mm-193 I have a matix, S, I want to write it in the form of S=v*v', where v is a column vector, v' is its transpose, hence is a row vector... How to decompose a matrix into such form, is there any technique can do this? > > I understand that not every matrix is decomposable into vector inner > product > v*v', but is there a technique to approximate the matrix by sum of a series of v*v'? For example, matrix M=v1*v1'+v2*v2'+v3*v3'+... > well M = v * v' will be symmetric in that M = M'. thus you will have trouble with non-symmetric matrices (for complex matrix, symmetric is hermitian symmetric). In the case that M is symmetric, try the singular-value decomposition (or, what amounts to nearly the same thing, an eigenvalue/eigenvector decomposition). be > non-symmetircal. In that case, I am actually looking for an optimal > approximation M=v1*v1'+v2*v2'+... in the minimum MSE sense... > Is there any technique for that? Break your matrix into two pieces, a symmetric and an anti-symmetric. M = S + A Given M, set S = (M+M')/2. Then 'nd A = M-S. You can't do anything about the A part, but the S is amenable to the SVD from which you do S = sigma1*v1*v1'+sigma2*v2*v2'+...+sigman*vn*vn[C apitalOTilde] where sigmak are positive real numbers and sigma1 >= sigma2 >= ... >= sigman with n being the size of M. You can roll the sigma into vectors by sigmak*vk*vk' = (sqrt(sigmak)*vk)*(sqrt(sigmak)*vk)'. -- === Subject: Re: (matrix analysis)how to convert a matrix to the inner product of two vectors? > I have a matix, S, I want to write it in the form of S=v*v', where v is a > column vector, v' is its transpose, hence is a row vector... > > How to decompose a matrix into such form, is there any technique can do > this? > I understand that not every matrix is decomposable into vector inner product v*v', but is there a technique to approximate the matrix by sum of a series of v*v'? For example, matrix M=v1*v1'+v2*v2'+v3*v3'+... > > well M = v * v' will be symmetric in that M = M'. thus you will have > trouble with non-symmetric matrices (for complex matrix, symmetric is > hermitian symmetric). In the case that M is symmetric, try the > singular-value decomposition (or, what amounts to nearly the same thing, > an eigenvalue/eigenvector decomposition). > Hi Johan, be non-symmetircal. In that case, I am actually looking for an optimal approximation M=v1*v1'+v2*v2'+... in the minimum MSE sense... Is there any technique for that? Break your matrix into two pieces, a symmetric and an anti-symmetric. M = S + A Given M, set S = (M+M')/2. Then 'nd A = M-S. You can't do anything about the A part, but the S is amenable to > the SVD from which you do > S = sigma1*v1*v1'+sigma2*v2*v2'+...+sigman*vn*vn[C apitalOTilde] where sigmak are positive real numbers and sigma1 >= sigma2 >= ... >= > sigman with n being the size of M. You can roll the sigma into > vectors by sigmak*vk*vk' = (sqrt(sigmak)*vk)*(sqrt(sigmak)*vk)'. > by the way, can I do the same trick for A as you've done for M? that's is, I take A1=(A+A')/2, then decompose it, .... go on and go on? === Subject: Re: (matrix analysis)how to convert a matrix to the inner product of two vectors? I have a matix, S, I want to write it in the form of S=v*v', where > v > is a column vector, v' is its transpose, hence is a row vector... How to decompose a matrix into such form, is there any technique > can > do this? > > I understand that not every matrix is decomposable into vector inner > product > v*v', but is there a technique to approximate the matrix by sum of a > series > of v*v'? > > For example, matrix M=v1*v1'+v2*v2'+v3*v3'+... > well M = v * v' will be symmetric in that M = M'. thus you will have trouble with non-symmetric matrices (for complex matrix, symmetric is hermitian symmetric). In the case that M is symmetric, try the singular-value decomposition (or, what amounts to nearly the same > thing, an eigenvalue/eigenvector decomposition). > > Hi Johan, > to be > non-symmetircal. In that case, I am actually looking for an optimal > approximation M=v1*v1'+v2*v2'+... in the minimum MSE sense... > > Is there any technique for that? Break your matrix into two pieces, a symmetric and an anti-symmetric. M = S + A Given M, set S = (M+M')/2. Then 'nd A = M-S. You can't do anything about the A part, but the S is amenable to the SVD from which you do S = sigma1*v1*v1'+sigma2*v2*v2'+...+sigman*vn*vn[C apitalOTilde] where sigmak are positive real numbers and sigma1 >= sigma2 >= ... >= sigman with n being the size of M. You can roll the sigma into vectors by sigmak*vk*vk' = (sqrt(sigmak)*vk)*(sqrt(sigmak)*vk)'. > by the way, can I do the same trick for A as you've done for M? > that's is, I take A1=(A+A')/2, then decompose it, .... go on and go > on? No, because A is anti-symmetric, i.e., A+A' = 0. -- === Subject: Re: (matrix analysis)how to convert a matrix to the inner product of two vectors? > I have a matix, S, I want to write it in the form of S=v*v', where v is a > column vector, v' is its transpose, hence is a row vector... > > How to decompose a matrix into such form, is there any technique can do this? > I understand that not every matrix is decomposable into vector inner product v*v', but is there a technique to approximate the matrix by sum of a series of v*v'? For example, matrix M=v1*v1'+v2*v2'+v3*v3'+... > > well M = v * v' will be symmetric in that M = M'. thus you will have > trouble with non-symmetric matrices (for complex matrix, symmetric is > hermitian symmetric). In the case that M is symmetric, try the > singular-value decomposition (or, what amounts to nearly the same thing, > an eigenvalue/eigenvector decomposition). > Hi Johan, be non-symmetircal. In that case, I am actually looking for an optimal approximation M=v1*v1'+v2*v2'+... in the minimum MSE sense... Is there any technique for that? Break your matrix into two pieces, a symmetric and an anti-symmetric. M = S + A Given M, set S = (M+M')/2. Then 'nd A = M-S. You can't do anything about the A part, but the S is amenable to > the SVD from which you do > S = sigma1*v1*v1'+sigma2*v2*v2'+...+sigman*vn*vn[C apitalOTilde] where sigmak are positive real numbers and sigma1 >= sigma2 >= ... >= > sigman with n being the size of M. You can roll the sigma into > vectors by sigmak*vk*vk' = (sqrt(sigmak)*vk)*(sqrt(sigmak)*vk)'. > Is your technique optimal in the MSE sense? Any proof? Michael. === Subject: Re: (matrix analysis)how to convert a matrix to the inner product of two vectors? I have a matix, S, I want to write it in the form of S=v*v', where > v > is a column vector, v' is its transpose, hence is a row vector... How to decompose a matrix into such form, is there any technique > can > do this? > > I understand that not every matrix is decomposable into vector inner > product > v*v', but is there a technique to approximate the matrix by sum of a > series > of v*v'? > > For example, matrix M=v1*v1'+v2*v2'+v3*v3'+... > well M = v * v' will be symmetric in that M = M'. thus you will have trouble with non-symmetric matrices (for complex matrix, symmetric is hermitian symmetric). In the case that M is symmetric, try the singular-value decomposition (or, what amounts to nearly the same > thing, an eigenvalue/eigenvector decomposition). > > Hi Johan, > to be > non-symmetircal. In that case, I am actually looking for an optimal > approximation M=v1*v1'+v2*v2'+... in the minimum MSE sense... > > Is there any technique for that? Break your matrix into two pieces, a symmetric and an anti-symmetric. M = S + A Given M, set S = (M+M')/2. Then 'nd A = M-S. You can't do anything about the A part, but the S is amenable to the SVD from which you do S = sigma1*v1*v1'+sigma2*v2*v2'+...+sigman*vn*vn[C apitalOTilde] where sigmak are positive real numbers and sigma1 >= sigma2 >= ... >= sigman with n being the size of M. You can roll the sigma into vectors by sigmak*vk*vk' = (sqrt(sigmak)*vk)*(sqrt(sigmak)*vk)'. > Is your technique optimal in the MSE sense? Any proof? How do you mean optimal? S can be built with no error. The best approximation of A by symmetric matrices is 0. If you mean can I drop terms in S = sigma1*v1*v1'+sigma2*v2*v2'+...+sigman*vn*vn[C apitalOTilde] and get the best MSE approximation with only n1 terms, then I think it is still optimal. I don't have a proof off-hand. -- === Subject: JSH: $100,000 US offer, Abel Prize One thing I've been fascinated by as I've considered replies to my posts is a loose group coordination between posters, as some try to post with math, and others just post various jibes, but all keep focused on pushing the false notion that my rather basic argument showing a problem with the de'nition of algebraic integers is wrong. Now I've tried to use money before, but failed as I think people in the math community understand that NONE of you can overcome that kind of coordinated effort, and even if you post your own version of my own argument, the rest of math society will shred you. However, the shortness and simplicity of the proof of the problem in core offers another solution--machine proof checking. That means I can make the offer of sharing at least $100,000 US with ONE person or one group that collaborates to produce the check by machine from the Abel Prize, assuming, of course that I win it. The bene't to me is that the Abel Prize is close to $800,000 US so what's a measly $100,000? And that way I know that some of you with the expertise may be brave enough to go against the group, for the money, knowing that all these attacks will fall apart when faced with a check by computer. Of course, I will at some point start working on the computer check myself, and in that case, I will happily give myself the money, assuming I win. That leaves any would be winners with me with the question, will mathematicians behave so badly as to try and deny me any math prizes? If so, don't worry, I'll 'nd private funds to replace, but again, that's assuming, so it's not a guarantee. I wish I could do better, but the situation is odd. I can see mathematicians, unless completely broken by the event, seeking revenge by trying to deny me any prizes, but then, there are lots of millionaires and billionaires out there, who will probably help out in that case. No guarantees, but hey, for those of you with the expertise, at least you can make math history, and potentially get serious cash in the process. Here's the current outline of the proof: PROOF OF CORE ERROR 1. Given P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) Variables: m, f, x, u E Ring of Algebraic integers P(m) is a polynomial with m the key variable. 2. Let P(m) = (a_1(m) x + uf)(a_2(m) x + uf)(a_3(m) x + uf) Variables: a_1, a_2, a_3, roots of cubic de'ned as follows. Cubic: a^3 + 3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m) Solving for roots of P(m) by setting x = -uf/a will give cubic. 3. Let P(m) = g_1(m) g_2(m) g_3(m) Variables: g_1, g_2, g_3 de'ned as follows g_1(m) = (a_1(m) x + uf), g_2(m) = (a_2(m) x + uf), g_3(m) = (a_3(m) x + uf). Tautological base: Change in terms independent of m happens independent of m. My point is that concentrating on terms independent of m, will give me results independent of m. So I get terms independent of m. 4. List of independent terms. Independent terms are found by setting m=0. Doing so with cubic gives: a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. (indices selected arbitrarily) Then from previous de'nitions: a. g_1 has value uf at m=0 indicating indepedent term uf b. g_2 has value uf at m=0 indicating independent term uf c. g_3 has value 3x + uf at m=0 indicating independent term 3x + uf So far so good, as now I have independent terms, which therefore will not change with m. CONDITION: f is coprime to 3, x and u This condition sets up the coprimeness results. PRIMARY ARGUMENT 5. Divide off f^2 from P(m). P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f Now I'm forcing a change by dividing off f^2, but again, I'm doing things independent of the value of m. 6. List resultant independent term P(0)/f^2 = u^2(3x + uf) Note that P(0)/f^2 is coprime to f given the condition above. which is P(0)/f^2 = u^2 g_3(0) which is P(0)/f^2 = g_1(0)/f g_2(0)/f g_3(0) Here the proof is basically complete as I've determined that f is a factor of two of the g's which may seem obvious, but remember, I'm setting up for a machine to check. Detail is necessary. a. g_1(0)/f is coprime to f b. g_2(0)/f is coprime to f c. g_3(0) is coprime to f Preliminary Finding: Independent terms of resultants are coprime to f. That sets up for the 'nale. 7. Backwards Theorem: Reverse use of distributive property proves g_1, g_2 have factor that is f. I 'nd it interesting that no comments from others have been made here at the heart of the proof. g_1(m) = a_1(m) x + uf g_2(m) = a_1(m) x + uf so to remove factor f from independent term uf, f must divide g_1(m) and g_2(m), from reverse use of distributive property: g_1(m)/f = a_1(m)/f x + u. That is, I found that the independent term changes by a factor of f when I divide P(m) by f^2, and I made that 'nding *independent* of m, so now I use it without concern about the value of m, i.e. in general. Then by a reverse use of the distributive property, it follows that a_1 and g_1 have f as a factor. (Note to readers: The point of using independent terms is that they are *independent* of the value of m.) 8. Veri'cation: Determine independent term of g_1(m)/f, by setting m=0. g_1(0)/f = a_1(0)/f x + u = u. Con'rmed factor f for g_1. Now maybe you see why posters have ignored the full outline in their replies. Determine independent term of g_2(m)/f, by setting m=0. g_2(0)/f = a_2(0)/f x + u = u. Con'rmed factor f for g_1. You see, there's no room for reasonable doubt. 9. Core error determination 10. Conclusion Core error, of course. James Harris === Subject: Re: JSH: $100,000 US offer, Abel Prize [snip hysterical rambling] > 9. Core error determination g's do not have a factor that is f in the ring of algebraic integers > goes. > 10. Conclusion Core error, of course. James Harris Wacky, isn't it? But, hey, it's just basic math. Yup, yup, yup! -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: $100,000 US offer, Abel Prize >One thing I've been fascinated by as I've considered replies to my >posts is a loose group coordination between posters, as some try to >post with math, and others just post various jibes, but all keep >focused on pushing the false notion that my rather basic argument >showing a problem with the de'nition of algebraic integers is wrong. Now I've tried to use money before, but failed as I think people in >the math community understand that NONE of you can overcome that kind >of coordinated effort, and even if you post your own version of my own >argument, the rest of math society will shred you. However, the shortness and simplicity of the proof of the problem in >core offers another solution--machine proof checking. That means I can make the offer of sharing at least $100,000 US with >ONE person or one group that collaborates to produce the check by >machine from the Abel Prize, assuming, of course that I win it. You really have no idea what an idiot you are. Supposing that you _were_ right about the Error. You think that would win an Abel prize? Tee-hee. >The bene't to me is that the Abel Prize is close to $800,000 US so >what's a measly $100,000? And that way I know that some of you with >the expertise may be brave enough to go against the group, for the >money, knowing that all these attacks will fall apart when faced with >a check by computer. Of course, I will at some point start working on the computer check >myself, and in that case, I will happily give myself the money, >assuming I win. That leaves any would be winners with me with the question, will >mathematicians behave so badly as to try and deny me any math prizes? If so, don't worry, I'll 'nd private funds to replace, but again, >that's assuming, so it's not a guarantee. Pretty exciting offer: If someone agrees with you they get $100,000, unless they don't. No reason anyone would see anything weasely about that. > I wish I could do better, >but the situation is odd. Heh-heh. Yes, it is. ************************ David C. Ullrich === Subject: Re: JSH: $100,000 US offer, Abel Prize > One thing I've been fascinated by as I've considered replies to my posts is a loose group coordination between posters, as some try to post with math, and others just post various jibes, but all keep focused on pushing the false notion that my rather basic argument showing a problem with the de'nition of algebraic integers is wrong. Now I've tried to use money before, but failed as I think people in the math community understand that NONE of you can overcome that kind of coordinated effort, and even if you post your own version of my own argument, the rest of math society will shred you. However, the shortness and simplicity of the proof of the problem in core offers another solution--machine proof checking. That means I can make the offer of sharing at least $100,000 US with ONE person or one group that collaborates to produce the check by machine from the Abel Prize, assuming, of course that I win it. You really have no idea what an idiot you are. Supposing that > you _were_ right about the Error. You think that would win an > Abel prize? Tee-hee. > The bene't to me is that the Abel Prize is close to $800,000 US so what's a measly $100,000? And that way I know that some of you with the expertise may be brave enough to go against the group, for the money, knowing that all these attacks will fall apart when faced with a check by computer. Of course, I will at some point start working on the computer check myself, and in that case, I will happily give myself the money, assuming I win. That leaves any would be winners with me with the question, will mathematicians behave so badly as to try and deny me any math prizes? If so, don't worry, I'll 'nd private funds to replace, but again, that's assuming, so it's not a guarantee. Pretty exciting offer: If someone agrees with you they get $100,000, > unless they don't. No reason anyone would see anything weasely about > that. > I wish I could do better, but the situation is odd. Heh-heh. Yes, it is. ************************ David C. Ullrich I had a feeling that it would come down to money eventually....How did I know? David Moran === Subject: Re: JSH: $100,000 US offer, Abel Prize linux) > I had a feeling that it would come down to money eventually....How did I > know? You're hardly prescient. The money angle is not new at all. See author: jstevh%40msn.com&hl=en&lr=&ie=UTF-8&selm= 3c65f87.0201251039.5eb94c4e ,---- | And make no mistake people, the gold rush just started with my offer | of $10,000 US. Eventually now, some people will 'gure out that | there's potentially much more money in all of this. | | [...] | | I'll just 'nish shooting down the objections currently being raised | on the newsgroup, and now for once I might actually have an audience | that realizes that I've done so, who won't just *claim* that other | people are right, even when they make egregious errors and | insupportable statements, because you see, they'll 'nally know what's | in it for them--cold, hard cash. `---- JSH even refers to this previous offer in his current deal. >Now I've tried to use money before, but failed as I think people in >the math community understand that NONE of you can overcome that kind >of coordinated effort, and even if you post your own version of my own >argument, the rest of math society will shred you. This is not a new low for James. It is an old one. -- Jesse Hughes We will run this with the same kind of openness that we've run Windows. Steve Ballmer, speaking about MS's new .Net project. === Subject: Re: JSH: $100,000 US offer, Abel Prize >The bene't to me is that the Abel Prize is close to $800,000 US so >>what's a measly $100,000? Pretty exciting offer: If someone agrees with you they get $100,000, >unless they don't. No reason anyone would see anything weasely about >that. What if more than eight people agreed with JH, wouldn't that mean he'd lose a sizable sum of money? Not that such a thing would ever happen, of course. === Subject: Re: JSH: $100,000 US offer, Abel Prize > >The bene't to me is that the Abel Prize is close to $800,000 US so >what's a measly $100,000? Pretty exciting offer: If someone agrees with you they get $100,000, unless they don't. No reason anyone would see anything weasely about that. What if more than eight people agreed with JH, wouldn't that mean he'd > lose a sizable sum of money? Not that such a thing would ever happen, > of course. I wouldn't rule it out. One can't believe impossible things, said Alice. I daresay you haven't had much practice, responded the Queen. When I was your age, I always did it for half-an-hour a day. Why, sometimes I've believed as many as six impossible things before breakfast! === Subject: Re: JSH: $100,000 US offer, Abel Prize Correction at 4. > 4. List of independent terms. > Independent terms are found by setting m=0. Doing so with cubic > gives: > a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. That should be a^3 - 3a^2 = 0. I will be working from the outline posted in this thread from here on out for the Abel Prize $100,000 Giveaway. I see it as a challenge to win the money for myself, and put the rest of you out in the cold, yet again. James Harris === Subject: Re: JSH: $100,000 US offer, Abel Prize > Correction at 4. > 4. List of independent terms. Independent terms are found by setting m=0. Doing so with cubic gives: a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. > That should be > a^3 - 3a^2 = 0. > I will be working from the outline posted in this thread from here on > out for the Abel Prize $100,000 Giveaway. > I see it as a challenge to win the money for myself, and put the rest > of you out in the cold, yet again. > James Harris As always, you're repeating yourself like a bad recording: never giving new information, constantly focussing on the same errors... Jeroen === Subject: Re: JSH: $100,000 US offer, Abel Prize Correction at 4. > 4. List of independent terms. > > Independent terms are found by setting m=0. Doing so with cubic > gives: > > a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. That should be a^3 - 3a^2 = 0. I will be working from the outline posted in this thread from here on out for the Abel Prize $100,000 Giveaway. I see it as a challenge to win the money for myself, and put the rest of you out in the cold, yet again. James Harris > As always, you're repeating yourself like a bad recording: never giving > new information, constantly focussing on the same errors... > Jeroen I blame Google for the mulitiple postings. In any event, I 'gure I'll have to put my proof in machine checkable format myself as I've had to do so much myself up to now, and that's more fun in any event. Trouble is, it could take me a couple of months to work everything out, so I'm back in layback mode. Math society is kind of funny. Now I'm having some amusement at how easy it was for so many of you to lie, and trash the very thing that should have been most important to you--mathematics. After all, I'm NOT a mathematician. I just have fun watching you run. Run mathematicians, RUN!!! I'm coming for you. It may take a few months, but I'll get that computer check and then your lives will be ended as you previously knew it. So run Barry Mazur! RUN, RUN! Run Andrew Granville! RUN, RUN! Run Ralph McKenzie! RUN, RUN!!! You *may* have a few months before I chase you down. And when I have the computer saying I'm right, your running will be over. James Harris === Subject: Re: JSH: $100,000 US offer, Abel Prize > > Correction at 4. > 4. List of independent terms. Independent terms are found by setting m=0. Doing so with cubic gives: a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. > > That should be > > a^3 - 3a^2 = 0. > > I will be working from the outline posted in this thread from here on > out for the Abel Prize $100,000 Giveaway. > > I see it as a challenge to win the money for myself, and put the rest > of you out in the cold, yet again. > > James Harris As always, you're repeating yourself like a bad recording: never giving new information, constantly focussing on the same errors... Jeroen I blame Google for the mulitiple postings. In any event, I 'gure > I'll have to put my proof in machine checkable format myself as I've > had to do so much myself up to now, and that's more fun in any event. Trouble is, it could take me a couple of months to work everything > out, so I'm back in layback mode. Math society is kind of funny. Now I'm having some amusement at how > easy it was for so many of you to lie, and trash the very thing that > should have been most important to you--mathematics. After all, I'm NOT a mathematician. I just have fun watching you run. Run mathematicians, RUN!!! I'm coming for you. It may take a few months, but I'll get that > computer check and then your lives will be ended as you previously > knew it. So run Barry Mazur! RUN, RUN! Run Andrew Granville! RUN, RUN! Run Ralph McKenzie! RUN, RUN!!! You *may* have a few months before I chase you down. And when I have the computer saying I'm right, your running will be > over. > James Harris James, I have a question for you. I've written programs in the past and I am wondering if you are going to supply the source code of your program along with the output? It could be easy to make a program to do what you want it to do, but it's pointless if it's not based on an explicit formula. David Moran === Subject: Re: JSH: $100,000 US offer, Abel Prize Correction at 4. > 4. List of independent terms. > Independent terms are found by setting m=0. Doing so with cubic > gives: > a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. That should be a^3 - 3a^2 = 0. I will be working from the outline posted in this thread from here on out for the Abel Prize $100,000 Giveaway. I see it as a challenge to win the money for myself, and put the rest of you out in the cold, yet again. James Harris === Subject: Re: JSH: $100,000 US offer, Abel Prize Correction at 4. > 4. List of independent terms. > Independent terms are found by setting m=0. Doing so with cubic > gives: > a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. That should be a^3 - 3a^2 = 0. I will be working from the outline posted in this thread from here on out for the Abel Prize $100,000 Giveaway. I see it as a challenge to win the money for myself, and put the rest of you out in the cold, yet again. James Harris === Subject: Re: JSH: $100,000 US offer, Abel Prize > Correction at 4. > 4. List of independent terms. Independent terms are found by setting m=0. Doing so with cubic gives: a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. > That should be > a^3 - 3a^2 = 0. > I will be working from the outline posted in this thread from here on > out for the Abel Prize $100,000 Giveaway. > I see it as a challenge to win the money for myself, and put the rest > of you out in the cold, yet again. Dream on, baby. Remember the prime counting challenge? Where you've been ABSOLUTELY SLAUGHTERED by me? When it is about boasting and lying, you are absolutely at the top of the chart. When it is about delivering, nobody has yet seen you delivering _anything_. === Subject: Re: JSH: $100,000 US offer, Abel Prize Correction at 4. > 4. List of independent terms. > Independent terms are found by setting m=0. Doing so with cubic > gives: > a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. That should be a^3 - 3a^2 = 0. I will be working from the outline posted in this thread from here on out for the Abel Prize $100,000 Giveaway. I see it as a challenge to win the money for myself, and put the rest of you out in the cold, yet again. James Harris === Subject: Re: JSH: $100,000 US offer, Abel Prize > Correction at 4. > 4. List of independent terms. Independent terms are found by setting m=0. Doing so with cubic gives: a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. That should be a^3 - 3a^2 = 0. I will be working from the outline posted in this thread from here on > out for the Abel Prize $100,000 Giveaway. I see it as a challenge to win the money for myself, and put the rest > of you out in the cold, yet again. > James Harris Why are you so rude? There's no need to be this rude. David Moran === Subject: Re: JSH: $100,000 US offer, Abel Prize === >Subject: JSH: $100,000 US offer, Abel Prize >One thing I've been fascinated by as I've considered replies to my >posts is a loose group coordination between posters, as some try to >post with math, and others just post various jibes, but all keep >focused on pushing the false notion that my rather basic argument >showing a problem with the de'nition of algebraic integers is wrong. Duh! Now I've tried to use money before, but failed as I think people in >the math community understand that NONE of you can overcome that kind >of coordinated effort, and even if you post your own version of my own >argument, the rest of math society will shred you. However, the shortness and simplicity of the proof of the problem in >core offers another solution--machine proof checking. That means I can make the offer of sharing at least $100,000 US with >ONE person or one group that collaborates to produce the check by >machine from the Abel Prize, assuming, of course that I win it. Can't work it out yourself, eh? Figures. The bene't to me is that the Abel Prize is close to $800,000 US so >what's a measly $100,000? And that way I know that some of you with >the expertise may be brave enough to go against the group, for the >money, knowing that all these attacks will fall apart when faced with >a check by computer. Of course, I will at some point start working on the computer check >myself, Sure you will. >and in that case, I will happily give myself the money, >assuming I win. That leaves any would be winners with me with the question, will >mathematicians behave so badly as to try and deny me any math prizes? Don't you have to have it published 'rst? If so, don't worry, I'll 'nd private funds to replace, but again, >that's assuming, so it's not a guarantee. I wish I could do better, >but the situation is odd. I can see mathematicians, unless completely broken by the event, >seeking revenge by trying to deny me any prizes, but then, there are >lots of millionaires and billionaires out there, who will probably >help out in that case. Hey, I just got an idea for a TV program. We'll have some guy pretend to be a millionaire and have him round up all the crackpots, like Harris, and have them compete for a million dollar prize. And on the last show when the best proof is selected, it will be revealed to be a hoax! Can you imagine how funny Harris would be if he really thought he had a chance to make money? Of course, to be selected, he'd have to convince the selection comittee, made up of, say, sci-math readers. No guarantees, but hey, for those of you with the expertise, at least >you can make math history, and potentially get serious cash in the >process. Here's the current outline of the proof: PROOF OF CORE ERROR > 1. Given P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > Variables: m, f, x, u E Ring of Algebraic integers > P(m) is a polynomial with m the key variable. > You are already in trouble here in step 1. You say that m is an algebraic integer. Then if P is a polynomial, P(m) is another algebraic integer, not a polynomial. Example: 5 is an algebraic integer. x^2 is a polynomial, and P(x) = x^2 is a polynomial function. P(5) is the algebraic integer 25. There is no such thing, unless you have previously de'ne it, as a key variable. You are not going to be able to get an automated proof without de'ning and using your terminology with much greater precision and consistency than you have so far. A good place to start would be for you to understand the distinction among a polynomial (as an element of a polynomial ring); a polynomial function; and an algebraic integer. === Subject: Re: $100,000 US offer, Abel Prize > One thing I've been fascinated by as I've considered replies to my > posts is a loose group coordination between posters, as some try to > post with math, and others just post various jibes, but all keep > focused on pushing the false notion that my rather basic argument > showing a problem with the de'nition of algebraic integers is wrong. Now I've tried to use money before, but failed as I think people in > the math community understand that NONE of you can overcome that kind > of coordinated effort, and even if you post your own version of my own > argument, the rest of math society will shred you. However, the shortness and simplicity of the proof of the problem in > core offers another solution--machine proof checking. That means I can make the offer of sharing at least $100,000 US with > ONE person or one group that collaborates to produce the check by > machine from the Abel Prize, assuming, of course that I win it. The bene't to me is that the Abel Prize is close to $800,000 US so > what's a measly $100,000? And that way I know that some of you with > the expertise may be brave enough to go against the group, for the > money, knowing that all these attacks will fall apart when faced with > a check by computer. Of course, I will at some point start working on the computer check > myself, and in that case, I will happily give myself the money, > assuming I win. That leaves any would be winners with me with the question, will > mathematicians behave so badly as to try and deny me any math prizes? If so, don't worry, I'll 'nd private funds to replace, but again, > that's assuming, so it's not a guarantee. I wish I could do better, > but the situation is odd. I can see mathematicians, unless completely broken by the event, > seeking revenge by trying to deny me any prizes, but then, there are > lots of millionaires and billionaires out there, who will probably > help out in that case. No guarantees, but hey, for those of you with the expertise, at least > you can make math history, and potentially get serious cash in the > process. Here's the current outline of the proof: PROOF OF CORE ERROR 1. Given P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) Variables: m, f, x, u E Ring of Algebraic integers P(m) is a polynomial with m the key variable. > 2. Let P(m) = (a_1(m) x + uf)(a_2(m) x + uf)(a_3(m) x + uf) Variables: a_1, a_2, a_3, roots of cubic de'ned as follows. Cubic: a^3 + 3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m) Solving for roots of P(m) by setting x = -uf/a will give cubic. > 3. Let P(m) = g_1(m) g_2(m) g_3(m) Variables: g_1, g_2, g_3 de'ned as follows g_1(m) = (a_1(m) x + uf), g_2(m) = (a_2(m) x + uf), g_3(m) = (a_3(m) x + uf). > Tautological base: Change in terms independent of m happens independent of m. > My point is that concentrating on terms independent of m, will give me > results independent of m. So I get terms independent of m. 4. List of independent terms. Independent terms are found by setting m=0. Doing so with cubic > gives: a^3 - 3a = 0, which gives a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. (indices selected arbitrarily) Then from previous de'nitions: a. g_1 has value uf at m=0 indicating indepedent term uf b. g_2 has value uf at m=0 indicating independent term uf c. g_3 has value 3x + uf at m=0 indicating independent term 3x + uf So far so good, as now I have independent terms, which therefore will > not change with m. CONDITION: f is coprime to 3, x and u This condition sets up the coprimeness results. PRIMARY ARGUMENT 5. Divide off f^2 from P(m). P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f Now I'm forcing a change by dividing off f^2, but again, I'm doing > things independent of the value of m. 6. List resultant independent term P(0)/f^2 = u^2(3x + uf) Note that P(0)/f^2 is coprime to f given the condition above. which is P(0)/f^2 = u^2 g_3(0) which is P(0)/f^2 = g_1(0)/f g_2(0)/f g_3(0) Here the proof is basically complete as I've determined that f is a > factor of two of the g's which may seem obvious, but remember, I'm > setting up for a machine to check. Detail is necessary. a. g_1(0)/f is coprime to f b. g_2(0)/f is coprime to f c. g_3(0) is coprime to f > Preliminary Finding: Independent terms of resultants are coprime to f. That sets up for the 'nale. 7. Backwards Theorem: Reverse use of distributive property proves > g_1, g_2 have factor that is f. I 'nd it interesting that no comments from others have been made here > at the heart of the proof. g_1(m) = a_1(m) x + uf g_2(m) = a_1(m) x + uf so to remove factor f from independent term uf, f must divide g_1(m) > and g_2(m), from reverse use of distributive property: g_1(m)/f = a_1(m)/f x + u. That is, I found that the independent term changes by a factor of f > when I divide P(m) by f^2, and I made that 'nding *independent* of m, > so now I use it without concern about the value of m, i.e. in general. Then by a reverse use of the distributive property, it follows that > a_1 and g_1 have f as a factor. (Note to readers: The point of using independent terms is that they > are *independent* of the value of m.) > 8. Veri'cation: Determine independent term of g_1(m)/f, by setting m=0. g_1(0)/f = a_1(0)/f x + u = u. Con'rmed factor f for g_1. Now maybe you see why posters have ignored the full outline in their > replies. Determine independent term of g_2(m)/f, by setting m=0. g_2(0)/f = a_2(0)/f x + u = u. Con'rmed factor f for g_1. You see, there's no room for reasonable doubt. 9. Core error determination g's do not have a factor that is f in the ring of algebraic integers > goes. > 10. Conclusion Core error, of course. James Harris Paying someone to agree with you...new all time low. No matter how you look at it, you are WRONG! David Moran === Subject: The Passing of a Mathematician Just today, in a perusal of some web sites that I visited on a routine basis, but could not visit for some time due to some problems with my Internet connection, I discovered that the world had lost an able and talented mathematician. I am referring to David Robbins, who passed away in September, at the age of 61, of pancreatic cancer. While the occasion of a man's death usually silences those who have unkind things to say about him, I suspect that there are some who felt that the world had lost the bene't of his talents before his death. This is because, although he had some work published in the open literature, much of his work was shrouded in the deepest secrecy, helping unravel mathematical problems connected with the codebreaking activities of the NSA. I do not share that view; I think the value of democracy, as we enjoy it in North America, Europe, and some other places, is obvious and great, and in devoting his life and talents to his country, he helped keep alive the kind of society in which other mathematicians can do their work in peace and in good conscience. But I suspect, too, that seeing many of their best students disappear into the NSA, some mathematics professors do bewail this aspect of the cost to humanity of national defense. Just as others, perhaps with somewhat less justi'cation, bewail the money spent on building bombs, in the belief that it could be used to grow more food. (Since, as Mark Twain noted, land is something of which they're not making any more, throwing more money at food production is likely only to drive up the price. Of course, that could still feed the hungry - giving more money to the right people, for example, could lead to less meat being eaten, increasing the ef'ciency of food production.) John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: The Passing of a Mathematician > [...] I think the value of democracy, as we enjoy > it in North America, Europe, and some other places, is obvious and > great, and in devoting his life and talents to his country, he helped > keep alive the kind of society in which other mathematicians can do > their work in peace and in good conscience. In which particular way does the NSA keep alive the value of democracy? Marc Olschok === Subject: Re: The Passing of a Mathematician > [...] I think the value of democracy, as we enjoy it in North America, Europe, and some other places, is obvious and great, and in devoting his life and talents to his country, he helped keep alive the kind of society in which other mathematicians can do their work in peace and in good conscience. > In which particular way does the NSA keep alive the value of democracy? Through helping defending the homeland of US, which in turn is the unique defender of democracy and human rights etc. etc. etc. in the world. M. K. Shen === Subject: Re: The Passing of a Mathematician === >Subject: The Passing of a Mathematician >Message-id: <3f89dd7a.248567@news.ecn.ab.ca >Just today, in a perusal of some web sites that I visited on a routine >basis, but could not visit for some time due to some problems with my >Internet connection, I discovered that the world had lost an able and >talented mathematician. I am referring to David Robbins, who passed away in September, at the >age of 61, of pancreatic cancer. While the occasion of a man's death usually silences those who have >unkind things to say about him, I suspect that there are some who felt >that the world had lost the bene't of his talents before his death. >This is because, although he had some work published in the open >literature, much of his work was shrouded in the deepest secrecy, >helping unravel mathematical problems connected with the codebreaking >activities of the NSA. I do not share that view; I think the value of democracy, as we enjoy >it in North America, Europe, and some other places, is obvious and >great, and in devoting his life and talents to his country, he helped >keep alive the kind of society in which other mathematicians can do >their work in peace and in good conscience. But I suspect, too, that seeing many of their best students disappear >into the NSA, some mathematics professors do bewail this aspect of the >cost to humanity of national defense. Just as others, perhaps with >somewhat less justi'cation, bewail the money spent on building bombs, >in the belief that it could be used to grow more food. Come on, you don't seriously believe that growing more food while the population remains unchecked is any kind of answer, do you? >(Since, as Mark >Twain noted, land is something of which they're not making any more, >throwing more money at food production is likely only to drive up the >price. Of course, that could still feed the hungry - giving more money >to the right people, for example, could lead to less meat being eaten, >increasing the ef'ciency of food production.) John Savard >http://home.ecn.ab.ca/~jsavard/index.html -- Mensanator Ace of Clubs === Subject: Euler books where can I 'nd english or german translations of some of euler's works in the internet? I can't afford to buy those. I know for instance that Euler's opera omnia is very expensive. Farncis === Subject: Re: Euler books > I can't afford to buy those. That's what libraries are for. They buy one copy, and then many people can read it. === Subject: Re: Euler books where can I 'nd english or german translations of some of euler's >works in the internet? >I can't afford to buy those. I know for instance that Euler's opera >omnia is very expensive. Farncis > I thought Dover, which is generally very reasonbly priced, published The Elements. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Euler books > I thought Dover, which is generally very reasonbly priced, published The Elements. That is by Euclid, an ancient Greek mathematician. Bob Kolker > === Subject: Re: Euler books > where can I 'nd english or german translations of some of euler's works in the internet? I can't afford to buy those. I know for instance that Euler's opera omnia is very expensive. Farncis > I thought Dover, which is generally very reasonbly priced, published > The Elements. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu I thought that The Elements was written by Euclid, unless of course Euler has a book titled The Elements that I am not aware of. I would around and see if you can 'nd something. It is possible that Cornell might have something online. (???) Lurch > === Subject: Re: Euler books > >>where can I 'nd english or german translations of some of euler's >works in the internet? >I can't afford to buy those. I know for instance that Euler's opera >omnia is very expensive. >>Farncis > >>I thought Dover, which is generally very reasonbly priced, published >>The Elements. >>-->Stephen J. Herschkorn herschko@rutcor.rutgers.edu >> >I thought that The Elements was written by Euclid, unless of course Euler >has a book titled The Elements that I am not aware of. I >around and see if you can 'nd something. It is possible that Cornell might >have something online. (???) Lurch > Oops. My apologies. I misread the original post, and my comment was not very helpful here. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: help me...my problem?? > sequence {An} A1 = 1 > A2 = 8 > An = root (An-1 * An-2) (n=3,4,5....) 'nd lim An (n->in'nite) --------------------------------- i want your warm advice. help me...please Hello. A hint for you. I tried this out on an Excel spreadsheet, and after just a few iterations, it was obvious that the answer was converging on an integer between 1 and 8. === Subject: Re: help me...my problem?? > sequence {An} A1 = 1 > A2 = 8 > An = root (An-1 * An-2) (n=3,4,5....) 'nd lim An (n->in'nite) --------------------------------- i want your warm advice. help me...please I don't know if it is warm or not, but ... Let B(n) = Log_2(A(n)). Then B(1) = 0 B(2) = 3 B(n) = (B(n-1) + B(n-2))/2 (#1) Then B(n) is in (B(n-2), B(n-1)) if n is odd, and B(n) is in (B(n-1), B(n-2)) is n is even. And B(n) - B(n-1) = (B(n-2) - B(n-1))/2. Then B(n) has limit as n --> inf. Let it be B. If you let n --> inf in #1, you only get B = (B + B)/2, that says nothing about B. Then from #1, 2B(n) = B(n-1) + B(n-2) > 2B(n) + B(n-1) = 2B(n-1) + B(n-2) = 2B(n-2) + B(n-3) = ... = 2B(2) + B(1) = 6 Then, when n --> inf, 2B + B = 3B = 6 > B = 2 and A = Lim(A(n), n, inf) = 2^2 = 4 If B(1) = p and B(2) = q, B = (p + 2q)/3 And if A(1) = r > 0, A(2) = s > 0, A(n) = sqrt(A(n-1)*A(n-2)) then A = cbrt((A(2))^2*A(1)) -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: help me...my problem?? In sci.math, hot-girl : > sequence {An} > A1 = 1 > A2 = 8 > An = root (An-1 * An-2) (n=3,4,5....) > 'nd lim An (n->in'nite) > --------------------------------- > i want your warm advice. > help me...please > You could try the sequence B1 = 0 B2 = 3 Bn = (Bn-1 + Bn-2) / 2, n > 2 and see where that goes, then relate Ai to Bi. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: help me...my problem?? > sequence {An} > A1 = 1 > A2 = 8 > An = root (An-1 * An-2) (n=3,4,5....) > 'nd lim An (n->in'nite) > Substitute some logarithm: Bn := log(An) to get the arithmetic mean instead of the geometric: Bn = (Bn-1 + Bn-2) / 2 Now solve this recurrence relation by another substitution (ok, this one's tricky): Cn := Bn - Bn-1. Towards which number do the Cn converge? How can you then calculate the limit of the Bn sequence? === Subject: Re: help me...my problem?? >sequence {An} A1 = 1 >A2 = 8 >An = root (An-1 * An-2) (n=3,4,5....) 'nd lim An (n->in'nite) --------------------------------- i want your warm advice. help me...please > Start by looking at Bn = 2log An -- Wim Benthem === Subject: how to differentiate the trace of a matrix with respect to a vector?(not hw prloblem) Please help me: how to differentiate tr[(A+B)*v*v'] with respect to column vector v, tr[(v*v')^2] with respect to column vector v(for this case, I guess since tr[(v*v')^2]=(tr[v*v'])^2, so I'd better 'nd Difftr[v*v'] with respect to v, am I right?) -Walala === Subject: Re: Is this newsgroup useless? <3F81A4DB.2364E9EE@ix.netcom.com> <3f85b353$16$fuzhry+tra$mr2ice@news.patriot.net> at 10:28 PM, Bart Goddard said: >3. This was meant to be humour, dumbass. *PLONK* -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Is this newsgroup useless? participant, I'm puzzled why this bothers you so much. The huge majority of posts on sci.math are about mathematics. I'm glad you said that, because I wanted to but hesitated to do so > because after all I'm one of the guys who's part of the problem > according to various posts here. That's my observation, as someone who plonks all the threads about supposed elementary proofs of the conjecture that Wiles turned into a theorem, but who actively looks for posts by some of the participants, including David Ullrich, who truth-squad those threads. There is plenty to read here, and plenty of it easy to read with the right newsreading software. > Seems to me that an even huger majority of the _threads_ are > about mathematics, making it even easier to 'lter out the crap > if you want. > It's easy to skip the other threads, assuming a moderately decent newsreader. Agreeing with the above, I do miss the pre-AOL days on most newsgroups. . . . -- Karl M. Bunday Christ has set us free. Galatians 5:1 Learn in Freedom (TM) http://learninfreedom.org/ kmbunday AT earthlink DOT net (preferred email address) === Subject: Re: topological terminology >So... homotope or homotop, which is the verb? Well, I know that Buckminster Fuller claimed that any noun can be verbed, but I would probably cringe if I saw that done to homotopy. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: topological terminology > So... homotope or homotop, which is the verb? > Pete > Use it in a sentence? > The space X homotoped into Y? > This avoids the question though. I've seen both homotope and homotop and >> the past tense of both would seem to be homotoped. I think homotope is >> more common, but that's also the one I use, so I'm biased. My point was that I never heard it used as a verb, so if you used it in >a sentence it would show something to me. I tend to be more interested in isotopy than homotopy. Here's the only sentence I could 'nd on the hard drive here at work (but I can look at home too if you really want a corpus): Now the process adequately sketched in Fig.~ref{handle 'gure} (swapping a collar of their common boundary from one handlebody to the other to shrink the transverse disks and make $Bcap bigcup_{k,r}D_{k+rnu}=emptyset$; shortening the $1$--handles around their shrunken transverse disks to make $Bcap bigcup_{k,r}eta_s(h1{k+rnu})=emptyset$; isotoping the handlebody decomposition with shrunken, ^^^^^^^^^ shortened $1$--handles back to the original one, and carrying $B$ along by the isotopy) lets $B$ acquire property~eqref{braid misses 1-handles} while preserving properties eqref{braid exterior to binding} and eqref{braid in Hl}. I try to avoid the in'nitive because isotop looks bad and isotope reeks of chemistry. This last objection doesn't apply to homotope, of course. Lee Rudolph === Subject: Re: Subspace question In , on >Is R2 considered a subspace of R3? No, but it's isomorphic to a subspace. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Subspace question >I'm a linear algebra student, and normally I would ask this to my TA, but he >is out to a conference. Is R2 considered a subspace of R3? I'm guessing no, because R2 cannot >contain the origin in R3 (0,0,0). Strictly speaking no, R^2 is not a subspace of R^3, because as you point out the elements of R^2 are not even elements of R^3. But a simple no here is probably not the whole story - people do sometimes talk about R^2 as a subspace of R^3 - when they do that they're speaking loosely, considering (x,y) and (x,y,0) as the same thing. (Also note that in some situations is a subspace of means is isomorphic to a subspace of - if that's what we mean then of course the answer is yes.) > ************************ David C. Ullrich === Subject: Cauchy sequences with Random Variables I am trying to do a problem on cauchy sequences with random variables, but don't know what exactly I need to prove. Show that the sequence of random variables X_n converges with probability one if and only if the random variables R_N de'ned by R_N = sup |X_m - X_n| satisfy R_N converges in probability to 0. (with m,n greater than or equal to N). This looks somewhat complicated if I use the de'nitions of almost sure convergence(of converges with probability one) and convergence in probability. Can I just merely prove that X_n is a cauchy sequence in the real numbers if and only if X_n converges? Steve === Subject: Slow Factoring Method (was: factoring to satis'ability) There has been a thread about converting factoring problems into Boolean Satis'ability problems. Let's call these FACTSAT problems. I want to describe a method for generating FACTSAT problems. I don't know much about factoring and much of this may be elementary. Start with a Boolean circuit that can multiply two 2-bit binary numbers. I use my own strange notation. A1 means the variable (A). A0 means (NOT(A)) A0B1 means (NOT(A) AND B) A1 + B0 means (A OR NOT(B)) x*y means x times y z = x * y de'ne z as ZYXW de'ne x as CA de'ne y as DB ZYXW = CA * DB W1 = B1A1 X1 = D1B0A1 + D1C0A1 + C1B1A0 + D0C1B1 Y1 = D1C1A0 + D1C1B0 Z1 = D1C1B1A1 I also want the inverse of this circuit. This is easily done using deMorgan's theorem. W0 = (B0+A0) X0 = (D0+B1+A0)(D0+C1+A0)(C0+B0+A1)(D1+C0+B0) Y0 = (D0+C0+A1)(D0+C0+B1) Z0 = (D0+C0+B0+A0) Let's factor 6 (base 10). We know that the output variables of the multiplication circuit must be Z0Y1X1W0. W0 = B0 + A0 X1W0 = D1B0A1 + C1B1A0 Y1X1W1 = D1C1B0A1 + D1C1B1A0 Z0Y1X1W1 = D1C1B0A1 + D1C1B1A0 Convert these satisfying assignments back into numbers: Z0Y1X1W1 = D1C1B0A1 + D1C1B1A0 ZYXW = CA * DB 0110 = (11*10) + (10*11) 6 = (3*2) or (2*3) We have found all 2-bit * 2-bit factors of 6. The factors 1*6 and 6*1 are not in this list because it takes three bits to represent 6. We would need a 3*3 FACTSAT to 'nd 6*1. There is a much easier way to genertate this FACTSAT. FACTSAT can be converted into a modular arithmetic problem, like many Boolean expressions. We are essentially taking a number, z, and calculating z mod 2^i for all i>0. Using 6 we see: 6 mod 2^1 = 0 6 mod 2^2 = 2 6 mod 2^3 = 6 Any factors of 6 must satisfy this same set of conditions: x*y mod 2 = 0 x*y mod 4 = 2 x*y mod 8 = 6 Consider a set of n*n bit multipliers. Start with a 1*1 multiplier. Since z mod 2 = 0 we know there are only three possible values for x and y: x = 0 or 1 y = 0 or 1 x*y mod 2 = 0 x*y = 0*0 or 1*0 or 0*1 We can reduce this list by only listing x*y only if x>=y. We need to remember that x*y actually represents two satisfying assignment, x*y and y*x, unless x=y. x*y mod 2 = 0 x*y = 1*0 or 0*0. Now, consider a 2*2 bit FACTSAT for 6. We can start with the solutions we found for the 1*1 FACTSAT. There are only four possible combinations for the two new bits we are adding to the FACTSAT. x*y = ?1*?0 01*00 mod 4 = 0 (1*0=0) 01*10 mod 4 = 2 (1*2=2) 11*00 mod 4 = 0 (3*0=0) 11*10 mod 4 = 2 (3*2=6) x*y = ?0*?0 00*00 mod 4 = 0 (0*0=0) 00*10 mod 4 = 0 (0*2=0) 10*00 mod 4 = 0 (2*0=0) 10*10 mod 4 = 0 (2*2=4) There are only two solutions such that: x*y mod 4 = 2 x*y = 2*1 or 3*2 I think these are called rings. This is the set of all pairs of non-negative integers x*y such that: (x >= y) and (x < 4) and (y < 4) and (x*y mod 2 = 0) and (x*y mod 4 = 2). Note that 2*1 =/= 6. Even though I have been calling this a FACTSAT, not every satisfying assignment is actually a factor of 6. What this formula really 'nds is the two low order bits of any factors of z where the two low order bits of z are equal to 10. For example, 10 base 10 = 1010 base 2. 10 mod 2 = 0 10 mod 4 = 2 x*y = 5*2 = 101*010 = ?10*?01 = 2*1 We can convert this modular arithmetic formula back into DNF: (Remember that x*y represents x*y and y*x if x=/=y). z mod 2 = 0 z mod 4 = 2 z = x*y = 2*1 or 1*2 or 3*2 or 2*3 X1W0 = 10*01 or 01*10 or 11*10 or 10*11 X1W0 = D1B0*C0A1 + D0B1*C1A0 + D1B1*C1A0 + D1B0*C1A1 X1W0 = D1B0A1 + C1B1A0 This is the same formula we got above. Say we want to 'nd any 50*50 bit factors of a 100 bit number. This would result in a FACTSAT with at least 100 variables. We could use the method above to create a 10*10 FACTSAT that is a sub-problem of the 50*50 FACTSAT. Since this 10*10 FACTSAT is in DNF, it really represents a list of satisfying assignments for the 10*10 sub-problem. The 10*10 sub-problem could be used to pre-process the 50*50 problem. The 50*50 problem could be reduced from 100 variables to 80 variables and a list of all satisfying assignments for the other 20 variables. I have written a slow factoring program that uses the method I described. This program is able to factor 62,062,883 = 7877 * 7879 in a few seconds. The program created a 13*13 bit FACTSAT with 2048 solutions for 62,062,883. This is a FACTSAT with 26 variables and about 4096 DNF clauses. This is a 6x6 bit (12 variable) FACTSAT for 1023: Factor 1023 Bit 0 : 1023 mod 2 = 1 1*1 Bit 0: 1 solution. Bit 1: 1023 mod 4 = 3 3*1 Bit 1: 1 solution. Bit 2: 1023 mod 8 = 7 7*1 5*3 Bit 2: 2 solutions. Bit 3: 1023 mod 16 = 15 15*1 13*11 9*7 5*3 Bit 3: 4 solutions. Bit 4: 1023 mod 32 = 31 31*1 29*11 27*13 25*23 21*3 19*5 17*15 9*7 Bit 4: 8 solutions. Bit 5: 1023 mod 64 = 63 63*1 61*43 59*13 57*55 53*35 51*5 49*47 45*27 41*39 37*19 --33*31 29*11 25*23 21*3 17*15 9*7 Bit 5: 16 solutions. Factor found. Russell - 2 many 2 count === Subject: Re: Slow Factoring Method > There is a much easier way to genertate this FACTSAT. > FACTSAT can be converted into a modular arithmetic problem, > like many Boolean expressions. > We are essentially taking a number, z, and calculating > z mod 2^i for all i>0. if z is a 2N bit number which is the product of two odd N-bit numbers, then for any odd N-bit number x there will exist another odd N-bit number y such that z = x * y mod 2^N This is because any odd residue modulo a power of two has a multiplicative inverse. So we just take y = (z * x^-1 mod 2^N). There are, as you may have noticed, 2^(N-1) odd residues modulo 2^N. So by the time you get through your least signi'cant half product, you will be tallying satisfying assignments whose number is exponential in the problem size. > Say we want to 'nd any 50*50 bit factors of a 100 bit number. > This would result in a FACTSAT with at least 100 variables. > We could use the method above to create a 10*10 FACTSAT > that is a sub-problem of the 50*50 FACTSAT. > Since this 10*10 FACTSAT is in DNF, it really represents a list > of satisfying assignments for the 10*10 sub-problem. > The 10*10 sub-problem could be used to pre-process the 50*50 problem. > The 50*50 problem could be reduced from 100 variables to 80 variables > and a list of all satisfying assignments for the other 20 variables. As Yoda might say, In combinatorics, free lunch, there is not, hmmmmm? Again, culling isn't going to occur until you are into the most signi'cant half of the output bits, and by that time, your list of satisfying assignments for the sub problem will total 2^49 in number. > Factor 1023 > Bit 1: 1 solution. ... > Bit 2: 2 solutions. ... > Bit 3: 4 solutions. ... > Bit 4: 8 solutions. ... > Bit 5: 16 solutions. The technical term for this is doubling. -- Eric Michael Cordian 0+ O:.T:.O:. Mathematical Munitions Division Do What Thou Wilt Shall Be The Whole Of The Law === Subject: Re: Slow Factoring Method > There is a much easier way to genertate this FACTSAT. FACTSAT can be converted into a modular arithmetic problem, like many Boolean expressions. > We are essentially taking a number, z, and calculating z mod 2^i for all i>0. if z is a 2N bit number which is the product of two odd N-bit numbers, > then for any odd N-bit number x there will exist another odd N-bit > number y such that z = x * y mod 2^N This is because any odd residue modulo a power of two has a multiplicative > inverse. So we just take y = (z * x^-1 mod 2^N). There are, as you may have noticed, 2^(N-1) odd residues modulo 2^N. So by the time you get through your least signi'cant half product, you > will be tallying satisfying assignments whose number is exponential in the > problem size. > I said it was slow. Say we want to 'nd any 50*50 bit factors of a 100 bit number. This would result in a FACTSAT with at least 100 variables. We could use the method above to create a 10*10 FACTSAT that is a sub-problem of the 50*50 FACTSAT. > Since this 10*10 FACTSAT is in DNF, it really represents a list of satisfying assignments for the 10*10 sub-problem. The 10*10 sub-problem could be used to pre-process the 50*50 problem. The 50*50 problem could be reduced from 100 variables to 80 variables and a list of all satisfying assignments for the other 20 variables. As Yoda might say, In combinatorics, free lunch, there is not, hmmmmm? Again, culling isn't going to occur until you are into the most > signi'cant half of the output bits, and by that time, your list of > satisfying assignments for the sub problem will total 2^49 in number. It's not pretty. I think you are forgetting that I am only interested in x and y where x < 2^10 and y < 2^10 in the 10*10 sub-problem. The DNF form will still have a lot of clauses. But I can solve the 10*10 sub-problem and its not as bad as it looks. Using my program it appears a 10*10 FACTSAT for an odd number with 20 or more bits usually has 256 solutions. That's not bad for a 20 variable SAT problem. Disk space is cheap. It is interesting that a FACTSAT reduced to the minimum number of variables requires so many clauses. A lot of the solutions will reduce in a minimal DNF. Factor 1023 > Bit 1: 1 solution. > ... Bit 2: 2 solutions. > ... Bit 3: 4 solutions. > ... Bit 4: 8 solutions. > ... Bit 5: 16 solutions. The technical term for this is doubling. True. But it shows there are only 16 (actually 32) possible solutions to a 12 variable SAT problem. A FACTSAT for an even number can grow even faster. Russell - 2 many 2 count === Subject: Re: Spivak's Calculus at 02:03 PM, lrudolph@panix.com (Lee Rudolph) said: >Have you looked at Apostol's _Mathematical Analysis_? An excellent book, with some marvelous exercises. Is it still in print? I'm not sure, though, that I would recommend it to a student without a certain amount of Mathematical maturity. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Spivak's Calculus Do you feel it would be worth purchasing the 1st edition? It is only >$20. >Compared with the near $100 for the 3rd. I am just looking for a gentle >introduction to Calculus/Analysis that I can use as self-teaching text. >For >my background, I 'nd Rudin too terse. >> Whoo-hoo. Gentle is not the word I'd have chosen 'rst, or tenth, >> to describe Spivak's _Calculus_. The problems, in particular, tend >> to be brutal. >> Have you looked at Apostol's _Mathematical Analysis_? Or, for that >> matter, Courant's calculus textbook? I have never seen Rudin's >> calculus text, just (many many years ago) his _Real and Complex >> Analysis_, and slightly more recently a monograph or two on some >> topics in several complex variables. But I don't think Spivak >> is notably less terse than Rudin, and I do think that both >> Apostol and Courant are probably less terse while being quite >> as rigorous as necessary. >> Lee Rudolph Lee, are you refering to Spivak's Calculus, or to his Calculus on manifolds ? He's referring to Calculus. I deduce this from two facts: (i) he _said_ he was referring to Calculus, (ii) everything he said about that book is absolutely true. >Lurch > ************************ David C. Ullrich === Subject: Re: Naive Q: Set theory, logic - which comes 'rst? <1xUcb.14254$O85.6040@pd7tw1no> <3f79e264$7$fuzhry+tra$mr2ice@news.patriot.net> <3f82ebb8$3$fuzhry+tra$mr2ice@news.patriot.net> >The point I was trying to make was that you have to distinguish >between provability in a speci'c system and provability in general, I studied Mathematics, not Metaphysics. My Mathematical training did not encompass provability in general; we had to prove things in speci'c axiom systems with speci'c rules of inference. G.9adel showed results concerning the axiom system that we call the integers, and that was what I claimed. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Naive Q: Set theory, logic - which comes 'rst? > The point I was trying to make was that you have to distinguish between provability in a speci'c system and provability in general, > I studied Mathematics, not Metaphysics. My Mathematical training did > not encompass provability in general; we had to prove things in > speci'c axiom systems with speci'c rules of inference. G.9adel showed > results concerning the axiom system that we call the integers, and > that was what I claimed. When mathematicians say that a theorem has been proven, they mean that its truth has been established beyond all doubt. That is not the same as provability in a speci'c formal system. The integers have more properties than can be captured by any axiom system. Therefore no axiom system deserves the name the integers. G.9adel's results were not about any speci'c axiom system, but about a large class of axiom systems. You said that there are statements about natural numbers that can neither be proven nor disproven, which may or may not be true, but does not follow from what G.9adel showed. G.9adel did show that in a given formal system for natural numbers there are statements about natural numbers that can neither be proven nor disproven. Mattias === Subject: Re: Naive Q: Set theory, logic - which comes 'rst? > at 01:50 PM, mattias_wikst71@hotmail.com (Mattias Wikstr?m) said: [You snipped the following:] > Personally I would say that the twin prime hypothesis is de'nitely either true or false, because the intuitive notion of the natural numbers determines that system of objects completely. > > Alas, Kurt G.9adel showed that to be false; there are statements about > natural numbers that can neither be proven nor disproven. [End of snipped text] G.9adel did not show this. He showed that for a given sound formal system in which the statements about natural numbers can be formulated, there are statements that can neither be proven nor disproven. > Google for G.9adel number. He represented statements as natural > numbers. The theorems he proved about statements were also theorems > about the integers representing them. Of course. The point I was trying to make was that you have to distinguish between provability in a speci'c system and provability in general, and that G.9adel showed nothing about the latter. Mattias === Subject: Re: Homeomorphism and boundaries <3f85d5c4$21$fuzhry+tra$mr2ice@news.patriot.net> at 11:12 PM, magidin@math.berkeley.edu (Arturo Magidin) said: >And is f a homeomorphism? That was part of the hypothesis above. Sorry, I missed that. I thought that the claim was that a 1-1, continuous and open map must preserve boundaries, and *that* is false. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Homeomorphism and boundaries Visiting Assistant Professor at the University of Montana. > at 11:12 PM, magidin@math.berkeley.edu (Arturo Magidin) said: >And is f a homeomorphism? That was part of the hypothesis above. Sorry, I missed that. I thought that the claim was that a 1-1, >continuous and open map must preserve boundaries, and *that* is false. Yes; an open, continuous, 1-1 map preservers interiors; a surjective, closed, continuous map preserves closures. Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Homeomorphism and boundaries Visiting Assistant Professor at the University of Montana. > at 11:12 PM, magidin@math.berkeley.edu (Arturo Magidin) said: >And is f a homeomorphism? That was part of the hypothesis above. Sorry, I missed that. I thought that the claim was that a 1-1, >continuous and open map must preserve boundaries, and *that* is false. Did your examples satisfy that condition? You mapped (0,2pi) to RxR by sending x to (0,x), but that map is not open, because (0,2pi) is an open set in the domain but f(0,2pi) is not an open set in RxR. It's not quite closed, either, but it would be if you go to [0,2pi]. Then you mapped (0,2pi) to C by sending x to exp(ix), and that map is also not an open map. Of course, if your domain changed to [0,2pi], compact, then it would be a closed continuous mapping, but it would not be injective. Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Homeomorphism and boundaries Visiting Assistant Professor at the University of Montana. > at 06:15 PM, magidin@math.berkeley.edu (Arturo Magidin) said: >f(int(A)) = f( union{B : B is open, B contained in A}) >> = union{ f(B): B is open, B contained in A} (since f is >>bijective) >> = union{ f(B): f(B) is open, f(B) contained in f(A)} >> = int(f(A)) No. Consider A=(0,2Pi) with the standard topoplgy, B=RxR with the >standard topoplgy. De'ne f: A -> B as f(x) =(0,x). Int(A) = A, but >Int(f(A)) is the null set. Just before the equations above, there was the following line: Thus, if f:X->Y is a homeomorphism, it follows that for all A, By removing it, you make it seem that I am saying something different; my equations applied only for the case of f a homeomorphism, which your example is not. >>f(clos(A)) = f( intersection{B: B closed, B contains A}) >> = intersection {f(B): B closed, B contains A} >> = intersection {f(B): f(B) closed, f(B) contains f(A)} >> = clos(f(A)). No. Consider A=(0,2Pi) with the standard topoplgy. De'ne f: A -> C as >f(x) = exp(i*x). cl(A)=A, but f(A) is not closed; it has the limit >point 1. No. Your f is not a homeomorphism. Arturo Magidin magidin@math.berkeley.edu === Subject: Re: oh...my god....genius very...very...very...thank you..sir very...very...very... you are perfect genius. === Subject: Re: oh...my god....genius > very...very...very...thank you..sir > very...very...very... > you are perfect genius. > Well, I practice a lot when I'm alone. -- Chris Henrich To succeed it is not enough to be stupid: you must also be well-mannered.' -Voltaire === Subject: Re: De Moivre formula at 02:42 PM, kyh <@hotmail.com> said: >I have some question about De Moivre formula. As soon as you start dealing with complex numbers, you need to understand that LN (natural log) is a multi-valued function, and that accordingly exponential expressions are ambiguous.I have been simply accepted that formula as : (cos t + i*sin t)^n = >cos(tn)+i*sin(tn) >and I regard it as true for every t and n. For n an integer. Beyond that you need to deal with the ambiguity. >but, in case t=Pi/6 and n=2, De Moivre formula doesn't hold. What makes you believe that? It's more generally useful to use the form exp(i*theta) = cos theta + i*sin theta, from which you can trivially derive the other. Note that I avoided the more conventional notation e^i*theta because exp is a well de'ned function not subject to the ambiguities of exponentiation. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: De Moivre formula > Ooops!!! > I misunderstood sin <=> cos for solving following exercise. : What is sum of natural #s n s.t. (sin t + i*cos t)^n=sin(nt)+i*cos(nt) ? Now, I see my misunderstanding point. > It is true for all integer n, s.t. n = 1 (mod 4). -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Antidiagonal, In'nity <3f79d8b6$5$fuzhry+tra$mr2ice@news.patriot.net> <3f7cafc2$11$fuzhry+tra$mr2ice@news.patriot.net> <3f82e66f$1$fuzhry+tra$mr2ice@news.patriot.net> <3f85c273$18$fuzhry+tra$mr2ice@news.patriot.net> at 09:58 PM, raf@tiki-lounge.com (Ross A. Finlayson) said: >Look at a probability distribution function. Which one? And what do you mean by probability distribution function? >Is it everywhere discontinuous? Depending on the de'nition, the question might not even have a meaning. >You don't have to §ip in'nitely many coins to tell if a real >number from [0,1) is less than 1/2, just one. >Consider x and y, independent variables, x goes to in'nity: y/x is >effectively zero or indeterminate, No. y/x is a function of x and y. The limit of y/x as x approaches in'nity is 0, regardless of the value of y. There is no indeterminism. >If y is a dependent variable of x, then, y/x is >not necessarily zero or indeterminate (divergent). Indeterminate is not the same as divergent. >I'm pretty sure that the elements within the hyperreals are the same >as those in the reals. You're pretty wrong. >Mathematics is a particularly tractable subject: Only if you actually study it instead of inventing unde'ned terms as you go along. >By the same token, those areas in which not all agree are the >signposts of areas where there is room for development, eg Euclid's >Fifth Postulate: Agree or Disagree. Disagree: there is no room for development of things know to be impossible or wrong. There is no room for attempting to express 2^.5 is a ratio of integers. There is no room for attempting to trisect an angle with compass and straightedge. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Antidiagonal, In'nity > I consider that there is not a next. I also consider that there is. Then all is revealed! -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: The physical universe: an abelian group? at 02:44 PM, snizpilbor@yahoo.com (Sniz Pilbor) said: >If we de'ne multiplication of two bits of matter/antimatter as >connecting them (a*b = a connected to b)... What does that mean? >ssociativity and commutativity are obvious, Not until you de'ne what you mean by connecting them, and maybe not even then. >inverse of a piece of matter is an equal piece of antimatter, That would imply that an electron connected to a positron is a vacuum. I'm having trouble 'guring out what kind of de'nition of connected to would lead to that result.What happened to the photons? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: The physical universe: an abelian group? > If we de'ne multiplication of two bits of matter/antimatter as > connecting them (a*b = a connected to b)... associativity and > commutativity are obvious, it certainly seems like closure is obvious > (how can you connect two pieces of matter/antimatter to obtain > something out of this universe?), the unit is vacuum and as for > inverse... inverse of a piece of matter is an equal piece of > antimatter, and vice versa. It seems to me like this very de'nitely > makes the physical universe an abelian group. Am i correct? No, I believe that some chemical reactions are noncommutative or nonassociative, but your idea is brilliant. > Something I was thinking about... if we do allow the physical > universe, ie the set of all matter and antimatter and the vacuum, to > be an abelian group, call it U, then it seems to me that if someone > were raised from birth percieving, instead of U, some set S which is > isomorphic to U, they would never ever be able to tell that they were > unnormal. (As an example: if, from birth, you saw everything upside > down. You would think it were perfectly normal and what's more, it > would be impossible to diagnose). Am i mistaken? No, I refer you to an interesting book that I enjoyed. It is called There are two errors in the the title of this book. It is about paradoxes similar to the book's title and has a chapter on different universes such as a Right/Left handed Universe or a uniform doubling of size. > What sort of subgroups of the physical universe could we come up with? > Perhaps, for example, the subgroup of all transparent > matter/antimatter? The subgroup of all matter/antimatter which does > not contain some basic element (an element that can't be created by > combining different elements)? Alot of things seem like candidates, > except that all these subgroups necessarily must contain the vacuum > (this rules out, for example, a subgroup of all organic, or inorganic, > material) The trivial subgroup, the vacuum. The set of all protons (or any === Subject: re: Topology of the Universe Jack, You say, In particular I do not as yet see any physical necessity for Saul-Paul's Ansatz that physical 3D expanding accelerating cosmic space is actually SU(2)/ID rather than the SU(2) shown in Fig 3... The idea that 3D cosmological space SU(2)/ID rather than SU(2) is the view of Luminet et al. in their *Nature* (9 October) paper. They say (p. 593-594): The Poincare dodecahedral space is a dodecahedral block of space with opposite faces abstractly glued together, so objects passing out of the dodecahedron across any face return from the opposite face. Light travels across the faces in the same way, so if we sit inside the dodecahedron and look outward across a face, our line of sight re-enters the dodecahedron from the opposite face. We have the ILLUSION [emphasis added] of looking into an adjacent copy of the dodecahedron. Yes, but I have a hard time understanding how to translate that clear formal statement into physical operational informal terms of what an Argonaut with Alcubierre warp drive /zpf 'eld super-technology would actually experience on a fast trip across space now when c/Ho ~ 10^28 cm? Suppose Kirk & Crew shoot across a distance ~ 10^28 cm in say one week proper time. Since they are, at all points on their worldline, on a local free-§oat timelike geodesic of their own making, Alcubierre points out that there is no necessary huge time-dilation twin effect as in global special relativity. That is, they can return in ~ 1 week of Earth time. So now the Mission is to forge straight ahead not to do a 180 turn around back to Earth. With your SU(2)/ID which Lou agrees with, they do so continuously and get back to Earth anyway! OK maybe it works that way. But maybe it does not? That is an empirical issue that cannot be described by pure mathematics alone. The issue here is how to tie the pure math to the actual experiences. Perhaps they do approach an edge or a wall in space that is a giant wormhole and they can see Earth straight ahead through the wormhole mouth. It's a big mouth of course of order c/Ho diameter. I suppose you would say, Well the mouth is so big, that there is no real operational difference in the two descriptions. Perhaps that is correct. How could we pin down the speci'c 'nite topology of SU(2)/ID of 120 spherical 3D snugly tileddodecahedrons for example. I mean suppose Kirk's ship is tracked by radar as an FTL object on what appears to us to be a spacelike worldline even though locally on the ship it is timelike geodesic. Will we see 120 images of the star ship at some point? Or something like that? We must be able to detect SU(2)/ID in some way for the Star Ship voyage for it to make physical sense and to tell it is not some other topology? face return from the opposite face and I have yet to understand physically what exactly that formal idea means. What does it mean for an object to pass out of any face the dodecahedron and to then return from the opposite face? The two faces have a co-moving separation c/H(t) does the object return at the same t as in a Star Gate or not? Also maybe we can have it return at t +- delta t for a time machine effect with some tweaking. What does the word face mean physically, or operationally, in Luminet's writing? That is the essence of my question of interpretation of the math here I suppose. Poincare dodecahedral space as they describe it is exactly the same as SU(2)/ID. These authors are quite aware of this equivalence as they show in the paper Cosmic microwave background constraints on multi-connected spherical spaces, which you can download from: http://xxx.lanl.gov/abs/astro-ph/0303580 On page 2 (of this 5 page paper), they say: The 'nite subgroups of S^3 are the cyclic groups Zn, the binary dihedral groups D*m, the binary tetrahedral, octahedral and icosahedral groups, respectively of order n, 4m, 24, 48 and 120. Now since you are looking for a physically signi'cant difference between SU(2)/ID and SU(2) tiled with 120 spherical dodecahedra, I should point out the fundamental signi'cance of the volume of the 3D space, which affects both the light travel times as well as the density of the space. As these authors say: In all cases the volume of the space S^3/G is the volume of the 3-sphere S^3 divided by the order |G| of the holonomy group. structures S^3/G, and that has to do with the A-D-E classi'cation of these spaces. This is because A-D-E Coxeter graphs classify at least 20 physically relevant mathematical objects. As I have mentioned previously, these include Coxeter (re§ection) groups, Lie algebras, gravitational instanton spaces, catastrophe bundles, 2D conformal 'eld theories, Heisenberg algebras, and much more. Agreed, how do we get physically testable ideas out of that math? That's a rhetorical question at this point. I don't expect you to have de'nite answers. I have started writing this up in a Word 'le, and will send it to you as soon as I 'nish it. Saul-Paul Cool. I am still not clear on how to physically interpret Luminet's Fig 3. === Subject: Re: deep holes of leech lattice Stephen: as I am sure that the answer might have been a bit overwhelming for the questioneer. But I don't know that much about lattice constructions either, so my mouth is still hanging open too. === Subject: Re: deep holes of leech lattice by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9C5Qvn22692; >Hello Is there any algorithm to calculate the deep holes of leech lattice? >Any software can do that? I tried MAGMA but MAGMA gave no result >I read SPLAG however I am not clear about it >I wanna have the whole collection of leech lattice deep holes >is there any way for me to calculate them? 24! = 620448401733239439360000 |C0(0)| = 24! / 74613 |Co(0)| = 8315553613086720000 A fundamental domain has |C0(0)| / g(P) holes of type P. number of holes Name vertices -------------------------------------------------------- 8292375 A(1)^24 48 60010496000 A(2)^12 36 12084312240000 A(3)^8 32 277185120436224 A(4)^6 30 60153020928000 D(4)^6 30 2406120837120000 A(5)^4 D(4) 29 7071048990720000 A(6)^4 28 21655087534080000 D(6)^4 28 19248966696960000 E(6)^4 28 32482631301120000 A(7)^2 D(5)^2 28 25665288929280000 A(8)^3 27 173240700272640000 D(8)^3 27 1385925602181120000 E(8)^3 27 103944420163584000 A(9)^2 D(6) 27 1039444201635840000 D(10) E(7)^2 27 346481400545280000 A(11) D(7) E(6) 27 159914492559360000 A(12)^2 26 1039444201635840000 D(12)^2 26 519722100817920000 A(15) D(9) 26 4157776806543360000 D(16) E(8) 26 692962801090560000 A(17) E(7) 26 831555361308672000 A(24) 25 4157776806543360000 D(24) 25 -------------------------------------------------------- 14717067522059208599 deep holes in a fundamental domain (data taken from table 23.1 and calculated from table 25.1a) In a fundamental domain, a 20 digit number of deep holes. Are you sure you want them all? Chapter 23 gives explicit MOG coordinates for some vertices surrounding a hole of each type. The center of a hole can be found by a weighted average of the vertices in any of the connected components of the Coxeter-Dynkin diagram of the hole. The weights to use are listed in 'gure 23.1. The meaning of the lines in the Coxeter-Dynkin diagram is shown in 'gure 23.3. The 4 by 6 boxes contain MOG coordinates. The box arrangement is easier to read than 24 coordinates listed in a row. In several examples, the coordinates of the center of the hole are given. Some vertices of a hole of type A(4)^6 are shown in 'gure 23.5. Theorem 12 gives the center at c = (1/5) (4,8,4,4, ... 6,2,2,2). Each of the six A(4) components is a pentagon, adding up to 30 vertices. As explained in 'gure 23.3, the 30 pentagon edge lengths are sqrt(6). There are 435 ways to choose 2 out of the 30 vertices. Of these, the 30 edges have length sqrt(6), and the 405 non-edges have length sqrt(4). The 30 vertices of this hole, as with any deep hole, are sqrt(2) away from the center. Chapter 11 explains the MOG coordinate system. Figure 4.12 shows the lattice generator matrix. Table 4.14 lists the number of vertices at distance sqrt(4), sqrt(6), sqrt(8), etc. from a given lattice vertex. (The table and 'gure references are to the 2nd edition of SPLAG.) === Subject: Re: Solving recurrence rel. a_2n = F(a_n) >>I am trying to 'nd a method for solving linear recurrence relations of >>the form: >> a_2n = F(a_n) >>For example, >> a_2n = c * a_n + d * a_n-1 + b * a_n-2 + 1 >>Could someone point me to literature that will help me? >>--john > Look up generating functions. Perhaps a good place to start is > http://mathworld.wolfram.com/GeneratingFunction.html. > Your example seems incomplete, however. What about the odd terms? I apologize! Yes, I forgot to say that the odd terms of the recurrence relation are also linear: a_2n+1 = G(a_n) For example, a_2n+1 = k * a_n - 1 F() and G() are linear functions. --john === Subject: Re: Solving recurrence rel. a_2n = F(a_n) > I am trying to 'nd a method for solving linear recurrence relations of > the form: > a_2n = F(a_n) > For example, > a_2n = c * a_n + d * a_n-1 + b * a_n-2 + 1 > Could someone point me to literature that will help me? > --john Look up generating functions. Perhaps a good place to start is http://mathworld.wolfram.com/GeneratingFunction.html. Your example seems incomplete, however. What about the odd terms? === Subject: binomial theorem -help wanted could anybody help out a novice with these 2 questions http://users.ncable.net.au/~luddite/15.htm and http://users.ncable.net.au/~luddite/23.htm any help or pointers would br greatly appreciated === Subject: Re: binomial theorem -help wanted > could anybody help out a novice with these 2 questions > http://users.ncable.net.au/~luddite/15.htm > and > http://users.ncable.net.au/~luddite/23.htm > any help or pointers would br greatly appreciated To show 0 = sum((-1)^k*C(n, k), k=0..n) look at (1 - 1)^n. To show, for even n, sum(C(n, 2*k), k=0..n/2) = sum(C(n,2*k-1), k=1..n/2) = 2^(n-1) write out the above sum. -- Paul Sperry Columbia, SC (USA) === Subject: Re: unprovability of the security of computational cryptography > In the recent paper Resource Bounded Unprovability of Computational Lower > Bounds (Okamoto & Kashima) the authors claim that the security of any > computational cryptographic scheme is unprovable in the standard notation of > the modern cryptography, .... Does it mean that the security is unprovable > even under plausible assumptions (P!=NP, NP!=coNP, etc)? > Your question could, it seems to me, interpreted in a number of ways: 1. Is the quote saying that ...? 2. Is the paper trying to say that ...? 3. Is it true that ...? My reply to the 'rst interpretation is: I do not know what it (the quote) is saying. Anybody cares to educate me on what notation is or is not standard? What would happen to the provability of security if we start using a different set of notations? Any Onon-computational' cryptographic scheme to enlighten here? Are there any cryptographic schemes that the quote is trying to exclude? QC? My reply to the second is: What is the relevance? A paper may try to say anything it wants to say. Whether what is said is in agreement with reality is quite another matter. My reply to the third is: No (preferrably with the words from even onward removed). I prefer the removal is because those words are not only a convolution but are also introducing something fallacious. plausible is not plausible to me at all. One should no longer take P != NP as an assumption. It is a FACT. One ought not assume NP != coNP any longer, because it is FALSE. Without even really proving the existence of one-way functions, one can, in an amusing way, show a formal proof that the following statement is false: Security of no cryptographic scheme is provable under the assumptions: A. P != NP and B. NP != coNP and C. etc. I can provide a pointer to a bit of formal stuff, from which you can trivially provide the proof (that the above statement is false). I can also give you the names of people who I think can. If they won't, it should not be the end of the world. Or perhaps formalism is not of interest to you. You may just want to get a sense of how a proof of P != NP would look like. For which I may be able to provide an informal, personal opinion. Or maybe, you just want to ask such a question to indicate that you are able ask such a question. Maybe, you are just tired of the whole issue. Let us assume that you lost interest in the old millennium even before it was declared a challenge for the new century. So why can't we all lose interest as you did? For no other than God's divine sake and for the sake of sacred mathematics, which being part of the divinity under the assumption that that is His profession, can we put a stop to this pornographic parade? Perhaps, we should change the challenge for the new century to: 1. Prove IP = PSPACE. (the Ogreat' difference between seeing truth and being convinced) 2. Prove one of the following: (NP upper boundary) A. NP - E is empty or B. NP - E is not empty or C. emptiness of NP - E is not decidable. Since the 'rst challenge is one-directional, I think one should even feel comfortable enough to offer a prize for it. === Subject: Re: unprovability of the security of computational cryptography > 1. Prove IP = PSPACE. (the Ogreat' difference between > seeing truth and being convinced). You'll be happy to hear that Shamir did this in 1991. A. Shamir, IP = PSPACE, in Proceedings, 31st Annual IEEE Symposium on Foundations of Computer Science, pages 11-15, 1990. === Subject: Re: unprovability of the security of computational cryptography > 1. Prove IP = PSPACE. (the Ogreat' difference between seeing truth and being convinced). > You'll be happy to hear that Shamir did this in 1991. > A. Shamir, IP = PSPACE, in Proceedings, 31st Annual IEEE Symposium on Foundations of Computer Science, pages 11-15, 1990. Yes, I have heard of that long ago, but I am de'nitely NOT happy. You think I propose something that trivial for the new century? :) === Subject: Re: Finishing argument, core error proven In sci.physics, The Ghost In The Machine : > Followups to sci.math. > In sci.physics, James Harris > > For me there have been two perspectives as I work to 'gure out how to >> explain the de'nition problem in mathematics with LOTS of opposition, >> and I wonder about mathematicians so dedicated to attacking an >> argument that is clearly correct. > I remind of that as I present what should 'nish their ability to >> distract, as I've seen a strange and dedicated effort to ignore the >> actual math, and simply toss up just about anything rather than face >> the truth. > All variables are in the ring of algebraic integers unless otherwise >> stated. > Let>> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > and let > R(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f > so P(m) = f^2 R(m). > Now consider > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > where the a's are given by the following cubic: > a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m). > Then it must be true that the following factorization exists > R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > where the b's are given by the following cubic: > b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m), > where a_3 = b_3, and at m=0, b_3 = 3. > An interesting problem for you. Setting m = 1, u = 1, f = 2, > one gets > P(1) = 28*x^3 - 36*x + 8, or > R(1) = P(1)/f^2 = 7*x^3 - 9*x + 2. > x=1 is clearly a root here. Dividing, we get > R(m) = (x-1) * (7*x^2 + 7*x - 2) > so the other two roots are 1/2 +/- sqrt(105)/14. > Your Oa-cubic' is given by C_a = a^3 + 9*a^2 - 28 > By a little empirical testing (translation: I got lucky :-) ), > a_1 = -2; this gives C_a = (a+2)(a^2 + 7*a - 14), > which factors further into (a+2)(a+7/2-sqrt(105)/2)(a+7/2+sqrt(105)/2). > So far, not too bad; a_1 is clearly divisible by 2 (-2/2 is an > algebraic integer). The other two roots, however, are not, as > (x+7/4-sqrt(105)/4)*(x+7/4+sqrt(105)/4) = x^2+7/2x-7/2 > Clearly, 7/4sqrt(105)/4 are not algebraic integers; > therefore 7/2sqrt(105)/2 are not divisible by 2. Aargh. Those should be -7/4sqrt(105)/4 and -7/2sqrt(105)/2 respectively. Not that it matters overly much. > Your Ob-cubic' is given by C_b = b^3 + 9*b^2 - 7. > It turns out this cubic is not zeroed by any of the following: > x = -2 > x = -1 > x = -1/2 > x = -7/2+sqrt(105)/2 > x = -7/4+sqrt(105)/4 > x = -7/8+sqrt(105)/8 > x = -7/2-sqrt(105)/2 > x = -7/4-sqrt(105)/4 > x = -7/8-sqrt(105)/8 > which therefore means I've found a counterexample to your > a_3 = b_3 hypothesis in your above statement. > (I don't know the actual roots of C_b. I'm not sure I need to. > Please verify the above computations, then 'x your proof, James.) > [rest snipped] > -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Finishing argument, core error proven > In sci.physics, The Ghost In The Machine R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > where the b's are given by the following cubic: > b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m), The given cubic is incorrect, as it is b^3+...+ 3(-1+mf^2)b^2+...- (m^3 f^4 - 3m^2 f^2 + 3m). > where a_3 = b_3, and at m=0, b_3 = 3. The statement that a_3 = b_3 in general is incorrect. An interesting problem for you. Setting m = 1, u = 1, f = 2, one gets P(1) = 28*x^3 - 36*x + 8, or R(1) = P(1)/f^2 = 7*x^3 - 9*x + 2. x=1 is clearly a root here. Dividing, we get R(m) = (x-1) * (7*x^2 + 7*x - 2) so the other two roots are 1/2 +/- sqrt(105)/14. Your Oa-cubic' is given by C_a = a^3 + 9*a^2 - 28 By a little empirical testing (translation: I got lucky :-) ), a_1 = -2; this gives C_a = (a+2)(a^2 + 7*a - 14), which factors further into (a+2)(a+7/2-sqrt(105)/2)(a+7/2+sqrt(105)/2). So far, not too bad; a_1 is clearly divisible by 2 (-2/2 is an algebraic integer). The other two roots, however, are not, as (x+7/4-sqrt(105)/4)*(x+7/4+sqrt(105)/4) = x^2+7/2x-7/2 Clearly, 7/4sqrt(105)/4 are not algebraic integers; therefore 7/2sqrt(105)/2 are not divisible by 2. And I assume you're trying to determine divisibility by *looking* at them? Here's where my example with x=2a, y=b, xy=2, where x, y and b are algebraic integers but Oa' is not applies. You can't *look* at those roots and tell because the sqrt() operator is not a number, and sqrt(105) is NOT a single number, but two, even if you *say* you wish to only take the positive root. So one of the numbers you obtained, based on some rather basic algebra, should have 2 as a factor, while the other does not. Readers should note the partial con'rmation with the one root where you *can* see that 2 is the factor, and remember that ALL reducible polynomials have shown this feature. That is, with actual examples, my predictions are correct. If I were wrong, then someone should be able to 'nd some reducible which proves it. Aargh. Those should be > -7/4sqrt(105)/4 > and > -7/2sqrt(105)/2 > respectively. Not that it matters overly much. > Your Ob-cubic' is given by C_b = b^3 + 9*b^2 - 7. It turns out this cubic is not zeroed by any of the following: x = -2 x = -1 x = -1/2 x = -7/2+sqrt(105)/2 x = -7/4+sqrt(105)/4 x = -7/8+sqrt(105)/8 x = -7/2-sqrt(105)/2 x = -7/4-sqrt(105)/4 x = -7/8-sqrt(105)/8 which therefore means I've found a counterexample to your a_3 = b_3 hypothesis in your above statement. The statement was wrong as, in general, a_3 does not equal b_3, while they are equal at m=0, besides I had to correct the cubic as well. (I don't know the actual roots of C_b. I'm not sure I need to. Please verify the above computations, then 'x your proof, James.) [rest snipped] It's 'xed. Oh, yeah, readers should focus on the question of the proper cubic as rationally, there IS a single cubic which will give b_1, b_2 and b_3. Before any poster wishes to challenge the argument, I suggest they try and 'nd an alternate *single* cubic to the one I gave. Yup, it's an intellectual challenge, as even understanding the math here requires that you rely extensively on mathematical logic, and not on trying to physically *see* it, as otherwise you're likely to get lost. James Harris === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. >> In sci.physics, The Ghost In The Machine > Then it must be true that the following factorization exists > R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > where the b's are given by the following cubic: > b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m), The given cubic is incorrect, as it is b^3+...+ 3(-1+mf^2)b^2+...- (m^3 f^4 - 3m^2 f^2 + 3m). >> where a_3 = b_3, and at m=0, b_3 = 3. >The statement that a_3 = b_3 in general is incorrect. I thought b_1=a_1/f, b_2=a_2/f, and that P(m) = (a_1x + uf) (a_2x + uf) (a_3x + uf) R(m) = P(m)/f^2. So (and I am not asserting anythign here are algebraic integers, this is just an equation as COMPLEX NUMBERS) R(m) = (1/f^2)(a_1x + uf)(a_2x + uf) (a_3x + uf) = (1/f)(a_1x + uf)*(1/f)(a_2x+uf)*(a_3x+uf) = ((a_1/f)x + u)((a_2/f)x+u)(a_3x+uf) = (b_1x + u)(b_2x+u)(a_3x+uf) = (b_1x + u)(b_2x+u)(b_3x+uf). I guess, if b_1x = -u, or b_2x = -u, then b_3 could be pretty much anything. But for MOST values, it will surely be the case that b_1x+u is nonzero, and b_2x+u is nonzero. Cancelling, we have a_3x + uf = b_3x+uf or a_3x = b_3x. Since it was stated that x is coprime to f, it cannot be that x is zero, so we get that a_3=b_3 in all cases, except perhaps for the cases where a_1 = -uf/x or a_2 = -uf/x. This will be true for almost all values of m, surely. So let us assume that we are in one of those values. Then b_3 must equal a_3, as above. So that means that b_3 is a root of the polynomial that de'nes the a's. Let us assume also that f is not a unit, so that claims about divisibility make sense (and as is the case in your applications). We have two cases: (1) The polynomial that de'nes the a's is irreducible over Q. Then any polynomial with integer coef'cients that has b_3 as a root must be a multiple of the polynomial that de'nes the a's. That polynomial is cubic, so if you have a cubic polynomial that de'nes the b's, and it is a multiple of the polynomial that de'nes the a's (also degree 3), then the two must be equal. So the b's are de'ned by the same polynomials that de'ne the a's. Which means either that b1=a1 and b2=a2, or else b1=a2 and b2=a1. But the 'rst would mean a1/f = a1, or a1=a1*f, which gives a1=0 and analogously a2=0; and the second would mean that a1/f = a2 and then b1 is still a multiple of f, which you said was not the case. (2) The polynomial that de'nes the a's is reducible over Q. Then you must have that the factor x-a3 factors out; otherwise, any polynomial which has b3 as a root and has integer coef'cients will also have one of a2 and a1 as a root, and that will lead to the same contradiction as above. In this case, it is possible to 'nd a cubic that has the bs as roots. Which means a3 is an integer. So we have three cases: (1) a_1=-uf/x or a_2=-uf/x; in that case, b3 can be anything. (2) If a_1 is not -uf/x and a_2 is not -uf/x, and the polynomial de'ning the a's is reducible over Q, with a factor of x-a3, so a3 is an integer. Then b3=a3, and it is possible that you can 'nd a (REDUCIBLE) cubic that de'nes the b's, by 'nding a qudratic that de'nes b1 and b2. (3) If a_1 is not -uf/x and a_2 is not -uf/x, and the polynomial de'ning the a's is irreducible over Q, then b3 must equal a3, and there is NO polynomial of degree 3 that has b1, b2, and b3 as roots. Unless you can show that we are always in case (1) and (2) (which we most de'nitely are not), then you cannot in general 'nd a cubic that de'nes b1, b2, and b3. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his 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 'gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde'nite 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 Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Finishing argument, core error proven In sci.math, James Harris >> In sci.physics, The Ghost In The Machine > Then it must be true that the following factorization exists > R(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > where the b's are given by the following cubic: > b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m), > The given cubic is incorrect, as it is > b^3+...+ 3(-1+mf^2)b^2+...- (m^3 f^4 - 3m^2 f^2 + 3m). > where a_3 = b_3, and at m=0, b_3 = 3. > The statement that a_3 = b_3 in general is incorrect. It doesn't matter, as I couldn't match any of the b's with the uncorrected cubic (even though I don't know what the roots are; I can tell the a's don't satisfy the cubic). I'll admit I'll now need to 'gure out where you're going with the corrected cubic. As it is, I still wonder how your system works, as the following example suggests problems with your logic: Let x^3 - 9 = 0. Then is it the case that two of the roots must be divisible by 3 in the realm of algebraic integers? Yes or no? > An interesting problem for you. Setting m = 1, u = 1, f = 2, > one gets > P(1) = 28*x^3 - 36*x + 8, or > R(1) = P(1)/f^2 = 7*x^3 - 9*x + 2. > x=1 is clearly a root here. Dividing, we get > R(m) = (x-1) * (7*x^2 + 7*x - 2) > so the other two roots are 1/2 +/- sqrt(105)/14. > Your Oa-cubic' is given by C_a = a^3 + 9*a^2 - 28 > By a little empirical testing (translation: I got lucky :-) ), > a_1 = -2; this gives C_a = (a+2)(a^2 + 7*a - 14), > which factors further into (a+2)(a+7/2-sqrt(105)/2)(a+7/2+sqrt(105)/2). > So far, not too bad; a_1 is clearly divisible by 2 (-2/2 is an > algebraic integer). The other two roots, however, are not, as > (x+7/4-sqrt(105)/4)*(x+7/4+sqrt(105)/4) = x^2+7/2x-7/2 > Clearly, 7/4sqrt(105)/4 are not algebraic integers; > therefore 7/2sqrt(105)/2 are not divisible by 2. > And I assume you're trying to determine divisibility by *looking* at > them? I can tell that the polynomial with integer coef'cients which is satis'ed by these roots does not have a_n = 1 or -1 as required by the de'nition of algebraic integer. The polynomial x^2 + 7/2x- 7/2 perverts to 2 * x^2 + 7x - 7, which clearly does not have algebraic integer roots because a_n = 2. > Here's where my example with x=2a, y=b, xy=2, where x, y and b are > algebraic integers but Oa' is not applies. > You can't *look* at those roots and tell because the sqrt() operator > is not a number, and sqrt(105) is NOT a single number, but two, even > if you *say* you wish to only take the positive root. sqrt(n) is a perfectly 'ne number, even an algebraic integer, for any integer n (including negative ones). The convention is that sqrt(n) is positive, sqrt(-n) is a multiple of Oi'. You may have also noticed that I'm considering both positive and negative values, so your objection isn't that much of an issue. Would you prefer I use ()^(1/2) instead? It doesn't make much difference. > So one of the numbers you obtained, based on some rather basic > algebra, should have 2 as a factor, while the other does not. > Readers should note the partial con'rmation with the one root where > you *can* see that 2 is the factor, and remember that ALL reducible > polynomials have shown this feature. > That is, with actual examples, my predictions are correct. So show one, with your corrected cubic. > If I were wrong, then someone should be able to 'nd some reducible > which proves it. > Aargh. Those should be > -7/4sqrt(105)/4 >> and >> -7/2sqrt(105)/2 > respectively. Not that it matters overly much. >> Your Ob-cubic' is given by C_b = b^3 + 9*b^2 - 7. > It turns out this cubic is not zeroed by any of the following: > x = -2 > x = -1 > x = -1/2 > x = -7/2+sqrt(105)/2 > x = -7/4+sqrt(105)/4 > x = -7/8+sqrt(105)/8 > x = -7/2-sqrt(105)/2 > x = -7/4-sqrt(105)/4 > x = -7/8-sqrt(105)/8 > which therefore means I've found a counterexample to your > a_3 = b_3 hypothesis in your above statement. > The statement was wrong as, in general, a_3 does not equal b_3, while > they are equal at m=0, besides I had to correct the cubic as well. > (I don't know the actual roots of C_b. I'm not sure I need to. > Please verify the above computations, then 'x your proof, James.) > [rest snipped] It's 'xed. Oh, yeah, readers should focus on the question of the > proper cubic as rationally, there IS a single cubic which will give > b_1, b_2 and b_3. > Before any poster wishes to challenge the argument, I suggest they try > and 'nd an alternate *single* cubic to the one I gave. > Yup, it's an intellectual challenge, as even understanding the math > here requires that you rely extensively on mathematical logic, and not > on trying to physically *see* it, as otherwise you're likely to get > lost. > James Harris -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Finishing argument, core error proven The more I have read about this, the more interested and confused I have gotten. Can someone give me a brief explanation of what this core error is, what the big deal it about is, and what the mathematical §aw about it is. Please don't give me dismissive he/they know(s) nothing arguements, I can evaluate mathematical claims myself. I have been at the recieving end of similar ridicule, much of it self-imposed, for trying to do things such as proving the countability of the reals or disproving Godel's Incompleteness Theorem. Once I saw the mathematics of my I was wrong, I gladly admitted that I was. If only in respect to the mathematical process, I would be very appreciative if someone showed what is being claimed and why it is wrong as simply as possible without simply saying to boils down to 0=1. While James may very likely be wrong, I doubt Godel was well recepted at 'rst. Steven Margolin === Subject: Re: Finishing argument, core error proven linux) > While James may very likely be wrong, I doubt Godel was well > recepted at 'rst. I've never heard that Goedel had to struggle for acceptance of his result. Does anyone who knows a bit more about the history of Goedel's incompleteness results care to comment? -- Jesse Hughes Well, if I can get [my proof of FLT accepted], then I hopefully get a book deal down the road, and maybe I get to go on OOprah'. James Harris, on the rewards of mathematical endeavours. === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. > While James may very likely be wrong, I doubt Godel was well >> recepted at 'rst. I've never heard that Goedel had to struggle for acceptance of his >result. Does anyone who knows a bit more about the history of >Goedel's incompleteness results care to comment? As far as I am aware, Goedel's work was hailed upon reception. So much so that he never had to get around to writing Part II of his paper, in which he was going to expand on some of the more technical parts of his argument; pretty much everyone agreed it cold be done. Legend has it that upon being told of the result Hilbert refused to believe it was true, but that once he read the proof he simply admitted it was correct. There were some confrontations regarding credit. Some years after the publication (about 10 or so, if memory serves), there were some published attacks, in which they claimed that Goedel had not actually proved the inconsistency or incompleteness of arithmetic, but had merely found another antinomy (i.e., another paradox/contradiction, like Russell's paradox, etc). Several people responded to the attack, ->incorrectly<- trying to explain Goedel's a response to the original attack that cleared up the matter to everyone's satisfaction. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his 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 'gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde'nite 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 Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Finishing argument, core error proven > While James may very likely be wrong, I doubt G.9adel was well recepted at 'rst. I've never heard that Goedel had to struggle for acceptance of his > result. Does anyone who knows a bit more about the history of > Goedel's incompleteness results care to comment? > I could be wrong about that particular example, but I am sure you can think of another person who was rejected at 'rst but later proved right. Galois, I think, was rejected by Gauss. > -- > Jesse Hughes > Well, if I can get [my proof of FLT accepted], then I hopefully get a > book deal down the road, and maybe I get to go on OOprah'. > James Harris, on the rewards of mathematical endeavours. === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. While James may very likely be wrong, I doubt G.9adel was well > recepted at 'rst. >> I've never heard that Goedel had to struggle for acceptance of his >> result. Does anyone who knows a bit more about the history of >> Goedel's incompleteness results care to comment? >I could be wrong about that particular example, but I am sure you can think >of another person who was rejected at 'rst but later proved right. Galois, >I think, was rejected by Gauss. No, Galois was encouraged by French mathematicians to whom he sent early versions of his stuff; he did not send anything to Galois that I am aware of. I think you are thinking of Abel, who sent his paper On the solution of the general quintic (or some such title) to Gauss, hoping Gauss would give him a recommendation for a position somewhere. Gauss never read it, because the title led him to believe that it was one of the many dozens of crank paper he used to receive with proposed general solutions of the quintic. Had Abel titled his paper On the impossibility of solving the general quintic, he might have had more luck. Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Finishing argument, core error proven Visiting Assistant Professor at the University of Montana. >The more I have read about this, the more interested and confused I have >gotten. Can someone give me a brief explanation of what this core error >is, what the big deal it about is, and what the mathematical §aw about it >is. James has a family of polynomial, parametrized by a variable m, which he factors in a speci'c way, into linear terms with algebraic integer coef'cients. He claims, as a consequence of an argument he provides, that the coef'cients will necessarily have certain divisibility conditions for any value of m. When his claim is checked for speci'c values of m, it turns out that some of his claims are provably wrong: the coef'cients do NOT have the divisibility properties he claims they should have. James has been unable to 'nd any error in the speci'c calculations. He is also sure his own argument about their divisibility is correct. He therefore concludes that there is something fundamentally wrong with the algebraic integers: they lead to a contradiction. He has gone on record as saying that what is wrong with the algebraic integers is their de'nition. As far as James's argument about the divisiblity conditions for any value of m, nobody that I am aware of has agreed to that argument. Most people who have looked at it have identi'ed at least one place where they state there is a gap or jump. James disagrees that there is either. So, everyone agrees that the speci'c calculations that disprove James's general claims are correct. As far as I know, only James agrees that his general claim is correct. Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Finishing argument, core error proven The more I have read about this, the more interested and confused I have gotten. Can someone give me a brief explanation of what this core error is, what the big deal it about is, and what the mathematical §aw about it is. James has a family of polynomial, parametrized by a variable m, which > he factors in a speci'c way, into linear terms with algebraic integer > coef'cients. He claims, as a consequence of an argument he provides, > that the coef'cients will necessarily have certain divisibility > conditions for any value of m. When his claim is checked for speci'c values of m, it turns out that > some of his claims are provably wrong: the coef'cients do NOT have > the divisibility properties he claims they should have. James has been unable to 'nd any error in the speci'c > calculations. He is also sure his own argument about their > divisibility is correct. He therefore concludes that there is > something fundamentally wrong with the algebraic integers: they lead > to a contradiction. He has gone on record as saying that what is > wrong with the algebraic integers is their de'nition. As far as James's argument about the divisiblity conditions for any > value of m, nobody that I am aware of has agreed to that > argument. Most people who have looked at it have identi'ed at least > one place where they state there is a gap or jump. James disagrees > that there is either. So, everyone agrees that the speci'c calculations that disprove > James's general claims are correct. As far as I know, only James > agrees that his general claim is correct. > Let me direct this to James: Look over your work with a fresh mind. Look at the errors people have pointed out. Go in with the mindset that perhaps you have found a core error, but you just want to micro inspect every detail of your proof. If you think that you are making a reasonable leap, prove a lemma. Call it the Harris Lemma if you like. However, you must be ready to accept the fact that you messed up somewhere. If it turns out you made a mistake, admit to it. You will not be thought of any less, and you might even be respected for it. Regardless, you will learn from your mistake. I am not a psychiatrist, but I have had similar experience. I remember in sixth grade, a friend of mine and I thought we had discovered something, quasirationals, like 1/49, that were rational but neither terminated nor Conjecture, he used something like FLT to prove that 1/49 must repeat within 49 digits, but Excel could only consider 30 digits. At that point I terminate/repeat in different bases and how long the cycles will be. We felt silly afterwards, but we learned a lot about number theory from it. James, I suggest you do the same. Remember it could take a hundred, a thousand, pages to prove the Riemann Hypothesis, and there could be debate about the validity of the proof, but a single number could disprove it one sentence. Steven Margolin > > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu > === Subject: Re: Finishing argument, core error proven Nntp-Posting-Host: hera.cwi.nl >> where the b's are given by the following cubic: >> >> b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m), > > The given cubic is incorrect, as it is > > b^3+...+ 3(-1+mf^2)b^2+...- (m^3 f^4 - 3m^2 f^2 + 3m). Can you tell me what ... means here? Can you write it down for us? As written it is not a polynomial. > (x+7/4-sqrt(105)/4)*(x+7/4+sqrt(105)/4) = x^2+7/2x-7/2 > > Clearly, 7/4sqrt(105)/4 are not algebraic integers; > therefore 7/2sqrt(105)/2 are not divisible by 2. > > And I assume you're trying to determine divisibility by *looking* at > them? No, by looking at the polynomial of which they are roots. -7/4+-sqrt(105)/4 are roots of 2x^2 + 7x - 7, so they are not algebraic integers. -- 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: Finishing argument, core error proven Nntp-Posting-Host: hera.cwi.nl > >> where the b's are given by the following cubic: > >> > >> b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m), > > > > The given cubic is incorrect, as it is > > > > b^3+...+ 3(-1+mf^2)b^2+...- (m^3 f^4 - 3m^2 f^2 + 3m). > > Can you tell me what ... means here? Can you write it down for us? > As written it is not a polynomial. I had a deja vu here. A long time ago James argued that hou could write: (x + y) = (sqrt(x) + ... + sqrt(y))^2 and called that a factorisation (not revealing what ... was). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Letters & Combinations Out of the 26 letters of the English alphabet we generate words containing 6 unique letters. So there are c(26,6)=230,230 such words. I asked to be provided with a list of 25, 6 letter words, in such a way that these words to intersect as many as possible of the 230,230 words, in at least 3 letter. I wish to rank the answers I received according with the total number of words they intersect. Is there any way, other than brute force, to calculate: 1) How many words are intersected by the list only once, 2) How many twice, 3) How many three times...... An examples of such a list is below: A B F K P X A B M S V Y A C G I N W A C J L R X A D K L O U A E H Q T Z B C D K S Z B E G L N W B H I J K V B I Q R T U C E O P U V C F H L M Q C H N P T Y D E I M N X D F G H R V D J P Q W Y E F I O Y Z E K R S W Y F J N Q S U G I L P S T G J K M O T G Q U X Y Z H O S U W X L T V W X Z M N O P R Z By a brute force algorithm I come up with the following for the above list. 229,942 words out of the 230,230 are intersected in at least 3 letters. 288 words are not intersected. This answers the ranking of the lists. I could revise my algorithm to calculate the questions 1,2 & 3 but does exist some other faster way to calculate the nunber of intersections? === Subject: Math History Posters Hello! I am a math teacher in Colorado Springs, CO. Three years ago I wanted math history posters to share with my students - such topics as, who came up with algebra and when? Or, what did Descartes contribute besides the coordinate plane? However, I could not 'nd any such items at the teacher stores. That is when I made my own. Other math teachers have since shown an interest in my posters. I am still creating more after school each day. There is so much to cover! Some categories that I have are: Regions - Egyptian, Greek, Mayan, and more. Eras - Medieval, Renaissance, 20th Century, and more. Topics - Algebra, Women in Mathematics, Calculating Machines. Mathematicians - Descartes, Euler, Turing and more. If you would like to see what I have done so far, please visit my online Math History Catalog. The URL is at http://www.mathisradical.com/Catalog.htm I'm not a professional sales person, just a math teacher who came up with a product that 'ts a need in a mathematics classroom. They are inexpensive and useful. I also have a free newsletter that includes lesson plans, biographies of mathematicians and Black Line Masters for your classroom. Kavon Rueter kavon@mathisradical.com http://www.mathisradical.com === Subject: Re: The Riemann Hypothesis > Is de Brange's proposed proof for the Riemann hypothesis acceptable I don't know, but I haven't heard about anyone accepting it. and what was the implications of the Riemann hypothesis Better error terms in pretty much any estimate involving prime numbers, for starters. There's tons of info on RH available in books & on the net, see for yourself. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The Riemann Hypothesis >> and what was the implications of the Riemann hypothesis > Better error terms in pretty much any estimate involving prime numbers, > for starters. There's tons of info on RH available in books & on the > net, see for yourself. On the other hand, few people (include me among the many) know what the present error terms are, so an improvement can hardly be earth-shaking news. If the generalised Riemann hypothesis holds then the Miller-Rabin test for primality takes polynomial time; but then again, since it only takes a few seconds in practice this is of purely theoretical interest. So I would say the proof of the Riemann hypothesis in itself would have very few direct consequences -- mainly because the reason it was introduced, to prove the Prime Number Theorem, was proved using weaker properties of the Riemann zeta function. -- Timothy Murphy e-mail: tim /at/ birdsnest.maths.tcd.ie (all email over 80k dispatched to /dev/null) tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: The Riemann Hypothesis by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9BGbUT06613; Is de Brange's proposed proof for the Riemann hypothesis acceptable and what was the implications of the Riemann hypothesis === Subject: Re: consecutive composite integers > ....If the conjecture is true, then I can prove > that the product of consecutive integers can never be a perfect power > of an integer. (I know -- it's already been proven but I can't 'nd > the proof and furthermore, I don't know the name attached to the > theorem.) Math Reviews sez, MR0376517 (51 #12692) Erd.9as, P.; Selfridge, J. L. The product of consecutive integers is never a power. Illinois J. Math. 19 (1975), 292--301. 10B05 (10A30) The authors prove that the product of two or more consecutive positive integers is never a power. That is, the Diophantine equation $(n+1)cdots(n+k)=x^l$ has no integer solution $x$ if $kgeq 2$, $lgeq 2$ and $ngeq 0$. Many special cases of this theorem have previously been discussed, beginning with Goldbach in 1724, who, in a letter to D. Bernoulli, treated the case $k=3,l=2$. The next important contribution was made by J. Liouville [J. Math. Pures Appl. (2) 2 (1857), 277--278], who proved that no solution exists if at least one factor $n+1,cdots,n+k$ is prime, or if $k>n-5$. Contributions up to 1918 are described in L. E. Dickson's book History of the theory of numbers, Vol. 2: Diophantine analysis, pp. 679--680, reprint, Chelsea, New York, 1966; MR 39 #6807b]. These deal mainly with the case $l=2$ and special values of $k$. For example, S. Narumi [T.99hoku Math. J. 11 (1917), 128--142; Jbuch 46, 214] proved that no solution exists if $l=2$ and $kleq 201$. A paper by R. Obl.87th [T.99hoku Math. J. 38 (1933), 73--92; Zbl 8, 196] contains a survey of later literature. The next important contribution was made by O. Rigge [Ninth Congress of Scandinavian Mathematicians (Helsinki, 1938 (Swedish)), pp. 155--160, Mercator, Helsinki, 1939; Zbl 21, 010] and by the 'rst author [J. London Math. Soc. 14 (1939), 194--198; MR 1, 4], who proved that no solution exists when $l=2$. Shortly thereafter the 'rst author and Rigge proved that for 'xed $l$ there are at most 'nitely many solutions. In 1940, the 'rst author and C. L. Siegel jointly proved that no solutions exist for $k>c$, where $c$ is an absolute constant, but that proof, which used the Thue-Siegel theorem, was never published. Later the 'rst author [Nederl. Akad. Wetensch. Proc. Ser. A 58 (1955), 85--90; MR 16, 797] gave a different proof of the existence of $c$ by an elementary method. An improvement of this method has now settled the problem in its fullest generality. In fact, the authors prove the following stronger result: Let $k,l,n$ be integers such that $kgeq 3$, $lgeq 2$ and $n+kgeq p^{(k)}$, where $p^{(k)}$ is the least prime $p$ satisfying $pgeq k$. Then there is a prime $pgeq k$ for which the power of $p$ dividing $(n+1)cdots(n+k)$ is not divisible by $l$. Reviewed by T. M. Apostol -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: consecutive composite integers >Examining a table of factors and primes, I found that for any sequence >of consecutive composite numbers there is always one integer that has >a prime factor larger than any other prime factor of any of the other >integers. Further, this prime is not raised to any power. > Thus the conjecture needs be limited to > maximal sequences of composite numbers longer than 1. Then, these should be counter-examples: 148992 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 97 148993 = 13 * 73 * 157 148994 = 2 * 23 * 41 * 79 148995 = 3 * 3 * 5 * 7 * 11 * 43 148996 = 2 * 2 * 193 * 193 740600 = 2 * 2 * 2 * 5 * 5 * 7 * 23 * 23 740601 = 3 * 3 * 19 * 61 * 71 740602 = 2 * 29 * 113 * 113 1474244 = 2 * 2 * 29 * 71 * 179 1474245 = 3 * 3 * 5 * 181 * 181 1474246 = 2 * 83 * 83 * 107 1674434 = 2 * 31 * 113 * 239 1674435 = 3 * 5 * 7 * 37 * 431 1674436 = 2 * 2 * 647 * 647 106891244 = 2 * 2 * 59 * 673 * 673 106891245 = 3 * 3 * 3 * 3 * 5 * 19 * 29 * 479 106891246 = 2 * 7 * 7 * 11 * 229 * 433 All these sequences are maximal, i.e. tightly bounded by two prime numbers. Helmut Richter === Subject: Re: Linear Algebra question on Vector/Subspaces/Spanning Sets sci group. >>Does the 'rst book cover all in the latter? > No, de'nitely not. They are about the same size (IIRC) but the second > one covers much more ground - assuming as it does a greater degree of > mathematical maturity (whatever that is :-). > Well, after cheaping out for a couple of weeks, I've ordered A primer on linear algebra for 45 bucks. It seems Linear algebra is still an active text at Berkeley and very popular. > I'd recommend also the classic Finite-Dimensional Vector Spaces by > Paul Halmos and for good exercises Linear Algebra (on the Schaum > series) by Seymour Lipschutz. > Cheap! And on the way. >>I have been downloading and printing out everything I can 'nd on >>geometric algebra. algebra' tells me >>I've got a lot to learn. > Yes. But I think you'll like it. > > Felix. Best, Dan. -- http://lakeweb.net http://ReserveAnalyst.com dbAtLakewebDotCom === Subject: Tricky integral: e^{x/x+2y} I'm supposed to integrate e^{x/x+2y} with limits 0 < x < 1-2y and 0 < y < 1/2. Could someone tell me what the concept involved here is? So far the course has mostly covered coordinate changes and div,grad, curl. I'm not seeing how the course material pertains to this integral. === Subject: Re: Tricky integral: e^{x/x+2y} > I'm supposed to integrate > e^{x/x+2y} > with limits 0 < x < 1-2y and 0 < y < 1/2. > Could someone tell me what the concept involved here is? So far the > course has mostly covered coordinate changes and div,grad, curl. I'm > not seeing how the course material pertains to this integral. similar, your integrand is e^(1+2*y) and your integral is trivially (1-2*y)*e^(1+2y) === Subject: Re: Tricky integral: e^{x/x+2y} Yes, I should include parenthesis. The integrand is e^{x/(x+2y)} > I'm supposed to integrate > e^{x/x+2y} > with limits 0 < x < 1-2y and 0 < y < 1/2. > Could someone tell me what the concept involved here is? So far the >> course has mostly covered coordinate changes and div,grad, curl. I'm >> not seeing how the course material pertains to this integral. similar, your integrand is e^(1+2*y) and your integral is trivially(1-2*y)*e^(1+2y) === Subject: Re: Tricky integral: e^{x/x+2y} > I'm supposed to integrate > e^{x/x+2y} > with limits 0 < x < 1-2y and 0 < y < 1/2. > Could someone tell me what the concept involved here is? So far the > course has mostly covered coordinate changes and div,grad, curl. I'm > not seeing how the course material pertains to this integral. > A simple substitution z=x+2y should suf'ce. HTH, Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de === Subject: How to 'nd maximum orthogonal vectors ? I have vectors of the form [a1, a2, a3, a4, ...] where a1, a2 etc can be either +1 or -1. For example, say I have a set S of 16 such vectors for [a1, a2, a3, a4]. I want to 'nd maximum orthogonal vectors from this set S where orthogonality is de'ned as a1*b1 + a2*b2 + ... = 0; [a1, a2, ...] and [b1, b2 ...] are vectors from the set S de'ned above. Is there a simple algorithm to solve this problem ? Or is there a similar known problem which I can refer ? Thanx, Ab. === Subject: Re: How to 'nd maximum orthogonal vectors ? > > I have vectors of the form [a1, a2, a3, a4, ...] where a1, a2 etc can be > either +1 or -1. For example, say I have a set S of 16 such vectors for [a1, a2, a3, a4]. > I want to 'nd maximum orthogonal vectors from this set S where orthogonality > is de'ned as a1*b1 + a2*b2 + ... = 0; [a1, a2, ...] and [b1, b2 ...] are > vectors from the set S de'ned above. > Is there a simple algorithm to solve this problem ? Or is there a similar known problem which I can refer ? you can look up Hadamard matrices for something like this... + + + + + + - - + - + - + - - + and similar for order any power of 2