mm-359 === Subject: : Re: Equalizing Golf to Baseball Re: B. Bonds test on distance of playing field Re: better to slow pitch in majors> (snipped)> Because Golf is similar to Baseball in that the ball tee-ed off is a ballat zero> speed and would simulate a self pitched baseball which is nearly zerospeed.> So, all I would have to do is compute a factor of a golf club compared toa> baseball bat. Compute another factor of the aerodynamics of a golf ball> compared to a major league baseball. Then one more factor of the muscles> used in baseball compared to the muscles used in golf.> Plugging in all those factors in an equation.Can you imagine how far a golf ball would go if it was fired toward thetee at 200 MPH and you could hit the dang thing? My God, that wouldbe spectacular! === Subject: : Sum (1/(N logN) [was Re: is this a solid proof for series?]> ... a divergent series needn't be satisfy the> condition that a_n > k / n^p for some p in [0, 1] - as an example> consider a_n = 1 / (n log n).I was led to this series when trying to construct a series which divergedmore slowly than the harmonic. On the face of it it seemed to diverge suchthat you needed to square the number of terms to increase the sum so far byln2. I was however unable to prove it. Is there a simple proof? === Subject: : Re: Sum (1/(N logN) [was Re: is this a solid proof for series?]> ... a divergent series needn't be satisfy the> condition that a_n > k / n^p for some p in [0, 1] - as an example> consider a_n = 1 / (n log n).I was led to this series when trying to construct a series which diverged> more slowly than the harmonic. On the face of it it seemed to diverge such> that you needed to square the number of terms to increase the sum so far by> ln2. I was however unable to prove it. Is there a simple proof?There is a simple method that handles quite a few series like this. If the elements of the series a(n) are positive and decreasing as they are here, then the sum of a(n) converges/diverges if and only if the sum of 2^k * a(2^k)converges/diverges. That sum is in your example the sum over 2^k * (1 / 2^k / log 2^k) = 1 / (k log 2)which is divergent. === Subject: : Re: Sum (1/(N logN) [was Re: is this a solid proof for series?]>... a divergent series needn't be satisfy the>>condition that a_n > k / n^p for some p in [0, 1] - as an example>>consider a_n = 1 / (n log n).> I was led to this series when trying to construct a series which diverged> more slowly than the harmonic. On the face of it it seemed to diverge such> that you needed to square the number of terms to increase the sum so far by> ln2. I was however unable to prove it. Is there a simple proof?Well, my proof would be the integral test.Sum ( i = 3 to n ) 1 / (n log n) >= Sum ( i = 2 to n ) Integral ( x = i- 1 to i ) 1 / (x log x) dx = Integral (x = 2 to n) 1 / ( x log x) dx = log (log n ) - c.So if you square n then you do indeed add log 2 to (a lower bound for) the sum.I suppose there are probably more elementary proofs, but the integral test is generally nice and handy so why not use it? :)David === Subject: : Re: Sum Of All IntegersIn sci.physics, J 891<386aaf52.0311020656.9936f73@ posting.google.com>:>> Note subject change. Followups to sci.math only.[snip for brevity]>> Pedant point: I'm not sure if the sum of integers is well-defined.>> Or one can simply observe that the partial sums>> S_0 = 0>> S_1 = 0 + 1>> S_2 = 0 + 1 - 1>> S_3 = 0 + 1 - 1 + 2>> S_4 = 0 + 1 - 1 + 2 - 2>> ...>> S_k = 0 + 1 - 1 + 2 - 2 ... + n>> where there are (k+1) terms, is 0 if k is odd but k/2 if k is even.[Accuracy warning, I am working from memory of my university days long> ago.]In any sequence, this series does not converge in the standard sense> at all. But even if it did, it is not surprising that you can> rearrange it and get different answers.A series is said to be absolutely convergent if the sum of the> absolute values of the terms also converges (not necessarily to the> same value). If the terms of an absolutely convergent series are> rearranged then it will still converge and to the same value. If the> terms of a convergent series are all positive (e.g. the power series> of exp(x) for any positive x) then the series is obviously absolutely> convergent. But there are many series which are not all positive that> are also absolutely convergent. The power series exp(x) for negative> x or sin(x) and cos(x) for any x are examples. The convergence of> exp(x) can be used to easily prove the convergence of these series.If a series converges but is not absolutely convergent, then it is> said to be conditionally convergent. These are odd beasts. You can> rearrange the terms to make them diverge, and more surprisingly, you> can rearrange the terms to arrive at any desired answer.The series 1 - 1/2 + 1/3 - 1/4 + 1/5 etc is an example of a> conditionally convergent series. It is easy to prove that it> converges but if you make all the terms positive then it diverges. So> you can rearrange it to come to any answer you like.For the meaning of the terms as I am using them and some more detail,> see http://mathworld.wolfram.com/AbsoluteConvergence.html and links on> that page. This is a great site.You are correct as far as *my* memory goes (I graduated in 1983) :-).In any event, good explanation; also, I've found mathworld.wolfram.comis chock full of stuff. -- #191, ewill3@earthlink.netIt's still legal to go .sigless. electron-dot-cloud are galaxies === Subject: : Re: B. Bonds test on distance of playing field Re: better to slow pitch > Too simple. In fact, you just flunked physics 101. With stuff likeTry getting a real full name, so you flunked what, second grade.When you make a silly assertion twice on the Internet that a 100mphfastball is a disadvantage to hitters compared to a 0 speed tossup hit. You mustthen be called either a troll or a dunce or both.Archimedes Plutonium, a_plutonium@hotmail.comwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies Too simple. In fact, you just flunked physics 101. With stuff likeTry getting a real full name, so you flunked what, second grade.When you make a silly assertion twice on the Internet that a 100mph> fastball is a disadvantage to hitters compared to a 0 speed tossup hit. You must> then be called either a troll or a dunce or both.Archimedes Plutonium, a_plutonium@hotmail.com> whole entire Universe is just one big atom where dots> of the electron-dot-cloud are galaxiesRecommended reading:The Physics of Baseball - Momentum at:http://library.thinkquest.org/11902/physics/momentum.htmlBat Weight, Swing Speed and Ball Velocity by Daniel A. Russell:http://www.kettering.edu/~drussell/bats-new/batw8. htmForce of a Bat on a Baseball:http://hypertextbook.com/facts/2000/ AlbertKlyachko.shtmlSpeed of the Fastest Pitched Baseball:http://hypertextbook.com/facts/2000/ LoriGrabel.shtmlHTHHP === Subject: : normal subgroup......thank you....alwayslet H : subgroup of group Glet N = intersection aHa^(-1) [a in G]show that N is normal subgroup of G------------------------------------i think..........any g in G, any n in N => gng^(-1) in N (i use)let n = ai*hj*ai^(-1) in Ng*ai*hj*ai^(-1)*g^(-1) = (g*ai) hj (g*ai)^(-1) in Num.............i have a doubt final result. (g*ai) hj (g*ai)^(-1) in Nhow do you think about it??advice ......please...... very much.... === Subject: : Re: normal subgroup......> thank you....alwayslet H : subgroup of group Glet N = intersection aHa^(-1) [a in G]show that N is normal subgroup of G------------------------------------> I think you are having problems working out a proof.Here are some tips that may help you write out a formal proof.First, start off by prefixing your proof with the phrase Proof:so that one can see where the statement of the theorem ends andyour proof begins.Second, restate the hypotheses that are given in the theorem.Third, restate the claim that you are going to prove.Thus, you would have here in your case the following:=== Proof: Let G be a group. Let H be a subgroup of G. Let N = intersection over a in G of aHa^(-1). Claim: N is normal subgroup of G===Note that your claim has two things to prove: 1) N is a subgroup of G. 2) N is normal in G.You are concentrating on just proving the second statement below.But, you must also give a reason why 1) is true.You can give a reason for why 1) is true by quoting a prior theorem.Thus, you add to your proof the following:=== N is a subgroup of G since .===> i think..........any g in G, any n in NHere you introduce any.My suggestion is that you should not work with *any* g or*all* g.Instead, you should work with a particular (but arbitrary) g.I find that this makes the proof less abstract. There isjustification for doing this if you are aware of mathematicallogic.Thus, you continue in your proof as follows:=== Let g be an element of G. Let n be an element of N.===Note that implicitly you are aiming at showing thatgng^(-1) is an element in N, which will show N isnormal in G by the definition of being normal.Thus, you will have proved the second statement ofyour claim.> => gng^(-1) in N (i use)Here you use the implication sign =>.I don't like having a string of implies in a proofsince it makes the proof statements unwieldly.However, this proof is not the best example ofshowing how to avoid using =>. So, I will pass ongoing into more detail on that.> let n = ai*hj*ai^(-1) in NHere, you lost the thread of what needs to be proved.Recall that you want to show that gng^(-1) is an element in Nfor the particular g and n that I am currently working with.You want to look at the definition of N: Let N = intersection over a in G of aHa^(-1).One method is to replace N by its definition.Thus, you want to prove: gng^(-1) is an element of the intersection over a in G of aHa^(-1).Now, how do you prove an element is in the intersection of severalsets? The standard way is to prove that the element is in each ofthe sets. That is, you are replacing working with all a in G byworking with a particular (but arbitrary) a in G. This againmakes it more concrete.Thus, you have the following:=== Let a be an element of G. Claim: gng^(-1) is an element of aHa^(-1).=== Note that now a, g, n are particular elements.> g*ai*hj*ai^(-1)*g^(-1) = (g*ai) hj (g*ai)^(-1) in NWhat is ai? What is hj? You haven't introduce them yet.Instead, you want to prove that gng^(-1) is an element of aHa^(-1).But, you want to use the information that you have stated abovefor the elements a, g, n. In particular, you want to use theinformation that n is an element of N and that N is thethe intersection over a in G of aHa^(-1). That is, by definitionof intersection, n is an element of aHa^(-1) for all a in G.Note that this last a is not the same a that we selectedearlier, since it is quantified by all. That is, you arefree to select whatever a you want in G and still haven is an element of aHa^(-1).Now, the heart of the proof is to choose an appropriate ain the statement For all a in G, n is in aHa^(-1) so thatyou can show that gng^(-1) is an element of aHa^(-1).Let me eliminate the ambiguity of the two a's in that lastparagraph. That is, you have the following situation.The heart of the proof is to choose an appropriate bin the statement For all b in G, n is in bHb^(-1) so thatyou can show that gng^(-1) is an element of aHa^(-1).You can work backwards to see what is the required choicefor b.> um.............i have a doubt final result. (g*ai) hj (g*ai)^(-1) in Nhow do you think about it??advice ......please...... very much....I realize that the above may be a little vague. But, I thinkit gives you some suggestions on how to approach a (formal)proof.-- Bill Hale === Subject: : Re: normal subgroup...... Adjunct Assistant Professor at the University of Montana.>thank you....alwaysI think you should spend more time on your own homework. You areasking the newsgroup for ->really easy<- problems. That leads me tobelieve that either you are not trying at all, (in which case, whyshould we do your homework for you?) or else you are trying andfailing (in which case, you should not be in the class you are in;these are ->easy<- problems, if you cannot do them, then you arefailing the class).>let H : subgroup of group G>let N = intersection aHa^(-1) [a in G]>show that N is normal subgroup of G>------------------------------------>i think..........>any g in G, any n in N => gng^(-1) in N (i use)>let n = ai*hj*ai^(-1) in N>g*ai*hj*ai^(-1)*g^(-1) = (g*ai) hj (g*ai)^(-1) in N>um.............i have a doubt final result. (g*ai) hj (g*ai)^(-1) in N>how do you think about it??Show thatg (intersection of aHa^{-1} [a in G] ) g^{-1}is the same asintersection of (gaHa^{-1}g^{-1} [a in G]). Then take it from there. === Subject: : Re: normal subgroup......>let H : subgroup of group G>let N = intersection aHa^(-1) [a in G]>show that N is normal subgroup of G> Show that> g (intersection of aHa^{-1} [a in G] ) g^{-1}> is the same as> intersection of (gaHa^{-1}g^{-1} [a in G]). Then take it from there.That leads to some general theorems.Arbitrary intersection of subgroups is a subgroup.Arbitrary intersection of normal subgroups is a normal subgroup.Now when H is subgroup, aHa^-1 is subgroup, thus N is subgroup.Now if H is normal subgroup, then aHa^-1 is normal subgroup.and N too would be normal. Indeed H = aHa^-1 = N.However H is just any subgroup... === Subject: : Re: normal subgroup...... Adjunct Assistant Professor at the University of Montana.>>let H : subgroup of group G>>let N = intersection aHa^(-1) [a in G]>>show that N is normal subgroup of G>> Show that>> g (intersection of aHa^{-1} [a in G] ) g^{-1}>> is the same as>> intersection of (gaHa^{-1}g^{-1} [a in G]). Then take it from there.>That leads to some general theorems.>Arbitrary intersection of subgroups is a subgroup.>Arbitrary intersection of normal subgroups is a normal subgroup.>Now when H is subgroup, aHa^-1 is subgroup, thus N is subgroup.>Now if H is normal subgroup, then aHa^-1 is normal subgroup.>and N too would be normal. Indeed H = aHa^-1 = N.>However H is just any subgroup...However what?Here's a further hint, then, for you: show that if g is fixed, then asa ranges over all elements of G, ga also ranges over all elements of G. === Subject: : Homegeneous relations...HelloCan anyone tell me if the following is a linear homogeneous recurrencerelation?a_n = a_(n-1) / nMy first reaction was to say not homogeneous because of the n. But does thatrule apply in this case or only in the form similiar to:a_(n-1) + n where n is totally independent ? === Subject: : Re: Homegeneous relations...> Can anyone tell me if the following is a linear homogeneous recurrence> relation?a_n = a_(n-1) / nYes, it is.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: : Re: Homegeneous relations...Content-transfer-encoding: 8bit> Hello> Can anyone tell me if the following is a linear homogeneous recurrence> relation?a_n = a_(n-1) / nMy first reaction was to say not homogeneous because of the n. But does that> rule apply in this case or only in the form similiar to:> a_(n-1) + n where n is totally independent ? Linear? Yes.Homogeneous? Yes.Constant coefficient? No.-- === Subject: : Re: algebra problem.... Adjunct Assistant Professor at the University of Montana.>group G : not cyclic group such that |G| = 27>1. show that G : not simple group>i don't know second problem.Clearly.Two hints:(a) Show that if g^a = e, then a divides |G|. (Easy)(b) Assume there is some g for which g^9 is not equal to e; what isthe order of g then?>um.................>advice please.......my doctor....One advice I would give you, given the number of requests for help youhave thrown into the group in the last few weeks, is to ask yourprofessor for more help or to drop the class. If I had a student whoneeds to ask as many questions on his or her homework as you do, andat such a level as your questions, I would definitely suggest to himor her to re-evaluate his/her commitment to the course, and possiblyjust drop out. === Subject: : Re: algebra problem....>group G : not cyclic group such that |G| = 271. show that G : not simple group2. show that any g in G, g^9 = e (e =identity)>i don't know second problem.Clearly.Two hints:(a) Show that if g^a = e, then a divides |G|. (Easy)Sigh; this is wrong as written, of course. Here, it should say Showthat if a is the ORDER of g, then a divides |G|. That is, a isassumed to be the smallest positive integer such that g^a=e., sans .sig === Subject: : Re: algebra problem....>>group G : not cyclic group such that |G| = 27>>1. show that G : not simple group>>I agree with everything that Arrturo says below. But I would be intriguedto know how you managed to prove that a group of order 27 is not simpleby using Sylow's Theorem.Derek Holt.>>2. show that any g in G, g^9 = e (e =identity)>>i don't know second problem.>Clearly.>Two hints:>(a) Show that if g^a = e, then a divides |G|. (Easy)>(b) Assume there is some g for which g^9 is not equal to e; what is>the order of g then?>>um.................>>advice please.......my doctor....>One advice I would give you, given the number of requests for help you>have thrown into the group in the last few weeks, is to ask your>professor for more help or to drop the class. If I had a student who>needs to ask as many questions on his or her homework as you do, and>at such a level as your questions, I would definitely suggest to him>or her to re-evaluate his/her commitment to the course, and possibly>just drop out.= === Subject: : Re: Groups isomorphic? Adjunct Assistant Professor at the University of Montana.>in message :>> Consider, for example, that there are nonabelian groups of exponent p>> and order p^3, so that p^3-1 elements have order p and one element has>> order 1.>Nitpick: This is only true for p odd. Of course o knows>this, but maybe it will save the OP from looking for a>non-abelian group of exponent 2 and order 8.You are absolutely right; I've been working in my research with groupsof exponent p and class 2, p and odd prime, and I seem to havedeveloped the bad habit of not specifying the prime is odd... === Subject: : Re: Set Exponentation Adjunct Assistant Professor at the University of Montana.>My understanding of taking sets to powers of other sets is really bad.>My professor briefly mentioned something about the number of all>possible functions that exist between the two sets, but he is so>MIND-NUMBING that I can't stay awake in his class . . .>Let A and B both be infinitely countable sets. What does A^B mean?>Let A be finite and B be countable. What does A^B meanIn general, if A and B are sets, then A^B is the set of all functionf:B->A. That is, A^B is the collection of all subsets S of BxA such that: (a) For every b in B there exists a in A such that (b,a) is in S;and (b) For every b in B, if (b,a) and (b,a') are both in S, then a=a'.>For example, we have the set 2^N, whose cardinality is equivalent to>that of the set of real numbers.2^N is the set of all functions from N to 2={0,1}; you can think of itas the set of all characteristic functions of subsets of N.> If 2^N is the set of all infinite>sequences of zeros and ones,Yes.> then I can see how it relates to Cantor's>Diagonalization. >But HOW can 2^N = {0,1}^N DESCRIBE the set of all infinite sequences>of zeros and ones?Because, by definition, a sequence is a map whose domain is N, thenatural numbers. Here, you are specifying the image, {0,1}. So anysequence of zeros and ones is a map from N to {0,1}, hence an elementof 2^N.>Which leads to more questions . . . >{0,1)^1 = ? I assume you mean {0,1}^1; since 1, as a set, is 1={0}, that means allfunctions from {0} to {0,1}. There are exactly two functions: the onemapping 0 to 0, and the one mapping 0 to 1. So{0,1}^1 = {0,1}^{0} = { {(0,0)}, {(0,1)} }.>{0,1}^{0,1} = ?This is 2^2; it is the collection of all functions from {0,1} to{0,1}. There are four such functions: everything to 0; everything to1; 0 to 0 and 1 to 1; 0 to 1 and 1 to 0.{0,1}^{0,1} = { {(0,0), (1,0)}, {(0,1),(1,1)}, } { {(0,0), (1,1)}, {(0,1),(1,0)}. }>{0,1}^n = ? (where n belongs to N) The collection of all functions from n={0,1,2,...,n-1} to {0,1}>Let A = 2 = {0,1}. Let B = 3 =>{0,1,2} What does A^B mean?The set whose elements are all functions from {0,1,2} to{0,1}. There's eight of them.> What does B^A mean? The set whose elements are all maps from {0,1} to {0,1,2}; there are 9of them.>I'm betting that A^B= B^A is false. That's correct. === Subject: : Re: Set ExponentationXevious grava .88 la saucisse et au marteau:> Let A and B both be infinitely countable sets. What does A^B mean?The set of the applications from B to A. To each element of B can beassigned an element in A. Therefore, the cardinality of this set isCard(A)^Card(B).> Let A be finite and B be countable. What does A^B mean?Same thing. > For example, we have the set 2^N, whose cardinality is equivalent to> that of the set of real numbers. If 2^N is the set of all infinite> sequences of zeros and ones, then I can see how it relates to Cantor's> Diagonalization. But HOW can 2^N = {0,1}^N DESCRIBE the set of all> infinite sequences of zeros and ones? When I look at the text, 2^N> means nothing to me. Which leads to more questions . . .> {0,1)^1 = ?The set containing the applications 1-> 1 and 1-> 0 (not veryinteresting applications).> {0,1}^{0,1} = ?The set containing the four applications:f1:0 -> 01 -> 0f2:0 -> 11 -> 0f3:0 -> 01 -> 1f4:0 -> 11 -> 1> {0,1}^n = ? (where n belongs to N)I let you imagine what it is.> Let A = 2 = {0,1}. Let B = 3 = {0,1,2}> What does A^B mean? What does B^A mean?Idem.> I'm betting that A^B = B^A is false.Yes. The first set contains 8 applications and the second one 9.-- Genji === Subject: : Re: Set ExponentationBy the way, what Le Roux calls an application is usually called a function in English. His point is correct. Given two sets A and B, by definition, A^B = {f in P(B x A): f is a function with domain B}. When both A and B are finite, card(A^B) = card(A)^card(B). In fact, that is true when either A or B are inifinite as well. One fact of interest: If 1 < card(A) <= card(C), 1 < card(B) <= card(C), and C is infinite, then card(A^C) = card(B^C). I am pretty sure one can demonstrate this without the axiom of choice; I am sure someone here will correct me if I am wrong.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: : Re: Set Exponentation grava .88 la saucisse et au marteau:> By the way, what Le Roux calls an application is usually called a > function in English. His point is correct. Given two sets A and B, Oops, sorry. That's because there is a difference between fonction andapplication in French, and I thought there was the same difference inEnglish.-- Genji === Subject: : Re: Set Exponentation> grava .88 la saucisse et au marteau:>> By the way, what Le Roux calls an application is usually called a >> function in English. His point is correct. Given two sets A and B, >>Oops, sorry. That's because there is a difference between fonction and>application in French, and I thought there was the same difference in>English.That's interesting. What's the difference? I have not seen application used this way in English, but I am not a professional mathematician.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: : Re: Set Exponentation> grava ? la saucisse et au marteau:>> By the way, what Le Roux calls an application is usually called a > function in English. His point is correct. Given two sets A and B, >>Oops, sorry. That's because there is a difference between fonction and>>application in French, and I thought there was the same difference in>>English.> That's interesting. What's the difference? I have not seen > application used this way in English, but I am not a professional > mathematician.Could it be the same distinction as between function and map?Marc === Subject: : Re: Set Exponentation grava .88 la saucisse et au marteau: > By the way, what Le Roux calls an application is usually called a >> function in English. His point is correct. Given two sets A and B, Oops, sorry. That's because there is a difference between fonction and>application in French, and I thought there was the same difference in>English. That's interesting. What's the difference? I have not seen > application used this way in English, but I am not a professional > mathematician. The main difference is nobody in America uses Eniglish Set theorists definition of function. And we're the only ones to do math applications, it's just the same old story for France. They act like they discovered DNA of something. But, since it's just another case of Napoleon fever, the only thing we can tell them is the same thing we tell Canadian doctors: Call us in morning, after you get rid of those Ruskie snivels. === Subject: : French vs. English (Was: Re: Set Exponentation)I do not think Nicolas will mind my sharing this e-mail:>>That's interesting. What's the difference? I have not seen >>application used this way in English, but I am not a professional >>mathematician.>>An application f:E->F is a fonction defined over the whole subset E.>log:R->R is a fonction (but valid only on R+*)>log:R+*->R is an application*Very* interesting. In English, the concept of function has a certain asymmetry to it. A function is a special type of relation. One of the axioms of a function is that, for each element in its domain (but not the range), there is at least one ordered pair in the function which has the element as its first coordinate. Apparently, the French fonction drops this requirement, so that if f is a fonction from A to B, neither projection must be the entire superset.I find the definition of application confirmed in the 1977 notes, Logique et Theories Axiomatiques, by J.L. Krivine. (I received these when I was taking an undergraduate logic/set theory course with G. Sabbagh at Jussieu (Paris VII) in 1980.) Interestingly, Krivine never defines the word fonction, though he uses it in reference to an application and as a subheading for the section where he introduces the concept.This is another interesting linguistic contrast between French and English mathematics. The other one I am thinking of is field (literally, champs) vs. corps (literally, body): In a field (corps commutatif), multiplication is by definition commutative; not commutativity is an essential feature of multiplication for English speaker; not so for French speakers. To this we may compare the lack of agreement for the definition of ring even in English alone. Some authors (notably Isaac) require a ring to contain a multiplicative identity; others do not.(BTW, as I was composing this, I received the following:>>That's interesting. What's the difference? I have not seen >>application used this way in English, but I am not a professional >>mathematician.>>After a Google research, I found that the translation of application>is map or mapping.>hth)-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: : Re: French vs. English (Was: Re: Set Exponentation)>>log:R->R is a fonction (but valid only on R+*)>>log:R+*->R is an applicationfunction and total function. Do mathematicians say that too?Martin === Subject: : Re: French vs. English (Was: Re: Set Exponentation)> This is another interesting linguistic contrast between French and> English mathematics. The other one I am thinking of is field> (literally, champs) vs. corps (literally, body): In a field> (corps commutatif), multiplication is by definition commutative; not> commutativity is an essential feature of multiplication for English> speaker; not so for French speakers.I don't think that follows at all.In so far as there is any reason for the difference in usage,I suspect that it reflects a difference in the feelor philosophy of the two languages.A person writing in English would feel uneasyat repeating commutative field 50 or 100 times --the natural tendency would be to find a shorthand form.The French, on the other hand, are perfectly happy as far as I can seeto repeat a compound phrase indefinitely.(Do the Germans actually prefer it?)-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ietel: +353-86-2336090, +353-1-2842366s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: : Re: French vs. English> This is another interesting linguistic contrast between French and>> English mathematics. The other one I am thinking of is field>> (literally, champs) vs. corps (literally, body): In a field>> (corps commutatif), multiplication is by definition commutative; not>> commutativity is an essential feature of multiplication for English>> speaker; not so for French speakers.I don't think that follows at all.> In so far as there is any reason for the difference in usage,> I suspect that it reflects a difference in the feel> or philosophy of the two languages.A person writing in English would feel uneasy> at repeating commutative field 50 or 100 times --> the natural tendency would be to find a shorthand form.The French, on the other hand, are perfectly happy as far as I can see> to repeat a compound phrase indefinitely.> (Do the Germans actually prefer it?)Depending on which is used more often, one may useSchiefkoerper and Koerper (cr skew-field)orKoerper and kommutativer KoerperAs you indicated above, it is mostly a matter of convenience.Marc === Subject: : Re: French vs. English> This is another interesting linguistic contrast between French and>> English mathematics. The other one I am thinking of is field>> (literally, champs) vs. corps (literally, body): In a field>> (corps commutatif), multiplication is by definition commutative; not>> commutativity is an essential feature of multiplication for English>> speaker; not so for French speakers. You can infer that, but most English writers don't even use the word commutative in the context of multiplication. If you don't use the phrase symmetric field, they take it for granted that you're a Russian spy, and shoot you. I don't think that follows at all.> In so far as there is any reason for the difference in usage,> I suspect that it reflects a difference in the feel> or philosophy of the two languages.A person writing in English would feel uneasy> at repeating commutative field 50 or 100 times --> the natural tendency would be to find a shorthand form.The French, on the other hand, are perfectly happy as far as I can see> to repeat a compound phrase indefinitely.> (Do the Germans actually prefer it?)Depending on which is used more often, one may useSchiefkoerper and Koerper (cr skew-field)orKoerper and kommutativer KoerperAs you indicated above, it is mostly a matter of convenience.Marc electron-dot-cloud are galaxies === Subject: : Re: new way of describing ellipse for math> Consider rotating the circle about a diameter, (which becomes the> major axis of your ellipse), while a perpendicular diameter becomes> the minor axis as the circle is projected on a stationary,> (non-rotating), plane paralel to the major axis.Still cannot picture your above. Regardless it is too complicated. I need animmediate and simple method.> The corners will not give you a unique elipse, but if you connect the> midpoints of the oposite sides and use them as major and minor axes,> you've got one.Sounds like the ellipse is then inscribed inside the rectangle. I want therectangle inscribed inside the circle and ellipse.I want to get away from axes. I want to connect a unique circle to a uniqueellipse using a unique rectangle. Every circle has a unique square inscribed.By throwing in a circle into the construction, that circle ought to allow me toget away without needing any axes or foci.If the expression that a ellipse is a squashed circle has any truth to it thensome such construction ought to exist.Archimedes Plutonium, a_plutonium@hotmail.comwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies === Subject: : Re: new way of describing ellipse for mathSounds like the ellipse is then inscribed inside the rectangle. I want the> rectangle inscribed inside the circle and ellipse.I want to get away from axes. I want to connect a unique circle to a unique> ellipse using a unique rectangle. Every circle has a unique square inscribed.By throwing in a circle into the construction, that circle ought to allow me to> get away without needing any axes or foci.If the expression that a ellipse is a squashed circle has any truth to it then> some such construction ought to exist.Archimedes Plutonium, a_plutonium@hotmail.com> whole entire Universe is just one big atom where dots> of the electron-dot-cloud are galaxies If you are looking for a unique elipse defined only by an inscribedrectangle, you are out of luck. Four points arranged as the corners ofa rectangle define a whole family of elipses including one circle. An elipse is not a 'squashed` circle, it is a rotated one. (Conic Sections - remember?) Whether you like it or not, I believe that the relationship you arelooking fordoes involve the rotation posted earlier. Consider the circle with its inscribed square. (You didn't say youhad trouble with that.) If you rotate this figure about one side ofthe square, and project it, you get a rectangle, (the square with oneset of opposite sides forshortened), and a unique circumscribedelipse, (the circle rotated).Hint: if you wanted to graph this elipse, you could multiply one setof ordinates from the graph of the circle with its center at theprojected intersection of the diagonals of the square by the cosine ofthe angle of rotation. This is High School stuff. If you have trouble with these concepts,you are wasting your time in attempting to understand, much lessmodify, string theory. === Subject: : Re: Factorial/Exponential Identity, InfinityFish, or cut bait.Is the dis-parity discrepancy contradictory? Why or why not?Ross === Subject: : Re: Math dependency logic REVISEDContent-transfer-encoding: 8bitA sure sign of the decline of sci.math/sci.logic presided> over by David Ullrich and the Boyz is that these NGs attract> loathsome mutants like you. > Is it lighten up time yet?One of the invariants of Usenet (like, for the past twenty years or so)is that, except on the relatively few groups that are moderated,*nobody* is in charge. Anybody who wants to can pipe up and sayanything.> Here one may shill for whatsoever and whosoever one pleases--> including that puke, Lyndon LaRouche--as long as one> fires off a few at JSH.> I don't read everything on sci.math; but my choices are informed inpart by morbid curiosity, so I am surprised to have missed the plugsfor Lyndon LaRouche. > Welcome to sci.math/sci.logic, where loathsome mutants gyre> and gimble in good company.Cool. Way cool. But how do you tell them apart? (Assuming you want to.)-- Chris HenrichMoral indignation is jealousy with a halo. -- H. G. Wells === Subject: : Re: Math dependency logic REVISED Adjunct Assistant Professor at the University of Montana. [.newsgroups trimmed.]>>[cut]> Pray give a *proof* that your> ring (which includes all possible values of a_1(x)/7 and a_2(x)/7 for> integer x) does not contain 1/7.>>I realize that you are trying to get Harris to see his error by>>asking him to prove that 1/7 is not in his ring. You are using>>this approach since Harris rejects concrete counterexamples,>>says valid definitions are flawed, and refuses to go into more>>detail when questioned on how one statement follows from another>>logically (instead, he gives metaphysical reasons why the>>questioned statement *must* be true).>>However, I feel that your approach gives a misleading idea of>>what constitutes a proof. Your approach seems to say that>>even after a proof has been presented, other things need also>>be considered and proved. My objection is that what if I failed>>to notice that I must give a proof that 1/7 is not in the ring?>>It doesn't appear to be needed in the proof given by Harris.>> You have a point, of sorts. The problem here is that James refuses to>> tell us what IS in his ring; he only tells us what is NOT in his ring.>Mathematics is about tracing out an argument from a truth using>logical steps to get to a conclusion which you then know MUST BE TRUE.>If you start with a truth, and use only logical steps to get to a>conclusion, it MUST BE TRUE.>I do that, which is how my work should be faced.Keep telling yourself that; someday maybe you'll believe it.>It amazes me how many of you seem to doubt that fact!!!I don't doubt it. I know for a fact it is not true that your workproceeds along only logical steps. When you finish with the case ofx=0, you perform a logical fallacy called a non-sequitur, when youclaim that the only way a factorization can occur is the way youclaim it occurs, and when you claim that something should be analgebraic integer. Neither of those two steps are justified by theprevious (mostly correct if also mostly irrelevant) work.>Instead posters try to find a backdoor.> His definition is that the object ring is the ring of numbers such>> that the only integers which are units are 1 and -1. He claims that it>> is more inclusive than the algebraic integers, but refuses to present>> a single element of the object ring which is not an algebraic integer,>> and refuses to explain why the integers themselves, which satisfy his>> definition, are not the object ring.>> His ->only<- objection to Dik's use of James's argument in a different>> polynomial is that the argument clearly does not apply, since in Dik's>> case it is easily provable that the conclusion would yield that 1/7 is>> in the ring where all these factorizations take place.>The argument I use relies on numbers like 7 being NUMBERS and not>variable, as well as on the distributive property: a(b+c) = ab + ac.Nonsense. That is only what you tell yourself because you cannotunderstand the objections raised to your work.You are assigning quasi-magical properties to the value of a functionat zero. You are so confused that you attack your own method when Iuse it.>Now I've pointed out before that what I have is short and basic, but>posters continually try to make endruns and obfuscate the facts.No, what you have is short and wrong. We've told you EXACTLY where youare wrong, but you continue to lie and claim nobody has every pointedout a line of your argument which is wrong.>> It is not unreasonable to request that James explain to us why this is>> also not the case in his argument, since he is introducing inverses of>> algebraic integer factors of 7 all over the place.>And I have explained it. To nobody's satisfaction but your own. That is not an explanation,that is a reply. And while every explanation and every answer is areply, not every reply is an answer and not every reply is anexplanation.You are very good at replying. But you have not provided answers orexplanations. At least, not answers or explanations that anybody findssatisfying, except for you.>I noticed something simple, which challenges the assumptions of>mathematicians, and rather than do what they're supposed to do: focus>on the argument itself; they're trying to cheat.>It's intellectual dishonesty. You *focus on the argument presented*.I have noticed that I've explained several times, by giving EXPLICITformulas, why your argument about what the constant terms imply forarbitrary values is incorrect, and why your claim that a factorizationmust be in some specific way is wrong.Rather than address those points, you either remove them and repeatyour argument, or you attack me personally.You do not focus on the argument presented, which is IN DIRECT REPLY,AND USING THE METODS OF, the argument ->you<- present.I also notice that you do not consider this to be intellectualdishonesty on YOUR part.Why is that? Is it just because, as always, you are being a hypocrite? power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: : Re: Math dependency logic REVISED Adjunct Assistant Professor at the University of Montana. [.newsgroups trimmed.]>In sci.physics, ><: [.snip.]>> His definition is that the object ring is the ring of numbers such>> that the only integers which are units are 1 and -1. He claims that it>> is more inclusive than the algebraic integers, but refuses to present>> a single element of the object ring which is not an algebraic integer,>> and refuses to explain why the integers themselves, which satisfy his>> definition, are not the object ring.>Uh....very dumb question, but, if a ring A contains a subring B,>aren't all the units of subring B also units of A?Yes; but the converse does not hold. That is, it is possible forsomething to be (a) an element of B; and (b) a unit in A' and (c) nota unit in B.>Subring B in this case is the ring of algebraic integers, which>contains many units: n - sqrt(n-1) and n + sqrt(n-1) being>some. In fact, any equation>x^n + a_{n-1} * x^{n-1} + ... + a_1 * x 1 = 0>(where n > 0 and the a_i are integers) will have unit roots,>as one can easily prove.>Therefore, an encompassing object ring of the algebraic>integers cannot have 1 and -1 as its only units, James specifies that the only INTEGERS which are units are 1 and -1;he does not say anything about any other elements which may beunits are 1 and -1 [emphasis added]. [.rest deleted.] power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: : Re: Math dependency logic REVISEDIn sci.math, <>:> [.newsgroups trimmed.]>In sci.physics, >><>: [.snip.]His definition is that the object ring is the ring of numbers such> that the only integers which are units are 1 and -1. He claims that it> is more inclusive than the algebraic integers, but refuses to present> a single element of the object ring which is not an algebraic integer,> and refuses to explain why the integers themselves, which satisfy his> definition, are not the object ring.>>Uh....very dumb question, but, if a ring A contains a subring B,>>aren't all the units of subring B also units of A?Yes; but the converse does not hold. That is, it is possible for> something to be (a) an element of B; and (b) a unit in A' and (c) not> a unit in B.Correct, of course. :-) For example, the ring of integers isa subring of the algebraic integers, but there are an infinitenumber of units in the algebraic integers, whereas the integershave only two.>Subring B in this case is the ring of algebraic integers, which>>contains many units: n - sqrt(n-1) and n + sqrt(n-1) being>>some. In fact, any equation>>x^n + a_{n-1} * x^{n-1} + ... + a_1 * x 1 = 0>>(where n > 0 and the a_i are integers) will have unit roots,>>as one can easily prove.>>Therefore, an encompassing object ring of the algebraic>>integers cannot have 1 and -1 as its only units, James specifies that the only INTEGERS which are units are 1 and -1;And he is quite correct in this case. But I think he'svery confused regarding factorization of algebraic integers.For example, 7^(2/3) is an algebraic integer. (It's nota unit, either.) x^3 - 49 can be rewritten (correctly)asx^3 - 49 = (x - 7^(2/3) * v1) * (x - 7^(2/3) * v2) * (x - 7^(2/3) * v3)where v1, v2, and v3 are units. v1 = 1,v2 = cos 2pi/3 + i sin 2pi/3, v3 = cos 4pi/3 + i sin 4pi/3.However, one cannot concludex^3 - 49 = (x - 7 * w1) * (x - 7 * w2) * (x - w3)where w1, w2, and w3 are algebraic integers. (Algebraic *numbers*,yes.)This is of course only a strawman, intended to be vaguely similarto JSH's proof, but it turns out his suffers from a similarproblem, at least in the examples I've seen thus far.> he does not say anything about any other elements which may be> units are 1 and -1 [emphasis added].I'll admit I am now curious as to what subrings of the algebraicintegers are such that the only units of the subring are 1 and -1.Or is James' object ring a superset?My brain hurts. :-)[rest deleted]-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: : Re: Math dependency logic REVISED> < Uh....very dumb question, but, if a ring B contains a subring A,> aren't all the units of subring A also units of B?>> Yes; but the converse does not hold. That is, it is possible>> for something to be a nonunit of A but a unit in B> Correct, of course. :-) For example, the ring of integers is> a subring of the algebraic integers, but there are an infinite> number of units in the algebraic integers, whereas the integers> have only two.That doesn't exemplify o's remark, which is that a nonunitin a ring A need not stay a nonunit in a ring B containing A.However, this is true in case A < B is an _integral_ ring extension;then any (non)unit in A remains the same in B. Integral extensionscannot alter whether an element is a unit (or not). This is afundamental property of integral extensions. Indeed, when thisproperty is generalized to the LYING-OVER (LO) or SURVIVAL propertyand further combined with the INCOMPARABILITY (INC) property, oneobtains a characterization of integral ring extensions, namelyTHEOREM For commutative rings R < T the following are equivalent(1) R < T is an integral extension(2) A < B has INC and LO for all A,B with R < A < B < T(3) A < A[u] has INC and LO for all A,B with R < A < T, all u in TRecall that one defines a ring extension R < T to beLYING-OVER (LO) if for any prime ideal P in R there existsa prime ideal Q in T lying over P, i.e. Q / R = P.INCOMPARABLE (INC) if two different primes P, Q in T with thesame contraction in R are incomparable: neither P < Q nor Q < PSURVIVAL if each proper (prime) ideal P of R survives in T,i.e. P doesn't explode to 1 in T: PT != TIn particular: r is a nonunit of R <=> (r) < 1, so if R < Tis survival then rT < 1 so r remains a nonunit in T.For details see my prior post [1], to which I've just posted a reply [2] containing much further detail, including (online) references.-Bill Dubuque[1] http://google.com/groups?selm=y8zr83t2fom.fsf% 40nestle.ai.mit.edu[2] http://google.com/groups?selm=y8zu15lm5am.fsf% 40nestle.ai.mit.edu === Subject: : Re: Math dependency logic REVISED Adjunct Assistant Professor at the University of Montana.>In sci.math, ><:>> [.newsgroups trimmed.]>In sci.physics, ><:>> [.snip.]>> His definition is that the object ring is the ring of numbers such>> that the only integers which are units are 1 and -1. He claims that it>> is more inclusive than the algebraic integers, but refuses to present>> a single element of the object ring which is not an algebraic integer,>> and refuses to explain why the integers themselves, which satisfy his>> definition, are not the object ring.>Uh....very dumb question, but, if a ring A contains a subring B,>aren't all the units of subring B also units of A?>> Yes; but the converse does not hold. That is, it is possible for>> something to be (a) an element of B; and (b) a unit in A' and (c) not>> a unit in B.>Correct, of course. :-) For example, the ring of integers is>a subring of the algebraic integers, but there are an infinite>number of units in the algebraic integers, whereas the integers>have only two.Sorry, but that is a BAD example of the situation I mentioned. In thissituation, B=integers; A=algebraic integers, but there is NO element xof A such that: (1) x is in B (i.e., is an integer) (2) x is a unit in A; and (3) x is not a unit in B.That is, the situation I mentionted does NOT occur in your example.What you gave was an example of a subring B of A, and element x of A,such that: (1) x is a unit in A; and (2) x is not in B (hence not a unit in B).A better example would be setting B=the integers, A = Z[1/2], the ringof all rationals that can be written as p/q with q a power of 2. Then,settting x=2, gives you an element of B which is a unit in A but notin B. power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: : Re: Math dependency logic REVISED Adjunct Assistant Professor at the University of Montana.>my question, or thing that always bugs me, is why i (and -i) are not>included as units, Not in cluded as units of ->what<-?James is specifying that no INTEGERS other than 1 and -1 should beunits; he does not say anything about what other sorts of units may beincluded. power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: : Re: Math dependency logic REVISEDIn sci.math, <>:>>[cut]> Pray give a *proof* that your> ring (which includes all possible values of a_1(x)/7 and a_2(x)/7 for> integer x) does not contain 1/7.>>I realize that you are trying to get Harris to see his error by>>asking him to prove that 1/7 is not in his ring. You are using>>this approach since Harris rejects concrete counterexamples,>>says valid definitions are flawed, and refuses to go into more>>detail when questioned on how one statement follows from another>>logically (instead, he gives metaphysical reasons why the>>questioned statement *must* be true).>>However, I feel that your approach gives a misleading idea of>>what constitutes a proof. Your approach seems to say that>>even after a proof has been presented, other things need also>>be considered and proved. My objection is that what if I failed>>to notice that I must give a proof that 1/7 is not in the ring?>>It doesn't appear to be needed in the proof given by Harris.There is some truth to what you say. On the other hand, James claims that a_1(x) and a_2(x) are divisible> by 7 in the object ring. The only property he has given on the> object ring is that no integers other than 1 and -1 are units. If his> claim that a_1(x)/7 and a_2(x)/7 are in the object ring is true,> then it ... must be the case that a_1(x) and a_2(x) are multiples of 7?Or what?:-)[rest snipped]-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: : Graph terminologyGiven a directed graph G = (V, U), where V is the set of vertices, andU is the set of arcs, I define H = (U, W), where the set of arcs U ofG becomes the set of vertices of H, and W is the set of arcs of H suchthat u = (u1, u2) is in H if and only if u1 = (v1, v2) and u2 = (v2,v3).I would like to know if there is a standard name for graph H. And isthere some symbolism (like H = G' for instance) that comes with that?Christophe === Subject: : Re: Region of Convergence of 1/zeta(s)>>there are half-planes of convergence and of absolute convergence,>>but they need not be the same; e.g. sum (-1)^n/n^s has convergence>>iff Re(s) > 0 and abolute convergence iff Re(s) > 1.Is it possible for the limit function to be analytic in a larger> right-half-plane than the half-plane of conditional convergence?Yes. This function is entire.A few follow-up questions;If a Dirichlet series does not contain any poles, does it convergeeverywhere?Related question; for all Dirichlet series that do have poles, does analyticcontinuation ALWAYS imply that one or more poles are cancelled ? Forexample, my understanding of analytic continuation of the series sumn^(-s) is that we multiply by (1 - 2^(1-s)), which has a zero thatcancels the pole at s = 1. This extends the ROC to re(s) > 0. We canthen divide by the same factor (1 - 2^(1-s)) to get back the samefunction we started with.I am just a lowly electrical engineer dabbling in the world of math,so if I have made a terminology error, please don't crucify me!Bob AdamsBob Adams === Subject: : Re: Region of Convergence of 1/zeta(s)> If a Dirichlet series does not contain any poles, does it converge> everywhere?I think you're asking the following: If the Dirichlet seriessum(a_n/n^s) converges for sufficiently large real part to a functionF, and F can be continued to an entire function, must the originalseries converge everywhere?An example (a_n = (-1)^n) has already been posted showing that theanswer is no. You may be interested to know that *if* all the a_nare nonnegative, then the answer is yes. This follows from a theoremof Landau which says that a Dirichlet series with nonnegativecoefficients cannot be continued to a region containing its realabscissa of convergence.For applications of this corollary, see Chapter VI of Newman'sAnalytic Number Theory, where it is employed in natural proofs ofthe nonvanishing of L-series on the line Re(s) = 1.Take care,Paul === Subject: : Re: Region of Convergence of 1/zeta(s)> If a Dirichlet series does not contain any poles, does it converge> everywhere?Hello Liz.I'm not sure what you mean by does not contain any poles. AsI pointed out sum_{n=1}^infinity (-1)^n/n^s is a Dirichlet seriesconverging iff Re(s) > 0. But its limit is the restrictionof an entire function (no poles anywhere is C). Doesthis Dirichlet series not contain any poles?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: : Re: How to find probability mass function of this?>>Suppose that we have two (fair) dice. Let>>X = the highest [sic - SJH] number on any of the dice,>>Y = the sum of both dice.>>What is the joint mass function here? >>There are 6*6 ways of throwing two dice. X can be any integer from 1>to 6. Given X, Y can be any integer from X+1 to 2X, but 2X only occurs>if both are the highest, while in the other cases either dice can be>the highest . So the formula becomes:>Prob(X=x,Y=y)> =2/36 if 1<=x<=6, x+1<=y<=2x-1> =1/36 if 1<=x<=6, y=2x > =0/36 otherwise.Next exercise:Let U and V be independent random variables each with Uniform(0,1) distribution. Determine the joint density of X = maj(U,V) and Y= U+V.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: : Re: How to find probability mass function of this?> So the formula becomes:> Prob(X=x,Y=y)> =2/36 if 1<=x<=6, x+1<=y<=2x-1> =1/36 if 1<=x<=6, y=2x > =0/36 otherwise. Next exercise:> Let U and V be independent random variables each with Uniform(0,1) > distribution. Determine the joint density of X = maj(U,V) and Y= U+V.Assuming that is a serious questionThe natural way of extending the previous thinking would be (for someconstant K)p(x,y)=2K if 01, or y2x and if the boundary needs a density then take account of the precisedefinition of the density of Uniform(0,1) at 0 and 1. === Subject: : Re: How to find probability mass function of this?>>So the formula becomes:>Prob(X=x,Y=y)> =2/36 if 1<=x<=6, x+1<=y<=2x-1> =1/36 if 1<=x<=6, y=2x > =0/36 otherwise.>>Next exercise:>>Let U and V be independent random variables each with Uniform(0,1) >>distribution. Determine the joint density of X = maj(U,V) and Y= U+V.>>Assuming that is a serious questionIt was meant as an exercise for Konrad (and other interested beginning students of probability).>The natural way of extending the previous thinking would be (for some>constant K)>p(x,y)>=2K if 0=0 outside the boundary[...]Not at all rigorous.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: : Re: How to find probability mass function of this? charset=iso-8859-1> Determine the joint density of X = maj(U,V) and Y= U+V.What is maj?-- KindlyKonrad-------------------------------------------------- -May all spammers die an agonizing death; have no burial places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be lazy--------------------------------------------------- === Subject: : Re: How to find probability mass function of this?>>Determine the joint density of X = maj(U,V) and Y= U+V.>>What is maj?the larger-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: : JSH: Factorization P(x) = 2(x(x+1)/2 + 1)Some people have gotten excited by their ability to find *different*factorizations than the ones I've been giving, where they end up withunits other than 1 or -1, and maybe an example will help some of you,so here's this post.Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1)which I pick because it's valid in the ring of integers, but NOTalways in the ring of algebraic integers.That, of course, is because every integer is either even or 1 awayfrom being even, so it works.Now then, what if someone wanted to question that fact, so they cameback at me with a *different* factorization that did somethingdifferent.Why would that be relevant?It wouldn't.You MUST FOLLOW A PROOF, and not second guess the process.Now I've given a proof, and mathematicians need to behave like peoplewho believe in what they're doing.And make no mistake, if you choose NOT to behave like mathematicians,then it seems to me that there's no reason for you to remainmathematicians, even if it is only in name.After all, fair is fair, and it's inhumane what you people are doingto me, basically punishing me for making discoveries. I haverights!!!James Harris === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1)In sci.math, James Harris<3c65f87.0311021757.2ecadddb@ posting.google.com>:> Some people have gotten excited by their ability to find *different*> factorizations than the ones I've been giving, where they end up with> units other than 1 or -1, and maybe an example will help some of you,> so here's this post.Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1)That's not a factorization, just a rewrite (although one couldquibble over the '2'). In this caseP(x) = (x - (-1/2 + i*(sqrt(7))/2)) * (x - (-1/2 - i*(sqrt(7))/2))is a factorization.which I pick because it's valid in the ring of integers, but NOT> always in the ring of algebraic integers.Erm, it's an abstract formalism/equation/rewrite; howcan it not be valid? Unless 2 = 0, which is only thecase in the field of integers mod 2, which isn't all thatinteresting. :-)That, of course, is because every integer is either even or 1 away> from being even, so it works.This is true.Now then, what if someone wanted to question that fact, so they came> back at me with a *different* factorization that did something> different.Why would that be relevant?It wouldn't.You MUST FOLLOW A PROOF, and not second guess the process.I'm assuming there's a proof to follow somewhere on theWeb; do you have a Web site? I for one want to see thewhole proof, ideally with the corrections others havementioned in this newsgroup.Now I've given a proof, and mathematicians need to behave like people> who believe in what they're doing.You've given a sequence of what you purport are logical deductions.Some of them are logical enough. However, others are jumps overcrevasses which I for one can't follow.I can only illustrate with one of my own strawman examples, however:x^3 - 49 is factorizable into (x - r1)(x - r2)(x - r3), butnone of the roots is divisible by 7 -- x divisible by 7 over thealgebraic integers, means that one can find an algebraicinteger y such that x = 7 * y. Since all the r's are productsof (-sqrt(3)/2 + i/2)^n * 7^(2/3) (n=0,1,2), it's clear thatthere is no such algebraic integer in this strawman case.Does your proof suffer from a similar flaw? I hope not.And make no mistake, if you choose NOT to behave like mathematicians,> then it seems to me that there's no reason for you to remain> mathematicians, even if it is only in name.After all, fair is fair, and it's inhumane what you people are doing> to me, basically punishing me for making discoveries. I have> rights!!!Personally, I prefer to critique your proof, not you.I can't say anything at all whether you smell good or bad,run around in your underwear (or not in your underwear),do weird things with a shishkebab, a chainsaw, and aSuperball, attend church, give to charity, drive around thecity with a large sign saying I [heart] MATH, etc. :-)Who really cares? We judge according to what one posts.It's all we have. :-)> James Harris-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1)> In sci.math, James Harris> :> Some people have gotten excited by their ability to find *different*> factorizations than the ones I've been giving, where they end up with> units other than 1 or -1, and maybe an example will help some of you,> so here's this post.Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1)That's not a factorization, just a rewrite (although one couldYou're showing troubling ignorance here considering how often you'veposted in my threads as if you know basic mathematics.In my example P(x) is factored into 2 and x(x+1)/2 + 1, and *both* areindeed factors.Possibly the symbols are confusing you, so here are some numbers:Let x=1, then P(1) = 4, and the factors are 2 and 2.It IS a factorization poster. You're just lost on basics.> quibble over the '2'). In this caseP(x) = (x - (-1/2 + i*(sqrt(7))/2)) * (x - (-1/2 - i*(sqrt(7))/2))is a factorization.Yes, it is a factorization, but it's not the only factorizationposter.Ok, let's get one thing straight: the discussions here are over*advanced* topics, so if you don't even know what a factorization is,you might want to sit back and just read without posting.And yes, when I see such posts I downgrade a poster.If I downgrade a poster far enough, I just don't worry about readingtheir posts, unless I start seeing evidence that the newsgroup as awhole is convinced by them in their confusion.James Harris === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Adjunct Assistant Professor at the University of Montana.>Some people have gotten excited by their ability to find *different*>factorizations than the ones I've been giving, where they end up with>units other than 1 or -1, and maybe an example will help some of you,>so here's this post.>Consider P(x) = x^2 + x + 2, which factors nicely as >P(x) = 2(x(x+1)/2 + 1)That's not a factorization; that's a rewriting.>which I pick because it's valid in the ring of integers, but NOT>always in the ring of algebraic integers.Nope. For every algebraic integer value of x, P(x) is also analgebraic integer. I think what you mean is that not every partial calculation is analgebraic integer; that is, x(x+1)/2 is not an algebraic integer forevery algebraic integer value of x; even though x(x+1)/2 is always aninteger for every integer value of x. power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Adjunct Assistant Professor at the University of Montana.>>Some people have gotten excited by their ability to find *different*>>factorizations than the ones I've been giving, where they end up with>>units other than 1 or -1, and maybe an example will help some of you,>>so here's this post.>>Consider P(x) = x^2 + x + 2, which factors nicely as >>P(x) = 2(x(x+1)/2 + 1)>That's not a factorization; that's a rewriting.Oh, I get it. You are factoring it as 2 times ((x)(x+1)/2) + 1.>>which I pick because it's valid in the ring of integers, but NOT>>always in the ring of algebraic integers.To be honest, I fail to see your point. Nobody is saying that youcannot factor your polynomial asP(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7).What people are saying is that this factorization does not yieldalgebraic integer factors whenever P(x) is irreducible; you claim itSHOULD. Are you claiming that your factorization above, which is validin algebraic integers when x=0, should also be valid for arbitraryalgebraic integer values of x? And if so, why do you think so? power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1)>[...]>After all, fair is fair, and it's inhumane what you people are doing>to me, basically punishing me for making discoveries. Huh??? The only punishment anyone's been giving you is tosay you're wrong and explain why. A minute ago you asked meWhat the does agreement matter? This is a little puzzling,since you regard disagreement as punishment.>I have>rights!!!Yes. You have the right to say what you want. And we all havethe right to say what we want about what you say.We'd even have the _right_ to say you were wrong if you wereactually _right_! But you're not, and you _don't_ have the rightto agreement from people about things you're wrong about,sorry.>James Harris************************ === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > Some people have gotten excited by their ability to find *different* > factorizations than the ones I've been giving, where they end up with > units other than 1 or -1, and maybe an example will help some of you, > so here's this post. > Consider P(x) = x^2 + x + 2, which factors nicely as > P(x) = 2(x(x+1)/2 + 1) > which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers.Can you give an example where x^2 + x + 2 is not divisible by 2 whenx is an algebraic integer?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1)> Some people have gotten excited by their ability to find *different*> factorizations than the ones I've been giving, where they end up with> units other than 1 or -1, and maybe an example will help some of you,> so here's this post.> Consider P(x) = x^2 + x + 2, which factors nicely as > P(x) = 2(x(x+1)/2 + 1)> which I pick because it's valid in the ring of integers, but NOT> always in the ring of algebraic integers.Can you give an example where x^2 + x + 2 is not divisible by 2 when> x is an algebraic integer?It's the FACTORIZATION Dik Winter, and if you can't grasp that point,you'll never get it.The factorization is valid in integers for any integer x, but not inthe ring of algebraic integers for any algebraic integer x.My point is that you have to focus on the *factorization* and itsvalidity in particular rings.Some factorizations will be valid in one ring, but not another.James Harris === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1)... > Consider P(x) = x^2 + x + 2, which factors nicely as > P(x) = 2(x(x+1)/2 + 1) > which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers. > Can you give an example where x^2 + x + 2 is not divisible by 2 when > x is an algebraic integer? > It's the FACTORIZATION Dik Winter, and if you can't grasp that point, > you'll never get it. > The factorization is valid in integers for any integer x, but not in > the ring of algebraic integers for any algebraic integer x.Oh, there are enough algebraic integers where it is valid in the ringof algebraic integers. For instance: x = sqrt(2). o gave anexample for which it is indeed not valid. > My point is that you have to focus on the *factorization* and its > validity in particular rings. > Some factorizations will be valid in one ring, but not another.Yes, and your factorisation P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)is in general not valid in the ring of algebraic integers. So whatare you trying to show?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1)> ...> Consider P(x) = x^2 + x + 2, which factors nicely as > P(x) = 2(x(x+1)/2 + 1)> which I pick because it's valid in the ring of integers, but NOT> always in the ring of algebraic integers.> Can you give an example where x^2 + x + 2 is not divisible by 2 when> x is an algebraic integer?> It's the FACTORIZATION Dik Winter, and if you can't grasp that point, you'll never get it.> The factorization is valid in integers for any integer x, but not in> the ring of algebraic integers for any algebraic integer x.Oh, there are enough algebraic integers where it is valid in the ring> of algebraic integers. For instance: x = sqrt(2). o gave an> example for which it is indeed not valid. > My point is that you have to focus on the *factorization* and its> validity in particular rings.> Some factorizations will be valid in one ring, but not another.Yes, and your factorisation> P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)> is in general not valid in the ring of algebraic integers. So what> are you trying to show?That's the point. I *prove* that if you have coprimeness between 7and 22 in the ring in which the factorization is valid, where 7 is NOTa unit (and neither is 22), then the constant terms of the factorsthat result from dividing P(x) by 49 *MUST* be coprime to 7.That result is then ring independent to the extent that it's notparticular to a particular ring, but to a particular property *of* thering.However, the ring of algebraic integers, which has the desiredcoprimeness property, does NOT always have a_1(x)/7 and a_2(x)/7 asmembers!It's a great result, which follows from some fascinatingly basicalgebra.Get it yet Dik Winter?James Harris === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Adjunct Assistant Professor at the University of Montana.>> ...>> Consider P(x) = x^2 + x + 2, which factors nicely as >> P(x) = 2(x(x+1)/2 + 1)>> which I pick because it's valid in the ring of integers, but NOT>> always in the ring of algebraic integers.>> Can you give an example where x^2 + x + 2 is not divisible by 2 when>> x is an algebraic integer?>> It's the FACTORIZATION Dik Winter, and if you can't grasp that point,>> you'll never get it.>> The factorization is valid in integers for any integer x, but not in>> the ring of algebraic integers for any algebraic integer x.>> Oh, there are enough algebraic integers where it is valid in the ring>> of algebraic integers. For instance: x = sqrt(2). o gave an>> example for which it is indeed not valid.>> My point is that you have to focus on the *factorization* and its>> validity in particular rings.> Some factorizations will be valid in one ring, but not another.>> Yes, and your factorisation>> P(x)/49 = (5 a1/7 + 1)(5 a2/7 + 1)(5 b3 + 22)>> is in general not valid in the ring of algebraic integers. So what>> are you trying to show?>That's the point. I *prove* that if you have coprimeness between 7>and 22 in the ring in which the factorization is valid, where 7 is NOT>a unit (and neither is 22), then the constant terms of the factors>that result from dividing P(x) by 49 *MUST* be coprime to 7.What you FAIL to realize, though, is that this DOES NOT MEAN that thefactorization isP(x)/49 = (5a1/7 + 1)(5a2/7 + 1)(5b3 + 22).What you get is that 49 = w_1(x)*w_2(x)*w_3(x), with w_i varying withthe values of x, andP(x)/49 = (( 5a1(x)+7)/w_1(x) - 1) + 1)* (((5a_2(x)+7)/w_2(x) - 1) + 1) * ((( 5b_3(x)+7)/w_3(x) - 22) + 22).That's IT. You are ASSUMING your conclusion when you claim thatw_1(x)=7 for all x.Remember: Given ANY function f(x) and ANY constant r, f(x) can bewritten as f(x)=g(x) + r, where r is constant, and g(x) is afunction. Given ANY function f(x), and any two values r and s, we canwrite f(x) = g(x) + c, where c is a constant, and g(r)=s: just letg(x) = f(x) - f(r) + sand let c = f(r)-s.Theng(x) + c = f(x)-f(r) + s + f(r) - s = f(x)and g(r) = f(r) - f(r) + s = s.So you can take the function 5a_1(x) + 7, divide it by ANY FUNCTIONr(x) WHICH SATISFIES r(0)=7, and then you can write(5a_1(x)+7)/r(x) = h_1(x) + 1where h_(1) = 0.Likewise, you can divide 5a_2(x)+7 by ANY FUNCTION s(x) whichsatisfies s(0)=7, and then you can write it as(5a_2(x)+7)/s(x) = h_2(x) + 1And you can divide 5b_3(x) + 22 by ANY FUNCTION t(x) that satisfiest(0)=1, and then you can write the result as(5b_3(x)+22)/t(x) = h_3(x) + 22.So, if you make ANY choice for which r(x)*s(x)*t(x) = 49, thefactorization will work; and if you make the choice in such a way thatr(x)*s(x)*t(x) = 49, r(x) divides a_1(x) and 7, s(x) divides a_2(x)and 7, and t(x) divides a_3(x) and 7 (all in the ring of algebraicintegers), then the factorization will work, and it will NOTNECESSARILY BE OF THE FORM YOU CLAIM IT ->MUST<- BE.You cannot conclude that, because you can ALWAYS write the function ash_1(x) + 1 with h_1(x) = 0, regardless of the value ofw_1(x). Likewise for w_2 and for w_3. And if you choose themCORRECTLY, then you get that all of them have values in the algebraicintegers.Get it, James S. Harris? power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) Adjunct Assistant Professor at the University of Montana.> Some people have gotten excited by their ability to find *different*> factorizations than the ones I've been giving, where they end up with> units other than 1 or -1, and maybe an example will help some of you,> so here's this post.Consider P(x) = x^2 + x + 2, which factors nicely as P(x) = 2(x(x+1)/2 + 1)which I pick because it's valid in the ring of integers, but NOT> always in the ring of algebraic integers.>Can you give an example where x^2 + x + 2 is not divisible by 2 when>x is an algebraic integer?Ah, I think I see now what James is trying to claim; that we can writeit as x(x+1)/2 + 1 times something...If x=i, then i^2+i+2 = -1+i+2 = 1+i; 1+i is a divisor of 2 and not aunit (in Z[i], its norm is 2, and if r is an algebraic integer unit,then it is a unit in the ring of integers of Q(r)); and it is a properdivisor of 2, since (1-i)(1+i) = 2 and 1-i is also not an algebraicinteger unit.If it were a multiple of 2, then 1+i would be an associate of 2, whichwould make 1-i into a unit, which is impossible. power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1)> Some people have gotten excited by their ability to find *different*> factorizations than the ones I've been giving, where they end up with> units other than 1 or -1, and maybe an example will help some of you,> so here's this post.> Consider P(x) = x^2 + x + 2, which factors nicely as > P(x) = 2(x(x+1)/2 + 1)> which I pick because it's valid in the ring of integers, but NOT> always in the ring of algebraic integers.Can you give an example where x^2 + x + 2 is not divisible by 2 when> x is an algebraic integer?I think James' point is that since x(x+1)/2 is not an algebraic integer forall algebraic integers x, writing the polynomial in that form implicitlyinvolves 'going into a larger ring'. On the other hand, if x is an integer,x(x+1) is always even, so x(x+1)/2 makes sense within the integers.But then, as so often before, this is a factorization of *numbers* writtenas if it were one of *polynomials*.-- Dave TaylorWhen I want your opinion, I'll ... I'll never want your opinion[BtVS] === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1)> Some people have gotten excited by their ability to find *different*> factorizations than the ones I've been giving, where they end up with> units other than 1 or -1, and maybe an example will help some of you,> so here's this post.> Consider P(x) = x^2 + x + 2, which factors nicely as> P(x) = 2(x(x+1)/2 + 1)> which I pick because it's valid in the ring of integers, but NOT> always in the ring of algebraic integers.> Can you give an example where x^2 + x + 2 is not divisible by 2 when> x is an algebraic integer?That's impossible because if you do a little mathematical induction, youfind that P(1)=4 and then evaluating for k+1, you get(k+1)^2+k+1+2k^2+2k+1+k+1+2k^2+3k+4So P(x) is even for all x.David Moran> -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,+31205924131> home: bovenover 215, 1025 jn amsterdam, nederland;http://www.cwi.nl/~dik/ === Subject: : Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1) > Some people have gotten excited by their ability to find *different* > factorizations than the ones I've been giving, where they end up with > units other than 1 or -1, and maybe an example will help some of you, > so here's this post. > Consider P(x) = x^2 + x + 2, which factors nicely as > P(x) = 2(x(x+1)/2 + 1) > which I pick because it's valid in the ring of integers, but NOT > always in the ring of algebraic integers. > Can you give an example where x^2 + x + 2 is not divisible by 2 when > x is an algebraic integer? That's impossible because if you do a little mathematical induction, you > find that P(1)=4 and then evaluating for k+1, you get > (k+1)^2+k+1+2 > k^2+2k+1+k+1+2 > k^2+3k+4(= (k^2 + k + 2) + 2(k + 1)) > So P(x) is even for all x.Yes, for the integers. Bit how about the algebraic integers?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: : Re: limitations on mutual anti-correlation?After all that about how I use Google to cross-post, Imindlessly posted this just to sci.math. So this willappear twice on sci.math, sorry.|> Second attempt at posting-- google pled server error thismorning.||Interesting to see how many people do post from Google.I do it when I decide to cross-post something, which aol'sown newsreader doesn't do. I sometimes think of getting abetter ISP, but it hasn't been a priority-- and using Googlefor just when I cross-post is ok. Doing it this way has theuseful feature of giving me a slight disincentive to joiningcross-posted threads. Quite a few of them cross-posted fromsci.physics seem very junky lately-- has sci.physics beengoing through a noisy phase lately?|Interesting learning problem: after the first time it happens and you|lose your careful reply, you get in the habit of copying your post|before hitting post. This behavior is reinforced so long as the|process fails regularly, but soon rejected as unnecessary overhead|after a winning streak ... until the next failure.||See, Google is like life. With mondo storage devices with file systems optimized forbig appends and deletes!I compose longer posts in notepad, then copy and paste, sokeeping a copy is nearly automatic. A fixed-width font makesit easier to avoid extra long lines and to compose ASCIIgraphics (see below). :-)|I also learned though that even though your post seems to be lost,|hitting refresh and then resend from the dialogue box trys the|post again ... complete with your text: you text is in some memory|buffer somewhere, even if you can't see it.I had just had to refresh a couple of previous pages, so Ifigured Google might actually be too busy or something....It also seems sometimes to claim it's failed, but then youfind it succeeded after all (d'oh) and the retry posted aduplicate.|But enough administrivia -- well, besides the annoying point that we|seem to have incommensurate line lengths; which is odd, consideringwe|are both apparently posting from Google.Yeah, I find I can compose something in notepad going allthe way out to 80 columns, cut and paste it into Google'swindow, and it looks fine, but gets posted with wrappedlines all over. Perhaps that's what the preview page is for.I think I'll reformat what you quoted from me, below, ifyou don't mind.[...]|> |This was the first and only intuitive example I came up with as an|> |example of an impossible putative correlation structure for 3random|> |variables ...||> it's a nice enough example of what it means to violate|> Bell's inequality that I've seen it used as an|>illustration, ||Oh. And I was _so_ proud of it. But are you sure seen it used is|not recall seeing Ed Green use it? My bag of tricks is so|small, and growing so slowly of late, I trot this one out everychance|I get.I thought I'd seen some other usenet poster use it, but Icould be wrong.|> in spite of not being the kind of|> violation of Bell's inequality allowed by quantum mechanics.||Well, I'm not sure what the difference in kind is.Just that quantum mechanics predicts the possibility ofmaking certain violations and not others. Quantum mechanicsdoesn't permit three variables to perfectly anticorrelatein pairs like that. For one thing, the correlation matrixis not positive semidefinite, and I just tried to show thatthe correlation matrix you get according to quantum mechanicsalways is.[...]|> |In the famous so-called Bell inequalities we consider 4 random|> |variables (of which only two may be observed at a time), and in|> effect|> |test whether our observations are compatible with maskedsimultaneous|> |observations of 4 random variables.|> |> One complication is that not all pairs taken from these|> four are distant from each other, so we lose our reason|> for expecting that the hypothetical actual values of the|> pair of variables can be measured independently of each|> other.||I don't follow you here. Probably there is some orthogonality to our|understanding -- since I normally find this -- but also I'm not even|sure if you really mean distant or if that's a typo for distinct.I'm saying that not only that, but some of the pairs ofvariables can't be observed at the same time. The originalposter had given a matrix and asked whether it was thematrix of covariances of a set of random variables. Wedon't get all the entries of a matrix from our data; we justget some of the entries, and have the question of whetherthose entries could be the covariances of a set of randomvariables.The criterion for that is in principle just whether it'spossible to fill in the remaining entries of the matrix insuch a way that it satisfies the condition: positivesemidefiniteness. It's a bit messy to determine that ingeneral (apparently), but I found a way to fill in certainmatrices, below, based on analogous features of quantummechanics.Typically, the experiment involves creating a pair ofthen plenty of theories would say that the one measurementaffects the result of the other measurement-- just byGenerally, theories that say making one measurementaffects the results of the other measurements can easilyget away with predicting just the same kind of correlationsas quantum mechanics does. That's the way some formulationsof quantum mechanics work, after all. Usually the reasonwhy a theory does not predict this kind of mutual influenceis because it doesn't predict such influences betweenmeasurements being made at a distance from each other.Ideally, the measurements should be made at a space-likeseparation (i.e., far enough apart and close enough to thesame time that a signal travelling at the speed of lightdoes not have time to get from the one measurement to theother one).|Anyway, in lose our reason ..., etc: this may touch on my|idiosyncratic approach, but our reason for expecting I think needbe|nothing more than an argument that, for a given class of theories,the|data would be compatible with a single joint distribution. IMO the|argument is often bollixed at this point. We don't have to assign|meaning, for example, to what we would have gotten if we measured|the remaining values: a common philosophical tongue twister.Sure. The derivation of the fact that the data is compatiblewith a joint distribution does typically use the fact thatthe simultaneous measurements we're making are being made onjoint distribution-- but it could easily just give somearbitrary other results caused by the two measurementsinterfering with each other. |> In these experiments these pairs of variables are usually|> also incompatible according to quantum mechanics, so we|> can't even ask what distribution of joint values of both|> variables it says we should get. ||Right. But then my take is ... which of course I'm happy to repeat| ... that ... scratch that. We don't even really care whatquantum|mechanics says we should get. The test stands alone ... ah, youjar|loose old rants from the attic ... as a test on a class of classical|theories. We are _not_ testing quantum mechanics, we are testing a|class of theories pretty much defined by the test itself.In a certain sense, I would say we are testing it, becausethe experiment we're trying to (re)design is one whichwould either falsify a bunch of other theories, or falsifyquantum mechanics. If the results come out as predicted byquantum mechanics, then whether they falsify another theorydepends solely on what the results are, and what the othertheory predicts they should be, of course.If all the measurements we use in the experiment arecompatible with each other, according to quantum mechanics,then quantum mechanics will predict results that come froma joint distribution of all the corresponding variables,which is no good. If some of them are incompatible accordingto quantum mechanics, then since space-like separated onesare compatible, it must be that some of our measurements arenot space-like separated. But, turning back to the classicaltheories, if two measurements are not space-like separated,it's easy to produce a theory in which the first measurementruins the second one.So we are stuck with designing an experiment where on eachrun we only make measurements that are separate in space,but where the measurements made on different runs areneither space-like separated nor compatible in quantummechanics' terms.[...]|I also liked to point out that this mysterious class of theories|would not seem very mysterious in some more prosaic situations: likea|smallish black box, e.g., which produced output at either end in the|from of flashing lights. A violation of a Bell-like inequality for|such a box would simply mean that our choice of which flashing lights|to observe at either end was somehow communicated to the oppositeend,|and influenced the outcomes there. If the box were, say, .3m on a|side, we wouldn't find this possibility very spooky!Indeed, if a science museum, say, were to make a quantummechanics exhibit with a box like that in it, one wouldtend to assume they'd just wired it up to work like that.Finding examples of theories lying squarely inside oroutside is straightforward.The mystery comes in only when you try to categorize allpossible explanations. It seems to be so easy to think onehas found all the ways that a theory can predict suchviolations only to find out that there are other verydifferent ones too.I think I've seen several times, for instance, a personreact to a description of one experiment by saying thatsuch experimental results would HAVE to be the result ofsome kind of signal being sent from one measuring stationto the other. I'm not sure I know of anyone who's come upwith a local explanation, one where no signal is sent,without knowing how quantum mechanics predicts it first.I find it especially hard to say how one should categorizesuch things as theories with quantum logic, or nonstandardprobability theory. Over here on sci.math :-) we can talkabout random variables safe in the knowledge that we meanfunctions on measure spaces (with the measure of the wholespace being 1), for which the usual laws can be rigorouslydeduced, but over there in sci.physics it seems like somekind of caution is in order.|> For other readers, it might be worth pointing out that thecorrelation|> matrix is positive semidefinite just when the covariance matrix is.||I wondered about that, and I must say the equivalence is plausiblebut|not obvious. It's not obvious to me that the weighting might somehow|shift a positive determinant to a negative one, or vice versa --|though of course derivation is the key to this difficulty. :-)A symmetric matrix M is positive definite when for eachnonzero vector v, we have v^T M v > 0, where v^T is thetranspose of v (v written as a row vector). The functionv^T M v is called the quadratic form associated with M.The 2 by 2 case is good enough to show why the equivalenceholds. Suppose the variables have standard deviations ofs1 and s2, and the correlation matrix is (a11 a12) (a21 a22).Then if v=(r s)^T, then v^T M v= a11r^2 + (a12+a21)rs + a22s^2. Now the covariance matrixM' is instead (s1^2*a11 s1s2*a12) (s1s2*a21 s2^2*a22)and v^T M' v = s1^2*a11*r^2 + s1s2(a12+a21)rs + s2^2*a22*s^2= a11(r*s1)^2 + (a12+a21)(r*s1)(s*s2) + a22(s*s2)^2, whichis just v'^T M v, where v' = (r*s1 s*s2)^T. So it's >0.To put it in words, the value of the quadratic form associatedwith the covariance matrix evaluated on v is the same as thevalue of the quadratic form associated with the correlationmatrix evaluated at a different vector, namely v with itscoordinates scaled by the standard deviations of the variables.The reverse also holds of course, when the standard deviationsare nonzero, by scaling by the reciprocals of the standarddeviations.[...]|> I tried to figure out whether or not this works. Apparently,|> it doesn't work because (and here's the punch line) quantum|> mechanics also predicts that the matrix is positive|> semi-definite! To put it another way, if the experiment|> works the way quantum mechanics says it will, we can partially|> fill in the entries of the correlation matrix with the values|> we can get directly from the data (and put 1's on the|> diagonal), and the resulting partial matrix is consistent|> with being filled out to being a positive semidefinite matrix.||I'm not sure what you're talking about ... it's quite possible there|is more than one matrix that we might think is the matrixassociated|with the system: but I'm fairly clear on this: if quantum mechanics|really does predict something which falls outside our nominal class|of theories, then there must be _some_ matrix of putative|correlations which _cannot_ be so filled out. Otherwise there would|be no discrimination between interesting outcomes, andno point.The experimental data includes more information than justthe correlations, however.Consider the version of the experiment in which there arethree axes 1, 2, and 3. One correlation matrix obtainedis A1 A2 A3 B1 B2 B3 A1 ( 1 ? ? 1 -1/2 -1/2) A2 ( ? 1 ? -1/2 1 -1/2) A3 ( ? ? 1 -1/2 -1/2 1 ) B1 ( 1 -1/2 -1/2 1 ? ? ) B2 (-1/2 1 -1/2 ? 1 ? ) B3 (-1/2 -1/2 1 ? ? 1 ).We don't get correlations for the ? entries because it wouldwhich as described above is not so helpful.Now for any random variables A1, A2, A3, B1, B2, and B3which have the correlations given above, the perfectcorrelations between A1 and B1, A2 and B2, and A3 and B3imply that those pairs are linearly related. That makesthem equivalent as far as taking correlations is concerned.So finding six such variables is equivalent to findingthree with a correlation matrix ( 1 -1/2 -1/2) (-1/2 1 -1/2) (-1/2 -1/2 1 ).There's no problem in doing so; if we let A1=B1=1, A2=A3=B2=B3=0 with probability 1/3, A2=B2=1, A1=A3=B1=B3=0 with probability 1/3, A3=B3=1, A1=A2=B1=B2=0 with probability 1/3,then we get just such a correlation matrix.But the measurements tell us that each of these variablesis 1 half of the time, not a third of the time!|It did occur to me right away there is some difficulty with positive|semi-definite vs. positive definite in discriminating between|cases. Because ... given some set of putative jointly distributed|random variables which illegal correlation structure, we could always|flesh them out with _more_ putative random variables, linear|combinations of the first set, such that the determinate of the|correlation matrix was zero. So zero just isn't good enough.If you extend a matrix which isn't positive semidefinite,however, the extension still is not positive semidefinite.If v^T M v < 0, and we extend M with more rows and columns,and extend v with 0's at the end, then v^T M v is still <0.[Much quotation deleted.]|You've lost me completely -- I need to go review/learn some linear|algebra. But just give me a precis of what you say you've just shown,|and how it relates to our discussion?First, I gave the example I describe a little differentlyabove, of hypothetical experimental results as predicted byquantum mechanics which violate Bell's inequality, togetherwith a non-Bell-violating way in which the same correlationscould be produced. This shows that the violation of Bell'sinequalities in this case is not detectable just by lookingat measured correlations between variables.Second, I sketched a proof that this is not an isolated case:the correlations you would get from any other experimentwhose results were consistent with quantum mechanics couldalso be reproduced by a joint distribution of randomvariables.It's crucial, of course, that the correlations are not allof the data we get from the experiment. The data as a wholeis incompatible with having a joint distribution, but thecorrelations can be generated by one.|Anyway, as I said, you've revived an old rant in my head, one which|include the coinage Rutherford test. Simply, although the wholeEPR|tradition culminating in Bell's work is or course intimate with|quantum mechanics, nothing in the final test need be or ought to beon|some irreducible level. As I said, we are _not_ testing quantum|mechanics, we are testing a class of theories best characterized as|... ta dum ... the class of theories tested for by investigation of|compatibility with a single joint distribution given a particular|experimental set-up -- which is a slightly less circular way ofsaying|the class of theories tested for by this test.||What we want to convince ourselves of ... which I've made no argument|of here for brevity ... is the equivalence of this test to the vaguer|but more motivated idea of sending out some undefined information|from a central source to some antipodes of our apparatus where,|interacting with local chance influence but not anymore backreacting|on the information which travelled to the other end of our apparatus,|produces results. Any theory/statistical observations _not_ in this|class is ... most cautiously and tautologically, but correctly ...|_not_ a result which is compatible with some undefined information|propagating outward from a source to two antipodes, not further|interacting except with local influence ... etc.||Anyway, I invoke Rutherford as somebody who would not have knownabout|qm, but could definitely understand the physical reasoning in ruling|out this class of theories based on some statical experimental|observations: and if we are really ruling this class of theories out|of nature, then we have better be able to build a black box ... avery|long baseline black box ... producing outcomes falsifying the|indicated class of models without any more need for special pleading|or interpretation. I'm not sure if any contemporary quantum optics|test meets this test or not.Every now and then I have a fantasy of bringing some wellrespected thinker from the past to the present day andtaking them on a little tour. We now have artificiallighting not requiring fire, see? Let me warn you about therather fast-moving oil-powered vehicles we now use, beforewe go out by the street. You are right, there are nomusicians there, there's a device in there for repeatingsounds. We have big machines that fly, and many of us havetravelled in them many times...!When it comes to these experiments, I sometimes imaginedescribing them to Einstein, Podolsky and Rosen. I thinkthey had a lot of the same motivation as Bell, and wouldhave enjoyed seeing how attempts at investigating the issuehad progressed experimentally.Keith Ramsay === Subject: : Re: limitations on mutual anti-correlation?|> Second attempt at posting-- google pled server error this morning.||Interesting to see how many people do post from Google.I do it when I decide to cross-post something, which aol'sown newsreader doesn't do. I sometimes think of getting abetter ISP, but it hasn't been a priority-- and using Googlefor just when I cross-post is ok. Doing it this way has theuseful feature of giving me a slight disincentive to joiningcross-posted threads. Quite a few of them cross-posted fromsci.physics seem very junky lately-- has sci.physics beengoing through a noisy phase lately?|Interesting learning problem: after the first time it happens and you|lose your careful reply, you get in the habit of copying your post|before hitting post. This behavior is reinforced so long as the|process fails regularly, but soon rejected as unnecessary overhead|after a winning streak ... until the next failure.||See, Google is like life. With mondo storage devices with file systems optimized forbig appends and deletes!I compose longer posts in notepad, then copy and paste, sokeeping a copy is nearly automatic. A fixed-width font makesit easier to avoid extra long lines and to compose ASCIIgraphics (see below). :-)|I also learned though that even though your post seems to be lost,|hitting refresh and then resend from the dialogue box trys the|post again ... complete with your text: you text is in some memory|buffer somewhere, even if you can't see it.I had just had to refresh a couple of previous pages, so Ifigured Google might actually be too busy or something....It also seems sometimes to claim it's failed, but then youfind it succeeded after all (d'oh) and the retry posted aduplicate.|But enough administrivia -- well, besides the annoying point that we|seem to have incommensurate line lengths; which is odd, considering we|are both apparently posting from Google.Yeah, I find I can compose something in notepad going allthe way out to 80 columns, cut and paste it into Google'swindow, and it looks fine, but gets posted with wrappedlines all over. Perhaps that's what the preview page is for.I think I'll reformat what you quoted from me, below, ifyou don't mind.[...]|> |This was the first and only intuitive example I came up with as an|> |example of an impossible putative correlation structure for 3 random|> |variables ...||> it's a nice enough example of what it means to violate|> Bell's inequality that I've seen it used as an|>illustration, ||Oh. And I was _so_ proud of it. But are you sure seen it used is|not recall seeing Ed Green use it? My bag of tricks is so|small, and growing so slowly of late, I trot this one out every chance|I get.I thought I'd seen some other usenet poster use it, but Icould be wrong.|> in spite of not being the kind of|> violation of Bell's inequality allowed by quantum mechanics.||Well, I'm not sure what the difference in kind is.Just that quantum mechanics predicts the possibility ofmaking certain violations and not others. Quantum mechanicsdoesn't permit three variables to perfectly anticorrelatein pairs like that. For one thing, the correlation matrixis not positive semidefinite, and I just tried to show thatthe correlation matrix you get according to quantum mechanicsalways is.[...]|> |In the famous so-called Bell inequalities we consider 4 random|> |variables (of which only two may be observed at a time), and in|> effect|> |test whether our observations are compatible with masked simultaneous|> |observations of 4 random variables.|> |> One complication is that not all pairs taken from these|> four are distant from each other, so we lose our reason|> for expecting that the hypothetical actual values of the|> pair of variables can be measured independently of each|> other.||I don't follow you here. Probably there is some orthogonality to our|understanding -- since I normally find this -- but also I'm not even|sure if you really mean distant or if that's a typo for distinct.I'm saying that not only that, but some of the pairs ofvariables can't be observed at the same time. The originalposter had given a matrix and asked whether it was thematrix of covariances of a set of random variables. Wedon't get all the entries of a matrix from our data; we justget some of the entries, and have the question of whetherthose entries could be the covariances of a set of randomvariables.The criterion for that is in principle just whether it'spossible to fill in the remaining entries of the matrix insuch a way that it satisfies the condition: positivesemidefiniteness. It's a bit messy to determine that ingeneral (apparently), but I found a way to fill in certainmatrices, below, based on analogous features of quantummechanics.Typically, the experiment involves creating a pair ofthen plenty of theories would say that the one measurementaffects the result of the other measurement-- just byGenerally, theories that say making one measurementaffects the results of the other measurements can easilyget away with predicting just the same kind of correlationsas quantum mechanics does. That's the way some formulationsof quantum mechanics work, after all. Usually the reasonwhy a theory does not predict this kind of mutual influenceis because it doesn't predict such influences betweenmeasurements being made at a distance from each other.Ideally, the measurements should be made at a space-likeseparation (i.e., far enough apart and close enough to thesame time that a signal travelling at the speed of lightdoes not have time to get from the one measurement to theother one).|Anyway, in lose our reason ..., etc: this may touch on my|idiosyncratic approach, but our reason for expecting I think need be|nothing more than an argument that, for a given class of theories, the|data would be compatible with a single joint distribution. IMO the|argument is often bollixed at this point. We don't have to assign|meaning, for example, to what we would have gotten if we measured|the remaining values: a common philosophical tongue twister.Sure. The derivation of the fact that the data is compatiblewith a joint distribution does typically use the fact thatthe simultaneous measurements we're making are being made onjoint distribution-- but it could easily just give somearbitrary other results caused by the two measurementsinterfering with each other. |> In these experiments these pairs of variables are usually|> also incompatible according to quantum mechanics, so we|> can't even ask what distribution of joint values of both|> variables it says we should get. ||Right. But then my take is ... which of course I'm happy to repeat| ... that ... scratch that. We don't even really care what quantum|mechanics says we should get. The test stands alone ... ah, you jar|loose old rants from the attic ... as a test on a class of classical|theories. We are _not_ testing quantum mechanics, we are testing a|class of theories pretty much defined by the test itself.In a certain sense, I would say we are testing it, becausethe experiment we're trying to (re)design is one whichwould either falsify a bunch of other theories, or falsifyquantum mechanics. If the results come out as predicted byquantum mechanics, then whether they falsify another theorydepends solely on what the results are, and what the othertheory predicts they should be, of course.If all the measurements we use in the experiment arecompatible with each other, according to quantum mechanics,then quantum mechanics will predict results that come froma joint distribution of all the corresponding variables,which is no good. If some of them are incompatible accordingto quantum mechanics, then since space-like separated onesare compatible, it must be that some of our measurements arenot space-like separated. But, turning back to the classicaltheories, if two measurements are not space-like separated,it's easy to produce a theory in which the first measurementruins the second one.So we are stuck with designing an experiment where on eachrun we only make measurements that are separate in space,but where the measurements made on different runs areneither space-like separated nor compatible in quantummechanics' terms.[...]|I also liked to point out that this mysterious class of theories|would not seem very mysterious in some more prosaic situations: like a|smallish black box, e.g., which produced output at either end in the|from of flashing lights. A violation of a Bell-like inequality for|such a box would simply mean that our choice of which flashing lights|to observe at either end was somehow communicated to the opposite end,|and influenced the outcomes there. If the box were, say, .3m on a|side, we wouldn't find this possibility very spooky!Indeed, if a science museum, say, were to make a quantummechanics exhibit with a box like that in it, one wouldtend to assume they'd just wired it up to work like that.Finding examples of theories lying squarely inside oroutside is straightforward.The mystery comes in only when you try to categorize allpossible explanations. It seems to be so easy to think onehas found all the ways that a theory can predict suchviolations only to find out that there are other verydifferent ones too.I think I've seen several times, for instance, a personreact to a description of one experiment by saying thatsuch experimental results would HAVE to be the result ofsome kind of signal being sent from one measuring stationto the other. I'm not sure I know of anyone who's come upwith a local explanation, one where no signal is sent,without knowing how quantum mechanics predicts it first.I find it especially hard to say how one should categorizesuch things as theories with quantum logic, or nonstandardprobability theory. Over here on sci.math :-) we can talkabout random variables safe in the knowledge that we meanfunctions on measure spaces (with the measure of the wholespace being 1), for which the usual laws can be rigorouslydeduced, but over there in sci.physics it seems like somekind of caution is in order.|> For other readers, it might be worth pointing out that the correlation|> matrix is positive semidefinite just when the covariance matrix is.||I wondered about that, and I must say the equivalence is plausible but|not obvious. It's not obvious to me that the weighting might somehow|shift a positive determinant to a negative one, or vice versa --|though of course derivation is the key to this difficulty. :-)A symmetric matrix M is positive definite when for eachnonzero vector v, we have v^T M v > 0, where v^T is thetranspose of v (v written as a row vector). The functionv^T M v is called the quadratic form associated with M.The 2 by 2 case is good enough to show why the equivalenceholds. Suppose the variables have standard deviations ofs1 and s2, and the correlation matrix is (a11 a12) (a21 a22).Then if v=(r s)^T, then v^T M v= a11r^2 + (a12+a21)rs + a22s^2. Now the covariance matrixM' is instead (s1^2*a11 s1s2*a12) (s1s2*a21 s2^2*a22)and v^T M' v = s1^2*a11*r^2 + s1s2(a12+a21)rs + s2^2*a22*s^2= a11(r*s1)^2 + (a12+a21)(r*s1)(s*s2) + a22(s*s2)^2, whichis just v'^T M v, where v' = (r*s1 s*s2)^T. So it's >0.To put it in words, the value of the quadratic form associatedwith the covariance matrix evaluated on v is the same as thevalue of the quadratic form associated with the correlationmatrix evaluated at a different vector, namely v with itscoordinates scaled by the standard deviations of the variables.The reverse also holds of course, when the standard deviationsare nonzero, by scaling by the reciprocals of the standarddeviations.[...]|> I tried to figure out whether or not this works. Apparently,|> it doesn't work because (and here's the punch line) quantum|> mechanics also predicts that the matrix is positive|> semi-definite! To put it another way, if the experiment|> works the way quantum mechanics says it will, we can partially|> fill in the entries of the correlation matrix with the values|> we can get directly from the data (and put 1's on the|> diagonal), and the resulting partial matrix is consistent|> with being filled out to being a positive semidefinite matrix.||I'm not sure what you're talking about ... it's quite possible there|is more than one matrix that we might think is the matrix associated|with the system: but I'm fairly clear on this: if quantum mechanics|really does predict something which falls outside our nominal class|of theories, then there must be _some_ matrix of putative|correlations which _cannot_ be so filled out. Otherwise there would|be no discrimination between interesting outcomes, andno point.The experimental data includes more information than justthe correlations, however.Consider the version of the experiment in which there arethree axes 1, 2, and 3. One correlation matrix obtainedis A1 A2 A3 B1 B2 B3 A1 ( 1 ? ? 1 -1/2 -1/2) A2 ( ? 1 ? -1/2 1 -1/2) A3 ( ? ? 1 -1/2 -1/2 1 ) B1 ( 1 -1/2 -1/2 1 ? ? ) B2 (-1/2 1 -1/2 ? 1 ? ) B3 (-1/2 -1/2 1 ? ? 1 ).We don't get correlations for the ? entries because it wouldwhich as described above is not so helpful.Now for any random variables A1, A2, A3, B1, B2, and B3which have the correlations given above, the perfectcorrelations between A1 and B1, A2 and B2, and A3 and B3imply that those pairs are linearly related. That makesthem equivalent as far as taking correlations is concerned.So finding six such variables is equivalent to findingthree with a correlation matrix ( 1 -1/2 -1/2) (-1/2 1 -1/2) (-1/2 -1/2 1 ).There's no problem in doing so; if we let A1=B1=1, A2=A3=B2=B3=0 with probability 1/3, A2=B2=1, A1=A3=B1=B3=0 with probability 1/3, A3=B3=1, A1=A2=B1=B2=0 with probability 1/3,then we get just such a correlation matrix.But the measurements tell us that each of these variablesis 1 half of the time, not a third of the time!|It did occur to me right away there is some difficulty with positive|semi-definite vs. positive definite in discriminating between|cases. Because ... given some set of putative jointly distributed|random variables which illegal correlation structure, we could always|flesh them out with _more_ putative random variables, linear|combinations of the first set, such that the determinate of the|correlation matrix was zero. So zero just isn't good enough.If you extend a matrix which isn't positive semidefinite,however, the extension still is not positive semidefinite.If v^T M v < 0, and we extend M with more rows and columns,and extend v with 0's at the end, then v^T M v is still <0.[Much quotation deleted.]|You've lost me completely -- I need to go review/learn some linear|algebra. But just give me a precis of what you say you've just shown,|and how it relates to our discussion?First, I gave the example I describe a little differentlyabove, of hypothetical experimental results as predicted byquantum mechanics which violate Bell's inequality, togetherwith a non-Bell-violating way in which the same correlationscould be produced. This shows that the violation of Bell'sinequalities in this case is not detectable just by lookingat measured correlations between variables.Second, I sketched a proof that this is not an isolated case:the correlations you would get from any other experimentwhose results were consistent with quantum mechanics couldalso be reproduced by a joint distribution of randomvariables.It's crucial, of course, that the correlations are not allof the data we get from the experiment. The data as a wholeis incompatible with having a joint distribution, but thecorrelations can be generated by one.|Anyway, as I said, you've revived an old rant in my head, one which|include the coinage Rutherford test. Simply, although the whole EPR|tradition culminating in Bell's work is or course intimate with|quantum mechanics, nothing in the final test need be or ought to be on|some irreducible level. As I said, we are _not_ testing quantum|mechanics, we are testing a class of theories best characterized as|... ta dum ... the class of theories tested for by investigation of|compatibility with a single joint distribution given a particular|experimental set-up -- which is a slightly less circular way of saying|the class of theories tested for by this test.||What we want to convince ourselves of ... which I've made no argument|of here for brevity ... is the equivalence of this test to the vaguer|but more motivated idea of sending out some undefined information|from a central source to some antipodes of our apparatus where,|interacting with local chance influence but not anymore backreacting|on the information which travelled to the other end of our apparatus,|produces results. Any theory/statistical observations _not_ in this|class is ... most cautiously and tautologically, but correctly ...|_not_ a result which is compatible with some undefined information|propagating outward from a source to two antipodes, not further|interacting except with local influence ... etc.||Anyway, I invoke Rutherford as somebody who would not have known about|qm, but could definitely understand the physical reasoning in ruling|out this class of theories based on some statical experimental|observations: and if we are really ruling this class of theories out|of nature, then we have better be able to build a black box ... a very|long baseline black box ... producing outcomes falsifying the|indicated class of models without any more need for special pleading|or interpretation. I'm not sure if any contemporary quantum optics|test meets this test or not.Every now and then I have a fantasy of bringing some wellrespected thinker from the past to the present day andtaking them on a little tour. We now have artificiallighting not requiring fire, see? Let me warn you about therather fast-moving oil-powered vehicles we now use, beforewe go out by the street. You are right, there are nomusicians there, there's a device in there for repeatingsounds. We have big machines that fly, and many of us havetravelled in them many times...!When it comes to these experiments, I sometimes imaginedescribing them to Einstein, Podolsky and Rosen. I thinkthey had a lot of the same motivation as Bell, and wouldhave enjoyed seeing how attempts at investigating the issuehad progressed experimentally.Keith Ramsay === Subject: : Re: JSH: Distributive property, math argument>Consider that if I have f(x) = x + 3, where 3 has itself as a factor>and f(x) has 3 as a factor, then x *must* have 3 as a factor.Then I would conclude that f(x)/3 is in the ring you are in, bydefinition of factor.>Now let's say x=7. Then, 7/3 must be in the ring.>See?Is this supposed to prove that 7/3 is an algebraic integer? To do so,you must start from the assumption that 10/3 is an algebraic integer,which it isn't. - Randy === Subject: : Re: JSH: Distributive property, math argument> ...> The start is simple enough, where the x shown is in the ring of> algebraic integers.> Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 Why does it not work with my polynomial? It's not your polynomial, of course, it's the *factorization* that you> use. At the start of the post I point out that if f(x) = x + 3, and f(x)> has 3 as a factor, then x must have it as a factor as well. That's *basic* Dik Winter, and if you accept that result, then at> least some things should be clear to you. Notice I didn't mention a ring for x, as the statement is true,> without regard to the commutative ring. yes, you did.> You said Consider f(x) = x+3, where x is an algebraic integer. Yup, you're right. I'd forgotten that quickly and didn't go back to> check before making the reply. In any event, maybe I should say that it doesn't matter what the ring> is. Consider that if I have f(x) = x + 3, where 3 has itself as a factor> and f(x) has 3 as a factor, then x *must* have 3 as a factor. Now let's say x=7. Then, 7/3 must be in the ring. See?No, I don't understand . You said it was the ring of algebraic integers> and that f(x), given as f(x) = x+3, has 3 as a factor. Then instead of> concluding that this places a *restricition* on the values that x can take> (i.e. x must be an algebraic integer that is divisible by 3) you choose x to> be 7 and conclude that therefore 7/3 =2.33333... must be in the ring of> algebraic integers.!!!!OOPS! In my previous post I have f(x) = 2((x+1)/2 + 1) when I shouldhave had f(x) = 2(x(x+1)/2 + 1).James Harris === Subject: : Re: JSH: Distributive property, math argument> ...> The start is simple enough, where the x shown is in the ring of> algebraic integers.> Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 Why does it not work with my polynomial? It's not your polynomial, of course, it's the *factorization* that you> use. At the start of the post I point out that if f(x) = x + 3, and f(x)> has 3 as a factor, then x must have it as a factor as well. That's *basic* Dik Winter, and if you accept that result, then at> least some things should be clear to you. Notice I didn't mention a ring for x, as the statement is true,> without regard to the commutative ring. yes, you did.> You said Consider f(x) = x+3, where x is an algebraic integer. Yup, you're right. I'd forgotten that quickly and didn't go back to> check before making the reply. In any event, maybe I should say that it doesn't matter what the ring> is. Consider that if I have f(x) = x + 3, where 3 has itself as a factor> and f(x) has 3 as a factor, then x *must* have 3 as a factor. Now let's say x=7. Then, 7/3 must be in the ring. See?No, I don't understand . You said it was the ring of algebraic integers> and that f(x), given as f(x) = x+3, has 3 as a factor. Then instead of> concluding that this places a *restricition* on the values that x can take> (i.e. x must be an algebraic integer that is divisible by 3) you choose x to> be 7 and conclude that therefore 7/3 =2.33333... must be in the ring of> algebraic integers.!!!!Nope. I'm saying that it doesn't matter what you might *say* the ringis, given f(x) = x+3, if you also have that f(x) has 3 as a factor and3 has itself as a factor, as then necessarily x has 3 as a factor,which means that if x=7, which IS an algebraic integer, it *still*must have 3 as a factor for the other statements to be true.Look at it as a logical argument:1. f(x) = x + 32. 3 has itself as a factor3. f(x) has 3 as a factor4. Therefore, x has 3 as a factor.You see, 1., 2., and 3., logically force 4. from the distributiveproperty, understand?Here's another example to help you out.Now consider the factorizationf(x) = 2((x+1)/2 + 1)where you'll notice that the factorization exists in the ring ofintegers, but does NOT always exist in the ring of algebraic integers.Think about what I said does NOT always exist and consider thepurpose of the use of words.James Harris === Subject: : Re: JSH: Distributive property, math argument> ...> The start is simple enough, where the x shown is in the ring of> algebraic integers.> Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 Why does it not work with my polynomial? It's not your polynomial, of course, it's the *factorization* that you> use. At the start of the post I point out that if f(x) = x + 3, and f(x)> has 3 as a factor, then x must have it as a factor as well. That's *basic* Dik Winter, and if you accept that result, then at> least some things should be clear to you. Notice I didn't mention a ring for x, as the statement is true,> without regard to the commutative ring. yes, you did.> You said Consider f(x) = x+3, where x is an algebraic integer. Yup, you're right. I'd forgotten that quickly and didn't go back to> check before making the reply. In any event, maybe I should say that it doesn't matter what the ring> is. Consider that if I have f(x) = x + 3, where 3 has itself as a factor> and f(x) has 3 as a factor, then x *must* have 3 as a factor. Now let's say x=7. Then, 7/3 must be in the ring. See?No, I don't understand . You said it was the ring of algebraic integers> and that f(x), given as f(x) = x+3, has 3 as a factor. Then instead of> concluding that this places a *restricition* on the values that x can take> (i.e. x must be an algebraic integer that is divisible by 3) you choose x to> be 7 and conclude that therefore 7/3 =2.33333... must be in the ring of> algebraic integers.!!!!Nope. I'm saying that it doesn't matter what you might *say* the ring> is, given f(x) = x+3, if you also have that f(x) has 3 as a factor and> 3 has itself as a factor, as then necessarily x has 3 as a factor,> which means that if x=7, which IS an algebraic integer, it *still*> must have 3 as a factor for the other statements to be true.Look at it as a logical argument:1. f(x) = x + 32. 3 has itself as a factor3. f(x) has 3 as a factor4. Therefore, x has 3 as a factor.You see, 1., 2., and 3., logically force 4. from the distributive> property, understand?Here's another example to help you out.Now consider the factorizationf(x) = 2((x+1)/2 + 1)where you'll notice that the factorization exists in the ring of> integers, but does NOT always exist in the ring of algebraic integers.> As you noted elsewhere, what you meant to say here was: f(x) = 2*(x*(x + 1)/2 + 1) This can be re-written as f(x) = x^2 + x + 3. Your statement is diametrically wrong. This polynomial does NOT factor with integer coefficients, but it DOES factor in thealgebraic integers: f(x) = (x - r1)*(x - r2), where r1 = (-1 + sqrt(-11))/2 and r2 = (-1 - sqrt(-11))/2,and of course both r1 and r2 are algebraic integers. The factorization that you gave is not a factorization inEITHER the integers or the algebraic integers (dividing by2 is the same as multiplying by 1/2). Are you totally losing it, or what ???> Think about what I said does NOT always exist and consider the> purpose of the use of words.> Uh ... right. Whatever you say, Einstein. It might also be worth noting that if x is an algebraicinteger, f(x) = x^2 + x + 3 is also: the set of A.I.'s isclosed under multiplication and addition. Nora B.James Harris === Subject: : Re: JSH: Distributive property, math argument Adjunct Assistant Professor at the University of Montana. [.snip.]>> Now consider the factorization>> f(x) = 2((x+1)/2 + 1)>> where you'll notice that the factorization exists in the ring of>> integers, but does NOT always exist in the ring of algebraic integers.> As you noted elsewhere, what you meant to say here was:> f(x) = 2*(x*(x + 1)/2 + 1)> This can be re-written as> f(x) = x^2 + x + 3.> Your statement is diametrically wrong. This polynomial does >NOT factor with integer coefficients, but it DOES factor in the>algebraic integers:> f(x) = (x - r1)*(x - r2), where> r1 = (-1 + sqrt(-11))/2 and> r2 = (-1 - sqrt(-11))/2,>and of course both r1 and r2 are algebraic integers.> The factorization that you gave is not a factorization in>EITHER the integers or the algebraic integers (dividing by>2 is the same as multiplying by 1/2).> Are you totally losing it, or what ???James is once again confusing a polynomial factorization and a numericfactorization; that is, the difference between the values of apolynomial always dividing the values of another polynomial, and apolynomial divind the other polynomial.Here, we have:F(x) = x^2 + x + 3.P(x) = 2Q(x) = (1/2)(x^2+x+3).Then F(x) = P(x)*Q(x) in the ring of all polynomials over Q. INADDITION, Q(x) is a function from Z to Z, and P(x) is a function fromZ to Z, with the properties that, AS FUNCTIONS, F(x) = P(x)*Q(x).That is, for each value of x, Q(x) divides F(x) in Z, which is adifferent statement from the polynomial Q(x) divides the polynomialF(x) in Z[x].Amazing, that after only about two years he finally realizes what hewas told so long ago: that a factorization of the VALUES of apolynomial does not always correspond to the a factorization of thePOLYNOMIAL. power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can criticize. A great many people are staggered to this extent, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan == === Subject: : Re: JSH: Distributive property, math argument Here's another example to help you out.Now consider the factorizationf(x) = 2((x+1)/2 + 1) That should be f(x) = 2(x(x+1)/2 + 1).> where you'll notice that the factorization exists in the ring of> integers, but does NOT always exist in the ring of algebraic integers.Think about what I said does NOT always exist and consider the> purpose of the use of words. James Harris === Subject: : Re: JSH: Distributive property, math argument> ...> The start is simple enough, where the x shown is in the ringof> algebraic integers.> Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 Why does it not work with my polynomial? It's not your polynomial, of course, it's the *factorization* thatyou> use. At the start of the post I point out that if f(x) = x + 3, andf(x)> has 3 as a factor, then x must have it as a factor as well. That's *basic* Dik Winter, and if you accept that result, then at> least some things should be clear to you. Notice I didn't mention a ring for x, as the statement is true,> without regard to the commutative ring. yes, you did.> You said Consider f(x) = x+3, where x is an algebraic integer. Yup, you're right. I'd forgotten that quickly and didn't go back to> check before making the reply. In any event, maybe I should say that it doesn't matter what the ring> is. Consider that if I have f(x) = x + 3, where 3 has itself as a factor> and f(x) has 3 as a factor, then x *must* have 3 as a factor. Now let's say x=7. Then, 7/3 must be in the ring. See? No, I don't understand . You said it was the ring of algebraicintegers> and that f(x), given as f(x) = x+3, has 3 as a factor. Then instead of> concluding that this places a *restriction* on the values that x cantake> (i.e. x must be an algebraic integer that is divisible by 3) you choosex to> be 7 and conclude that therefore 7/3 =2.33333... must be in the ring of> algebraic integers.!!!!> Nope. I'm saying that it doesn't matter what you might *say* the ring> is, given f(x) = x+3, if you also have that f(x) has 3 as a factor and> 3 has itself as a factor, as then necessarily x has 3 as a factor,> which means that if x=7, which IS an algebraic integer, it *still*> must have 3 as a factor for the other statements to be true.Perhaps I misunderstood the example as in the original discussion of whichthis is an overly simplified example it was obvious that f(x) indeed hadsuch a factor.What I don't understand is why you think (in either the ring of integers oralgebraic integers), that given f(x) = x+3, you can simply *assert* as youdid that f(x) has 3 as a factor without investigating whether for variousvalues of x it indeed does so.In particular, for your example of x = 7, f(7) = 10, which does not have 3as a factor, nor does 7 have 3 as a factor.> Look at it as a logical argument:> 1. f(x) = x + 3> 2. 3 has itself as a factor> 3. f(x) has 3 as a factor> 4. Therefore, x has 3 as a factor.> You see, 1., 2., and 3., logically force 4. from the distributive> property, understand?No, given 1, 2 and 3, you can say:4. Therefore x is constrained to only those values that have 3 as a factorOR you can say:3. IF x has 3 as a factor THEN4. f(x) has 3 as a factorbut you can't turn it around the other way as you didThe distributive property: a(y+z) = ay + bz,applied to your equation is: 3(x/3 +1) = 3(x/3) + 3(1)It says nothing at all about x. Pick an x, any x. Pick 7. It does *not*have 3 as a factor although the distributive property is upheld.KeithK> Here's another example to help you out.> Now consider the factorization> f(x) = 2((x+1)/2 + 1)> where you'll notice that the factorization exists in the ring of> integers, but does NOT always exist in the ring of algebraic integers.> Think about what I said does NOT always exist and consider the> purpose of the use of words.> James Harris === Subject: : Re: JSH: Distributive property, math argument> ...> The start is simple enough, where the x shown is in the ring> of> algebraic integers.> Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 Why does it not work with my polynomial? It's not your polynomial, of course, it's the *factorization* that> you> use. At the start of the post I point out that if f(x) = x + 3, and> f(x)> has 3 as a factor, then x must have it as a factor as well. That's *basic* Dik Winter, and if you accept that result, then at> least some things should be clear to you. Notice I didn't mention a ring for x, as the statement is true,> without regard to the commutative ring. yes, you did.> You said Consider f(x) = x+3, where x is an algebraic integer. Yup, you're right. I'd forgotten that quickly and didn't go back to> check before making the reply. In any event, maybe I should say that it doesn't matter what the ring> is. Consider that if I have f(x) = x + 3, where 3 has itself as a factor> and f(x) has 3 as a factor, then x *must* have 3 as a factor. Now let's say x=7. Then, 7/3 must be in the ring. See? No, I don't understand . You said it was the ring of algebraic> integers> and that f(x), given as f(x) = x+3, has 3 as a factor. Then instead of> concluding that this places a *restriction* on the values that x can> take> (i.e. x must be an algebraic integer that is divisible by 3) you choose> x to> be 7 and conclude that therefore 7/3 =2.33333... must be in the ring of> algebraic integers.!!!! Nope. I'm saying that it doesn't matter what you might *say* the ring> is, given f(x) = x+3, if you also have that f(x) has 3 as a factor and> 3 has itself as a factor, as then necessarily x has 3 as a factor,> which means that if x=7, which IS an algebraic integer, it *still*> must have 3 as a factor for the other statements to be true. Perhaps I misunderstood the example as in the original discussion of which> this is an overly simplified example it was obvious that f(x) indeed had> such a factor.What I don't understand is why you think (in either the ring of integers or> algebraic integers), that given f(x) = x+3, you can simply *assert* as you> did that f(x) has 3 as a factor without investigating whether for various> values of x it indeed does so.It's a logical point, and perhaps a subtle one.If you have f(x) = x + 3, that f(x) has 3 as a factor, and you havethat 3 has itself as a factor, then it *must* be true that x has 3 asa factor.There is no investigation as it's logically forced.Now you may wish to challenge that f(x) has 3 as a factor, or that 3has itself as a factor, but the logical argument itself is notchallengeable.It's like:1. All dogs wear shoes.2. Ralph is a dog.3. Therefore Ralph wears shoes.The logical argument is ok, it's premise 1. that's faulty.So your investigation has to do with the premises, not with theconclusion. > In particular, for your example of x = 7, f(7) = 10, which does not have 3> as a factor, nor does 7 have 3 as a factor.Yes it does, in a ring where 7 is a unit.For instance, in reals, 3(7/3) = 7, so 3 is a factor of 7.> Look at it as a logical argument: 1. f(x) = x + 3 2. 3 has itself as a factor 3. f(x) has 3 as a factor 4. Therefore, x has 3 as a factor. You see, 1., 2., and 3., logically force 4. from the distributive> property, understand? No, given 1, 2 and 3, you can say:> 4. Therefore x is constrained to only those values that have 3 as a factorNo. You can't go to the end, you have to go to the beginning.> OR you can say:> 3. IF x has 3 as a factor THEN> 4. f(x) has 3 as a factor> but you can't turn it around the other way as you didYes I can. It's a *logical* argument.> The distributive property: a(y+z) = ay + bz,> applied to your equation is: 3(x/3 +1) = 3(x/3) + 3(1)It says nothing at all about x. Pick an x, any x. Pick 7. It does *not*> have 3 as a factor although the distributive property is upheld.KeithKIt does in the field of reals.James Harris === Subject: : Re: JSH: Distributive property, math argumentNoting your correction: > Now consider the factorization > f(x) = 2((x+1)/2 + 1)Corrected to: > f(x) = 2(x(x+1)/2 + 1) > where you'll notice that the factorization exists in the ring of > integers, but does NOT always exist in the ring of algebraic integers.Lessee. f(x) = 2(x(x + 1)/2 + 1) = x(x + 1) + 2 = x^2 + x + 3.What is the factorisation in the integers? In the algebraic integers thereis a ready factorisation for each and every algebraic integer x.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: : Re: JSH: Distributive property, math argument>> It's amazing how easily people can question very basic concepts when>> the stakes are high, so here I am again to help you out by reminding>> you just how basic the math argument I've been giving recently is.>> Consider f(x) = x+3, where x is an algebraic integer.>> Now then, if f(x) has 3 as a factor, then x *must* have 3 as a factor>> as well.>> That follows easily enough from the distributive property:>> a(b+c) = ab + ac>> and for quite a few months, I've noticed people willing to debate it,>> as if that property can just go away at a whim.>> That's not mathematics, and many of you are showing a puzzling lack of>> rationality by fighting mathematics itself, rather than just accepting>> the truth.>> After all, it's not like the math is mad at you, or that there's some>> conspiracy to screw you over as mathematical truth is just true.>> If you believed something else in the past that was wrong, it was just>> wrong.>> Now here's the argument, where those people questioning it have>> basically been questioning the distributive property.>> The start is simple enough, where the x shown is in the ring of>> algebraic integers.>> Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078>> which you'll notice has a constant term that is 1078.>> Well moving things around with P(x) gives you>> P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3>> which is a deliberate form to allow me to factor P(x), so that I have>> P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)>> where the a's are roots of>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).>> To use the distributive property as I wish I need to have constant>> terms.>> And it *appears* that the constant terms for the three factors are all>> 7, but that can't be right, as the constant term of P(x) is 1078.>> So I need to do another step, and analysis offers the simple technique>> of setting x=0 to pull out the constant terms, so setting x=0, I find>> that>> P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)>> as the cubic defining the a's at x=0 is>> a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and>> a_2(0) to equal 0, which leaves a_3(0) with a value of 3.>> So let a_3(x) = b_3(x) + 3, to keep indices matched. Then I have>> P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)>> P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)>> and now my constant terms work out correctly.>> Now I can use the distributive property, as P(x) has 49 as a factor of>> each term in>> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078>> so P(x) has 49 as a factor, so I can divide by 49, and dividing 1078>> by 49 gives me 22, as the new constant term.>> Well that means that>> P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22)>> is the only way that 49 can divide out and keep the constant terms>> matching, in what is a simple exercise dependent on the distributive>> property.>> For the following below, I am going to define g1, g2, and g3 as:>> g1 = 5*a_1 + 7>> g2 = 5*a_2 + 7 and>> g3 = 5*b_3 + 22 = 5*a3 + 7.>> OK, you have that P(x)/49 has constant term 22, and indeed>> g3 = 5*b_3 + 22 has constant term 22 also. It is almost irresistible>The proof shows what *must* be the case, which is how math works.>Human emotion is irrelevant.>> to conclude that the only way you can factor P(x)/49 is in >> the form>> P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22).>> As you say, the constant terms multiply together perfectly>> to give 22. Plus 22 is coprime to 7.>> But it's possible, for any given value of x, that 5*b3 + 22 is>> NOT coprime to 7. For example, what if b3 = 4 ?>I focus on constant terms because they are constant, so it hardly>matters about changes from the value of x. This is such a blatantly ridiculous statement! I said thateven though 22 is coprime to 7, it is not necessarily true that 5*b3 + 22 is coprime to 7. I gave you the example ofb3 = 4, for which 5*b3 + 22 = 5*4 + 22 = 20 + 22 = 42,which is actually DIVISIBLE by 7. Then you come back and say: I focus on constant terms because they are constant, so it hardly matters about changes from the value of x. Have you completely lost your marbles ? You knowthat b3 is dependent on x. What happens if for a givenvalue of x, b3 = b3(x) = 4 ???? How have you eliminatedthat possibility ? Not to mention the *many* other waysthat 5*b3 + 22 could be non-coprime to 7 in the algebraicintegers. In view of this example your statement, and especiallythe phrase hardly matters is simply senseless. Clearly you are just plain assuming that if the constant term g(0) of a function g(x) is coprime to 7, then g(x) is coprime to 7 for all x. This is so obviously false, yet you seem unable to grasp it. Why?>The proper approach is to look over the proof, step-by-step, and if>you see some step that bugs you, ask me about it. Which is exactly what I did. Your answer is not justinadequate; it is totally ridiculous! Plus you deleted completely my main objection! Your impliedoffer to respond when we ask about some step that bugs usis clearly insincere. You just go through the motions of flapping your lips and claim you have responded. It's dishonest.>I'm not interested in a lot of time wasted on tangents and wild goose>chases. It appears more like you are not interested in answeringsubstantive objections in a substantive way. What I broughtup is *central* to your argument, not a tangent or a wild goosechase. It is right down the middle. You are evading. Nora B.>James Harris === Subject: : Lp norm exerciseI'm trying to do one exercise of Rudin's Real and Complex Analysischapter on Lp spaces: Show that if 0 < m(E) < infty, 0 < ||f||_infty< infty then a_{n+1}/a_n -> ||f||_infty where a_n = int_E |f|^n dm,with Esubset R^k, m lebesgue measure.One can easily show lim sup a_{n+1}/a_n <= ||f||_infty ... but theother side is giving me some trouble ... Any hints ?nojb. === Subject: : Re: Lp norm exercise>I'm trying to do one exercise of Rudin's Real and Complex Analysis>chapter on Lp spaces: Show that if 0 < m(E) < infty, 0 < ||f||_infty>< infty then a_{n+1}/a_n -> ||f||_infty where a_n = int_E |f|^n dm,>with Esubset R^k, m lebesgue measure.>One can easily show lim sup a_{n+1}/a_n <= ||f||_infty ... but the>other side is giving me some trouble ... Any hints ?Hint: if n is large enough nearly all of a_n comes from the integral over those x for which |f(x)| is very near ||f||_infty.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: : Re: Sum of seriesTrishia Rose> Chintan M. Shah> (SNIP)> I wanted to know how to get the formula for sum of series.> (SNIP) The result can be proves by Mathematical Induction but how to actuallyget> the RHS needed to make use of the proof. (SNIP) For the special case where the exponent is 1, theres a remarkably> elementary demonstration (though not really rigorous) which I'm> amazed was not included in the mathworld link....Here's another bit that MathWorld missed. The polynomials S_p defined byS_p(n)=sum{k=1 to n}k^psatisfy this recursion relation (when we extend them to polynomial functionson R, so that they can be integrated):S_{p+1}(n) = (p+1) int{t=0 to n} S_p(t)dt + knwhere the constant k is such that S_{p+1}(1) = 1. Explicitly,k = 1 - (p+1) int{t=0 to 1} S_p{t}dt.I've never seen exactly this formula in print or on the web.LH === Subject: : Re: Sum of series> Later on he was>correcting his dad's checking and banking accounts at age 12. :*)Any reasonably intelligent 12-year-old would be able to do that.But Gauss did it when he was 3.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: : Re: Sum of series charset=iso-8859-7 Robert Israel > Later on he was>correcting his dad's checking and banking accounts at age 12. :*)> Any reasonably intelligent 12-year-old would be able to do that.Maybe _you_ were able to do that at age 12, who we know were reasonablewith math, but I still can't do it at age 40 :*)> But Gauss did it when he was 3.Yes. Didn't recall the source right. It is indeed 3.By the way, my source indicates sum(i,i=1..100), for the kid Gaussassignment, contrary to what Ignatio seems to quote. I assume his is a morerecent source, so I will go with that, although it seems to me a littleunreasonable for a teacher of elementary school to have given such a task tothe kids.> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2--Ioannis Galidakishttp://users.forthnet.gr/ath/jgal/------------------- -----------------------Eventually, _everything_ is understandable === Subject: : finding the extremity of a function of a matrixAs we know, for a function y = f(x), the extremities can be found byletting dy/dx = 0.Now I have a scalar function, J = f(X), where J is a scalar and X is aNxN matrix, firstly, D = (partial J/partial X(n1,n2)) leads to an NxNmatrix. Secondly, setting D to zero and to find the extremity of the X.Does it make sense?.Lin. S. === Subject: : Re: finding the extremity of a function of a matrixS. Lin grava .88 la saucisse et au marteau:> As we know, for a function y = f(x), the extremities can be found by> letting dy/dx = 0.> Now I have a scalar function, J = f(X), where J is a scalar and X is a> NxN matrix, firstly, D = (partial J/partial X(n1,n2)) leads to an NxN> matrix. Secondly, setting D to zero and to find the extremity of the X.> Does it make sense?Yes, as you simply consider f as a function from R^{N^2} to R andcompute its gradient.-- Nicolas === Subject: : Product of RealsIs it possible to determine what the uncountable product of all realnumbers in the interval [0.5, 1.5] is? Intuitively it seems like 1,but is this concept studied in general anywhere (e.g. the concept ofuncountably infinite products or sums of real numbers)? === Subject: : Re: Product of RealsX-ID: G56AbsZ6ZeqGmu6-y+wCKHfp96ae0Lom12c1ivDURsTvo-3+mCGPZH> Is it possible to determine what the uncountable product of all real> numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1,> but is this concept studied in general anywhere (e.g. the concept of> uncountably infinite products or sums of real numbers)?Lookup the term summable family. This is a concept which allows definitionof convergence for any numbers a_i with i in I where I is an arbitraryindex set. By applying exp-log-formalism (transform your product into asum) you get the analogue theory for products.A necessary condition for such a family to converge is that its support (thelargest set J subset I such that a_j <> 0 forall j in J) is countable -this means that your product cannot converge.Because equally well than your reasoning I could argue that it converges toany other real number <> 1. Cf. Riemann's rearrangement theorem.-- reverse my forename for mail! === Subject: : Re: Product of Reals>Is it possible to determine what the uncountable product of all real>numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1,>but is this concept studied in general anywhere (e.g. the concept of>uncountably infinite products or sums of real numbers)?I could argue that it's going to be zero, if it's defined (which it isn't).Write the above product as:(Product)_{0.5<=x<=1} (x * 1-x)Thus 0.5 times 1.5, 0.7 times 1.3, 0.9 times 1.1, etc. (This takes careof every product except for the middle term one, so multiply the answer byone when you're finished if it makes you feel better). Each term in this product is strictly less than one, and there are an infinite number of them. Hence the product is zero.Doug === Subject: : Re: Product of Reals> Each term in this product is strictly less than one, and there are an > infinite number of them. Hence the product is zero.Actually (1/2)*(3/4)*(7/8)*(15/16)*... > 0. === Subject: : Re: Product of Reals> Is it possible to determine what the uncountable product of all real> numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1,> but is this concept studied in general anywhere (e.g. the concept of> uncountably infinite products or sums of real numbers)?No, no, and no. It's possible to define something that works as the *average* of all the real numbers in the interval, and that's the integral of x over the interval, divided by the length of the interval. That works out, no surprise, to 1. That's average in the sense of arithmetic mean. There's also an average in the sense of geometric mean, which is something like what do want a product to do, and that's e^A, where A is the arithmetic mean of log x. So: exp ( integral from .5 to 1.5 of log x dx). I'll let you work it out.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: : Re: Product of RealsIs it possible to determine what the uncountable product of all real> numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1,> but is this concept studied in general anywhere (e.g. the concept of> uncountably infinite products or sums of real numbers)?No, no, and no. It's possible to define something that works as the *average* of > all the real numbers in the interval, and that's the integral of > x over the interval, divided by the length of the interval. That > works out, no surprise, to 1. That's average in the sense of arithmetic mean. There's also an > average in the sense of geometric mean, which is something like > what do want a product to do, and that's e^A, where A is the > arithmetic mean of log x. So: exp ( integral from .5 to 1.5 of log x dx). I'll let you work it out.What does this number represent? Can you point me to a referencewhich discusses this number? I got that that the above expressionequals 3^(1.5)/(2e) which is approximately equal to .955877. I guessa better question is what does the geometric mean of two numbersrepresent, and then I can perhaps apply this analogy to the aboveexpression. === Subject: : Re: Product of Reals> I guess a better question is what does the geometric mean of two > numbers represent, and then I can perhaps apply this analogy to the > above expression.The geometric mean is to multiplication what the arithmetic mean is to addition. Suppose shares of toolshed.com multiplied by 4 in value during 1999 and by 9 in 2000. The geometric mean of 4 and 9 is 6, and the net effect is the same as if the shares had multiplied in value by 6 each year.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: : Re: Product of RealsNobody grava .88 la saucisse et au marteau:> equals 3^(1.5)/(2e) which is approximately equal to .955877. I guess> a better question is what does the geometric mean of two numbers> represent, and then I can perhaps apply this analogy to the above> expression.The arithmetic mean c of a and b (a<= b) is such that a,c and b are 3consecutive terms of an arithmetic sequence, that is to says c-a = b-cThe geometric mean c of a and b is such that a,c and b are 3 consecutiveterms of a geometric sequence, that is to say c/a = b/c (if a<>0).hth-- Nicolas === Subject: : Re: Product of Reals>Is it possible to determine what the uncountable product of all real>numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1,>but is this concept studied in general anywhere (e.g. the concept of>uncountably infinite products or sums of real numbers)?> No, such a product is not well-defined.Note that we can convert this to a question of sums by taking logarithms. First (in contradistinctiion to this example) consider a sum of uncountably many positive terms. Then, for some positive integer k, uncountably many of these terms which exceed 1/k. Hence, the sum is infinite.In the case at hand, after taking logarithms, there are uncountable many negative terms and uncountaby many infinite terms. Hence, you are trying to compute the difference infinity - infinity.And why would you think this product is 1 in the first place?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: : Re: Product of RealsIs it possible to determine what the uncountable product of all real>numbers in the interval [0.5, 1.5] is? Intuitively it seems like 1,>but is this concept studied in general anywhere (e.g. the concept of>uncountably infinite products or sums of real numbers)?> No, such a product is not well-defined.Note that we can convert this to a question of sums by taking > logarithms. First (in contradistinctiion to this example) consider a > sum of uncountably many positive terms. Then, for some positive integer > k, uncountably many of these terms which exceed 1/k. Hence, the sum > is infinite.In the case at hand, after taking logarithms, there are uncountable many > negative terms and uncountaby many infinite terms. Hence, you are > trying to compute the difference infinity - infinity.And why would you think this product is 1 in the first place?I guess I was thinking of an average, which is of course a sum not aproduct. But how about this (which I'm sure is still wrong). What ifyou consider the infinite product of the numbers in the intervalA=[2/3,3/2]. Then is this product 1? I would attempt to prove thisas follows:Let x be in A. If x is 1 then it contributes nothing to the product,otherwise assume 2/3 <= x < 1. Then 1 < 1/x <= 3/2. So 1/x is in A. Likewise if 1 < x <= 3/2, then 1/x is in A. So for any number x in A,1/x is in A. Hence the product of all of them is 1.Or is there some problem with this? === Subject: : Re: Product of Reals>I guess I was thinking of an average, which is of course a sum not a>product. But how about this (which I'm sure is still wrong). What if>you consider the infinite product of the numbers in the interval>A=[2/3,3/2]. Then is this product 1? I would attempt to prove this>as follows:>Let x be in A. If x is 1 then it contributes nothing to the product,>otherwise assume 2/3 <= x < 1. Then 1 < 1/x <= 3/2. So 1/x is in A. >Likewise if 1 < x <= 3/2, then 1/x is in A. So for any number x in A,>1/x is in A. Hence the product of all of them is 1.Assume for simpleness that we are only concerned with rationals andthat the infinite sum of rationals on a closed interval iswell-defined. They are countable, so we can order them into a productwhere the product of two consecutive rationals equals 1. Then, takingthe logarithm of this product gives an alternating series whose termsapproach zero, so the series converges conditionally (to 0, ofcourse).However, if we change the ordering of the series and go back to ouroriginal product we can obtain a different value for the product. Butwe know that multiplication of rationals is supposed to commutate, sowe have run into a contradiction.I'm not sure either reasoning even works for reals. But you can seehow it's easy to reach very confusing results by assuming such aproduct is well-defined.>Or is there some problem with this?No problem that doesn't already exist with infinite sums. === Subject: : help me proofI'm a student in the mathematics department of College of NaturalSciences, Vietnam National University, Ho Chi Minh CityI have a problem with my homework Can professor help me proof ?: Assume E ={ (a1 ,a2 ,....,an , ... ) ; a1 ,a2 ,....,an , ... arecomplex numbers}Define vector addition and scalar multiplication on E such that every x,y in E ,every p is complex number x+y=(x1 +y1 ,x2 +y2 ,....,xn +yn , ... ) p .x= (px1 ,px2 ,....,pxn , ... ) such that E is a vector space.Proof no exist a norm on E such that if every sequence x(n) ofelements in E converge to 0 with this norm then limn->infinite x_m(n)=0with m=1,2,3......... thank professor very much ,professor can send to me by emailmy email : nguyencaotrongduy2@yahoo.com === Subject: : Re: Parabolic, Elliptical, and Hyperbolic Geometry.> How did these three geometries get their names? What does it have do to> with the conics by the same names?Not much. Elliptic means not enough, hyperbolic means too much. Elliptic geometry doesn't have enough parallel lines, hyperbolic has too many. Parabolic geometry is new to me.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: : Re: Parabolic, Elliptical, and Hyperbolic Geometry.>> How did these three geometries get their names? What does it have do to>> with the conics by the same names?>Not much. Elliptic means not enough, hyperbolic means too much. >Elliptic geometry doesn't have enough parallel lines, hyperbolic >has too many. >Parabolic geometry is new to me.According to http://members.aol.com/jeff570/mathword.html :The terms HYPERBOLIC GEOMETRY, ELLIPTIC GEOMETRY, and PARABOLICGEOMETRY were introduced by Felix Klein (1849-1925) in 1871 in .86berdie sogenannte Nicht-Euklidische Geometrie (On so-callednon-Euclidean geometry), reprinted in his Gesammelte mathematischeAbhandlungen I (1921) p. 246 (Ken Pledger and Smart, p. 301). I have not read that reference, but a likely reason for theterminology is that the 2nd-order local approximation of a surface inR^3 with positive, negative, or zero Gaussian curvature is anellipsoid, hyperboloid, or paraboloid (or plane), respectively.John Mitchell === Subject: : Re: Parabolic, Elliptical, and Hyperbolic Geometry.>Not much. Elliptic means not enough, hyperbolic means too much. >Parabolic geometry is new to me.That must be the one Goldilocks sat in. === Subject: : Re: Homological algebra> Can someone suggest a good book on Homological algebra??I'm not sure where you are starting from, but I have been> reading Hilton and Wu's _A_Course_in_Modern_Algebra_. The> final chapter introduces homological techniques (Ext and Tor).> This book has much to recommend it, but for now let me just> say that it is a reasonably thin book that starts out nice> and slow--groups--and concentrates basically on> concepts which will lead to homological techniques, without> much else. For example, there is a chapter on category theory> including a discussion of abelian categories and adjoint functors.> Projective and injective modules are discussed at length.The major problem with this book is that it is currently being> offered by in the Wiley Classic's Library at approximately $100,> which I suspect gives them a healthy profit margin. However, copies> are available on line.Best wishes,> MikeHomological Algebra and that it was a very clear exposition. I didn'tget very far into it, but that was not the fault of the book. Algebraic topology teachers are usually pretty clear that thehomological machinery is slick, gets the results out, and is not atall intuitive. It takes time and searching (I have been told) tounderstand the history and intuition behind the modern developments.Achava === Subject: : Re: No Perfect Cuboid Exists: - Proof> See also - http://www.bearnol.pwp.blueyonder.co.uk> ****************************************a^2+b^2=d^2> b^2+c^2=e^2> c^2+a^2=f^2> a^2+b^2+c^2=g^2Smallest solution hcf(a,b,c)=1a^2==b^2==c^2==0,1 [mod 4]therefore a^2+b^2+c^2==0 [mod 4]> => a^2==b^2==c^2==0 [mod 4]> => a==b==c==0 [mod 2]> => hcf(a,b,c)>=2> therefore g==1 [mod 2] (1)but hcf(a,b,c)=1> => only one (max) even> therefore all a,b,c odd [from (1)]therefore all d,e,f even> => d^2==e^2==f^2==0 [mod 4]> => d^2+e^2+f^2==0 [mod 4]but d^2+e^2+f^2=2.g^2> and g^2==1 [mod 4]> => d^2+e^2+f^2==2 [mod 4]=> contradiction, from which result follows*****************************************People a lot brighter than you and I have been working on this problem for decades with no success. Do you really believe that there is any chance that there is a proof this short and using nothing beyond the 1st week of a number theory course and that all those bright people have missed it? You are wasting your time.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: : Re: Making Star Trek Real> sarfatti@pacbell.net says...> Note some minor typographical corrections to the macro-quantum > geometrodynamical equations for the metric engineering of Star Gate Time > Travel Machines and Weightless FTL Warp Drives that underlie the physics > of UFO super-technology that I was put to work on in 1954 at the age of > 14 in a USG Project headed by Eugene Mc Dermott a co-founder of Texas > Instruments and a high level Defense Intelligence Honcho since WWII.Can you say that in one breath?Sarfatti has yet to learn that merely writing an equation (we won'targue the correctness of the equation, we'll assume they are correct)gives absolutely no clue as to how one goes about building devicesthat engineer the metric to conform to those equations.Check it out. Section 19.3 in MTW's GRAVITATION describes Mass andAngular momentum of a fully relativistic source. Even when all thenormal conservation laws are assumed to apply (and we don't know thisfor sure since matter doesn't occur at such densities and velocitiessimultaneously (fast moving things tend to fly apart)) there are someinteresting frame dragging properties of fact moving masses - asvonStockum showed. But where the hell do you get an infinitely longmass that's several hundreds of kilometers in diameter (think of areally big and dense string of spaghettie) whose density approachesthat of a neutron star and which is spinning at half light-speed? *That's* de minimus in providing a means to change the metric ininteresting ways. Assuming of course there aren't changes that mustbe made when and if we actually go to engineer the metric. That is,without a priori knowledge its not certain things will go our way. Sure, fully relativistic masses seem to have some interestingproperties, now. Enthusiasts like Sarfatti would have us believe thatour ignorance means there may be even more interesting things possiblethan anyone believes. Unfortunately, in the absence of sureknowledge, the universe may end up being less interesting than anyonebelieves as well. Which is just another way of saying, that despitepossibilities that may occur on the hairy edge of reality - we justdon't know what's possible. This includes the possibility that verylittle in the way of Star Trek Engineering is possible, just as verylittle in the way of Witchcraft is possible.Making our future in space a reality involves making what is knowngenrally available to the business world in an environment wheremissile and nuclear proliferation are non-issues. This is a tallorder. But one that may be brought about by a successful conclusionof our war on Terror. Given that, we also need to create a businessand economic environment more favorable to business development. Thisinvolves allowing businesses and people to own assets and resourcesoff-world, and to develop those assets without undue interference fromterrestrial governments and without undue taxation.Given these preconditions we can clearly show that investment in spacepropulsion will decrease the cost of momentum - and as this costdecreases we will see the following occur; (1) suborbital ballistic high-value package and personell delivery (2) orbiting satellites (comsats, spysats, navsats, weathersats, etc) (3) large orbiting satellites (powersats) (4) very large interplanetary spacecraft (asteroid capture,factorysats)then, the same development arc at far lower prices (5) suborbital ballistic low-value package and personnel delivery (6) orbiting satellites (space homes) (7) large orbiting satellites (full integration of Earth/Earth orbit) (8) very large interplanetary spaceraft (mobile spacehomes)then, flight to the stars at about 1/3 light speed; (9) interstellar space homesthen, once a number of colonies are set up surrounding sol, we canimagine research along the lines proposed by Sarfatti - basically,we'll collide shaped pieces of Iron-56 at 1/3 light speed - in aneffort to find out more about the metric, and see how we may engineerit to do some interesting things. Things like time travel, tappingzero point energy, and gravity drives and so forth.Once you have these things, even given the limitations of existingtime machine designs (you cannot visit a time before the first machineit built, or after the last machine ceases to function) its possibleto create an effective superluminal travel (assuming all the physicswe think we know remains unchanged under extreme conditions) - here'show that works; (a) Before you leave make a time violating region that lets youtravel to any point in time the region exists - do this by establishing aring of tiny black holes, preferrably by tapping the zero point energyof the vacuum in some way. (b) Then, use your gravity drive, and zero point energy generator tofly off at very high accelerations to a distant locale, say 10,000 lightyears away. (c) Very little shiptime later, but 10,000+ years later, arrive atyour destination. Explore, and so forth. Then return. (d) Arrive back at your starting point 20,000+ years later, whileonly a few days pass on the ship due to very high speeds most of theway. (e) Use the time machine to travel back in time to the point justafter your depe. For everyone's sanity exit the causalityviolating region so that the local time is synchronized with your shiptime. (You may send radio telescope messages back and forth this wayas well, to that everything is synchronized between the two frames. Thereply messages from the moving ship are picked up by the time machinein the future and routed, like internet packets, to the right timeexit, based on time stamps. Thus establishing instantaneous messaging)This is very much like Star Trek - but has nothing to do with the'science' of Star Trek. Anymore than the flying in an airplane has todo with the 'science' of flying by Witchcraft. They're both talkingabout flight - but one achieves it while the other does not. === Subject: : Re: Making Star Trek RealIf someone was dieing and wanted to see somewhere we can show them byusing the holodeck. If we can travel to other worlds will the warbetween human will it stop? === Subject: : Re: WMD in Iraq , UFO & Doomsday Fact or Fiction?http://www.amazon.com/gp/reader/0471265179/ref=sib_dp_ pt/102-9493486-1170531#reader-linkCheck out the reality of our foreign policy before making such insaneand foolish statements as those made here.The link above is to ALL THE SHAH'S MEN: An American Coup and theRoots of Middle East Terror. === Subject: : EvoGraph: a package for graph synthesis by genetic programmingX-AUTHid: hujianjuDear all:I am glad to announce that a package for evolving generic graphs andnetworks using genetic programming is available for download. http://www.egr.msu.edu/~hujianju/evograph/evograph.htmEvoGraph is a package for graph synthesis by genetic programming. It isdeveloped based on strongly-typed lilgp, originally developed at GARAGe, MSUand later patched by Sean Luke with the strongly-typed feature. Basically,EvoGraph provides a set of GP functions and terminals along with a C++ graphlibrary for evolving arbitrary graphs. The same technique is widely used indevelopmental genetic programming for evolving electric circuits, bondgraphs, neural networks.A benchmark problem: the wireless access point configuration problem isintroduced in this package. This problem requires simultaneous search ofboth topology and parameters, the same as in circuit synthesis. And therecould be a series of variations of this problem and the problem is alsoscalable in terms of problem size.This package is still in its starting stage and there is no complete or verydetailed documents. But it is easy to use and understand after readingrelated reference papers.Jianjun Hu (George)Genetic Algorithm Research & Application Group (GARAGe)Department of Computer Science & EngineeringMichigan State University hujianju@msu.eduWeb: www.egr.msu.edu/~hujianjuPhone: 517-355-3796(o) === Subject: : Re: Cardinal Arithmetic> Let k be a cardinal number. Let 0 = null and 1 = {0}.> Show that k^0 = 1 and k^1 = k for all k.There is only one function with domain of empty set andthat is the empty function, which is the same as theempty set. Thus k^0 = 1.Let K be a set with cardinality k. Thus k^1 = |{ {(0,x)} | x in K }|and { {(0,x)} | x in K }bijects with K by {(0,x)} -> x === Subject: : Re: Leibniz & Newton> Also, the dy/dx notation is a little misleading in that it makes you> think Fraction!> Actually, it's the explications of it, that are.> It's not difficult at all to come up with an explication that matches> what Leibnitz intended [... followed by an explication...]>Hi Alfred!> for your very detailed post. May I use your examples in my>classroom? I'm very impressed with your ability.Just to a web search on infinitesimals. It's all over theplace. === Subject: : Re: Leibniz & Newton>There is also Chandrasekhar's book Newton's _Principia_ for the>Common Reader, Oxford University Press, in which the most important>parts of Newton's great book are rewritten using modern notation and>terminology.I actually did that with Maxwell's treatise, rewriting a large chunkof it in the boldface, 3-D vector notation. Entire sections meltedaway into nothingness, with about a 50% size reduction and proportionateincrease in transparency. === Subject: : Re: Leibniz & Newton charset=iso-8859-1> It's not difficult at all to come up with an explication that matches> what Leibnitz intended and is perfectly consistent (and no, I'm NOT> talking about Non-Standard analysis here).Consider the algebra given by:> A = (x + d y: x, y in R; d^2 = 0 },If d^2 = 0, and the analysis is *not* non-standard, then how can d differ from 0? === Subject: : Re: Leibniz & Newton>> and is perfectly consistent (and no, I'm NOT talking about Non-Standard>> analysis here).>If d^2 = 0, and the analysis is *not* non-standard, ... which has nothing to do with anything that was said that you'rereplying to...http://search.yahoo.com/search?p=How To Search The Webhttp://search.yahoo.com/search?p=Non-Standard Analysishttp://search.yahoo.com/search?p=Wikipedia>then how can d differ from 0?*Sigh.*http://search.yahoo.com/search?p=Associative Linear Algebrashttp://search.yahoo.com/search?p=Vector Spaceshttp://search.yahoo.com/search?p=Nilpotent Algebrashttp://search.yahoo.com/search?p=Finitely Presented Algebrashttp://search.yahoo.com/search?p=Ideals+ Infinitesimalshttp://search.yahoo.com/search?p=Hypercomplex Numbers+Infinitesimals === Subject: : Re: Leibniz & Newton for your help, Mark. :)Kavon>> and is perfectly consistent (and no, I'm NOT talking about Non-Standard>> analysis here).>If d^2 = 0, and the analysis is *not* non-standard,> ... which has nothing to do with anything that was said that you're> replying to...> http://search.yahoo.com/search?p=How To Search The Web> http://search.yahoo.com/search?p=Non-Standard Analysis> http://search.yahoo.com/search?p=Wikipedia>then how can d differ from 0?> *Sigh.*> http://search.yahoo.com/search?p=Associative Linear Algebras> http://search.yahoo.com/search?p=Vector Spaces> http://search.yahoo.com/search?p=Nilpotent Algebras> http://search.yahoo.com/search?p=Finitely Presented Algebras> http://search.yahoo.com/search?p=Ideals+Infinitesimals> http://search.yahoo.com/search?p=Hypercomplex Numbers+Infinitesimals === Subject: : Re: Leibniz & Newton charset=Windows-1252I second that..r.e.s.> for your help, Mark. :)> Kavon>> and is perfectly consistent (and no, I'm NOT talking aboutNon-Standard>> analysis here).>If d^2 = 0, and the analysis is *not* non-standard, ... which has nothing to do with anything that was said that you're> replying to... http://search.yahoo.com/search?p=How To Search The Web> http://search.yahoo.com/search?p=Non-Standard Analysis> http://search.yahoo.com/search?p=Wikipediathen how can d differ from 0? *Sigh.* http://search.yahoo.com/search?p=Associative Linear Algebras> http://search.yahoo.com/search?p=Vector Spaces> http://search.yahoo.com/search?p=Nilpotent Algebras> http://search.yahoo.com/search?p=Finitely Presented Algebras> http://search.yahoo.com/search?p=Ideals+Infinitesimals> http://search.yahoo.com/search?p=Hypercomplex Numbers+Infinitesimals === Subject: : How to prove that sqrt(2) exist?I.e., how to prove that there exists a real number xsuch that x^2 = 2?I'm getting started with the rigourous side of maths(as a hobby -- being an engineer, with several yearsof experience, you understand that I haven't reallyneeded it), and really enjoyed reading a proof thatsqrt(2) can not be a rational number.However, the justification they give is that fromPythagoras' theorem, looking at a triangle withsides 1 and 1, then the hypothenuse is such thatits square is 2. I don't find that convincing atall (I mean, sure, intuitively yes -- I wouldn'teven need that evidence to convince me -- I havealways accepted that sqrt(2) exists, just because,well, it has to be there, somewhere between 1.4and 1.5 ... You know... engineers!! :-))But anyway, they don't really prove that sqrt(2) isa real number: they prove that there is no rationalnumber p/q such that (p/q)^2 is 2.So, I'm curious: how can one prove that sqrt(2)exists and it is a real number?!-- === Subject: : Re: How to prove that sqrt(2) exist?> I.e., how to prove that there exists a real number x> such that x^2 = 2?> I'm getting started with the rigourous side of maths> (as a hobby -- being an engineer, with several years> of experience, you understand that I haven't really> needed it), and really enjoyed reading a proof that> sqrt(2) can not be a rational number.> However, the justification they give is that from> Pythagoras' theorem, looking at a triangle with> sides 1 and 1, then the hypothenuse is such that> its square is 2. I don't find that convincing at> all ...If we allow that real numbers are in one-to-onecorrespondence with points on the x-axis, thenwe can mark off sqrt(2) on this real number line,using compass and straightedge. Voila. There it is.Or do you think this line has holes or gaps? === Subject: : Re: How to prove that sqrt(2) exist?>I.e., how to prove that there exists a real number x>such that x^2 = 2?>So, I'm curious: how can one prove that sqrt(2)>exists and it is a real number?[The first phrasing is better. Whatever existence means in a mathematical discussion, it isn't really productive to say sqrt(2) existsand then try to prove it's real; for example, in the set of integersmodulo 7, sqrt(2) certainly exists -- it's equal to 3 -- but Iwouldn't say it's a real number...]You've already gotten some fine answers. Just to add another one:there are several models you can use for what real numbers _are_,but the key property which all of them share, which distinguishesthe real numbers from the rational numbers, is this: every nonemptyset of real numbers which has any upper bound at all will have asmallest upper bound. (It's reasonable to be skeptical of thisassertion but it can be proved once you do provide a model ofwhat the real numbers are. Or, you can simply take it as an axiomand see what its consequences are.)Once you buy in to this axiom, you can easily prove that there existreal solutions to all kinds of problems. In the case of the squareroot of 2, you look at the set S of all real numbers x with x^2 <= 2.It's certainly not empty (e.g. x=0) and it's easily seen to haveupper bounds (e.g. whenever x^2 <= 2, we certainly have x < 2; x=3/2also serves as an upper bound for S, as does x= 17/12, etc.)So by the axiom, there is a _least_ upper bound L. The fact that it'san upper bound turns out to be enough to show that L^2 <=2 ; the factthat it's smaller than all the other upper bounds is enough to proveL^2 >=2 ; so in fact L^2 equals 2 on the nose.This same example shows that the set of rational numbers does NOTenjoy the least-upper-bound property. The set S has many _rational_upper bounds, as noted above. But there is no smallest one (for everyrational upper bound there's another smaller rational number whichis also an upper bound for S).dave === Subject: : Re: How to prove that sqrt(2) exist?In sci.math, Moreno:I.e., how to prove that there exists a real number x> such that x^2 = 2?I'm getting started with the rigourous side of maths> (as a hobby -- being an engineer, with several years> of experience, you understand that I haven't really> needed it), and really enjoyed reading a proof that> sqrt(2) can not be a rational number.However, the justification they give is that from> Pythagoras' theorem, looking at a triangle with> sides 1 and 1, then the hypothenuse is such that> its square is 2. I don't find that convincing at> all (I mean, sure, intuitively yes -- I wouldn't> even need that evidence to convince me -- I have> always accepted that sqrt(2) exists, just because,> well, it has to be there, somewhere between 1.4> and 1.5 ... You know... engineers!! :-))But anyway, they don't really prove that sqrt(2) is> a real number: they prove that there is no rational> number p/q such that (p/q)^2 is 2.So, I'm curious: how can one prove that sqrt(2)> exists and it is a real number?!> --> I'm not sure exists works in this context, as no numbersexist physically, be they the natural number 1, therational number 8/17, the radical sqrt(2), transcendentalpi, or the complex number 3 + 4i.However, presumably what Dedekind did (I'd have to look)is assume that one defines an entity using two sets over Q(the lower set having no greatest member), and show thatthese setpairs could be manipulated in straightforwardways to define addition, subtraction, multiplication, anddivision, over the real numbers (which contain, of course,the rational numbers as a subfield).In the case of sqrt(2), one simply uses S1={x in Q:x^2<2}and S2={x in Q:x^2>=2}. It turns out S2 has no leastmember, either, in this case.Addition is easy. Subtraction may require additionof the negative. Multiplication requires some care,because of silly behavior around 0. Division may requiremultiplication of the reciprocal.I see a potential problem with Dedekind cuts, if onlybecause one has to define the sets carefully lestone inadvertantly. In fact, you can probably see theerror I've made already; a more proper definition mightbeS1={x in Q: x < 0} union {x in Q: x >= 0 and x^2 < 2}S2={x in Q: x >= 0 and x^2 >= 2}which makes S1 into the desired form, as opposed to anopen interval centered around 0.A Cauchy sequence as suggested by Mathworld.wolfram.comcan also be used. Basically, it's a sequence (in thiscase, over Q) such thatlim (min(m,n)->+oo) d(a_m,a_n) = 0where d(a,b) is a metric over Q (the typical one is abs(a-b)).Or one can go the slightly silly route: sqrt(2) exists because(sqrt(2)^2) = 2; this treats the problem as a rather abstractbut somewhat useful formalism. For example, (sqrt(2) - 1)^2= sqrt(2)^2 - 2 * sqrt(2) * 1 + 1 = 3 - 2 * sqrt(2).Or one goes the engineering route, which you've stated youdon't like: sqrt(2) = 1.4142; pi = 3.1416; e=2.71828.http://mathworld.wolfram.com/DedekindCut.htmlhttp:// mathworld.wolfram.com/CauchySequence.htmlhttp:// mathworld.wolfram.com/Metric.html-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: : Re: How to prove that sqrt(2) exist?> In sci.math, Moreno> :I.e., how to prove that there exists a real number x> such that x^2 = 2?I'm getting started with the rigourous side of maths> (as a hobby -- being an engineer, with several years> of experience, you understand that I haven't really> needed it), and really enjoyed reading a proof that> sqrt(2) can not be a rational number.However, the justification they give is that from> Pythagoras' theorem, looking at a triangle with> sides 1 and 1, then the hypothenuse is such that> its square is 2. I don't find that convincing at> all (I mean, sure, intuitively yes -- I wouldn't> even need that evidence to convince me -- I have> always accepted that sqrt(2) exists, just because,> well, it has to be there, somewhere between 1.4> and 1.5 ... You know... engineers!! :-))But anyway, they don't really prove that sqrt(2) is> a real number: they prove that there is no rational> number p/q such that (p/q)^2 is 2.So, I'm curious: how can one prove that sqrt(2)> exists and it is a real number?!> --> I'm not sure exists works in this context, as no numbers> exist physically, be they the natural number 1, the> rational number 8/17, the radical sqrt(2), transcendental> pi, or the complex number 3 + 4i. All, surreal numbers exist physically, since it's well-known and well-documented that they were invented by idiots. You are right about 1 though, since nobody has yet proven than it exists at all.However, presumably what Dedekind did (I'd have to look)> is assume that one defines an entity using two sets over Q> (the lower set having no greatest member), and show that> these setpairs could be manipulated in straightforward> ways to define addition, subtraction, multiplication, and> division, over the real numbers (which contain, of course,> the rational numbers as a subfield).In the case of sqrt(2), one simply uses S1={x in Q:x^2<2}> and S2={x in Q:x^2>=2}. It turns out S2 has no least> member, either, in this case.Addition is easy. Subtraction may require addition> of the negative. Multiplication requires some care,> because of silly behavior around 0. Division may require> multiplication of the reciprocal.I see a potential problem with Dedekind cuts, if only> because one has to define the sets carefully lest> one inadvertantly. In fact, you can probably see the> error I've made already; a more proper definition might> beS1={x in Q: x < 0} union {x in Q: x >= 0 and x^2 < 2}> S2={x in Q: x >= 0 and x^2 >= 2}which makes S1 into the desired form, as opposed to an> open interval centered around 0.A Cauchy sequence as suggested by Mathworld.wolfram.com> can also be used. Basically, it's a sequence (in this> case, over Q) such thatlim (min(m,n)->+oo) d(a_m,a_n) = 0where d(a,b) is a metric over Q (the typical one is abs(a-b)).Or one can go the slightly silly route: sqrt(2) exists because> (sqrt(2)^2) = 2; this treats the problem as a rather abstract> but somewhat useful formalism. For example, (sqrt(2) - 1)^2> = sqrt(2)^2 - 2 * sqrt(2) * 1 + 1 = 3 - 2 * sqrt(2).Or one goes the engineering route, which you've stated you> don't like: sqrt(2) = 1.4142; pi = 3.1416; e=2.71828. You're right, only mathematicans go the engineering route. Engineers go the route: sqrt(2)=5 - $3.4858, for every time you ask another stupid question.http://mathworld.wolfram.com/DedekindCut.html> http://mathworld.wolfram.com/CauchySequence.html> http://mathworld.wolfram.com/Metric.html === Subject: : Re: How to prove that sqrt(2) exist?> Or one can go the slightly silly route: sqrt(2) exists because> (sqrt(2)^2) = 2You can always define it that way (for instance, the wayyou define the imaginary unit). But that does not mean thatthat number you just defined is in the set of real numbers.In particular, I could arbitrarily come up with such definitionfor the rational numbers: I could simply say: sqrt(2) is therational number X such that X^2 is 2. I could claim that thenumber exists because I defined it, and that means that itexists. You can, however, prove that if such number exists,it is not a rational number (and thus, not an integer or naturalnumber either).The question remains -- when trying to be rigourous, at leastthe way I'm understanding it, as a beginner in the rigourousside of mathematics, you ask yourself the question: how do Ireally know that there is such number? How do I know that itwill not be the case that whatever number I can come up with,its square will be either greater than 2, or less than 2?> Or one goes the engineering route, which you've stated you> don't like: sqrt(2) = 1.4142; pi = 3.1416; e=2.71828.Well, yes. There's the practical side to it, which is whatI have all my life lived with -- it's not like now I want tobecome an actual mathematician; but I'm enjoying that sideof maths, and am trying to adapt my mind to work in theskeptical mode.-- === Subject: : Re: How to prove that sqrt(2) exist?>I.e., how to prove that there exists a real number x>such that x^2 = 2?>I'm getting started with the rigourous side of maths>(as a hobby -- being an engineer, with several years>of experience, you understand that I haven't really>needed it), and really enjoyed reading a proof that>sqrt(2) can not be a rational number.>However, the justification they give is that from>Pythagoras' theorem, looking at a triangle with>sides 1 and 1, then the hypothenuse is such that>its square is 2. I don't find that convincing at>all The fact that you don't find that convincing is agood thing, from a purely mathematical point of view.There are various ways to show that a real numberexists - the simplest is first to give a rigorousdefinition of continuous function, prove thetheorem that if a < b, f is continuous, f(a) < 0and f(b) > 0 then f(x) = 0 for some x betweena and b, and then apply this theorem withf(x) = x^2 - 2, a = 0, b = 2.There are a lot of details there - you can findproofs in many calculus books (especiallyold calculus books - they used to include alot more proofs). Take the calculus book youused in school and look for the IntermediateValue Theorem...> (I mean, sure, intuitively yes -- I wouldn't>even need that evidence to convince me -- I have>always accepted that sqrt(2) exists, just because,>well, it has to be there, somewhere between 1.4>and 1.5 ... You know... engineers!! :-))>But anyway, they don't really prove that sqrt(2) is>a real number: they prove that there is no rational>number p/q such that (p/q)^2 is 2.>So, I'm curious: how can one prove that sqrt(2)>exists and it is a real number?>!>************************ === Subject: : Re: How to prove that sqrt(2) exist?> The fact that you don't find that convincing is a> good thing, from a purely mathematical point of view.:-)> There are various ways to show that a real number> exists - the simplest is first to give a rigorous> definition of continuous function, prove the> theorem that if a < b, f is continuous, f(a) < 0> and f(b) > 0 then f(x) = 0 for some x between> a and b, and then apply this theorem with> f(x) = x^2 - 2, a = 0, b = 2.I actually thought about this (not in this muchdetail, but what I mean is that I thought of thecontinuity issue, but then I kind of dismissed it,since I thought the very notion of continuityrelies on the existence of numbers... So, thefunction has a chance at being continuous becausethe values exist -- how could I use that to provethat a value exists such that f(x) = 0?What I find curious is that this continuity issueseems to rely on the fact that between any twodistinct real numbers, there is at least one otherreal number -- but this fact is not a sufficientcondition for the existence of such number forwhich f(x) = 0: it is also true for rationalnumbers that between any two distinct rationalnumbers, there is at least one other rationalnumber.(at this point, I'm guessing my doubt becomessort of philosophical? Something akin to howdo we prove that 0 exists? or how do weprove that 1 is greater than 0?, or that sortof thing?)!-- === Subject: : Re: How to prove that sqrt(2) exist?>> The fact that you don't find that convincing is a>> good thing, from a purely mathematical point of view.>:-)>> There are various ways to show that a real number>> exists - the simplest is first to give a rigorous>> definition of continuous function, prove the>> theorem that if a < b, f is continuous, f(a) < 0>> and f(b) > 0 then f(x) = 0 for some x between>> a and b, and then apply this theorem with>> f(x) = x^2 - 2, a = 0, b = 2.>I actually thought about this (not in this much>detail, but what I mean is that I thought of the>continuity issue, but then I kind of dismissed it,>since I thought the very notion of continuity>relies on the existence of numbers... So, the>function has a chance at being continuous because>the values exist -- how could I use that to prove>that a value exists such that f(x) = 0?>What I find curious is that this continuity issue>seems to rely on the fact that between any two>distinct real numbers, there is at least one other>real number -- but this fact is not a sufficient>condition for the existence of such number for>which f(x) = 0: it is also true for rational>numbers that between any two distinct rational>numbers, there is at least one other rational>number.It's a long story. Start with that calculus book,as I suggested.>(at this point, I'm guessing my doubt becomes>sort of philosophical? Something akin to how>do we prove that 0 exists? or how do we>prove that 1 is greater than 0?, or that sort>of thing?)>!>************************ === Subject: : Re: How to prove that sqrt(2) exist?> I.e., how to prove that there exists a real number x> such that x^2 = 2?I'm getting started with the rigourous side of maths> (as a hobby -- being an engineer, with several years> of experience, you understand that I haven't really> needed it), and really enjoyed reading a proof that> sqrt(2) can not be a rational number.However, the justification they give is that from> Pythagoras' theorem, looking at a triangle with> sides 1 and 1, then the hypothenuse is such that> its square is 2. I don't find that convincing at> all (I mean, sure, intuitively yes -- I wouldn't> even need that evidence to convince me -- I have> always accepted that sqrt(2) exists, just because,> well, it has to be there, somewhere between 1.4> and 1.5 ... You know... engineers!! :-))But anyway, they don't really prove that sqrt(2) is> a real number: they prove that there is no rational> number p/q such that (p/q)^2 is 2.So, I'm curious: how can one prove that sqrt(2)> exists and it is a real number?!> --> The set of rationals q, such that q^2 > 2, defines a Dedekind cut, x, whose square is neither less than nor greater than 2. === Subject: : Re: How to prove that sqrt(2) exist?>I.e., how to prove that there exists a real number x>such that x^2 = 2?...>So, I'm curious: how can one prove that sqrt(2)>exists and it is a real number?It depends on which flavour of definition of real number you'reusing. Using Dedekind cuts it's fairly simple: you just split the rationals into {x: x <= 0 or x^2 < 2} and {x: x >= 0 and x^2 > 2}.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: : Re: Injective Mapping of Finite Set into Itself|Say I have a finite set S, and a mapping f: S -> S, which I know is|injective. How can I prove that f is surjective?By induction on the number of elements of f. If S is empty,then f is surjective because there are no elements thatneed to be mapped to. If S has n+1 elements, and theclaim is true for all sets with n elements, then let x bean element of S and consider the function g(y) = f(x) if f(y)=x and g(y) = f(y) if f(y)<>x.The function g is from S-{x} to S-{x}: if f(y)=x then bythe injectivity of f, one can't have f(x)=x also. Also gis injective, as g(y1) and g(y2) for y1<>y2 are eitherf(y1)<>f(y2) or f(x)<>f(y2) or f(y1)<>f(x); the casef(y1)=f(y2)=x does not occur because f is injective.So by induction g is surjective. This implies thatthe image of f on S contains S-{x}. If for some yin S-{x}, f(y)=x, then the image of f also containsx, and we are done. If for no y in S-{x} does f(y)=x,then g(y)=f(y) for all y in S-{x}, and f maps S-{x}surjectively to S-{x}. If so, then f(x) can't be anelement of S-{x}, because each of them is animage of an element of S-{x} under f, and f isinjective. So f(x)=x and we are done.Keith Ramsay === Subject: : Campbell's theorem for filtered point processesI am looking for an english reference for the Campbell theorem forstationary filtered point processes (not only Poisson). I have only areference inKoenig/Schmidt: Einfuehrung in Punktprozesse,have a reference which is easily accessible everywhere.-- Karl BreitungRemove nospam to get my reply address === Subject: : Re: Replacing sum of gaussians by a single one?Given a function f(x) constructed by a linear combination of two > gaussians:> f(x) = a*exp(-(x-m1)^2/(2*s1*s1)) + b*exp(-(x-m1)^2/(2*s1*s1))I assume that should be f(x) = a*exp(-(x-m1)^2/(2*s1*s1)) + b*exp(-(x-m2)^2/(2*s2*s2))> where m denotes mean and s denotes standard deviation. If the two > gaussians are close enough, such that replacing their superposition by> a single gaussian gives tolerable error, what are the optimal parameters> (c,m3,s3) for minimizing the error?> Formally:f^(x) = c*exp(-(x-m3)^2/(2*s3*s3))> Error = integral { (f^(x)-f(x))^2 dx} ,x from -inf to +inf.After spending a few hours, I computed the error analytically and then> tried to make its partial derivatives (with respect to c,m3 and s3) > equal to zero. This resulted in a very complicated non-linear equation> that I couldn't solve. Do you know any solution for (c,m3,s3) ? If> the optimal solution is too complicated, even a good sub-optimal is> helpful.The simplest approach to approximating a mixture of variables by asingle variable is to give the approximating variable the correctmean and variance. Rewrite the true density asf(x) = ((p1/s1)*exp(-(x-m1)^2/(2*s1*s1))+ (p2/s2)*exp(-(x-m2)^2/(2*s2*s2)))/sqrt(2*Pi),where p1 + p2 = 1. The mean of the mixture is m3 = p1*m1 + p2*m2.The variance is s3^2 = p1*(s1^2 + m1^2) + p2*(s2^2 + m2^2) - m3^2.The normal density with the same mean and variance isg(x) = exp(-(x-m3)^2/(2*s3*s3))/(s3*sqrt(2*Pi)). === Subject: : Re: Replacing sum of gaussians by a single one?> Given a function f(x) constructed by a linear combination of two > gaussians:> f(x) = a*exp(-(x-m1)^2/(2*s1*s1)) + b*exp(-(x-m1)^2/(2*s1*s1))where m denotes mean and s denotes standard deviation. If the two > gaussians are close enough, such that replacing their superposition by> a single gaussian gives tolerable error, what are the optimal parameters> (c,m3,s3) for minimizing the error?Well, mean square error makes life difficult in this case.If f(x) is normalized (so that int_{-infty}^{infty} f(x) dx = 1)and the approximation (call it g(x)) is also assumed to be normalized,then you can consider using the cross entropy, -int f(x) log g(x) dxas the goodness of fit instead of mean square error.If g(x) is a single Gaussian bump, setting its mean equal to themean of f(x) and its variance equal to the variance of f(x)minimizes the cross entropy. (You can verify this by computingderivatives of the cross entropy and setting derivatives to zero.)If f(x) is not normalized, you can of course compute a scalingfactor to make it normalized, and divide g(x) by the same factor.Offhand, I don't know if that yields the same result as minimizingthe cross entropy with respect to the mean, variance, and scalingfactor; I don't even know if cross entropy is really meaningfullyapplied to nonnormalized functions.For what it's worth,Robert Dodier--Far better an approximate answer to the right question, which is oftenvague, than an exact answer to the wrong question, which can always bemade precise. -- John W. Tukey === Subject: : Re: Replacing sum of gaussians by a single one?Could try making it linear.write m2=m1+delta mand m3 = m1 + delta Mditto s1this may work for the two initial gaussians being very close because whenthey are the same you know the answer.> [Hossein could not access to this group and asked me to send it. So> this is Hossein's message.]> Hello> Given a function f(x) constructed by a linear combination of two> gaussians:> f(x) = a*exp(-(x-m1)^2/(2*s1*s1)) + b*exp(-(x-m1)^2/(2*s1*s1))> where m denotes mean and s denotes standard deviation. If the two> gaussians are close enough, such that replacing their superposition by> a> single gaussian gives tolerable error, what are the optimal parameters> (c,m3,s3) for minimizing the error?> Formally:> f^(x) = c*exp(-(x-m3)^2/(2*s3*s3))> Error = integral { (f^(x)-f(x))^2 dx} ,x from -inf to +inf.> After spending a few hours, I computed the error analytically and then> tried to make its partial derivatives (with respect to c,m3 and s3)> equal to zero. This resulted in a very complicated non-linear equation> that I couldn't solve. Do you know any solution for (c,m3,s3) ? If> the> optimal solution is too complicated, even a good sub-optimal is> helpful.> Regards> --Hossein Mobahi === Subject: : Re: Real Roots of PolynomialThank you for the replies...Maybe i was a bit unclear as to what i was asking.I know that we can apply the Descartes' Sign Rule to determine themax number (upper limit) of positive and negative real roots of apolynomial. But from this we cannot (always) determine if a polynomialof degree n has n real roots.Now we can apply Sturms Theorem to determine the exact number ofdistinct real roots of a polynomial on an interval(a,b). However wemust limit ourselves to an interval here.So is there any specific condition whereby given a polynomial ofdegree n we can always state that, given this specific conditionholds, all its roots are real?Maybe what i am asking is rather silly/uneducated and i am missingsome basic points here, if so please tell :)>hi,>could anyone tell me what conditions are neccessary in order for a>polynomial to have real roots only?Look for Sturm sequences in an old Theory of equations> book, or elsewhere. They tell you exactly how many> real roots a real polynomial has. === Subject: : Re: Real Roots of Polynomial> Thank you for the replies...Maybe i was a bit unclear as to what i was asking.I know that we can apply the Descartes' Sign Rule to determine the> max number (upper limit) of positive and negative real roots of a> polynomial. But from this we cannot (always) determine if a polynomial> of degree n has n real roots.Now we can apply Sturms Theorem to determine the exact number of> distinct real roots of a polynomial on an interval(a,b). However we> must limit ourselves to an interval here.It is also possible, using synthetic division, to determine that no real roots are larger than a given value or no real roots smaller than a given value. There are other ways, too, of finding bounds on real roots. Thexe, together with Sturm's Theorem, do it all. === Subject: : Re: Real Roots of PolynomialNow we can apply Sturms Theorem to determine the exact number of> distinct real roots of a polynomial on an interval(a,b). However we> must limit ourselves to an interval here.Yes, but one can effctively find a number M such that all zeroesof the polynomial satisfy |z| < M. Apply Sturm to (-M,M).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: : Re: Real Roots of Polynomial> Now we can apply Sturms Theorem to determine the exact number of> distinct real roots of a polynomial on an interval(a,b). However we> must limit ourselves to an interval here.Yes, but one can effctively find a number M such that all zeroes> of the polynomial satisfy |z| < M. Apply Sturm to (-M,M).Yes obviously we can set our range to (-infinity, +infinity).But that is just a method for finding out the number of real zeros.What are the conditions that guarantee, given a polynomial of degreen, that the polynomial has n real zeros?For example given polynomial of degree 2, ax^2 + bx + c, The condition that, b^2 - 4ac > 0guarantees that this polynomial shall have 2 real zeros.Does there exist such a condition that holds for polynomials ofarbitary degree? === Subject: : Re: Real Roots of PolynomialFor example given polynomial of degree 2, > ax^2 + bx + c, > The condition that, > b^2 - 4ac > 0> guarantees that this polynomial shall have 2 real zeros.Does there exist such a condition that holds for polynomials of> arbitary degree?no === Subject: : Re: A 'basic' topology question about interiors Of interest perhaps are these formulas:When U subset A subspace S cl_A (U) = A / cl_S (U) int_A (U) = A / int_S (SA / U)/ / , intersect union---- === Subject: : Re: A 'basic' topology question about interiors> Of interest perhaps are these formulas:> When U subset A subspace S> cl_A (U) = A / cl_S (U)> int_A (U) = A / int_S (SA / U)Thank you! These are most helpful formulae.Ben Scott === Subject: : Re: Graduate algebra book> What's a good book for a first year graduate algebra course?> Something with alot of emphasis on factorization, polynomial rings,> fields, PIDs, Galois Theory, and of course all the more basic topics> of algebra as well like groups, ideals, integral domains, etc.That's graduate algebra?There's a recent book by Joe Rotman Advanced Modern Algebrawhich does all that, and more.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: : Re: Graduate algebra bookWhat's a good book for a first year graduate algebra course?> Something with alot of emphasis on factorization, polynomial rings,> fields, PIDs, Galois Theory, and of course all the more basic topics> of algebra as well like groups, ideals, integral domains, etc.That's graduate algebra?There's a recent book by Joe Rotman Advanced Modern Algebra> which does all that, and more.Uhh... Yeah. I mean do you feel like the topics I mentioned shouldbe covered at the undergraduate level and not at the graduate level? If so, then why do both Lang and Hungerford treat all of the abovetopics in detail in their GTM books? Granted, of course groups,ideals, and integral domains should be covered at the undergraduatelevel, but should free groups, finitely generated abelian groups, andgalois theory? Maybe if you go to Harvard or MIT. === Subject: : Re: Graduate algebra book> What's a good book for a first year graduate algebra course?> Something with alot of emphasis on factorization, polynomial rings,> fields, PIDs, Galois Theory, and of course all the more basic topics> of algebra as well like groups, ideals, integral domains, etc.>> That's graduate algebra?>> There's a recent book by Joe Rotman Advanced Modern Algebra>> which does all that, and more.Uhh... Yeah. I mean do you feel like the topics I mentioned should> be covered at the undergraduate level and not at the graduate level?> If so, then why do both Lang and Hungerford treat all of the above> topics in detail in their GTM books? Granted, of course groups,> ideals, and integral domains should be covered at the undergraduate> level, but should free groups, finitely generated abelian groups, and> galois theory? Maybe if you go to Harvard or MIT.All the listed topics can (and should) be taught in any good first coursein abstract algebra. Anyone lacking such fundamental algebraic knowledgewould be poorly prepared for graduate-level studies.-Bill Dubuque === Subject: : Re: Graduate algebra book> What's a good book for a first year graduate algebra course?> Something with alot of emphasis on factorization, polynomial rings,> fields, PIDs, Galois Theory, and of course all the more basic topics> of algebra as well like groups, ideals, integral domains, etc.>> That's graduate algebra?>> There's a recent book by Joe Rotman Advanced Modern Algebra>> which does all that, and more.Uhh... Yeah. I mean do you feel like the topics I mentioned should> be covered at the undergraduate level and not at the graduate level?> If so, then why do both Lang and Hungerford treat all of the above> topics in detail in their GTM books? Granted, of course groups,> ideals, and integral domains should be covered at the undergraduate> level, but should free groups, finitely generated abelian groups, and> galois theory? Maybe if you go to Harvard or MIT.All the listed topics can (and should) be taught in any good first course> in abstract algebra. Anyone lacking such fundamental algebraic knowledge> would be poorly prepared for graduate-level studies.-Bill DubuqueAre you serious? If so, maybe I'm just a complete idiot. But are youtelling me that Galois theory can and should be taught in a firstcourse in abstract algebra? Granted, all the topics I mentioned(groups, rings, ideals, integral domains, fields, PIDs, factorization)should be mentioned in a first course on abstract algebra, but howmuch detail can you possibly go into in such a short amount of time? Should free abelian groups, sylow theory, and solvable groups beincluded in this introductory course? How about Splitting Fields andGalois groups? If you say yes then you've lost your noodle, howeverthese still fall under the category of group theory or fieldtheory. I said in my original post that I'll have finishedhungerford by the time I start reading the new book, so when I referto group theory obviously I'm not talking about the definition of agroup, or a homomorphism, although many books will probably mentionthat anyway. === Subject: : Re: Graduate algebra book>> What's a good book for a first year graduate algebra course?>> Something with alot of emphasis on factorization, polynomial rings,>> fields, PIDs, Galois Theory, and of course all the more basic topics>> of algebra as well like groups, ideals, integral domains, etc.>> That's graduate algebra?>> There's a recent book by Joe Rotman Advanced Modern Algebra>> which does all that, and more.Uhh... Yeah. I mean do you feel like the topics I mentioned should> be covered at the undergraduate level and not at the graduate level?I don't feel anything, but all of these were covered in the firsttwo years of my undergraduate degree.> If so, then why do both Lang and Hungerford treat all of the above> topics in detail in their GTM books? Why did the publisher put them in the graduate texts series?> Granted, of course groups,> ideals, and integral domains should be covered at the undergraduate> level, but should free groups, finitely generated abelian groups, and> galois theory? Maybe if you go to Harvard or MIT.I went to neither.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: : Re: Graduate algebra book> What's a good book for a first year graduate algebra course?>> Something with alot of emphasis on factorization, polynomial rings,>> fields, PIDs, Galois Theory, and of course all the more basic topics>> of algebra as well like groups, ideals, integral domains, etc.>> That's graduate algebra?>> There's a recent book by Joe Rotman Advanced Modern Algebra>> which does all that, and more.Uhh... Yeah. I mean do you feel like the topics I mentioned should> be covered at the undergraduate level and not at the graduate level?I don't feel anything, but all of these were covered in the first> two years of my undergraduate degree.Well it certainly isn't a fact that they should be covered at theundergraduate level, hence it's a matter of opinion. Either you feel1) that they should be covered at the graduate level, 2) that theyshould be covered at the undergraduate level, or 3) you have noopinion or can't decideIf 3) were the case, you would not respond with such a snotty remarkasThat's graduate algebra?because you would have no opinion.Thus one can only conclude that you either feel 1) or you feel 2).In your opinion, what topics SHOULD be covered in a first yeargraduate algebra course? And what textbook is suitable for such acourse? === Subject: : Re: Graduate algebra book>> I don't feel anything, but all of these were covered in the first>> two years of my undergraduate degree.Well it certainly isn't a fact that they should be covered at the> undergraduate level, hence it's a matter of opinion. Either you feel> 1) that they should be covered at the graduate level, 2) that they> should be covered at the undergraduate level, or 3) you have no> opinion or can't decideIf 3) were the case, you would not respond with such a snotty remark> assnotty? that's abuse.> That's graduate algebra?because you would have no opinion.Thus one can only conclude that you either feel 1) or you feel 2).In your opinion, what topics SHOULD be covered in a first year> graduate algebra course? And what textbook is suitable for such a> course?Don't be asinine. I was merely expressing my surprisethat such elementary topics should be considered as graduate.And I did suggest a book.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: : Re: Graduate algebra book> I don't feel anything, but all of these were covered in the first> two years of my undergraduate degree.That's Europe. In America, undergraduates study liberal arts, andsqueeze in only a little of their specialty in the first two years. === Subject: : Re: Graduate algebra bookI would also recommend Dummit & Foote's book Abstract Algebra. It coversa lot of material. And I have often heard this book praised by graduatestudents. Check it out!Lurch> What's a good book for a first year graduate algebra course?> Something with alot of emphasis on factorization, polynomial rings,> fields, PIDs, Galois Theory, and of course all the more basic topics> of algebra as well like groups, ideals, integral domains, etc.> Something that has lots and lots of _good_ exercises? Hungerford is> good, but I'll have finished most of the exercises by the time I want> to start doin the exercises in this new book. Lang is probably good,> but there don't seem to be very many exercises. === Subject: : How to solve this differential equation?I've encountered such a D.E.,a(x,y)(dx/dt)^2+b(x,y)(dx/dt)(dy/dt)+c(x,y)(dy/dt)^2=f(t) where x=x[t],y=y[t] and a(x,y),b(x,y),c(x,y),f(t) are all knownfunctions.How to derive the equation in x,y,t?e.g. dx/dt+dy/dt=t => x+y=(1/2)t^2+const. (dx/dt)^2+(dy/dt)^2=t => plane curve with arc lengths(t)=(1/2)t^2+const, t is the parameter.Any comments or suggestions? === Subject: : Re: Time complexity of algorithms>> for that :) That was a nice tutorial and it definitely>> helped me understand the reasoning behind algorithms.>> I realize this might not be the best place to post such a question but>> since it basically involved math (what doesn't :)) and was independant>> of computer languages, I thought someone might be able to help me out.>> There still remains one problem.. how do I reason with the algorithm>> that I presented?>// part of code example by Rick> i := 1> while i <= n and A[i] = 0 do> i := i + 1> endwhile> return 1 + HelpMe(A, n-i)>First, the code is wrong (that could be part of the confusion). If A[i]<>0>for some i, then your while-loop is endless.This while loop is obviously NOT endless for any combination of values inA, as i is always increased, and the while loop stops if i>n.-- Wim Benthem === Subject: : Re: Time complexity of algorithms> for that :) That was a nice tutorial and it definitely>> helped me understand the reasoning behind algorithms.>> I realize this might not be the best place to post such a questionbut>> since it basically involved math (what doesn't :)) and was independant>> of computer languages, I thought someone might be able to help me out.>> There still remains one problem.. how do I reason with the algorithm>> that I presented?// part of code example by Rick> i := 1> while i <= n and A[i] = 0 do> i := i + 1> endwhile> return 1 + HelpMe(A, n-i)First, the code is wrong (that could be part of the confusion). IfA[i]<>0>for some i, then your while-loop is endless. This while loop is obviously NOT endless for any combination of values in> A, as i is always increased, and the while loop stops if i>n.How stupid. Of course it ends. I misunderstood the meaning of it. It stopsat the first nonzero entry of A (instead of just not doing the i:=i+1).The complexity is lowered by this. I think it'll be O(n) now, but take acloser look (I don't have much time now).- Arthur> -- > Wim Benthem === Subject: : Re: Time complexity of algorithms a lot Arthur! That explains it :) Just wondering, are there any good links/websites that explain how to form recurrence relationships from looking at code/algorithm? Rick === Subject: : Re: normal form of matrices over PID> Let R = Z[x], where x is 1/2 times> 1 + sqrt{-19}. I'm looking for a> 1 by 2 matrix over R that is not> equivalent by any sequence of> elementary column operations to> a matrix of the form (z 0). Is> (x+1 x-1) such a matrix? (Any other> would be fine!)I'm sure there is one: whether your example is one, I don't know.You should consult this paperSwan, Richard G.Generators and relations for certain special linear groups.Advances in Math. 6 1971 1--77 (1971).I don't have this to hand, so the following is based on vague memories.You are really asking if SL(2,R) = E(2,R) where E(2,R) is the subgroupof SL(2,R) generated by elementary matrices (1 0 /a 1) and (1 a/0 1).If R is Euclidean the answer is yes. If R is not Euclidean, even ifit's a Dedekind domain of class number 1 then then answer may be no.Swan analyses the situation by looking at an action of SL(2,R)(R imaginary quadratic) on hyperbolic 3-space. By finding a fundamentaldomain you can read off generators and relations, and see that the elementary matrices are inadequate as generators.Ther connection with your problem is that if (a b) is the top rowof a matrix in SL(2,R) but outside E(2,R), this is notequivalent under elementary column operations to (1 0) (despite a andb being coprime).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: : How do you integrate the function f(x) = x/(tanx) [0,pi/2]?How do you integrw do you integrate the function f(x) = x/(tanx) ?please send substitutes, etc. -- Georg Goerg === Subject: : Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]?> How do you integrw do you integrate the function f(x) = x/(tanx) ?Please forget my previous reply. It is wrong, as A. N. Niel pointedout. Your function is bounded, of course.Best regardsJose Santos === Subject: : Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]?> How do you integrw do you integrate the function f(x) = x/(tanx) ?The behaviour around 0 is like 1/x; therefore, the integral of f(x)in ]0,Pi/2] is equal to +oo.Best regards,Jose Santos === Subject: : Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]?> How do you integrw do you integrate the function f(x) = x/(tanx) ?The behaviour around 0 is like 1/x; therefore, the integral of f(x)> in ]0,Pi/2] is equal to +oo.Best regards,Jose SantosI think you are in error. === Subject: : Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]?> How do you integrw do you integrate the function f(x) = x/(tanx) ?> please send substitutes, etc. Maple does it in terms of a polylogarithm. Suggesting that theindefinite integral is not an elementary function. === Subject: : Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]?> How do you integrw do you integrate the function f(x) = x/(tanx) ?> please send substitutes, etc. > Maple does it in terms of a polylogarithm. Suggesting that the> indefinite integral is not an elementary function.Mathematica does similarly, giving x*Log[1 - E^((2*I)*x)] - (I/2)*(x^2 + PolyLog[2, E^((2*I)*x)])for the indefinite integral. However, based on the title of the thread,perhaps Georg really wants the definite integral from x = 0 to Pi/2. Thevalue of that is Pi/2*Log[2].David Cantrell === Subject: : Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]? How do you integrw do you integrate the function f(x) = x/(tanx) ?> please send substitutes, etc. Maple does it in terms of a polylogarithm. Suggesting that the> indefinite integral is not an elementary function.Mathematica does similarly, giving x*Log[1 - E^((2*I)*x)] - (I/2)*(x^2 + PolyLog[2, E^((2*I)*x)])for the indefinite integral. However, based on the title of the thread,> perhaps Georg really wants the definite integral from x = 0 to Pi/2. The> value of that is Pi/2*Log[2].> Try the substitution u= tan x to reduce it to standard nonelementary form.> David Cantrell === Subject: : Re: simple question about subspaces| at 08:46 PM, kramsay@aol.com (KRamsay) said:| | >Deriving it from the group properties does not indicate a way to| >eliminate references to 0 from the usual axioms for vector spaces.| | It's precisely because of the axioms that it is derived. A vector| space is defined as an ordered set with certin properties.Not necessarily ordered.| There are| various ways to do it, but some of them involve redundant| constituents.Okay, but what about how to eliminate the references to 0 in thedefinition without using an additional axiom?small trick one could use: I considered mentioning that, plus the fact that one could eliminate the references to 0 in the definition by having the axiom (x+(-x))+y=y and so on.I had explained previously why it was I wanted to get ridof references to 0 in the axioms. One answer to the originalposter's question was that the empty subspace is ruled outby the definition which accords with the definition ofvector space by conventions of universal algebra. I wantedto elaborate on this explanation a bit. The reason why theuniversal algebra definition rules out the empty set as asubstructure is that in the universal algebra treatment,the 0 is considered a 0-ary function, and closure under thisfunction means including 0 in the structure.Since it's possible to tweak the axioms slightly, so thatthey still fit into the framework of universal algebra, butlack these references to zero (and are equivalent aside fromallowing the empty set as a structure), just following theconventions of universal algebra isn't *quite* a completeexplanation of what the original poster asked about. We arein fact going the usual route and requiring the presence of0 in the structure for other considerations like elegance.It's not an attempt to improve on the axioms at all; it'san exploration of why we don't go down this other road.Now, for such a purpose, just lumping together the axiomsreferring to 0 under the rubric of is an abelian groupunder + does absolutely nothing. It appears as though youconsider doing such a lumping-together eliminates thereference to 0 in a manner worth mentioning, however.You replied: You don't need it; you can derive it from the group properties. | >The usual group axioms contain references to the identity | >(0 for the additive group),| | A group is nomally defined as an ordered pair (E,*) satisfyning| certain properties. One of those properties is that there exists an| identity element. Defining a group as (E,e,*) would be redundant.This helps me get rid of the references to the identity inthe axioms how?Imagine, if you will, that someone has been giving a briefexplanation of how it's possible to build a car without usingspark plugs to ignite the fuel.Someone pipes up, But you don't need to use this sort oftrick. Just use an engine. | >and if you figure out a set of axioms for groups which don't, [speaking hopefully] So you've also figured out some wayto build the engine without needing spark plugs?| And if your grandmother had wheels she'd be a wagon. Please don't| bother rebutting things that I never claimed.I said nothing about building an engine without using sparkplugs! | >because the references to 0 in the usual axioms for vector spaces| >are in the axioms for the space as an additive group.But the spark plugs normally are part of the engine.| Nu, so what else is new?Who gives a !|There is no need to include 0 as an explicit|element of the vector space precisely because the group axioms require|that it exist.Just use the engine! The engine will ignite and burn thefuel for you!|Similarly, there is no need to explicitly require|uniqueness, because the group axioms already suffice to prove it.| | >After some thought, I suspect you mean to say that we can derive| >(x+(-x))+y=y from the other axioms for VECTOR SPACES, not GROUPS.Perhaps you mean something other than the engine? |No. What I mean to say is that the group axioms are sufficient to|prove the existence and uniqueness of 0 and the existence and|uniqueness of -x for any x.||>but I don't see how to get from there|>to 0*v being an additive identity without using the existence of|>additive inverses (in some form).||Then it's a good thing that I didn't suggest that you don't need|is a group and on top of that make the inverse one of the constituent|operations, since the group axioms already ensure that it exists and|is unique.No, just use the engine, like I keep telling you!| >So unless I'm missing something here,| | Quite a bit.Former professors these days, boy, they just don't listen,do they?| To start out, you use the constants 0 and the operator| - without defining them.This is commonplace. Some authors phrase the axiom as sayingthat for each x there exists a y such that..., either assumeor prove that it's unique and then introduce the notation-x for this y. Others introduce a function unary - andgive identities for it. (The latter is what fits in withuniversal algebra, which we had been discussing.)|More important, the vector space axioms|include being a group under +, which your system is not; there is an|identity element but there are elements for which no inverse exists.Perhaps you think I've given at some point what I think areall the axioms of a vector space. I was addressing those inthe audience who would not need to have a list provided tothem. The details of the remaining axioms are irrelevantto the question of how to rephrase the ones involving 0 soas not to refer to 0.| >So please be less terse.I might have said instead: kindly explain how what you aresaying makes anything at all unnecessary for any meaningfulpurpose whatsoever.| A vector space is defined as an orderedYou don't mean ordered.| set containing some| combination of sets, some of them being functions, satisfying certain| conditions. There are various ways in which this can be done; the| definitions are formally equivalent. Some of the ways in which it can| be done include elements that are not present in orther ways. In| particular, any definition involving either the 0 vector or the unary| negation operator as elements of the ordered list is equivalent to| another definition that does not include them.| | Thus, If I defined[1] a vector space V to be an ordered set| (K,V_E,V_0,V_+,V_-,V_*) and gave the obvious axioms, that would be| equivalent to defining it as V'=(K,V_E,V_+,V_*) with the obvious| axioms. Similarly, it would be equivalent to V''=(K,V_G,V_*).| | [1] I'm using the notation V_E for the elements of V, V_0 for the null| vector, V_+ for vector addition, V_G for the group (V_E,V_+), etc.Great.Do me a favor, and the next time I'm trying to explainhow it is possible to do something, and you have somepiece of nearly content-free trivia, please phrase itmore like this: This does absolutely nothing to help youaccomplish what you're trying to do here, but....Keith Ramsay === Subject: : Re: grasshopper consciousness?NOVA | The Elegant Universe | PBS> http://www.pbs.org/wgbh/nova/elegant/A Theory of Everything?Some physicists believe string theorymay unify the forces of natureby Brian Greene physicists have identified -- electrons, neutrinos, quarks, and so on -- are the letters of all matter. Just like their linguistic counterparts, they appear to have no further internal substructure. String theory proclaims otherwise. According to with even greater precision -- a precision many orders of magnitude beyond our present technological capacity -- we would find that each is not pointlike but instead consists of a tiny, one-dimensional loop. contains a vibrating, oscillating, dancing filament that physicists have named a string. [...]http://www.pbs.org/wgbh/nova/elegant/everything.html _________Viewpoints on String TheorySteven Weinberghttp://www.pbs.org/wgbh/nova/elegant/ view-weinberg.html The Elegant Universe homepage [For more on a final theory, see A Theory of Everything?.] > AL-JILWAH (The Revelation)> The Hunger Site:> http://www.thehungersite.com/> ~o~> Jeff Mishlove> Thinking Allowed Productions> URL: http://www.thinking-allowed.com/> ~o~> What we may Learn from Future Physics Experiments> Andris Skuja, University of Maryland> Professor Andris Skuja> Web Page: http://www.physics.umd.edu/hep/> http://www.physics.umd.edu/people/faculty/skuja.html> C. SCOTT LITTLETON> http://www.oxy.edu/~yokatta/home.htm Any sufficiently advanced technology is> indistinguishable from magic> --Sir Arthur C. Clarke> Einstein's gravitons and Tony Smith's conformal gravitons.> Tony Smith's conformal gravitons> http://www.innerx.net/personal/tsmith/SegalConf.html> Conformal Groups are related to Moebius Transformations SPECIAL RELATIVITY PROHIBITS SPACELIKE CAUSATION> http://www.mindspring.com/~cerebroscopic/Angelidis.html Sloan Digital Sky Survey (SDSS) The First Detailed Full Sky Picture> of the Oldest Light in the Universe.> http://map.gsfc.nasa.gov/m_mm.html The Wilkinson Microwave Anisotropy Probe (WMAP)> })))> MSU University News> [...]> Cornish, Spergel and Starkman have posted> their findings on the Internet> ()*> where other scientists can evaluate them.> Ultimately, Cornish said, it's the scientific> community that will decide. The most amazing thing about this whole story is the> fact that we can go out there and try and figure out> the shape of the universe, Cornish said. ... Regardless> of what the results are, I think it's great for people to> know this is even a possibility. > * Constraining the Topology of the Universe> http://arxiv.org/abs/astro-ph/0310233 1 parsec = 3.2 light years distance> 1 Megaparsec (Mpc) = 3.2 light years distance times 1,000,000> 1 Gigaparsec (Gpc) = 3.2 light years distance times 1,000,000,000> 24 Gpc = 24 times 3.2 light years times one billion (size)> 76,800,000,000 Light Years!!> [Uhhhmmm... is that radius or diameter?]> Our universe, so they say, is 13.7 billion years old since the Big> Bang...> [[Cosmology Tutorial:> ]]> Wilkinson Microwave Anisotropy Probe> WMAP: http://map.gsfc.nasa.gov/> Data Analysis http://lambda.gsfc.nasa.gov/> So what the hell are WE doing here??!!> 0101010101010101010101010101010101010101> A high resolution foreground cleaned CMB map from WMAP> http://arxiv.org/abs/astro-ph/0302496 Prof. Max Tegmark> http://www.hep.upenn.edu/~max/wmap.html Brane Worlds, the Subanthropic Principle and the> Undetectability Conjecture by Beatriz Gato-Rivera> http://www.unknowncountry.com/mindframe/opinion/?id=96> Generally ...http://xxx.arxiv.cornell.edu/find/physics/1/au:+Gato_Rivera _B/0/1/0/all/0/1 Dark Matter, Extra Dimensions Related And Possibly Detectable> http://www.spacedaily.com/news/cosmology-03o.html Neuroblast Wormhole Behaviorism, a.k.a.> The Portal, to Dodecahedron Head> http://pw1.netcom.com/~mthorn/mt90.htm Meanwhile... n a n o t u b e s . . .> A New Alternative To Depleted Uranium?!http://popularmechanics.com/science/space/2002/7/ going_up/print.phtml The promise of inexpensive access to space is so important> to the human race that we are ready to meet these challenges> head on. Viewed in one way, the space elevator will be the> largest civil engineering project ever attempted,> Laubscher said. [...] Crawling to space: ... [N]ew work in China that suggests carbon nanotubes> can be fused together, without need of a matrix material.> [...]http://www.space.com/businesstechnology/technology/space_ elevator_030917.html Space Elevator, Nanotubeshttp://www.google.com/search?hl=en&ie=ISO-8859-1&q= Space+Elevator%2C+Nanotubes Aumakua _ Uhane _ Unihipilihttp://www.google.com/search?hl=en&lr=&ie=ISO-8859-1& q=Aumakua%2C+Uhane%2C+Unihipili Electromagnetic Mind Field> http://pw1.netcom.com/~mthorn/magnmind.htm>