. radio-telescope 1 <- synchronization clocks -> radio-telescope 2. Length of VLBI basis. |<-------------------------- {snip} ----------------------------->|./^ /^...^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ {snip} ^ ^ ^ ^ ^ ^ ^ ^ ^.| | | | | | | | | | | | | | | | | | | | | | | |. Noise microwave radiation[snip]> Alexsandr's contention is that since the detectors are situated all over> the Earth, and their detectors are absolutely synchronized (by later> comparisons of data streams from each detector), their detectors are absolutely synchronized (by later comparisons of data streams from each detector), With one stipulation, detectors will be synchronized physically ALMOUST absolutely, but detectors will be never synchronized absolutely precisely from a mathematical point of view.Always there is so-called a residual frequency of an interference .----------------------------------------------------------- ----------Hydrogenous atomic clock 1 & hydrogenous atomic clock 2 are physical Newtonian time in detector 1 & detector 2 accordingly.The atomic clocks are physical Newtonian time in each detector accordingly. The physical Newtonian time in the detector 1 is unpredictable differs from physical Newtonian time in the detector 2 . This phenomenon is termed as instability of atomic clocks. 1. rough synchronization of atomic clocksFor a tentative rough synchronization of atomic clocks transportable caesium frequency standard will be customaryly utillized.2. precise synchronization of atomic clocks in REMOTE POINTSThe most precise synchronization of atomic clocks in REMOTE POINTS is reached at deriving of VIRTUAL VLBI INTERFERENCE in the digital COMPUTER. I.e. Computer Data for tuning of a maximum of an interference we can utillize for PRECISE synchronization of atomic clocks in REMOTE POINTS in which one records of a signal from remote space object was yielded. === = ==Here fraud with synchronization of clocks in SR suffers complete experimental crash! >;^)))))) === === The instability of atomic clocks relative to each other is is very small, but this one ALWAYS exists.There are also other physical reasons of residual frequency of an interference .+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ +The trajectory to ?al rout of a method of synchronization of clocks in the theory of relativity uncloses a new method of synchronization of atomic clocks leaning on technology of VLBI interferometry.++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++++ > Sometimes claimants misquote or exaggerate to further their> own agendas. It is best to keep an open opinion until you have heard> from both sides of any story. -|Tom|-> Tom Van Flandern - Washington, DC - see our web site on replacement> astronomy research at http://metaresearch.org > Sometimes claimants misquote or exaggerate to further their> own agendas. It is best to keep an open opinion until you have heard> from both sides of any story. -|Tom|-> Tom Van Flandern - Washington, DC - see our web site on replacement> astronomy research at http://metaresearch.org === I think I have ?ed the problem. Please download the latest version from mywebsitehttp://www.dcproof.comPlease note that IE4 or later is required to view the help screen. You don'thave to be online, however, to view the help screen.Dan> Download a free beta version of my new proof checking software, DC Proof> 1.0> at> http://www.dcproof.com/> (for MS Windows 95 and later)> It is a learning aid to teach the fundamentals of logic and proof.After> entering a premise or assumption, the user can select rules of inference> for Logic, Set Theory and Number Theory from easy-to-use, drop-down menus.It> should be impossible to write an invalid proof in this system! Feedbackis> immediate with each new line.> A tutorial with examples and exercises demonstrating the main featuresof> DC> Proof is included along with the user manual. (Click the Help** Button.)> DC Proof is interesting from both pedagogical and theoretical> perspectives.> The axioms for set theory, for example, are a simpli?d version of the> standard ZF model. Yet they are robust enough to avoid the known> contradictions of naive set theory, and, I believe, to develop much of> modern-day mathematical theory.> Please let me know what you think of my new system.> Dan Christensen> Toronto, Canada> Neither the Help menu nor the Help** button produced any help. Here iswhat> got installed...> C:Program FilesDC Proofasyc?t.dll> C:Program FilesDC Proofcomcat.dll> C:Program FilesDC ProofCOMDLG32.OCX> C:Program FilesDC ProofDC Proof.chm> C:Program FilesDC ProofDC Proof.exe> C:Program FilesDC ProofDefaultDir.bin> C:Program FilesDC Proofe.gif> C:Program FilesDC Proofhhctrl.ocx> C:Program FilesDC Proofmsvbvm60.dll> C:Program FilesDC Proofoleaut32.dll> C:Program FilesDC Proofolepro32.dll> C:Program FilesDC ProofRecentFiles.bin> C:Program FilesDC ProofRICHTX32.OCX> C:Program FilesDC Proofstdole2.tlb> C:Program FilesDC Proofunins000.dat> C:Program FilesDC Proofunins000.exe> -- > Clive Tooth> http://www.clivetooth.dk> === >How do I use the sinh, cosh functions on my calculator? I'm not sure what>buttons to use. I have a Casio fx-991MS.> when all else fails, read the instructions! === Your posting this from Outlook Express V6. So you MUST have access to theWindows calculator.Make sure you havew it in scienti? mode. You will see across the top acheckbox which says hyp - click this on and the sin function becomes thesinh function etc. Note also you can select radians or degrees - the defaultis degrees which is not so usual for hyperbolic calculations. So key in3.1415Note that you can copy and paste into any windows apps, includingIf you also have the Of?e suite, have a look at Excel. It does hyperbolicsjust ?e, and lets you do lots of other stuff with the data (plotting.documenting calculations, database if you want).> How do I use the sinh, cosh functions on my calculator? I'm not sure what> buttons to use. I have a Casio fx-991MS.> === > Your posting this from Outlook Express V6. So you MUST have access to the> Windows calculator.> Make sure you havew it in scienti? mode. You will see across the top a> checkbox which says hyp - click this on and the sin function becomes the> sinh function etc. Note also you can select radians or degrees - thedefault> is degrees which is not so usual for hyperbolic calculations. So key in> 3.1415The concept of radians or degrees does not apply to the xxxh functions - youwill get the same (correct) answer in each case. Most Casio's have a hypkey which changes the next trig function to the corresponding hyperbolic.--Paul V. S. TownsendInterchange the alphabetic elements to reply === >How do I use the sinh, cosh functions on my calculator? I'm not sure what>buttons to use. I have a Casio fx-991MS.>Either the functions are present on your fx-991MS or not. If not present, youcould use the identity expression for the hyperbolic sine and cosine, andinclude the low-level pencil and paper process with the calculator, or composeyour calculations ON the calculator. Actually, I'm not familiar with yourcalculator; is that a graphing-type calculator? G C === >How do I use the sinh, cosh functions on my calculator? I'm not sure what>buttons to use. I have a Casio fx-991MS.> Either the functions are present on your fx-991MS or not. If notpresent, you> could use the identity expression for the hyperbolic sine and cosine, and> include the low-level pencil and paper process with the calculator, orcompose> your calculations ON the calculator. Actually, I'm not familiar with your> calculator; is that a graphing-type calculator?Nope, it's a standard scienti? calculator. I think it's one of the Casio'sbetter calculators of this type. === >How do I use the sinh, cosh functions on my calculator? I'm not sure what>buttons to use. I have a Casio fx-991MS.> Either the functions are present on your fx-991MS or not. If not> present, you> could use the identity expression for the hyperbolic sine and cosine, and> include the low-level pencil and paper process with the calculator, or> compose> your calculations ON the calculator. Actually, I'm not familiar with your> calculator; is that a graphing-type calculator?> Nope, it's a standard scienti? calculator. I think it's one of the Casio's> better calculators of this type.Use sinh(x) = (1/2)(exp(x) - exp(-x)) and cosh(x) = (1/2)(exp(x) + exp(-x)).[ exp(x) = e^x ]-- Paul SperryColumbia, SC (USA) === In sci.math, R3769Hey 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's 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>He can still go for the $20K, not to mention the $30K, the $50K,the $75K, the $100K, the $150K, or the $200K. I should mentionthat the $200K involves factoring a 2,048-bit number.I'm wondering how big JSH thinks that is. :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === > I ?d it dif?ult to put a litotic spin on ...Now you're sounding like Jack Sarfatti.Gib === > I'm NOT a mathematician.> However, I'm talking about two papers in rather fundamental areas of> mathematics.--ils duces d'Enron!http://larouchepub.com === In sci.math, Gerry Myerson> 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.>Depends on the width of the margin... :-)-- #191, ewill3@earthlink.net -- and the book he's reading :-)It's still legal to go .sigless. === In sci.math, Doug Norris:>prove if x is an irrational number then sqroot of x is irrational??>Any rational number has a ?ite number of digits> You mean, like 1/3 = 0.3333333333333333...?> DougIt's probably worth proving this is in?ite:Let 1/3 = .Q_1 Q_2 Q_3 ... Q_n ...where Q_i are integers from 0 to 9, in the usual fractional expansion.It's obvious thatQ_1 = ?0^1 / 3)for example. We also de?e a sequence R_i whereR_1 = 10 % 3 (or 10 mod 3) = 1and set things up such thatQ_{n+1} = ?_n * 10 / 3)R_{n+1} = R_n * 10 mod 3Since 10 mod 3 = 1, all of the R's will be 1. Therefore, all ofthe Q's will be 3.This proof is adapted from a long division algorithm that most peoplelearn in childhood. I'll admit I'm not entirely certain how todo it more rigorously although one could do the following.De?e a sequence of positive integers M_i, N_i for i > 0such thatM_i / 10^i <= 1/3 <= N_i / 10^iand M_i is as large and N_i is as small as possible.ThereforeM_i <= 10^i/3 <= N_iSince 10^i/3 is never an integer (10 mod 3 = 1, if 10^n mod 3 = 1then 10^(n+1) mod 3 = 10 mod 3 = 1), we let M_i = ?0^i/3)and N_i = ceil(10^i/3), M_i != N_i, and the expansion is thereforein?ite. (I can't use this proof to show it repeats.)This result is obviously generalizable; as long as therational number p/q is such that q has any primes otherthan 2 and 5, the expansion will be in?ite.One can of course explore things such as base 12 or 16. 10has such a paucity of prime factors. :-)In base 12, for instance, 1/3 = 0.4. Of course 1/5 = .249724972497...and 1/7 = .186A35186A35186A35...-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === In sci.math, Jeroen Boschma<3FD6D35E.8FF57477@fel.tno.nl>: I bet you already knew that cos pi = -1 and such. But I was shocked to ?d,> just now, that there appears to be a _vegetable_ in the domain of cos x.> (Yes, you read that correctly. Obviously this is a silly pseudo-mathematical> post, but then you could tell that from the thread's title.)> What is that vegetable? What is the value of cos x there?> The vegetable is lettuce.> According to my dictionary:> cos lettuce = romaine> Yup, and when Jesus comes, its the end of mathematics because he'll wash> away our sin's...This thread is de?itely a tangent...> Jeroen-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === >?st letters and the ?st two of the person I am writing the program >with, and on the same time the essence of Karma really ?s the game >(the sum of your actions and so on), but Iago is almost better...> And taken, AFAIK :-) === > Kolmogorov complexity, IIRC. Kolmogorov introduced the idea of> describing the complexity (or entropy, or information content) of a> sequence as the minimum length of an algorithm needed to generate it.> Of course, this de?ition is relative to a chosen way of describing> algorithms, you can derive many interesting facts about Kolmogorov> complexity that are true for *any* way of describing algorithms.> DaleThe Kolmogorov-Chaitin de?ition of complexity is in opencontradiction to System Theory, Cellular Automata experience andmathematics history.The complexity of a system results in the interaction and type o?teraction of its parts, not in the length of their description.Think in the smallness of Conway's algorithm to generate the highcomplex LIFE's game.Or the randomness adquired by certain symetric curves, only bychanging the position of a variable in its parametric equations.Or the algorithm to generate the digits of 1/3125 that is longer,(bytwo characters), than the algorith to generate the digits of theperiod of 1/97, but clearly, the last sequence is more complex thanthe former.The examples can continue to in?ite.L.Rodriguez === > |Perhaps we should post here de?itions of in?ite integer sequences,> |then rank our sequences by their randomness and by their simplicity> |of description somehow.> |> |The goal of the contest would be to post the sequence with the highest> |(amount of randomness) divided by, say, (the length of its decription> |in some language).> |> |By randomness, perhaps I mean the minimum order of a polynomial needed> |to duplicate the sequence for a ?ed number of terms (say, 100).> |> |There might be a much better de?ition of randomness. Feel free to> |post any de?ition you feel might be better.> if you don't ?d a better de?ition of randomness then my sequence is> f(n) = 2^n.n! is one character shorter ( === => Well, the last time I didn't follow something you suggested I was> applying Fubini to the wrong double integral. So maybe I'm 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?> You are right. It is wrong. I fell in the trap. Joel sent me just the idea with the identity, maybe he thought to use it differently, though I don't see how.Anyway, you have at least to agree that it was a nice idea :-)Ciprian Pop === > Well, the last time I didn't follow something you suggested I was> applying Fubini to the wrong double integral. So maybe I'm 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?>You are right. It is wrong. I fell in the trap. Joel sent me just the >idea with the identity, maybe he thought to use it differently, though I >don't see how.>Anyway, you have at least to agree that it was a nice idea :-)Oh, I'd agree to that even if I didn't have to. Seems sortof inspired by Fourier analysis, expressing |f| in terms of theFourier transform chi of the distribution of f. A very natural thingto try, since the distribution of f(x) + f(y) is just a convolution,which becomes a product under the Fourier transform -if only Re(chi^2) <= Re(chi) we'd be set...>Ciprian Pop****David C. Ullrich === One of the more persistent posters in replying to me has been DikWinter, a person who also has several webpages up about some of my oldwork, which I notice when I do Google searches.However, he not only seems to lack a throrough mathematicalunderstanding, he often screws up on basics, and I've yet to see anyother poster correct him.In my experience, posters on sci.math are labeled crank NOT for beingwrong, but for upsetting other posters!Now then I'll give a simple example to show one of Winter's mistakesin reasoning to show you exactly what I mean.For a while I posted about the factorization(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) = a_3(x) - 3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).Notice you have three factors of the polynomial:(5 a_1(x) + 7), (5 a_2(x) + 7), and (5 b_3(x) + 22)where if x=0, a_1(0) = a_2(0) = b_3(0) = 0.Dik Winter adamantly argued that 49 divided those factors as afunction of x, so that you had something like w_1(x) w_2(x) w_3(x) = 49where the w's are functions that vary as x varies, so thefactorization, after dividing both sides by 49 is((5 a_1(x)+ 7)/w_1(x))((5 a_2(x) + 7)/w_2(x))((5 b_3(x) + 22)/w_3(x))= 300125 x^3 - 18375 x^2 - 360 x + 22where he was always rather vague about the w's, and in fact never gavethem.That's a classic crank strategy--being vague about the details--buthere it turns out that Winter is easily foiled by considering constantterms.You see, you now have the factors((5 a_1(x)+ 7)/w_1(x)),((5 a_2(x) + 7)/w_2(x)), and((5 b_3(x) + 22)/w_3(x))which *look* terribly complicated, and Winter probably feels that he'ssafe with his assertion with expressions that are hard to resolve.However, now let me consider the possibility that the w's arealgebraic integer functions and that the factors shown are algebraicinteger functions for which I'll use f's.Then I havef_1(x) = ((5 a_1(x)+ 7)/w_1(x)),f_2(x) = ((5 a_2(x) + 7)/w_2(x)), andf_3(x) = ((5 b_3(x) + 22)/w_3(x))and letting x=0, I have((5 a_1(0)+ 7)/w_1(0))((5 a_2(0) + 7)/w_2(0))((5 b_3(0) + 22)/w_3(0))= 300125 0^3 - 18375 0^2 - 360 (0) + 22sof_1(0) = 7/w_1(0) = 1,f_2(0) = 7/w_2(0) = 1, andf_3(0) = 22/w_3(0)= 22.and it follows that w_1(0) = 7, w_2(0) = 7, and w_3(0) = 1. Now using g's to again separate out constant terms as the constantterms were separated out in my original factorization, I havef_1(x) = g_1(x) + 1, f_2(x) = g_2(x) + 1, and f_3(x) = g_3(x) + 22.So now I have(g_1(x) + 1)(g_2(x) + 1)(g_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22where the g's all equal 0, when x=0, which compares nicely with theoriginal.Now then you might suppose that there exist g's that will in generalbe algebraic integers, but in fact, they do not exist.I would ?d it intriguing if anyone is willing to dispute that fact.James Harris === [snip]> So now I have> (g_1(x) + 1)(g_2(x) + 1)(g_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where the g's all equal 0, when x=0, which compares nicely with the> original.> Now then you might suppose that there exist g's that will in general> be algebraic integers, but in fact, they do not exist.> I would ?d it intriguing if anyone is willing to dispute that fact.What fact? You have made an unsupported assertion about the properties ofthe ?g's without evidence or proof. Post your proof and THEN ask if anyonewishes to dispute it.--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 === > One of the more persistent posters in replying to me has been Dik> Winter, a person who also has several webpages up about some of my old> work, which I notice when I do Google searches.> However, he not only seems to lack a throrough mathematical> understanding, he often screws up on basics, and I've yet to see any> other poster correct him.> In my experience, posters on sci.math are labeled crank NOT for being> wrong, but for upsetting other posters!> Now then I'll give a simple example to show one of Winter's mistakes> in reasoning to show you exactly what I mean.> For a while I posted about the factorization> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> where b_3(x) = a_3(x) - 3 and the a's are roots of> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).> Notice you have three factors of the polynomial:> (5 a_1(x) + 7), (5 a_2(x) + 7), and (5 b_3(x) + 22)> where if x=0, a_1(0) = a_2(0) = b_3(0) = 0.> Dik Winter adamantly argued that 49 divided those factors as a> function of x, so that you had something like> w_1(x) w_2(x) w_3(x) = 49> where the w's are functions that vary as x varies, so the> factorization, after dividing both sides by 49 is> ((5 a_1(x)+ 7)/w_1(x))((5 a_2(x) + 7)/w_2(x))((5 b_3(x) + 22)/w_3(x))> => 300125 x^3 - 18375 x^2 - 360 x + 22> where he was always rather vague about the w's, and in fact never gave> them.> That's a classic crank strategy--being vague about the details--but> here it turns out that Winter is easily foiled by considering constant> terms.> You see, you now have the factors> ((5 a_1(x)+ 7)/w_1(x)),> ((5 a_2(x) + 7)/w_2(x)), and> ((5 b_3(x) + 22)/w_3(x))> which *look* terribly complicated, and Winter probably feels that he's> safe with his assertion with expressions that are hard to resolve.> However, now let me consider the possibility that the w's are> algebraic integer functions and that the factors shown are algebraic> integer functions for which I'll use f's.> Then I have> f_1(x) = ((5 a_1(x)+ 7)/w_1(x)),> f_2(x) = ((5 a_2(x) + 7)/w_2(x)), and> f_3(x) = ((5 b_3(x) + 22)/w_3(x))> and letting x=0, I have> ((5 a_1(0)+ 7)/w_1(0))((5 a_2(0) + 7)/w_2(0))((5 b_3(0) + 22)/w_3(0))> => 300125 0^3 - 18375 0^2 - 360 (0) + 22> so> f_1(0) = 7/w_1(0) = 1,> f_2(0) = 7/w_2(0) = 1, and> f_3(0) = 22/w_3(0)= 22.> and it follows that w_1(0) = 7, w_2(0) = 7, and w_3(0) = 1.> Now using g's to again separate out constant terms as the constant> terms were separated out in my original factorization, I have> f_1(x) = g_1(x) + 1, f_2(x) = g_2(x) + 1, and f_3(x) = g_3(x) + 22.> So now I have> (g_1(x) + 1)(g_2(x) + 1)(g_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where the g's all equal 0, when x=0, which compares nicely with the> original.> Now then you might suppose that there exist g's that will in general> be algebraic integers, but in fact, they do not exist.> I would ?d it intriguing if anyone is willing to dispute that fact.> James HarrisHmm sounds like you James. We've been wanting to get speci? informationfrom you, but you never answer those requests. You should be ashamed fordragging these people's reputations through the mud. You are a crank becauseyou are unwilling to learn and get mad when anyone says you are wrong. YOUACT LIKE A 5 YEAR OLD KID!-- David MoranChief MeteorologistOklahoma Storm Team === (g_1(x) + 1)(g_2(x) + 1)(g_3(x) + 22) => 300125 x^3 - 18375 x^2 - 360 x + 22> where the g's all equal 0, when x=0, which compares nicely with the> original.> Now then you might suppose that there exist g's that will in general> be algebraic integers, but in fact, they do not exist. It would be a very odd factorization in which any one of the g's is a constant not functionally dependent on x, so why should anyone expect such functions to be algebraic integers? Unless his understanding of mathematics was as erratic as JSH's.What they can be is homogeneous linear polynomials, each an algebraic number times x. It only then makes any sense to ask whether those constant multipliers can be algebraic integers (or reciprocals of algebraic integers). But JSH does not demnstrate that he has enough sense. === > One of the more persistent posters in replying to me has been Dik> Winter, a person who also has several webpages up about some of my old> work, which I notice when I do Google searches.> However, he not only seems to lack a throrough mathematical> understanding, he often screws up on basics, and I've yet to see any> other poster correct him.Dik T. Winter is probably the one person posting on sci.math with the most experience in the implementation of mathematical algorithms. Look, James, my prime counting implementation beat yours by a factor of 1000. You are just lucky that Dik hasn't felt any urge to trounce you even more. === Algebraic toplogy (AT herein) seems to be a good way of formalizingprotocols in distributed systems (DS herein) (such as decision problems like'consensus'). I am interested in learning more, however AT is a huge ?ldand I am only interested in learning the parts directly related todistributed computing. Can anyone suggest a book, that a) assumes noknowledge of algebraic toplogy b) assumes no more than undergraduate levelmath - ie: calculus, linear algebra, basic geometry, ability to do proofs,etc... c) focused on showing how AT can be utilized to solve DS problems anddoes not get into non-DS related aspects of AT (unless they are requiredbackground for understanding DS related AT topics).I have found many introductions on the net, but they seem to assume at leastbasic knowledge of topology, homotopy, and other topics I am not familiarwith, so I think a full book dedicated to the subject sounds more feasibleas a basis for learning AT for DS. But any links or online books you mayknow of will be of great help as well.l8r, Mike N. Christoff === Arrrghh.... this calculator is really annoying me.I've put my fx-991MS calculator into CMPLX mode, and I can get it tocalculate and show me, for example, the real and imaginary parts of 1 - 3i(simple I know but it de?itely works). However I can't get it to show methe real and imaginary parts of, for example, 1 - sqrt3i. It says MathERROR. This also happens with *any* other complex numbers calculationinvolving roots or powers.Am I missing something fundamental about this particular type of complexnumber?There's almost no point in the complex numbers functionality implemented inthis calulator if it can only work with simple complex numbers. Couldsomeone with the same or similar Casio calculator tell me if the samehappens on theirs? I speci?ally chose this calculator because it was themost expensive of it's type in the shop so I'm gonna be annoyed if thisdoesn't work.If I could get this to work it would help me a lot - normally I would be?e about calculating by hand - but I've got a multiple choice exam nextweek and having the calculator there and fully functioning would be a bighelp.Any suggestions? === >Arrrghh.... this calculator is really annoying me.>I've put my fx-991MS calculator into CMPLX mode, and I can get it to>calculate and show me, for example, the real and imaginary parts of 1 - 3i>(simple I know but it de?itely works). However I can't get it to show me>the real and imaginary parts of, for example, 1 - sqrt3i. It says Math>ERROR. This also happens with *any* other complex numbers calculation>involving roots or powers.If you mean (sqrt(3))*i, then you can almost certainly overcome theproblem by putting in some parentheses.If you mean sqrt(3i) then you are up against a restriction on theoperations allowed on your calculator. (Note that sqrt ismulti-valued.)>Am I missing something fundamental about this particular type of complex>number?>There's almost no point in the complex numbers functionality implemented in>this calulator if it can only work with simple complex numbers. Could>someone with the same or similar Casio calculator tell me if the same>happens on theirs? I speci?ally chose this calculator because it was the>most expensive of it's type in the shop so I'm gonna be annoyed if this>doesn't work.>If I could get this to work it would help me a lot - normally I would be>?e about calculating by hand - but I've got a multiple choice exam next>week and having the calculator there and fully functioning would be a big>help.>Any suggestions?> === I'm searching a conformal mapping that transform a rectangle in a ringbounded by two circles with the same center, i.e., a disc with acircle hole with the same center.The horizontal edges of the rectangle must transform in the two circle(with the same center), while the two vertical edges must coincide atthe end of the mapping and they must be transformed in a verticalradius line going from the inner to the outer circles.MAy anybody help me? === > I do admit that if enough people think that I'm just some nut, then I> have to wonder.We know you're a crank, Harris. People like you waste space insci.skeptic all the time.Why don't you go put on your tinfoil hat and invent a perpetual motion machine?-jcr === Hey, haven't been here in a while...Anyway, Thursday and Friday I was working on a problem, basically,given a fuciton and a continuous interval of that function, I wantedto ?d the length of its curve. I eventually came up with this:The length of the curve f(x) on the continuous curve [a,b] is theintegral taken from a to b of the squareroot of the sum of thederivitive of the function, squared, plus one.Bobby Simione === > The length of the curve f(x) on the continuous curve [a,b] is the> integral taken from a to b of the squareroot of the sum of the> derivitive of the function, squared, plus one.This formula for arc length is probably in every elementary calculus text. === >Consider : X_1, X_2, ... is a sequence of independent random variables with>?ite variances and a common distribution F such that F(0) = 0. Now>consider the product>Z_n = product (from k = 1 to n) of { (2(X_k)^2) / ( (X_{k-1})^2 +>(X_{k-2})^2 ) }What are X_0 and X_{-1}? However, you could start from 3.>In order to analyze Z_n, I'd like to talk about 2E[(X_n)^2] and>E[1/((X_{n-1})^2 + (X_{n-2})^2]. What is the relationship between these two>expectation? Is one always bigger than the other?The product is the same asE[1/((X_{n-1})^2 + (X_{n-2})^2)]*E((X_{n-1})^2 + (X_{n-2})^2)],which is strictly greater than 1 unless the distribution is trivial.>Then ?ally I'd like to show that Z_n converges with probability one to a>random variable that is ?ite with probability one.The only ?ite number this could possibly converge to is0, and this does happen, at least if the logarithm has a?ite mean. Only the last term involves X_n. The logarithm, suitable normalized, will be asymptoticallynormally distributed if the logarithm of the denominatorhas ?ite variance. One can apply the m-dependent centrallimit theorem. In that case (ln(Z_n) - n*A)/sqrt(n)is asymptotically normal with ?ite variance. But on theweaker condition that the logarithm has a ?ite mean, themean of the logarithm of the denominator is greater thanthat of the numerator, and the strong law of large numberswill have the product at least converging weakly to 0.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue UniversityX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCC1Fs02845; === > ...>am doing. in the end i am trying to show that Aut(Z16) is isomorphic>to Z2xZ8, and I dont think I want to use the fact that I know all>groups of order 8.> ...If Z16 stands for the cyclic group with 16 elements, then its rather simple to determine Aut(Z16): choose a generator, for example 1. Then an automorphism f is uniquely determined by prescribing the image f(1). This image must be another generator of Z16. So there are as many automorphisms as generators of Z16. Less than 16 by the way - so Aut (Z16) cannot be isomorphic to Z2xZ8.HX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCC1Ed02782; === In the ?st posting, I overlooked the trivial case of n = 1.The parts enclosed with *** may be skipped.*** ****I put trivial above in parenthesis denote a paradox:i) It is trivial... if indeed * is an associative.ii) If it were trivial, it shouldn't taken me so long to realize that this case must be handled differently. ...which I'm sure has it's origin in two interpretations of trivial.This all reminds me of a scene from the humorous movie The Princess Bride- where a warrior giant losesin a ?ht with a little man. To save his pride,he sais afterwards as an excuse I guess I havebeen accustomed to ?hting gangs of men at a time for so long that I forgot how to ?ht a single man!* *Here is the original posting with thecorrection:Perhaps there's another name for this besides what I'm calling it here.Def. Let X be a set. Call *: X times X -> X an associative if (a*b)*c = a*(b*c) for all a,b,c in X.Let Z be the whole numbers. Then normal multiplicationand ab = a+b-ab (ab normal multiplication) are associatives.Earlier, I had asked if the set of all associatives ofa given set had a particular name (still not sure): Let As(Z) be such a set on Z.Every x in N (naturals) has a unique representation asx =(x_1)(x_2)...(x_n) wherex_1, x_2, ...,x_n are (positive) primes and x_1 <= x_2 <= ... <= x_n (normal multiplication)If x in Z and x < 0, then x also has a (unique positive) primerepresentation asx =-(x_1)(x_2)...(x_n) wherex_1, x_2, ...,x_n are prime and x_1 <= x_2 <= ... <= x_n I would like to show that * in As(Z) where * is de?ed as:For x,y > 1 x*y = [(-1)^j]xy = [(-1)^j](x_1)(x_2)...(x_n)(y_1)(y_2)...(y_m) = [(-1)^j](z_1)(z_2)...(z_{n+m}) = [(-1)^j]zwhen (x_1)(x_2)...(x_n) (y_1)(y_2)...(y_m) (z_1)(z_2)...(z_{n+m}) are the (unique) prime representations of x, y and z (=xy) respectively and j is the minimum number of shifts necessary to order (x_1)(x_2)...(x_n)(y_1)(y_2)...(y_m)into (z_1)(z_2)...(z_{n+m}).If x or y = 0, then de?e x*y = 0.De?e 1*x = x = x*1 and (-1)x = -x = x(-1)If x < 1 or y < 1, then do the multiplication exactly as ifboth were positive, but substitute j with (j+k) werek = 1 <-> (x < 0 and y > 0) or (x > 0 and y < 0)k = 2 <-> x,y < 0.Notes: Whether the shift is to the left or the right does not play a role, i.e. j is the same number in either case. We cannot replace For x,y > 1 above with For x,y > 0; i.e., the rules for 1 must be de?ed differently for the law of association to hold (which they were). Otherwise, we would have, for example: (3*1)*1 = (-3)*1 = 3 != -3 = 3*1 = 3*(1*1) In the ?st equation above, 3 was shifted once to the right... -> j = 1 This does not happen when 1 is replaced by any other whole number, for example 2: (3*2)*2 = (-6)*2 = (-(2)(3))*2 = (2)(2)(3) = 12 and 3*(2*2) = 3*4 = [(-1)^2](2)(2)(3) = 12 In the second equation, j = 2 since each 2 was shifted once to the left to order (3)(2)(2) into (2)(2)(3) so that there were two total shifts. Alternatively, 3 was shifted twice to the right.Other examples:(3*10)*14 = (3*((2)(5))*14 = (-(2)(3)(5))*14 = (-(2)(3)(5))*((2)(7)) = -(2)(2)(3)(5)(7) = -420Reason: in the 2nd equation, one shift was necessary; 2 was shifted once to the left. in the 4th equation, two shifts were necessary; 2 was shifted twice to the left. 3*(10*14) = 3*((2)(5)*(2)(7)) = 3*(-(2)(2)(5)(7) = -(2)(2)(3)(5)(7) = -420Reason: in the 2nd equation, 2 was shifted once to the left. alternatively, 5 was shifted once to the right. in the 3rd equation, 3 was shifted twice to the right.We can now assign each x in N a parity according to:x in P^{+} <-> (x*x)/(xx) = 1x in P^{-} <-> (x*x)/(xx) = -1Ex. 10 in P^{-}, 70 in P^{-}, but 560 in P^{+}.Let R be the reverse operator, i.e.if x = (x_1)(x_2)...(x_n)= (x_1)*(x_2)*...*(x_n) is the prime representation of x, then Rx := (x_n)*...*(x_2)*(x_1)A few properties and conjectures (in the order of decreasing obviousness, as I see it now):i) RRx = xii) x*(Rx) in P^{+}iii) x in P^{+}, y in P^{-} -> x*y in P^{-} x in P^{-}, y in P^{+} -> x*y in P^{-} x in P^{-}, y in P^{-} -> x*y in P^{+} x in P^{+}, y in P^{+} -> x*y in P^{+}iv) x*y = y*x <-> x and y of same parityv) x*(y+z) = x*y + x*z <-> x*y and x*z of same parityC. DementX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCEOH512889; === thanksX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCEOHD12885; === I read your witten in mathematics and i think it is very nice in this time ....But u don't know where you are intrest in mathematics,i'am intrested in ring theory and make my search now in goldie ring..Or give me the adress of someone who intersting in this subject. thanks,X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCEOJL12935; === > However, I always questioned if saying that |N| < |R|> has any more meaning than saying that irrational numbers> are green.>The irrationals are green, and justi?bly so IMO, out of envy of the>rationals.> thats obviously because they are purple!> Dude...take off those colored glasses... I> see them clearly as Chartreuse.... with pink spots...>Now somewhat seriously:>Besides being a mathematician, I am also a musician. As such, I am aware>that some people experience synesthesia. The composer Scriabin was famous>for this, experiencing certain color sensations when he heard certain>pitches. I would not be greatly surprised if there were some people who>experience certain color sensations when contemplating certain numbers.>Does anyone know of any instances of such?>DavidGreetings fellow musician!Well, I don't see colors every time I see a number. But personally, the following list (expect for the number 1)has been completely obvious to me ever since I was about 14.In fact, even though I haven't thought about it directly for years, I would say the list probably seems more obvious to me now than ever.1 don't know, white?2 red 3 yellow4 green5 blue6 a yellowish orange7 brown8 black 9 yellowBy the way, there was a storyabout synesthesia on the BBC a few years back-I believe it centered around people who saw colorswhenever the smelled certain smells.C. Dement === [snip]>Besides being a mathematician, I am also a musician. As such, I am aware>that some people experience synesthesia. The composer Scriabin was>famous for this, experiencing certain color sensations when he heard>certain pitches.Greetings!Your ?al comment below made me think again about Scriabin. I may well bemistaken, but I seem to think now that he also may have associated varioussmells with pitches and colors.>I would not be greatly surprised if there were some>people who experience certain color sensations when contemplatingcertain numbers. Does anyone know of any instances of such?> Greetings fellow musician!> Well, I don't see colors every time I see a number. But> personally, the following list (expect for the number 1)> has been completely obvious to me ever since I was about 14.> In fact, even though I haven't thought about it directly for> years, I would say the list probably seems more obvious> to me now than ever.Fascinating!> 1 don't know, white?> 2 red> 3 yellow> 4 green> 5 blue> 6 a yellowish orangebecause 6 = 2*3 perhaps?> 7 brown> 8 black> 9 yellowbecause 9 = 3*3?Are numbers which you recognize as being divisible by 3 always at leasttinged with yellow?David> By the way, there was a story> about synesthesia on the BBC a few years back-> I believe it centered around people who saw colors> whenever the smelled certain smells.> C. DementX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCFtrR20081; === >[snip]>Every sequence of subsets of N convergent to N, isn't>uniformly convergent to N unless the sequence is eventually N.>Every sequence of subsets of R convergent to R, isn't uniformly>convergent to N unless the sequence is eventually dense.What is R? If it is a complex normed space, could I please see a proof of this second statement?C. DementX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCEOH312881; === >I am having a terrible time understand a mathematical derivation in this>computer graphics paper:><http://www.vis.uni-stuttgart.de/~roettger/data/ Papers/TERRAIN.PDFI know this is asking a lot, but if someone could go to page 4, and see if>they understand how the constant K is inferred, I would very much appreciate>it. I have been trying to ?ure this out with no success and I have posted>on several graphics forums, but no one seems to be able to ?ure it out.>Here are some speci? questions regarding the paper:>confused for the number 1):>Thus the following condition must hold for the decision variable f1 of an>adjacent block in order to limit the level differences:>(4) f1 < f2 <=> ln1 / (d*d21) < ln2 / [(d/2) * d22]>----------------->Ok, I think he is saying that if we are going to subdivide f2, which is a>level lower than f1, then we must also subdivide f1. But, couldn't the>relationship also be f1 <= f2, that will still ensure they are both>subdivided, right?>Also, the original equations of f is>(3) f = ln / [d*C*max(c*d2, 1)>How is it that he infers (4) from (3). I can see that we can cancel ?C',>but I do not follow how he simpli?d the max. function.>He continues:>For a point of view falling inside the rectangular region Equation (3) [f <>1] is always satis?d, since ln1 / d is always less than the minimum>resolution C.>----------------->Now when he says for a point of view falling inside the region, does he mean>the view point (camera position) lies inside the region, or does he mean the>region is inside the viewing frustum? I also do not follow how he says ln1>/ d < C.>He continues:>Outside this region the value of the fraction ln1 / (2*ln2) is bounded by>1/2 and the constant K:>1/2 < ln1 / (2*ln2) < K (C > 2)>K = L1 / (2*L2) = C / [2*(C - 1)]>---------------->I'm totally lost here. First of all, from the ?ure in the paper, I cannot>make out what ln1 / (2*ln2) means geometrically. Perhaps if I knew where he>obtained that relationship I would understand the rest; speci?ally how it>is bounded. Moreover, I do not understand how he obtains the relationship K>= L1 / (2*L2) = C / [2*(C - 1)] from that.>Did you contact author?X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCEOIX12918; === [snip]>Every sequence of subsets of N convergent to N, isn't>uniformly convergent to N unless the sequence is eventually N.>Every sequence of subsets of R convergent to R, isn't uniformly>convergent to N unless the sequence is eventually dense.>You claim A_n ->* A iff>A_n subset A forall n and for all p > 0, p in R exists q in N>for all a in A for all n > q exists b in A_n: d(a,b) < p>For all n, let A_n = D be a dense subset of R.>Let A be any set with D subset A subset R.>Then A_n ->* A for all such A's.>Your uniform convergence, and hence your convergence doesn't have unique>limits. Moreover, a sequence can even converge to different sets with>different cardinalities.Sorry for taking so long to respond... hadsome other stuff I was working on for a while.Remember the following part of the ?st message in this series of posting? **>You have a problem here, because a limit is not unique, if A' is a >subset of A, then A_n -> A implies A_n -> A'. My response:I hope I covered your concern in my ?st Re:, where the following was added to the initial question: ... A_n subset A for all n should also be required (or somethingsimilar like there exists an m such that for all n > m: A_n subset A). That way two sequences A_n and B_n approaching A willalso approach each other... Under these conditions, it appears the limit is almost unique, i.e., it is unique if the set A' subset A your arerefering to above is closed in the metric topology of X; a little proof:If A_n -> A and A_n -> A', then (by de?ition)A_n subset A for all n and A_n subset A' for all n. If A' is a proper subset of A, then there must exist an elementx which is in A but not in A'. Thus, there exists a sequence (x_n) converging to x in the metric topology of X withx_n in A_n for all n. Since A_n subset A' for all n, x_n in A' for all n. Since A' is closed, x in A', a contradiction.** *Thus, it was already known that the limit was notunique if the A_n's were not additionally requiredto be closed. If, for all n, we let A_n = D be a dense subset of R.and let A be any set with D subset A subset R.Then A_n ->* A for all such A's. But if, as above,we again assume A_n closed we have no problem- since thenit follows that A_n = D = A = X. C. DementX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCFe9v19025; === > Do negative numebrs exist in the real world or are they just a purely> symbolical and mathematical construct?>Do positive numebrs exist in the real world or are they just a purely>symbolical and mathematical construct?A like that question. Two bad (pun intended) you weren't myprofessor.Towards the end of my ?st semester at the university,while everyone else was learning about the integral, Ifelt betrayed because I kept thinking:Well, if it all depends on the naturals/reals and yourinitiation into mathematics is to sware an oath of faith to them (-> axioms), why haven't I ever seentwo things exactly alike? In addition (this time no pun intended),I wondered if the 2 in 2 bears on the hillwas really the same 2 as in 2 kilometers/hour?, etc.Although I probably wasn't thinking about then, this questiondoes have relevence in physics (in thermodynamics, for example,see extensive/intensive quantities).So, in my reasoning of the time, 2 was already strange enough;I never dared consider even stranger numbers, such as i.Consequently, I got behind really fast and had to play a lotof catch up, later.However, i seems no more strange to me today than 2.I think it's because I see i as part of agroup (for example, the quaternions). Sure,this group has 8 elements- but 8 as part of a mathematical set is not the same as 8 objects in nature:If someone had only said to me back then Look, you might not have ever seen two thingsexactly alike, but you have seen a lot of two thingswhich were almost alike... nature is only concernedwith the almost (by the way, has anyone seen two things exactlydifferent?). Therefore, the natural numbers can only beapproximately applied to nature (thus, there is no harm done in accepting the axioms as truths). There were also some philosophical questions that Ikept trying to apply to math. For example True and False. In my mind, I thought statement X false -> X cannot (physically) exist. That led me to strange conclusions like Everything anybody sais must be right,because if were wrong it couln't exist. But what they said*did* exist... In a freshman seminar speech, I even gotan oscilloscope and hooked it up to a microphone to show this.I stated into the microphone: 2+2 = 5 and said See, the waveson the screen? It really exists! It's not wrong!C. Dement X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCFmKt19390; === > Def.> Let X be a set. Call *: X times X -> X an> associative if (a*b)*c = a*(b*c) for all a,b,c in X.>(X,*) is a semi-group. 2x2 matrices is an example of>a noncommunative monoid, ie semi-group with identity.> Let Z be the whole numbers. Then normal multiplication> and ab = a+b-ab (ab normal multiplication) are associatives.>a#b = a+b - ab is know as symmetric difference in set theory.>Symmetric differences of a boolean ring produce a boolean algebra.>, I'm allergic to permutations.where I hopefully corrected a mistake.)Hey, after two months time it's funny that you write me Re: convergence on a space with no topology)C. DementX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCJ2R003492; === I am trying to do some research on alcuin's sequence, but all the references that I can ?d quote the terms of the sequence as the coef?ients in the Maclaurin expansion of [(1-x^2)(1-x^3)(1-x^4)]^-1. I ?d this surprising, as I believe the sequence to have been named after Alcuin of York (735-804) who lived many years before Maclaurin expansions were discovered. This sequence also gives the number of different triangles that have integral sides and perimeter n. I would think that it is much more likely that this is what Alcuin discovered, and years later someone found that the above expansion acted as a generating function. That is, of course, if the Alcuin who the sequence is named after is Alcuin of York and not someone else with the same name.I would be most grateful if you could either con?m or refute this, or point me in the direction of any references.Many thanksTonyX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCJ2R203504; === >What with the war on terror, and Canada's puny military and corrupt>police force, the only way to insure peace and safety for that>country, is for it to be annexed by America.>It is the only way foward for Canada aka Canuckistan.Maybe it would be more appropriate if Canada annexed the U.S.U.S. has a greater population, but Canada has a greater land area.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCL80P13965; === X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBCLLiM15116; === >prove if x is an irrational number then sqroot of x is irrational??? PROOF:> Assume that> (a)x is irrational;> (b)x^(1/2) is p/q for some natural numbers p, q.> Put x^(1/2) = p/q> ln(x) = 2ln(p/q))> x = exp(ln((p/q)^2))> = (p/q)^2.> But this contradicts (a). This completes the proof.>Is there some reason you're using logarithms and exponentials>in this problem? I mean, it's not incorrect, but it's truly>irrelevant to what you want to show.>Why not prove the contrapositive:> If sqrt(x) is rational, then x is rational.>or more simply put:> If y is rational, then y^2 is rational.>???>A word to the wise:> If you use things (e.g., concepts, functions,> constructions, calculations) that are not> required in, or relevant to, an argument, it> shows that you don't understand what the argument> is about.>I suppose that principle may apply more generally than>within mathematics.>DaleActually, I realized the unnecessity of logarithm, and put anotherthread. Please see the thread (the one at the bottom). ButI can not see any logical ? the argument.Contrapositive works if and only if the original argument works:)X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBD3OxR09540; === >I tried to solve another task on this issue today and failed. {/-(>First of all, I wanted to know if there is a general approach to solve>tasks of the form: You have given a fraction of the form P(x)/Q(x).>There P(x) and Q(x) are polynoms. You have to express this fraction>in the form sum(a_k*(m*x+n)^k, k, 0, in?ity). ?>[m, n are constants and a_k represents k constants.]>If somebody knows a general approach or a website (understandable>for a beginner) there this is explained then please tell me. (I somehow>must understand this matter.)>Anyway, this was my task today:>###>task:>Express 14 / ( 8*(x^2-1) ) as an in?ite sequence of the form:>sum(a_k*(2-x)^k, k, 0, in?ity).>____________>|my attempt(s):>14 / ( 8*(x^2-1) ) = (7/4) * ( 1 / ( (x-1)(x+1) ) )>partial fraction decomposition (Do I name it right?):1 = A*(x+1) + B(x-1)>if x = 1> => A = 1/2>if x = -1> => B = -1/2>==> (7/4) * ( 1 / ( (x-1)(x+1) ) )>= (7/4) * ( 1/( 2*(x-1) ) - 1/( 2*(x+1) ) )>= (7/8) * ( 1/(x-1) - 1/(x+1) ) ***>Attempt N 1:>*** = (7/8) * ( -1/(1-x) - 1 / ( 1-(-x) ) )>= (7/8) * ( -sum( x^k, k, 0, in?ity) - sum( (-x)^k, k, 0,>in?ity) )>= (7/8) * ( -sum( x^k - (-1)^k * x^k, k, 0, in?ity) )>= (7/8) * ( -sum( ( 1 - (-1)^k ) * x^k, k, 0, in?ity) ) = ?>I didn't know how to transform this to the demanded form.>Attempt N 2:>*** = (7/8) * ( 1/(1-2+x) + 1/(1-2-x) )>= (7/8) * ( 1 / ( 1 - (2-x) ) + 1 / ( 1-(2+x) ) )>= (7/8) * ( sum( (2-x)^k, k, 0, in?ity ) +> sum( (2+x)^k, k, 0, in?ity ) ) = ?>I didn't know how to simplify (2-x)^k + (2+x)^k, so>that I get (..)*(2-x)^k {the demanded form}.>###>That's it. It would be very nice if you told me a general>approach for these tasks.>Karl I don't know that it is very general but with something like14 / ( 8*(x^2-1) ) , I would think about the geometric series. A series of terms of the form ar^n converges for -1< r< 1 andhas sum a/(1-r). I can force 14/(8*(x^2-1)) into that form byrewriting it is (-14/8)(1-x^2) so that I need to take a= -14/8= -7/4 and r= x^2. That means a geometric sum of the formSum((-7/4)(x^2)^n) or (-7/4)(1+ x^2+ x^4+ ... + x^(2n)+ ... ).X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBD7ZlZ26134; === >
Anyway, I see no reason to stop
there. Since>the system already uses deci- and
centi-,>why not use all metric fractional pre?es?
This>would give: >decillion =3D 10^33>centillion =3D
10^303>millillion =3D 10^3003 =3D 10^(1000^1 + 3)
>micrillion =3D 10^3000003 =3D 10^(1000^2 + 3)>nanillion
=3D 10^(1000^3 + 3) >picillion =3D 10^(1000^4 +
3)>femtillion =3D 10^(1000^5 + 3)>attillion =3D 10^(1000^6
+ 3)>zeptillion =3D 10^(1000^7 + 3)>yoctillion =3D
10^(1000^8 + 3) >Uh, oh... ran out of pre?es! :-( =A0 BTW,
the>American version of that last one
has>1,000,000,000,000,000,000,000,003 zeroes,>and the
British version (10^(1,000,000^8))
has>1,000,000,000,000,000,000,000,000,000,000>
000,000,000,000,000,000 zeroes (somewhat>dif?ult to write
out in full). >Ooops... that should read:>decillion =3D
10^33>centillion =3D 10^303>millillion =3D 10^3003 =3D
10^(3*1000^1 + 3) >micrillion =3D 10^3000003 =3D 10^(3*1000^2
+ 3)>nanillion =3D 10^(3*1000^3 + 3) >picillion =3D
10^(3*1000^4 + 3)>femtillion =3D 10^(3*1000^5 + 3)>attillion
=3D 10^(3*1000^6 + 3)>zeptillion =3D 10^(3*1000^7 +
3)>yoctillion =3D 10^(3*1000^8 + 3) >vecillion =3D
10^(3*1000^9 + 3)>xonillion =3D 10^(3*1000^10 +
3)>dekacontillion =3D 10^(3*1000^11 + 3)>vigincontillion =3D
10^(3*1000^21 + 3)>trigincontillion =3D 10^(3*1000^31 +
3)>quadragincontillion =3D 10^(3*1000^41 +
3)>quinquagincontillion =3D 10^ (3*1000^51 +
3)>sexagincontillion =3D 10^ (3*1000^61 +
3)>septuagincontillion =3D 10^ (3*1000^71 +
3)>octogincontillion =3D 10^ (3*1000^81 +
3)>nonagincontillion =3D 10^ (3*1000^91 + 3)>hectillion =3D
10^ (3*1000^101 + 3)X-Received: (from approve@localhost) by
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>This is
computing.>For a = 0 to d> FOR b = a To d> FOR c = b TO d> IF
a+b+c = d THEN PRINT a,b,c> next> next>nextThere is a more
ef?ient solution, if the order of the numbers doesn't
matter. In that case, we can constrain a <= b <= c and
reason as follows: -We want all solutions for a + b + c = d
where a <= b <= cthese conditions are equivalent (after a few
lines of simple algebra) to 0 <= a <= d/3 a <= b <= (d-a)/2 c
= d-b-aSo we can get away withfor a = 0 to d/3 for b = a to
(d-a)/2 c = d-b-a print a, b, c end forend forX-Received:
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Sorry to post such a straightforward
problem... This is not a homework problem; I genuinely need
the answer for an MSc. research project I am working on.
Unfortunately I have not done any real maths in about 10
years and I am too rusty to ?ure this out.Here's the
problem: -Given three random variables A, B and C, and
givenH(A|B) = 0 (i.e. A is just a deterministic function of
B)proveH(B|C) >= H(A|C) I can see intuitively that it *has*
to be true for discrete distributions because you can't know
less about a function of B than you know about B, but I can't
produce a formal proof to that effect. (That's how rusty I
am!)Also, I really want the inequality to be true of
continuous distributions as well. But my intuitive reasoning
doesn't work there because they have that irritating property
of entropy not being invariant under relabelling, i.e.
H(f(X)) <> H(X).Can anyone help me? I would be extremely
grateful.Simon McGregorsm66blue.sussexpink.ac.ukRemove
colours from the preceding line and replace ?st . with
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Derivation of
these algorithms can be found
fromhttp://vsg.cape.com/~pbaum/date/date0.htmand their
referenced pages.>
I understand the formula for
converting a (Gregorian) date to a Julian day>number. I have
been unable to convince myself that the following
algorithm>converts a Julian day number to a (proleptic)
Gregorian date, for ALL>positive JDN's:>Given n, a modi?d
JDN:>day = n + 2432046;>c = (day + 36524)*4/146097 - 1;>day
-= c*146097/4;>y = (day + 365)*4/1461 - 1;>day -=
y*1461/4;>year = 100*c + y - 4800;>month = (5*day -
3)/153;>day -= (153*month + 2)/5;>if ((month += 3) > 12) {>
month -= 12;> year++;>}>I understand the second section,
computing the month and day, and adjusting>the month and
year from the computational convenience of beginning
years>with March. My problem is the computation of the
(unadjusted) year. I>recognize that the temporary values c
and y are the century and two-digit>year, respectively.
But how to derive this formula, or to prove that it
is>correct?>Eric Farmer>
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===
 Sub: Unsolvabve Geometrical Problems
Squaring a given circle is unsolvable . There are volumes of
pages on the subject which I have read - from Archimedes to
SrinivasaRamanujan. Well! I have found out a solution. Could
not believe it ? My logic is very simple.Pi is an
irrational number. So is square root of 2. There is a simple
geometrical construction to solve for square root of 2.
Similarly I have developed a simple geometrical construction
to solve for pi. My question is Where to send it? Is there
any Maths Forum where I can present my papers and answer
questions of the experts? 
===
> Sub: Unsolvabve Geometrical
Problems > Squaring a given circle is unsolvable . There are
volumes of pages on the subject which I have read - from
Archimedes to SrinivasaRamanujan.> Well! I have found out a
solution. Could not believe it ? My logic is very simple.>
Pi is an irrational number. So is square root of 2. There
is a simple geometrical construction to solve for square
root of 2. Similarly I have developed a simple geometrical
construction to solve for pi.> My question is Where to send
it? Is there any Maths Forum where I can present my papers
and answer questions of the experts? > I suggest that you
place your result on a Web page and post the URLhere. Then
anyone who's interested could take a look.Rickp.s. Your
keyboard has a Return or Enter key (or perhaps both).Learn
where it is and use it from time to time. My
newsreaderdoesn't wrap your lines and I have this thing
abouthorizontal scrolling. 
===
> Sub: Unsolvabve
Geometrical Problems > Squaring a given circle is unsolvable
. And its unsolvabilty has been proven.>There are volumes of
pages on the>subject which I have read - from Archimedes to
SrinivasaRamanujan.And none of them proved it. So what?>
Well! I have found out a solution. No, you haven't.>Could not
believe it ? Belief is irrelevant. It's been proven
impossible.>My logic is very simple.And wrong.>Pi is an
irrational number. It's also transcendental.>So is square
root of 2. But the square root of 2 is _not_ transcendental.
Look up the difference.>There is a simple>geometrical
construction to solve for square root of 2. Nobody said that
was impossible.>Similarly I have>developed a simple
geometrical construction to solve for pi.If you constructed
it, what you have is _not_ pi, because pi cannot
beconstructed.> My question is Where to send it? The circular
?e.>Is there any Maths Forum where I can>present my papers
and answer questions of the experts? Sure, right here on
sci.math. But instead of crowing about your proof, tryasking
forhelp in understanding why it's wrong.> May I request you
to kindly guide me in this regard. Sorry, I wouldn't be able
to ?d your fallacy even though I know it
exists.>--MensanatorAce of Clubs
===
> (axiom-schema) is
introduced...along with power set...pairing is> redundant,
they advocate keeping it because it is elementary and>
essential for the sequential development and even for a
scheme that> shuns replacement, pairing makes sense. I think
they are suggesting> that set theory even without replacement
is strong enough for almost> all practical purposes. What do
you think?>Oh, I see, I'm used to the shunning replacement
version, because I'm>used to GBN set theory. So I think of
replacement+comprehension as>just GBN's comprehension, but of
course, that isn't quite right, and>this is an example why.>I
believe that in GBN set theory, pairing *is* required, is it
not?Yes, because the NBG axioms do not have
propositionalfunctions. Functions are de?ed only by ordered
pairs.-- This address is for information only. I do not claim
that these viewsare those of the Statistics Department or of
Purdue University.Herman Rubin, Department of Statistics,
Purdue UniversityX-Received: (from approve@localhost) by
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>
I
understand the formula for converting a (Gregorian) date to a
Julian day>number. I have been unable to convince myself that
the following algorithm>converts a Julian day number to a
(proleptic) Gregorian date, for ALL>positive JDN's:>Given n,
a modi?d JDN:>day = n + 2432046;>c = (day + 36524)*4/146097
- 1;>day -= c*146097/4;>y = (day + 365)*4/1461 - 1;>day -=
y*1461/4;>year = 100*c + y - 4800;>month = (5*day -
3)/153;>day -= (153*month + 2)/5;>if ((month += 3) > 12) {>
month -= 12;> year++;>}>I understand the second section,
computing the month and day, and adjusting>the month and
year from the computational convenience of beginning
years>with March. My problem is the computation of the
(unadjusted) year. I>recognize that the temporary values c
and y are the century and two-digit>year, respectively.
But how to derive this formula, or to prove that it
is>correct?>Eric Farmer>

===
> achievement> What
achievement? Taking a set of irrelevant and badly> taught
courses and getting grades in them? Learning to> study for
exams and forget the material afterward?>Learning to show up
for work every day and eat up the nonsense the boss>dishes
out.>An important achievement.For a non-intellectual type of
job.-- This address is for information only. I do not claim
that these viewsare those of the Statistics Department or of
Purdue University.Herman Rubin, Department of Statistics,
Purdue University
===
>I have never reviewed material for
exams when I took the>course.> I like to say that a test
which can be studied for is not worth> giving (or
taking).>Been to a medical doctor lately? I'm glad mine
studied for their exams.Why? The important part here, as
elsewhere, is beingable to reason and understand. It is
seeing the symptoms,and what else may be odd. We will
eventually have theanalysis of the symptoms being done
probabilistically,as they should be now, and the patient
informed of theresult, and also of the distribution of
bene?s and adverse consequences of the available treatments.
Most medicine is not emergency room.-- This address is for
information only. I do not claim that these viewsare those of
the Statistics Department or of Purdue University.Herman
Rubin, Department of Statistics, Purdue University
===
> All
degrees, credentials, etc., should be by comprehensive>
examinations ALONE. Anything else is
anti-educational.>Written by someone with a degree which is a
result of a long paper.I did have a moderately long
examination, and would even now, decades later, be willing to
take an evenmore thorough comprehensive examination. -- This
address is for information only. I do not claim that these
viewsare those of the Statistics Department or of Purdue
University.Herman Rubin, Department of Statistics, Purdue
University
===
> I would not go THAT far, but I object to the
use of degrees> and other such credentials unless they can
provide reasonable> assurance that the person has a good
share of the important> knowledge and ability.>I know of no
graduate program which relies on a BA alone as its>admissions
criterion. So indeed, they all work pretty hard to make>sure
one has gotten that knowledge and ability.Having been
involved is selecting graduate students for along time, I see
no way to do this, although the attemptis made. Even a very
high ?st class ?st from anIndian university, and usually a
?st class is strong,can fail to have it.-- This address is
for information only. I do not claim that these viewsare
those of the Statistics Department or of Purdue
University.Herman Rubin, Department of Statistics, Purdue
University
===
> All degrees, credentials, etc., should be
by comprehensive> examinations ALONE. Anything else is
anti-educational.> Perhaps the appropriate method depends on
the degree. For example, one> could imagine that a degree in
music composition could have a ?al> hurdle where the
candidate composes a musical composition... A Ph.D. in>
mathematics could have a ?al hurdle where the candidate
produces> original theorems. And so on.Signi?ant original
theorems.Jon Miller
===
> All degrees, credentials, etc.,
should be by comprehensive> examinations ALONE. Anything
else is anti-educational.> Perhaps the appropriate method
depends on the degree. For example, one> could imagine that
a degree in music composition could have a ?al> hurdle
where the candidate composes a musical composition... A Ph.D.
in> mathematics could have a ?al hurdle where the candidate
produces> original theorems. And so on.>Signi?ant original
theorems.This is what a thesis is SUPPOSED to be, but often
is not.-- This address is for information only. I do not
claim that these viewsare those of the Statistics Department
or of Purdue University.Herman Rubin, Department of
Statistics, Purdue University
===
>|Why is that unfortunate?
Certainly it should not be oppressively>|dif?ult, but
neither should it be so easy as to allow
unquali?d>|applicants to get in.> Indeed.> Some people
here may not be aware of how much of a problem> it can be
when someone enters a graduate program without> really being
ready to bene? from it the way they want to. No kidding. I
just spent 3 years as head of the admissionscommittee for
regular freshmen. How DO you get it through parents heads
that declining admission totheir unprepared/untalented child
is doing them a huge favor?That admitting someone who has a
99% probability of failingall their ?st semester courses,
(most of which will benon-credit remedial courses anyway) is
tantamount to ?ancialaid fraud? That jumping into a
university setting that is overtheir head and failing, rather
than tech school or two-year collegeand succeeding, is a major
emotional blow to the kid, from which he will never recover?
That if we admit their kid, the only difference at the end of
three months is that he will have anabysmal academic record
that will PREVENT him from admission toa more suitable school
AND they will be out several thousanddollars for their
trouble?Why don't they just walk around bad neighborhoods at
night yelling please mug me and club my kid around the head
andshoulders! thereby leaving me and the taxpayers out of
theirstupidity?Bart
===
> In other words, it is simultaneously
true that both the number of balls > in the basket and the
number of balls removed grow arbitrarily large as > time
approaches noon.Agreed.> It makes no sense to talk about what
is in the basket at noon, as time> is outside the realm of
your thought experiment (your conclusion 3),> which is not
physically realizable.I think we all agree that this problem
cannot be done physically (atleast in our universe). Actions
cannot continue in?ity long (as theuniverse will die before
we are done), actions cannot be performed atfaster than the
speed of light, nor do we have the required in?iteamount of
matter to create these balls. Any attempt to answer thismust
be modelled with a different discipline.Set Theory seems like
the natural one, since it routinely handlesin?ite sized sets,
performing countable unions and differences. Heck, none of
these examples involves sets that are even uncountable! 
===
>
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 de?e the set Sn as follows:> Sn = Sn+1 U
{ 10n+1, 10n+2, ..., 10n+10 }  {10n+1}.Oops, that
formulation was way off. That should be: S0 = Empty set Sn+1
= Sn U { 10n+1, 10n+2, ..., 10n+10 }  {n+1} for n>0Boy, was
that sloppy of me. :-/ (Fortunately, it hadn't diminishedthe
spirited discussion.)Jonathan HoyleGene Codes
Corporation
===
> 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 de?e the set Sn as
follows:> Sn = Sn+1 U { 10n+1, 10n+2, ..., 10n+10 } 
{10n+1}.What the heck is that?> Let S = lim n->inf (Sn).> If
you agree with the above as a mathematical description of the>
problem, then Solution #2 is the only correct answer, for the
reasons> given.> Solution #1 is incorrect because it
confuses Card(lim n->inf Sn) with> lim n->inf Card(Sn).>
Solution #3 is avoided because no laws of physics are
involved in this> redescription, and these set operations are
de?ed on countable> unions.> Jonathan Hoyle> Gene Codes
CorporationSo you say no balls are left? Certainly the balls
are removed one at atime.What was the last ball removed if
there are no balls left? I would like toknow what natural
number (that's the way they are labeled) is on that
ball.Certainly you can tell me that number right
away.
===
>No, you have a basket ?led with some number of
indistinguishable >balls. At each epoch, you reach in and
remove one. You can even shake >them up ?st, if you like. I
think would most say that, after the nth >epoch, there are 9n
indistingushable balls in the basket. At noon, there >are
in?itely many indistinguishable balls in the basket.If we agree to map this to Set Theory, you can't 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 ?ite step, it absolutely>matters which ones you
remove at lim Sn. Calling them>indistinguishable doesn't
change this fact. Constantly removing the>ones that ?st 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 in?ite number.>For newcomers to this thread, I
repeat the problem under discussion.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.
How many balls are in the basket at noon (= 12)?I strongly
suspect that those who disagree with the analysis in the ?st
paragraph above (a contribution of mine) are being
disingenuous, though very likely not consciously so. Had I
presented this problem *before* any mention of a problem with
labels, would you really have disagreed?As I stated before,
this is all a question of modelling. Balls and baskets are
*not* mathematical objects. Though one will probably not be
so explicit, one would draw my stated conclusion above in the
following sense:The state at epoch n (i.e., at 12- 2^(1-n)) is
represented completely by the number a_n of balls in the
basket after the transaction at epoch n. To be painfully
explicit, a_n is determined bya0 = 0a_(n+1) = a_n + 10 -1 for
n in N.Since lim a_n = in?ity, we say there are an in?ite
number of balls in the basket at noon.>A much more simpli?d
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 in?ite>sized set. If you do not specify which ones you
take out, you cannot>determine the answer.>There are no
labels on the balls, hence no such thing appears in this
model. What do you say aboutProblem 6. You have a tub with
in?ite capacity. At time 12-2^(1-n), you add 10 liters of
water to the tub, then you scoop out and discard 1 liter. How
much water is in the tub at noon?Now let us look at explicit
models forProblem 2. 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. How many
balls are in the basket at noon?A possible model (suggested
by JH) is the following: Represent the state at epoch n as
S_n = the set of numbers which appear as labels on the balls
after the transaction at the nth epoch.S0 = {}S_(n+1) = (S_n
U {10n-9, 10n-8, ..., 10n})  {n}Since lim S_n = {}, one
concludes that the basket is empty at noon. One also notes
card(S_n) = a_n, which grows arbitrarily large.However, I am
not convinced that this is necessarily *the* correct model
for this situation, for two reasons.1) The situation is not
physically realizable, and2) No purpose for the answer is
speci?d. (After all, my ?ld is operations research!) I.e.,
what is the real question we are trying to answer? Noon is
outside of the realm of the experiment described. For
example, regardless of what answer we give to this ?titious,
essentially nonmathematical question, it is still true that no
?ite-capacity basket will ever hold all the balls which are
supposed to be in there simultaneously at some epoch.Because
of reason 2, it is not clear that the same model as in
problem 1 will not suf?e. What if the labels are invisible
(e.g., in white against a white background)?>[...]>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 ?ite. There are
an>in?ite number of times each ball is removed and an
in?ite number>of times it is put back (albeit with a new
label)...a classic in?ity>minus in?ity 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 ?ite>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.>De?e T_n as the set of numbers that were the initial
labels of the balls in the basket after the transaction at
epoch n.T_0 = 0T_(n+1) = (T_n U {10n-9, 10n-8, ..., 10n}) 
{10n}(T_n) is an increasing sequence of sets whose union is N
 10N. So there are an in?ite number of remaining in the
basket. How do you justify the difference between the two
answers?>[...]>As I have pointed out before, the very
interesting subtlety here is >dealing with the concept of
in?ity. I do not agree with JH's >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-de?ed 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 subtly
here I think is not in?ity 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's not even possible to
describe. >It seems to me that this problem has a solution
only within the ?ld>of mathematics, where in?ity
(particular within sets) is very well>de?ed.>I admit to
redundancy in the following remarks.The dif?ulty is one of
precise modelling of a physically unrealizable process (I
think we agree here) without interaction with the initial
poser. (There's the OR appearing again!) All of the
mathematical objects a_n, S_n, and T_n in the above
discussion are very precisely de?ed. Yet it remains a matter
of discussion which of these represents the *nonmathematical*
phenomenon (balls and baskets) to a suf?ient extent. I
disagree with the last quoted sentence.Perhaps we will have
to agree to disagree. I wonder what others think? (I have
cc'd this two a few other participants in this thread.) Do
you maintain that there is one correct answer here?-- Stephen
J. Herschkorn herschko@rutcor.rutgers.edu
===
>No, you have a
basket ?led with some number of indistinguishable >balls.
At each epoch, you reach in and remove one. You can even
shake >them up ?st, if you like. I think would most say
that, after the nth >epoch, there are 9n indistingushable
balls in the basket. At noon, there >are in?itely many
indistinguishable balls in the basket.>If we agree to map
this to Set Theory, you can't 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 ?ite step, it absolutely>matters which ones you
remove at lim Sn. Calling them>indistinguishable doesn't
change this fact. Constantly removing the>ones that ?st
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 in?ite number.> For newcomers to
this thread, I repeat the problem under discussion.> 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. How many balls are in the basket at
noon (= 12)?> I strongly suspect that those who disagree with
the analysis in the > ?st paragraph above (a contribution of
mine) are being disingenuous, > though very likely not
consciously so. Had I presented this problem > *before* any
mention of a problem with labels, would you really have >
disagreed?Whether the balls are distinguishable or not, they
have distinctidentities. If you ? your attention on any one
ball in the bucket, tothe exclusion of all others, you will
?d that with probablity 1, thatball will be removed before
noon (assuming random selection at eachstep).> As I stated
before, this is all a question of modelling. Balls and >
baskets are *not* mathematical objects. Though one will
probably not be > so explicit, one would draw my stated
conclusion above in the following > sense:> The state at
epoch n (i.e., at 12- 2^(1-n)) is represented completely by >
the number a_n of balls in the basket after the transaction at
epoch > n. To be painfully explicit, a_n is determined by> a0
= 0> a_(n+1) = a_n + 10 -1 for n in N.No, because this model
fails to capture the fact that some balls areremoved. Exactly
the same state would describe an experiment in which 9balls
are added at each step and none are removed.> Since lim a_n =
in?ity, we say there are an in?ite number of balls > in the
basket at noon.Not if we want to be precise, we don't. The
function describing the number ofballs at time t is
obviously discontinuous at a great many points. Whyshouldn't
t=0 (noon) be one of them?> Problem 6. You have a tub with
in?ite capacity. At time 12-2^(1-n), > you add 10 liters of
water to the tub, then you scoop out and discard 1 > liter.
How much water is in the tub at noon?Any given molecule is
eventually removed with probability 1. The totalnumber of
molecules is countable. Therefore, with probability 1, the
tubis empty at noon.> Now let us look at explicit models for>
Problem 2. 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. How many balls
are in the basket at noon?> A possible model (suggested by
JH) is the following: Represent the > state at epoch n as S_n
= the set of numbers which appear as labels on > the balls
after the transaction at the nth epoch.> S0 = {}> S_(n+1) =
(S_n U {10n-9, 10n-8, ..., 10n})  {n}> Since lim S_n = {},
one concludes that the basket is empty at noon. One > also
notes card(S_n) = a_n, which grows arbitrarily large.>
However, I am not convinced that this is necessarily *the*
correct model > for this situation, for two reasons.> 1) The
situation is not physically realizable, and> 2) No purpose
for the answer is speci?d. (After all, my ?ld is >
operations research!) I.e., what is the real question we are
trying to > answer? Noon is outside of the realm of the
experiment described. You have ?e apples at 11:55. I take
away two at 11:59. How manyapples do you have at noon?Do you
claim that noon is outside the realm of the
experimentdescribed?> For > example, regardless of what
answer we give to this ?titious, > essentially
nonmathematical question, it is still true that no >
?ite-capacity basket will ever hold all the balls which are
supposed > to be in there simultaneously at some
epoch.Depends on the sizes of the balls.> Because of reason
2, it is not clear that the same model as in problem 1 > will
not suf?e. What if the labels are invisible (e.g., in white >
against a white background)?The labels don't matter, unless we
use them to describe a particularorder in which the balls are
removed. In that case, it doesn't matterwhether the labels
can be seen or not, since the problem statement saysthe balls
are removed in a certain order. It doesn't say how. We
couldhave a very long tube in which balls are added at one
end and removedfrom the other in FIFO order.>[...]>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?Since this version has the
same ball being added and removed an in?itenumber of times,
I don't see how to decide where that particular ballmight be
when we ?ish. Rob Johnson would probably say the
functionfails to converge.-- Dave SeamanJudge Yohn's mistakes
revealed in Mumia Abu-Jamal
ruling.
===
>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?> Since this version has the same ball being
added and removed an in?ite> number of times, I don't see how
to decide where that particular ball> might be when we ?ish.
Rob Johnson would probably say the function> fails to
converge.That's an excellent point, Dave. I lost sight of the
fact that wemust look at the actual set element being
reinserted, regardless o? Labels are helpful as a
convenience for most of the otherproblems given, but in this
case following the labels does notfollow the balls since
each ball can have a different label atdifferent
times.Although I had ?st claimed that in Problem #4 the
container isempty, given Dave's argument, I would now say
that it is more likeProblem #1 and indeterminant.Jonathan
HoyleGene Codes Corporation