mm-4769 === Subject: Re: Linear Algebra: Another Question to do with inverses... I see how this disproves the original statement... I should have realized that I should be looking for a case where it is invertible, and then disproving the original statement rather than trying to prove it... If the question was instead: if there is a non-invertible square matrix B that satisfies B^3 - 2B + I = 0? I would just factor out a B to get B(-B^2 + 2I) = I, which shows that B is never non-invertible... so the statement would be false? I know I asked that question earlier, I just want to make sure I understand it correctly... If I can write the equation in this form BX=I, then B will never be non-invertible, as the only case where B can = 0 is when B is the zero matrix... I'm I understanding this correctly? === Subject: Re: Linear Algebra: Another Question to do with inverses... > me... sorry for not understanding before. I see how this disproves the original statement... I > should have realized that I should be looking for a > case where it is invertible, and then disproving the > original statement rather than trying to prove it... If the question was instead: > if there is a non-invertible square matrix B that > satisfies B^3 - 2B + I = 0? I would just factor out a B to get B(-B^2 + 2I) = I, > which shows that B is never non-invertible... so the > statement would be false? I know I asked that > question earlier, I just want to make sure I > understand it correctly... Yes, that's correct. > If I can write the > equation in this form BX=I, then B will never be > non-invertible, Yes. > as the only case where B can = 0 is > when B is the zero matrix... I'm I understanding this > correctly? > No. B does not need to be the zero matrix to be non-invertible. There only has to exist a vector x not equal to the null vector with B*x=0. > was very confused! Best wishes Torsten. === Subject: Re: Linear Algebra: Another Question to do with inverses... Understanding this now :) === Subject: I need solution manual of heat exchangers selection,rating, and thermal design(second edition Sadik kakac and hongtan liu).I am waiting your answer.Please hurry up...I will give you coast of it with paypall. I need solution manual of heat exchangers selection,rating, and thermal design(second edition Sadik kakac and hongtan liu).I am waiting your answer.Please hurry up...I will give you coast of it with paypall. === Subject: JSH: Explaining Pell's Equation Oddly enough a clue that seals the explanation to Pell's Equation came up in arguments about my general solution to binary quadratic Diophantine equations, where it's spectacularly and to me, satisfyingly simple. Essential to the argument is the simple relation that given x^2 + Dy^2 = F, it is also true that (x-Dy)^2 + D(x+y)^2 = F*(D+1) which allows you to generate a series of equations. But notice what happens with F=1, D=-3: 1. x^2 - 3y^2 = 1 2. (x+3y)^2 - 3(x+y)^2 = (-2) 3. (4x+6y)^2 - 3(2x + 4y)^2 = (-2)^2 and the 4 divides off!!! So that's 1. x^2 - 3y^2 = 1 2. (x+3y)^2 - 3(x+y)^2 = (-2) 3. (2x+3y)^2 - 3(x + 2y)^2 = 1. So you can start with x=1, y=0, and get 1. 1^2 = 1 2. (1)^2 - 3(1)^2 = (-2) 3. (2)^2 - 3(1)^2 = 1 and thereby get the first non-trivial solutions!!! Key here is the requirement that -D must be a quadratic residue modulo (D+1)^j where j is a natural number giving the level into the series you are at minus 1, and if THAT IS NOT POSSIBLE, then factors of (D+1)^j must divide off and if all its factors divide off you can get a non-trivial solution. So for D=-3, j=2, when (D+1)^j divides off completely, but for 7, you're looking at (-6)^j, and notice that factors of 2 routinely divide off, but the 3 is a hardy bugger, which hangs in there for some time, so you would look for when 7 is not a quadratic residue modulo 3^j (um, can someone give that answer? When is 7 not a quadratic residue modulo 3^j?). So you really have to look by prime factors of D+1. But then you have the explanation for why for some values of D you get solutions quickly while for others it can take a while as the mathematics is looking for when -D is NOT a quadratic residue modulo all the prime factors of D+1 raised to some even number power of j, so it dutifully goes along until it finds that situation and it can take a while. Cool. I like explanations. So that number theoretic structure I call a Diophantine supermap is the key to understanding Pell's Equation and all its behavior. Yeah, I knew it was important. Cool. James Harris === Subject: Re: JSH: Explaining Pell's Equation > Essential to the argument is the simple relation that given x^2 + Dy^2 > = F, it is also true that (x-Dy)^2 + D(x+y)^2 = F*(D+1) which allows you to generate a series of equations. But notice what > happens with F=1, D=-3: 1. x^2 - 3y^2 = 1 2. (x+3y)^2 - 3(x+y)^2 = (-2) 3. (4x+6y)^2 - 3(2x + 4y)^2 = (-2)^2 and the 4 divides off!!! So that's 1. x^2 - 3y^2 = 1 2. (x+3y)^2 - 3(x+y)^2 = (-2) 3. (2x+3y)^2 - 3(x + 2y)^2 = 1. So you can start with x=1, y=0, and get 1. 1^2 = 1 2. (1)^2 - 3(1)^2 = (-2) 3. (2)^2 - 3(1)^2 = 1 and thereby get the first non-trivial solutions!!! I tested this experimentally for all D from 2 to 199, inclusive, starting each with 1^2-D*0^2=1 and going until the first square term on the left exceeded 200 decimal digits. Results: Solutions were found, quickly, for D = 2, 3, 5, and none were found for any other D. For 2 and 3, the first solutions occur at j = 2, where j is the number of iterations. At D = 5, the first solution is at j = 6. For each of these, solutions continue to occur for higher values of j. Before hitting my 200 digit limit for x, 262 solutions are found for D = 2, 230 are found for D = 3, and 65 are found for D = 5. With the 200 digit limit, the program is able to test up to j = 374 for D = 6, and up to j = 170 for D = 199. The above is based on starting with 1^2 - D*0^2 = 1, and iterating, as JSH did in his example, looking for even j where (D-1)^j | x^2 and (D-1)^j | y^2. Out of curiosity, I tried one more thing. I took the sequence of solutions that are found by the iterate-and-divide procedure, and tries using THOSE solutions to start the iteration, instead of starting with (1,0). For all three D that worked with JSH's method, I tried one more thing. Let S0 = (1,0). Let S1 = the first (x,y) found by the JSH chain, and let S2 be the second solution, and so on. I tried starting the iteration with S1 instead of S0. That also gives solutions--the same solutions. Starting with S0, we get S0, S1, S2, ..., and starting with S1, we get S1, S2, ..... One might expect that this continues, but it does not. Starting the iteration with S2 does not give any more solutions before I hit my 200 digit arbitrary limit. Same starting with S3, S4, .... The above are just observations based on experiment. I have not tried to see if there is any good theoretical reason for apparently 2, 3, and 5 begin the only D amenable to JSH's method, and for S0 and S1 being good starting points for iteration, and S2, S3, ... being useless starting points. -- --Tim Smith === Subject: Re: JSH: Explaining Pell's Equation posting-account=Z3AipgkAAABkoMfyNwddSxsYhXHi5CDt CLR 1.1.4322; InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Essential to the argument is the simple relation that given x^2 + Dy^2 > = F, it is also true that (x-Dy)^2 + D(x+y)^2 = F*(D+1) which allows you to generate a series of equations. But notice what > happens with F=1, D=-3: 1. x^2 - 3y^2 = 1 2. (x+3y)^2 - 3(x+y)^2 = (-2) 3. (4x+6y)^2 - 3(2x + 4y)^2 = (-2)^2 and the 4 divides off!!! So that's 1. x^2 - 3y^2 = 1 2. (x+3y)^2 - 3(x+y)^2 = (-2) 3. (2x+3y)^2 - 3(x + 2y)^2 = 1. So you can start with x=1, y=0, and get 1. 1^2 = 1 2. (1)^2 - 3(1)^2 = (-2) 3. (2)^2 - 3(1)^2 = 1 and thereby get the first non-trivial solutions!!! I tested this experimentally for all D from 2 to 199, inclusive, > starting each with 1^2-D*0^2=1 and going until the first square term on > the left exceeded 200 decimal digits. Results: Solutions were found, quickly, for D = 2, 3, 5, and none were > found for any other D. For 2 and 3, the first solutions occur at j = 2, > where j is the number of iterations. At D = 5, the first solution is at > j = 6. For each of these, solutions continue to occur for higher values of j. > Before hitting my 200 digit limit for x, 262 solutions are found for D = > 2, 230 are found for D = 3, and 65 are found for D = 5. With the 200 digit limit, the program is able to test up to j = 374 for > D = 6, and up to j = 170 for D = 199. The above is based on starting with 1^2 - D*0^2 = 1, and iterating, as > JSH did in his example, looking for even j where (D-1)^j | x^2 and > (D-1)^j | y^2. Out of curiosity, I tried one more thing. I took the sequence of > solutions that are found by the iterate-and-divide procedure, and tries > using THOSE solutions to start the iteration, instead of starting with > (1,0). For all three D that worked with JSH's method, I tried one more thing. > Let S0 = (1,0). Let S1 = the first (x,y) found by the JSH chain, and > let S2 be the second solution, and so on. I tried starting the iteration with S1 instead of S0. That also gives > solutions--the same solutions. Starting with S0, we get S0, S1, S2, > ..., and starting with S1, we get S1, S2, ..... One might expect that this continues, but it does not. Starting the > iteration with S2 does not give any more solutions before I hit my 200 > digit arbitrary limit. Same starting with S3, S4, .... The above are just observations based on experiment. I have not tried > to see if there is any good theoretical reason for apparently 2, 3, and > 5 begin the only D amenable to JSH's method, and for S0 and S1 being > good starting points for iteration, and S2, S3, ... being useless > starting points. -- > --Tim Smith- Hide quoted text - - Show quoted text - Tim, see my various posts on this subject. You need 1-D to be a low power of 2 for this to work. I am not 100 percent sure about 8 and up being excluded, but I am pretty sure that they are. Brain still not quite functioning today. Achava === Subject: Re: JSH: Explaining Pell's Equation posting-account=Z3AipgkAAABkoMfyNwddSxsYhXHi5CDt 1.1.4322; PeoplePal 3.0),gzip(gfe),gzip(gfe) so you would look for when 7 is not a quadratic residue modulo > 3^j (um, can someone give that answer? When is 7 not a quadratic > residue modulo 3^j?). > When pigs fly? When hell freezes over? How about never. Since 7 is a quadratic residue of 3, it must also be a quadratic residue of all powers of 3. That works for any odd prime. It fails for the prime 2. 3, 5, and 7 are quadratic residues mod 2. 3 and 7 are not quadratic residues mod 4. None of them are quadratic residues mod 8. I believe that any number congruent to 1 mod 8 (which is the only quadratic residue mod 8) is also a square mod any power of 3 but I am so sleepy that I don't think that I should be held to this (see my most recent post to Mina World which has a real howler among other errors). At any rate, this stuff is pretty easy to prove and can be found in a great many books with titles resembling Elementary Number Theory. So I am thinking that it is pretty rare that your method will turn out to be useful. It is not uncommon that our ideas don't work. They have to be refined or replaced by better ones. Achava === Subject: Re: Explaining Pell's Equation > Oddly enough a clue that seals the explanation to Pell's Equation came This was solved a hundred year ago. === Subject: Re: Explaining Pell's Equation > Oddly enough a clue that seals the explanation to Pell's Equation came > This was solved a hundred year ago. So? He's not claiming to be the first to solve it. He's claiming that he has a *better* way to solve it. -- --Tim Smith === Subject: Re: Explaining Pell's Equation <48fd40c2$1@news.x-privat.org> posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Oddly enough a clue that seals the explanation to Pell's Equation came > This was solved a hundred year ago. Longer than that, goes back to Lagrange. I'm looking at an alternate explanation where quadratic residues play a crucial rule. Specifically the length of the cycle until continued fractions give a solution should be related to j where -D is NOT a quadratic residue modulo (D+1)^j where j is a even number, and -D cannot be made a quadratic residue modulo (D+1)^j/n^2, where n is any non-unit natural number other than (D+1)^j, when x^2 + Dy^2 = 1. That can look complicated but the explanation for the detail for D=-7. Then you've looking at (-6)^j, and very quickly find that 7 is not a quadratic residue for any even j's, but IS when you divide off all powers of 2. For instance 7 is NOT a quadratic residue 36, but IS a quadratic residue modulo 9. So the question is, how does j with the conditions noted above correlate with the length of the cycle for a continued fraction solution? James Harris === Subject: Re: Explaining Pell's Equation > Oddly enough a clue that seals the explanation to Pell's Equation came > up in arguments about my general solution to binary quadratic > Diophantine equations, where it's spectacularly and to me, > satisfyingly simple. trivial. Hacker Slang besides you subbed in some cheat numbers (show it is not general case) divide off divide off, your HS algebra teacher was a dork, for sure > and if all its factors divide off you can get a non-trivial > solution. So for D=-3, j=2, when (D+1)^j divides off divides off again! you so funny! James Harris Dorkus, try googling something else really old, and present it here as your discovery again. === Subject: Re: Explaining Pell's Equation Dude. Seriously. What the . === Subject: Re: JSH: Correction to solving binary quadratic Diophantine equations posting-account=Z3AipgkAAABkoMfyNwddSxsYhXHi5CDt 1.1.4322; PeoplePal 3.0),gzip(gfe),gzip(gfe) > A poster noted a problem with my original theory that I thought fully > covered binary quadratic Diophantine equations as I stupidly assumed > coprimeness between x and y given an equation like x^2 - Dy^2 = F when F can have a square as a factor. That poster gave the example of x^2 - 3y^2 = 4, but it turns out in > any case with a natural number n, not a unit, given x^2 - Dy^2 = n^2 where D is a natural number not a unit, solutions must exist as you > can reduce to (x/n)^2 - D(y/n)^2 = 1 > How do you know that both x and must be divisible by n? If you are going for integer solutions you have not fixed any flaw this way. I don't know what the problem you are trying to address might be, but this is clearly not the way to fix it unless you can show that x and y must be divisible by n or you are secretly going for rational rather than integer solutions. Achava === Subject: Re: JSH: Correction to solving binary quadratic Diophantine equations posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) A poster noted a problem with my original theory that I thought fully > covered binary quadratic Diophantine equations as I stupidly assumed > coprimeness between x and y given an equation like x^2 - Dy^2 = F when F can have a square as a factor. (In the following, I am using x^2+Dy^2 = F as the equation we want to > solve, since that is the one used in your paper). The same problem can arise when F does not have a square factor, since > your paper asserts that -D is a quadratic residue mod F*(D+1)^j. For Dividing off squares is for the form x^2 + Dy^2 = F. the equation x^2-3y^2=1, your paper asserts that 3 must be a quadratic > residue mod 2^j if the equation is solvable. 3 is a residue mod 2, so you are fine at j = 1 > 3 is not a residue mod 4, so this fails at j = 2 So you divide off 4. 3 is not a quadratic residue mod any positive power of 2 other than 2. That poster gave the example of x^2 - 3y^2 = 4, but it turns out in > any case with a natural number n, not a unit, given x^2 - Dy^2 = n^2 where D is a natural number not a unit, solutions must exist as you > can reduce to (x/n)^2 - D(y/n)^2 = 1 so the problem is with square factors, which have to be divided off to > force coprimeness which I stupidly assumed. But other than that the theory is ok. Easy fix. I boringly solved > generally solved binary quadratic Diophantine equations. So every time we hit 4, we have to divide it off? That means that if we > start with x^2 - 3y^2 = 1, and build your chain, our chain ends up being > a short loop, not a huge number theoretic structure. Nope. With D=-2, you have an even shorter loop since it's -1^j. All that happens is you divide off 4 at every other step, so given 1. x^2 - 3y^2 = 1 2. (x+3y)^2 - 3(x+y)^2 = (-2) 3. (4x+6y)^2 - 3(2x + 4y)^2 = (-2)^2 you can see where the 4 gets divided off easily enough. So that's 1. x^2 - 3y^2 = 1 2. (x+3y)^2 - 3(x+y)^2 = (-2) 3. (2x+3y)^2 - 3(x + 2y)^2 = 1 which is kind of neat. Notice then you can start with x=1, y=0, and > get 1. 1^2 = 1 2. (1)^2 - 3(1)^2 = (-2) 3. (2)^2 - 3(1)^2 = 1 so you can trace out the full series in this way, and it's still > infinitely large. So hey there's an interesting discovery there! With x^2 - Dy^2 = 1 if D is not a quadratic residue modulo (1-D)^j, then you can divide > off square factors (which must be there) until you have a modulus for > which D is a quadratic residue. If you have to divide off (1-D)^j itself, you can find the first non- > trivial solution as I showed above. Always. There should be a correlation between the length of the cycle when continued fractions give a solution and j when that happens that (1- D)^j fully divides off. So I'd guess that x^2 - 3y^2 = 1, gives a very quick answer with continued fractions. while for x^2 - 7y^2 = 1 it'd take longer, where I'm still curious about for what natural number j is 7 NOT a quadratic residue modulo 3^j. If you can also compare that to the continued fraction solution that would be nice as well. James Harris === Subject: Re: JSH: Correction to solving binary quadratic Diophantine equations posting-account=Z3AipgkAAABkoMfyNwddSxsYhXHi5CDt 1.1.4322; PeoplePal 3.0),gzip(gfe),gzip(gfe) > A poster noted a problem with my original theory that I thought fully > covered binary quadratic Diophantine equations as I stupidly assumed > coprimeness between x and y given an equation like x^2 - Dy^2 = F when F can have a square as a factor. (In the following, I am using x^2+Dy^2 = F as the equation we want to > solve, since that is the one used in your paper). The same problem can arise when F does not have a square factor, since > your paper asserts that -D is a quadratic residue mod F*(D+1)^j. For Dividing off squares is for the form x^2 + Dy^2 = F. the equation x^2-3y^2=1, your paper asserts that 3 must be a quadratic > residue mod 2^j if the equation is solvable. 3 is a residue mod 2, so you are fine at j = 1 > 3 is not a residue mod 4, so this fails at j = 2 So you divide off 4. 3 is not a quadratic residue mod any positive power of 2 other than 2. That poster gave the example of x^2 - 3y^2 = 4, but it turns out in > any case with a natural number n, not a unit, given x^2 - Dy^2 = n^2 where D is a natural number not a unit, solutions must exist as you > can reduce to (x/n)^2 - D(y/n)^2 = 1 so the problem is with square factors, which have to be divided off to > force coprimeness which I stupidly assumed. But other than that the theory is ok. Easy fix. I boringly solved > generally solved binary quadratic Diophantine equations. So every time we hit 4, we have to divide it off? That means that if we > start with x^2 - 3y^2 = 1, and build your chain, our chain ends up being > a short loop, not a huge number theoretic structure. Nope. With D=-2, you have an even shorter loop since it's -1^j. All that happens is you divide off 4 at every other step, so given 1. x^2 - 3y^2 = 1 2. (x+3y)^2 - 3(x+y)^2 = (-2) 3. (4x+6y)^2 - 3(2x + 4y)^2 = (-2)^2 you can see where the 4 gets divided off easily enough. So that's 1. x^2 - 3y^2 = 1 2. (x+3y)^2 - 3(x+y)^2 = (-2) 3. (2x+3y)^2 - 3(x + 2y)^2 = 1 which is kind of neat. Notice then you can start with x=1, y=0, and > get 1. 1^2 = 1 2. (1)^2 - 3(1)^2 = (-2) 3. (2)^2 - 3(1)^2 = 1 so you can trace out the full series in this way, and it's still > infinitely large. So hey there's an interesting discovery there! With x^2 - Dy^2 = 1 if D is not a quadratic residue modulo (1-D)^j, then you can divide > off square factors (which must be there) until you have a modulus for > which D is a quadratic residue. If you have to divide off (1-D)^j itself, you can find the first non- > trivial solution as I showed above. Always. There should be a correlation between the length of the cycle when > continued fractions give a solution and j when that happens that (1- > D)^j fully divides off. So I'd guess that x^2 - 3y^2 = 1, gives a very quick answer with > continued fractions. while for x^2 - 7y^2 = 1 it'd take longer, where I'm still curious about for what natural > number j is 7 NOT a quadratic residue modulo 3^j. It never is as I explained in a post earlier tonight. By the way, D will always be a quadratic residue mod 1-D as it is congruent to 1. That means that if 1-D is not a power of 2, then D will never be a quadratic residue of (1-D)^j. I am even sleepier now than when I it shouldn't be too hard to work out tomorrow. At this point your method is in need of some new ideas except possibly for the case where D = 1 - 2^n for some n. Achava > If you can also compare that to the continued fraction solution that > would be nice as well. > === Subject: Re: One more question about inverses.... > A(A + 2A^-1) = A^2 + 2I, goes nowhere that I can > see. What I did was: > took A + 2A^-1 = A^-1 (A^2 + 2I), > then tried to find a case where A^2 + 2I = 0 (correct > me if I'm wrong, but this would be because any matrix > that carries a transformation to zero is singular, > hence non-invertible). > Using the matrix A = > [1 -3 > 1 -1] > as an example, I obtain A^2 = > [-2 0 > 0 -2] > and then A^2 = -2I, so this would be a condition > where A^2 + 2I = 0. Therefore, I could conclude that > the original statement is false and A + 2A^-1 is not > invertible... correct? Yes, it's correct. But why not directly checking that A+2*A^(-1)=0 for A as defined above ? B^3 - 2B + I = (B - I)(B^2 + B - I) = 0 If B - I is invertible, then ... B^3 - 2B + I = (B - I)(B^2 + B - I) = 0 > Then said if B=I, then B - I = 0 and 0 = 0. > Therefore, this would be a condition where B-I > carries the transformation to zero, and is therefore > singular. So then, I could say that the original > statement is true, that the matrix (B-I) cannot be > invertible... Did what I do make sense; I'd appreciate any input! see the other thread. Best wishes Torsten. === Subject: Re: One more question about inverses.... <1768821.1224512585697.JavaMail.jakarta@nitrogen.mathforum.org Please don't over clip the reply. Keep essential context, especially the problem statement. Otherwise the thread can vanish into the memory hole. > A(A + 2A^-1) = A^2 + 2I, goes nowhere that I can see. What I did was: > took A + 2A^-1 = A^-1 (A^2 + 2I), > then tried to find a case where A^2 + 2I = 0 (correct me if I'm wrong, > but this would be because any matrix that carries a transformation to > zero is singular, hence non-invertible). > Using the matrix A = > [1 -3 > 1 -1] > as an example, I obtain A^2 = > [-2 0 > 0 -2] > and then A^2 = -2I, so this would be a condition where A^2 + 2I = 0. > Therefore, I could conclude that the original statement is false and A + > 2A^-1 is not invertible... correct? No way, I don't get it. In addition without the problem statement in your reply, I'm not going to take a second look at it and see if I can solve the problem or even get started with it. Nor am I going review the thread to refreash myself what the problem be. Haven't the time for it. > B^3 - 2B + I = (B - I)(B^2 + B - I) = 0 If B - I is invertible, then ... B^3 - 2B + I = (B - I)(B^2 + B - I) = 0 That's what I said. > Then said if B=I, then B - I = 0 and 0 = 0. By what right have you to assume B = I? > Therefore, this would be a condition where B-I carries the > transformation to zero, and is therefore singular. So then, I could say > that the original statement is true, that the matrix (B-I) cannot be > invertible... > Huh? I don't get it. Assume B - I is intervable. What do you get? In addition restore the problem statement. === Subject: Re: A cute little problem with logs : A student emailed me a problem, and I thought it might be fun to > share. Notation: log_p(q) means the base-p logarithm of q. Given: log_a(x) = c and log_b(x) = d. > Find: log_ab(x), the log of x in base (a*b), in terms of c and d. I'll post my solution in a day or two, but I'm curious if someone > else comes up with something more elegant. I'll post my answer now... > cd/(c+d) Passerby's, but I've enjoyed reading the others. One of the things I find endlessly delightful about math is that for all but the most trivial problems there are usually several solutions. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Shikata ga nai... === Subject: Re: A cute little problem with logs One of the things I find endlessly delightful about math is that for > all but the most trivial problems there are usually several > solutions. And with any luck they all give the same answer. -- He is not here; but far away The noise of life begins again And ghastly thro' the drizzling rain On the bald street breaks the blank day. === Subject: Vector algebra and Geometry help Demonstrate , using the definition of the dot product, that the length of the vectors a and b are equal if an only if the diagonals of the quadrilateral OAQB are orthogonal. A Q O B its meant to be a rhombus Hint: Use that if (OA-OB) is perependicualr to (OA+OB) then OA = OB and vice versa. P.S dont tell me the answer just need some help of what to do help much appreciated === Subject: Re: Vector algebra and Geometry help <25053963.1224603387587.JavaMail.jakarta@nitrogen.mathforum.org>, Ad > Demonstrate , using the definition of the dot product, that the length of the > vectors a and b are equal if an only if the diagonals of the quadrilateral > OAQB are orthogonal. A Q > O B its meant to be a rhombus > Hint: Use that if (OA-OB) is perependicualr to (OA+OB) then OA = OB and vice > versa. P.S dont tell me the answer just need some help of what to do help much appreciated I can't see any diagram so I'm guessing the problem is what I think it is. I'm going to use for the dot product of u and v. One diagonal is a + b; the other is a - b. Use the fact that a + b is orthogonal to a - b iff = 0 and that and are the squares of the lengths of a and b respectively. -- Paul Sperry Columbia, SC (USA) === Subject: Solution manual to Cost Accounting A Managerial Emphasis by Charles T. 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For example, if one takes the integral of xcos(x)dx and lets u = x and dv = cos(x)dx then would the integral take the form: x[sin(x) + c] - integral[sin(x) + c]dx ??? my uni lecture notes seem to say that the integral will take the form: xsin(x) - integral(sin(x))dx Obviously a constant is produced when taking the integral of sin(x)dx, but my question is, if we really do let dv = cos(x)dx and wish to find v, then to find v, it seems we must indeed take the integral of both sides to get v + c1 = sin(x) + c2 which can be written as v = sin(x) + c3 where c3 = c2 - c1 So when finding v from dv when doing integration by parts do we take some sort of pseudo integral (i.e. an integral in which the arbitary constant is not included) or are my lecture notes wrong? Any help would be greatly appreciated. === Subject: Re: Integration by parts - where does the constant go? >When finding v from dv when doing integration by parts, is a constant included in the expression for v? For example, if one takes the integral of xcos(x)dx >and lets u = x and dv = cos(x)dx then would the integral take the form: x[sin(x) + c] - integral[sin(x) + c]dx ??? my uni lecture notes seem to say that the integral will take the form: xsin(x) - integral(sin(x))dx Obviously a constant is produced when taking the integral of sin(x)dx, but my question is, if we really do let dv = cos(x)dx and wish to find v, then to find v, it seems we must indeed take the integral of both sides to get v + c1 = sin(x) + c2 which can be written as v = sin(x) + c3 where c3 = c2 - c1 So when finding v from dv when doing integration by parts do we take some sort of pseudo integral (i.e. an integral in which the arbitary constant is not included) or are my lecture notes wrong? Any help would be greatly appreciated. As people have been saying, you can use any v with the given dv. It doesn't matter which constant you add to v. For example, | | | u dv = u(v+c) - | (v+c) du | | | = uv + uc - cu - | v du | | = uv - | v du | Rob Johnson take out the trash before replying === Subject: Re: Integration by parts - where does the constant go? G. A. Edgar a .8ecrit : > There is a little paradox you can get this way. Suppose we want to > compute J = integral (1/x) dx. Integrate by parts: u = 1/x, dv = dx, > v = x, to get J = integral(1/x) dx = (1/x) x - integral x (-1/x^2) dx > = 1 + integral (1/x) dx = 1 + J. That is: J = 1 + J or 0 = 1. > No problem: when you use the notation integral without any lower or upper bound it means that J is not a number nor a function but an equivalence class of functions modulo the equivalence relationship: f-g is constant on every interval. -- Fatal === Subject: Re: Integration by parts - where does the constant go? >There is a little paradox you can get this way. Suppose we want to >compute J = integral (1/x) dx. Integrate by parts: u = 1/x, dv = dx, >v = x, to get J = integral(1/x) dx = (1/x) x - integral x (-1/x^2) dx >= 1 + integral (1/x) dx = 1 + J. That is: J = 1 + J or 0 = 1. This is not really a problem because the constants of integration for the two integrals don't need to be the same. However, this is a nice proof that 1 = 0; perhaps a bit more subtle than the division by 0 proofs. Rob Johnson take out the trash before replying === Subject: Re: Integration by parts - where does the constant go? westeez a .8ecrit : > When finding v from dv when doing integration by parts, is a constant included in the expression for v? Just chose the primitive you want, i.e. add every constant you wish. Once you've put a lower and an upper bound to your integral, these additive constants vanish. -- Fatal === Subject: Re: Dice Rolling Problem > I need some help with a problem. The first part > is: How many results are possible if n identical dice > are > rolled and each of 1,2,3,4,5,6 appears at least > once? My answer: X1+X2+X3+X4+X5+X6=n > where Xi >= 1 and Yi >= 0 > Xi=Yi+1 > Y1+Y2+Y3+Y4+Y5+Y6+6=n > Y1+Y2+Y3+Y4+Y5+Y6=n-6 > C(n-6,1) The second part: How many results are possible if 5 white dice, 10 > red > dice and 20 blue dice are rolled? It's not a probability problem. It's a counting > problem. I'm not sure how to go about it because > if > I put 5 in for n, I get C(-1,1) which is not > possible. I need some help with a problem. The first part > is: How many results are possible if n identical dice > are > rolled and each of 1,2,3,4,5,6 appears at least > once? My answer: X1+X2+X3+X4+X5+X6=n > where Xi >= 1 and Yi >= 0 > Xi=Yi+1 > Y1+Y2+Y3+Y4+Y5+Y6+6=n > Y1+Y2+Y3+Y4+Y5+Y6=n-6 > C(n-6,1) > Let A_i be the set of possible roles of n dice in > which number i does not appear. > Then you have to count the number of elements of > union_{i=1}^{6} A_i to get the number of > outcomes where at least one of the six > numbers 1,...,6 is missing. > The number of outcomes of rolling n dice where > only k of 6 possible numbers appear is C(n+k-1,k-1). Then, by the inclusion-exclusion-principle, the > answer > to your question is C(6,0)*C(n+6-1,6-1) - (C(6,1)*C(n+5-1,5-1) - > C(6,2)*C(n+4-1,4-1) + C(6,3)*C(n+3-1,3-1) - > C(6,4)*C(n+2-1,2-1) + C(6,5)*C(n+1-1,1-1)) A good exercise is solving this problem > under the condition that all dice > are considered of a different colour. > I now see that in case that the dice are identical the problem is much easier. The number of ways to distribute n identical balls on 6 urns such that each urn contains at least one ball is C(n-6+6-1,6-1) = C(n-1,5). (So the above sum should come out to be C(n-1,5)) > The second part: How many results are possible if 5 white dice, 10 > red > dice and 20 blue dice are rolled? It's not a probability problem. It's a counting > problem. I'm not sure how to go about it because > if > I put 5 in for n, I get C(-1,1) which is not > possible. > Best wishes Torsten. === Subject: Re: two persons betting game... >Two persons, A and B, each initially have 0 dollars. >>At each run, the winner will get $1, and the loser get $0, and the >>first person who gets $10 is the grand winner. >>At each run, one of them serves as a dealer, and when A serves as the >>dealer, he has 0.7 winning probability, whereas when B serves as the >>dealer, he has 0.6 winning probability. >>Suppose A is the first to be the dealer, >>he can choose to set the rules of the dealership from the following >>two options: >>(1) They take turns in serving as the dealer; >>(2) The winner from the previous run will continue to serve as the >>dealer in the subsequent run. >>Which rule should A choose? >>Any thoughts? The choice of rule doesn't matter since with either rule, the > probability that A wins is exactly 2753639317149973 / 3906250000000000 If I were A, I would choose (1). It is true that the chances of being the grand winner are the same with either rule, but the first rule has the advantage that when A loses, he comes away with more money. Expected dollar amount: (1) 35499016540353643 / 3906250000000000 ~= 9.09 (2) 34631020084339843 / 3906250000000000 ~= 8.87 Scott -- Scott Hemphill hemphill@alumni.caltech.edu This isn't flying. This is falling, with style. -- Buzz Lightyear === Subject: Re: Grid Of Revolution Game posting-account=dGiPYgkAAABSJ3xUlNLViQdT0h489hR6 AppleWebKit/523.10.3 (KHTML, like Gecko) Version/3.0.4 Safari/523.10,gzip(gfe),gzip(gfe) Highest score wins. I'll take first turn if n is odd. I don't care which turn I get when n is > even. For n even: divide the board in half, horizontally. For every cell in the > top half there is a corresponding cell in the bottom half. We write these > as [X,Y] with X and Y the colors they get. > At the end of the game, there are 4 possibilities: > [A,A],[B,B],[A,B],[B,A] > let's assume these occur m_aa, m_bb, m_ab, m_ba times. This means there > are 2*m_aa + m_ab + m_ba cells with color A, and 2*m_bb + m_ab + m_ba > cells with color B. We know that the total number of cells with color A > 2*m_bb. This means that, when rotating 180 degrees, both players will get > the same amount of point. In the case where n is odd, the first player should place an A in the > middle. He will play a B on the rest of the board, and will finally in > the last half-turn place another A somewhere. For the board, excluding > the middle cell, the above reasoning still holds, but the first player > will get another point for the middle cell, so will certainly win. Either way, it's not much of a game, i.m.h.o. robby Don't forget; the top grid can be rotated about many points, not just rotated about the center point of the grid. Your right; both players get the same score IF both players rotate about the middle point. Here is an example (4-by-4 board) of a colored grid that may help illustrate what is going on here. ABBB BAaA ABBB AABA Either player who rotates about the center point gets 4 points. On the other hand, if a player rotates about the square (the a) that is in the 2nd row and 3rd column, then player 1 would get 6 points, player 2 would get 3 points. (The left-most column and the bottom-most row do not line up with any squares of the other grid, if we rotate about the a.) Leroy Quet === Subject: Re: Grid Of Revolution Game > Here is an example (4-by-4 board) of a colored grid that may help > illustrate what is going on here. ABBB > BAaA > ABBB > AABA Either player who rotates about the center point gets 4 points. On the > other hand, if a player rotates about the square (the a) that is in > the 2nd row and 3rd column, then player 1 would get 6 points, player 2 > would get 3 points. > (The left-most column and the bottom-most row do not line up with any > squares of the other grid, if we rotate about the a.) Leroy Quet Wooops. I have things switched around. Rotating about the 'a', player 1 gets 3 points, and player 2 gets 6 points. Sorry. Leroy Quet === Subject: Re: Grid Of Revolution Game posting-account=dGiPYgkAAABSJ3xUlNLViQdT0h489hR6 AppleWebKit/523.10.3 (KHTML, like Gecko) Version/3.0.4 Safari/523.10,gzip(gfe),gzip(gfe) Okay. What matters, as far as the difference between the scores of player 1 and player 2, is simply the differences between the number of color A's and the number of color B's in the rectangle of overlap between the two grids. So, dispense with the rotation and the duplicate grid drawn on tracing paper. Let's simplify things. After the grid is filled in, each player chooses a rectangle of the grid where two adjacent sides of the rectangle match up with two adjacent sides of the grid. (This corresponds to the area of overlap under the previous rules.) Player 1 gets added to his/her score the number of squares of color A in the rectangle. Player B gets added to her/his score the number of squares of color B in the same rectangle. Then the other player chooses a rectangle, and the number of squares of color A are added to player 1's score, and the number of squares of color B are added to player 2's score. A much simpler game. Leroy Quet === Subject: Re: Grid Of Revolution Game the rectangle. Player B gets added to her/his score the number of > squares of color B in the same rectangle. Then the other player chooses a rectangle, and the number of squares of > color A are added to player 1's score, and the number of squares of > color B are added to player 2's score. A much simpler game. Aha. Still, not such an interesting game: I'll go second and be able to match the first player's score. If n is odd player A has an advantage and the game is not fair (there will be one more A than B in the grid), I'll assume n is even. When going second, if player A puts an 'A' at position x_1,y_1 and a B at position x_2,y_2, I play 'B' at position n+1-x_1, n+1-y_1 and 'A' at n+1- x_2,n+1-y_2 . In case x_2 = n+1-x_1, y_2 = n+1-y_1; player A is helping me, I just place another such move. Whatever rectangle player A takes to maximize his score, I take the mirror image and get equal score. robby === Subject: Re: Grid Of Revolution Game posting-account=dGiPYgkAAABSJ3xUlNLViQdT0h489hR6 AppleWebKit/523.10.3 (KHTML, like Gecko) Version/3.0.4 Safari/523.10,gzip(gfe),gzip(gfe) >... > Aha. Still, not such an interesting game: I'll go second and be able to > match the first player's score. If n is odd player A has an advantage and > the game is not fair (there will be one more A than B in the grid), I'll > assume n is even. >... You are right. I have made up much better games in the past (games where the just-match-your-opponent's-moves strategy won't necessarily work). Forget this game. I am removing it from my game blog. (http://gamesconceived.blogspot.com/) Leroy Quet === Subject: Re: lie group generator posting-account=IBUqVwoAAADepmzxVr9iEYD5Z0A483SY Gecko/20070530 Fedora/1.5.0.12-1.fc5 Firefox/1.5.0.12,gzip(gfe),gzip(gfe) On 22 Oct, 14:19, magya_bl...@yahoo.com why is a connected lie group generated by neighborhood of the Show that a neighbourhood of the identity in any topological group G generates an open subgroup of G. Victor Meldrew I don't believe it! === Subject: Re: lie group generator > Show that a neighbourhood of the identity in any > topological group G generates an open subgroup of G. Is true that any open subset of a topological group generates an open subgroup? It seems to me that such a subgroup is just a union of translations of the original generating set, and I guess unions of open sets are open. === Subject: Re: lie group generator >> Show that a neighbourhood of the identity in any >> topological group G generates an open subgroup of G. Is true that any open subset of a topological group > generates an open subgroup? Yes. That's very easy to prove. > It seems to me that such a subgroup is just a union of > translations of the original generating set, and I guess > unions of open sets are open. You guess? What do you think that open set means, in the first place? Jose Carlos Santos === Subject: Re: lie group generator > Show that a neighbourhood of the identity in any > topological group G generates an open subgroup of G. >> Is true that any open subset of a topological group >> generates an open subgroup? Yes. That's very easy to prove. > It seems to me that such a subgroup is just a union >> of translations of the original generating set, and I >> guess unions of open sets are open. Then it seems that a connected topological group is generated as a group by any open neighborhood. This seems a little strange. It seems that for a Lie group at least, the group generated by an open neighborhood of the identity is connected, but that an open neighborhood can generate a disconnected Lie group. Do Lie groups have the property that some group acts on their connected components? It seems like one can take representative elements from each component, and look at what subgroups are generated by the open neighborhoods containing it. Can one actually just take single representatives from each connected component? Perhaps the group acting is the quotient group by the component of the identity? This seems somewhat strange to me, but probably because of the mixture of finite and infinite groups. Must the quotient group be discrete? It seems hard to picture a continuous set of connected components. === Subject: Re: lie group generator > Then it seems that a connected topological group is > generated as a group by any open neighborhood. This > seems a little strange. Strange but true. > It seems that for a Lie group at least, the group > generated by an open neighborhood of the identity > is connected, No, since a neighbourhood of the identity may contain points not in the same connected component as the identity. > Do Lie groups have the property that some group acts > on their connected components? It seems like one can > take representative elements from each component, and > look at what subgroups are generated by the open > neighborhoods containing it. Can one actually just > take single representatives from each connected component? In a Lie group G (indeed in any topological group) the component H containing the identity is an open normal subgroup. Therefore we can consider the quotient G/H which is also a topological group, and is completely disconnected. There are interesting completely disconnected topological groups (not Lie groups) for instance the additive group Z p of the p-adic integers. Victor Meldrew I don't believe it! === Subject: Re: lie group generator questions, but they stray from Lie groups to topological groups, which I think covers a lot of weird things. > Then it seems that a connected topological group is > generated as a group by any open neighborhood. >This seems a little strange. Strange but true. It seems that for a Lie group at least, the group > generated by an open neighborhood of the identity > is connected, No, since a neighbourhood of the identity may contain > points not in the same connected component as the > identity. even circular. The identity component is a subgroup, and so if I take a neighborhood contained in it, then the subgroup generated by it is contained in it. I think this clears it up for me though. The key really is that the identity component is a normal subgroup, and the other components are cosets, so this gives the action. >>... > In a Lie group G (indeed in any topological group) > the component H containing the identity is an open > normal subgroup. Therefore we can consider the quotient > G/H which is also a topological group, and is > completely disconnected. .. >> Is the quotient group discrete? Ah, so completely disconnected, but not necessarily discrete. In a Lie group (over the reals or complexes), is a completely disconnected group discrete? > There are interesting completely disconnected > topological groups (not Lie groups) for instance > the additive group Z_p of the p-adic integers. of the identity is just the identity (zero), but the quotient group does not act very tamely on it. I seem to remember that most p-adic things were completely disconnected. Is there an easy way to have the component of the identity be a little bigger with the quotient still Z_p? I think Z_p as abelian group is pretty special (pure injective I think, but the projective-like properties are really the ones to worry about). I guess I want a normal subgroup of something like the real numbers, but that is an injective abelian group so the sequence would split if abelian. Is that just ok? Is the identity component of R x Z_p just R x 1? Are there less trivial examples? There are no nontrivial actions of Z_p on R^n, right? Seems unlikely to get a central nonsplit extension of Z_p by R. I think that rules out anything obvious to me at least. === Subject: Re: lie group generator since there is so much fun, can i ask another one? If H is a lie subgroup of a lie group G. Then Lie(H)={X in Lie(G) such that exp(tX) is in H}. In case we assume exp(tX) is in H, then is the proof simply that exp(tX) is a one parameter subgroup of H hence X is in Lie(H)? I think there is a flaw in this thinking, but don't see where. On Oct 22, 11:50am, Jack Schmidt groups, which I think covers a lot of weird things. Then it seems that a connected topological group is > generated as a group by any open neighborhood. >This seems a little strange. Strange but true. It seems that for a Lie group at least, the group > generated by an open neighborhood of the identity > is connected, No, since a neighbourhood of the identity may contain > points not in the same connected component as the > identity. even circular. The identity component is a subgroup, > and so if I take a neighborhood contained in it, then > the subgroup generated by it is contained in it. I think this clears it up for me though. The key > really is that the identity component is a normal > subgroup, and the other components are cosets, so > this gives the action. >... > In a Lie group G (indeed in any topological group) > the component H containing the identity is an open > normal subgroup. Therefore we can consider the quotient > G/H which is also a topological group, and is > completely disconnected. > .. >> Is the quotient group discrete? Ah, so completely disconnected, but not necessarily > discrete. In a Lie group (over the reals or complexes), > is a completely disconnected group discrete? There are interesting completely disconnected > topological groups (not Lie groups) for instance > the additive group Z p of the p-adic integers. of the identity is just the identity (zero), but the > quotient group does not act very tamely on it. I seem to remember that most p-adic things were > completely disconnected. Is there an easy way to > have the component of the identity be a little bigger > with the quotient still Z p? I think Z p as abelian group is pretty special (pure > injective I think, but the projective-like properties > are really the ones to worry about). I guess I want > a normal subgroup of something like the real numbers, > but that is an injective abelian group so the sequence > would split if abelian. Is that just ok? Is the identity component of R x Z p > just R x 1? Are there less trivial examples? There are no nontrivial > actions of Z p on R^n, right? Seems unlikely to get a > central nonsplit extension of Z p by R. I think that > rules out anything obvious to me at least. === Subject: Re: lie group generator > since there is so much fun, can i ask another one? If H is a lie > subgroup of a lie group G. Then Lie(H)={X in Lie(G) such that exp(tX) > is in H}. To be more precise: Lie(H) is the set {X in Lie(G) | for all real _t_, exp(tX) is in H}. > In case we assume exp(tX) is in H, then is the proof simply > that exp(tX) is a one parameter subgroup of H hence X is in Lie(H)? I > think there is a flaw in this thinking, but don't see where. of all one-parameter subgroups of H *or if* you defined Lie(H) in some other way and then proved that it can be identified with the set of all one-parameter subgroups of H. Jose Carlos Santos === Subject: Re: lie group generator <17886658.1224701443045.JavaMail.jakarta@nitrogen.mathforum.org> <6mar8vFd9dinU1@mid.individual.net> posting-account=epuBzwkAAADSPMLj47hQogB-dyIt9wW_ rv:1.9.0.3) Gecko/2008092414 Firefox/3.0.3,gzip(gfe),gzip(gfe) > of all one-parameter subgroups of H *or if* you defined Lie(H) in some > other way and then proved that it can be identified with the set of all > one-parameter subgroups of H. I assumed that there is a bijection between Lie(H) (tangent space of H at identity) and the set of one parameter subgroups of H. When I looked at John Lee's book Introduction to Smooth Manifolds, on page 532 corollary 20.14, it uses a more involved argument with foliations to show that tangent at identity is in Lie(H) (as defined above) which i thought was not necessary (unless the assumption that exp(tX) is in H by itself does not imply that it is a one-parameter subgroup of H). === Subject: Re: lie group generator >> of all one-parameter subgroups of H *or if* you defined Lie(H) in some >> other way and then proved that it can be identified with the set of all >> one-parameter subgroups of H. I assumed that there is a bijection between Lie(H) (tangent space of H > at identity) and the set of one parameter subgroups of H. Well... yes there is. You don't have to assume it. The canonical bijection is (where OPG(H) = {one-parameter subgroups of H}) OPG(H) -----> Lie(H) f |--> f'(0). > When I > looked at John Lee's book Introduction to Smooth Manifolds, on page > 532 corollary 20.14, it uses a more involved argument with foliations > to show that tangent at identity is in Lie(H) (as defined above) which > i thought was not necessary (unless the assumption that exp(tX) is in > H by itself does not imply that it is a one-parameter subgroup of H). Of course it does. Jose Carlos Santos === Subject: Re: lie group generator Show that a neighbourhood of the identity in any > topological group G generates an open subgroup of G. >> Is true that any open subset of a topological group >> generates an open subgroup? Yes. That's very easy to prove. > It seems to me that such a subgroup is just a union of >> translations of the original generating set, and I guess >> unions of open sets are open. You guess? What do you think that open set means, in the first place? > Jose Carlos Santos I would guess that the Mr. Schmidt was doing the Usenet version of thinking out loud. I suppose that might be termed thinking out type, if it weren't for the exceptionally clumsy wording. Dale === Subject: Re: number elements in this factor ring days. My association with the Department is that of an alumnus. Z_3[x] / has how many elements? Since elements in every coset of ideal will have the same last three >coefficients Huh? >I think answer is by product rule, 3*3*3 (because each >coefficient can have 3 possible values, 0, 1, 2). Some help please. If F is a field (as Z_3 is), and p(x), q(x) are any two polynomial with q(x) nonzero, then there exist unique polynomial a(x) and b(x) such that p(x) = q(x)a(x) + b(x), and b(x)=0 or deg(b)All, I was able to follow the derivation of the formula for the Cissoid of >Diocles, which for a circle of radius 1/2 is: y^2 = x^3/(1-x) When I label the intersection point of the line y = -2x + 2 with y^2 = >x^3/(1-x) as P, the slope at the point equals the cube root of 2. I am not able to see intuitively why this is the case. Can someone explain this in an intuitive way? Diana M. Assuming I didn't make some dumb mistake, the slope at P is approximately 2.88988 Why do think the slope should equal the cube root of 2? === Subject: Re: Cissoid of Diocles I have defined point D = (0, 2), point O = (0, 0), point A = (1, 0). I have defined a circle of radius 1/2 centered at (1/2, 0). I am defining the point P to be the intersection of the line from D to A, [y = -2x +2], and the cissoid, [y^2 = x^3/(1-x)] I am trying to see intuitively why a line through the origin and P, line OP, intersects the line x = 1 at the cube root of 2. Since I know that the line OP intersects the line x = 1 at the cube root of 2, the slope of P would be the cube root of 2. If I have made an error, please correct me. Diana === Subject: Re: Cissoid of Diocles posting-account=75Rr6AoAAADNxXCCWzGrlkVXGpmaLNsm 2.0.50727),gzip(gfe),gzip(gfe) Well, I got some help with this and solved it by doing the following: First, I needed to solve for x with the two equations: 1) y = -2x + 2 2) y^2 = x^3/(1-x) Substitute the right side of equation 1) into the left side of equation 2). (-2x+2)^2 = x^3/(1-x) ... 4(1-x)^3 = x^3 ... x = 2^(2/3)/(2^(2/3)+1) y = 2/(2^(2/3)+1) Slope of OP = y/x = 2^(1-(2/3)) = 2^(1/3) So, line OP intersects the line x=1 at 2^(1/3) === Subject: Re: Cissoid of Diocles >Well, I got some help with this and solved it by doing the following: First, I needed to solve for x with the two equations: 1) y = -2x + 2 >2) y^2 = x^3/(1-x) Substitute the right side of equation 1) into the left side of >equation 2). (-2x+2)^2 = x^3/(1-x) >... >4(1-x)^3 = x^3 >... >x = 2^(2/3)/(2^(2/3)+1) >y = 2/(2^(2/3)+1) Slope of OP = y/x = 2^(1-(2/3)) = 2^(1/3) So, line OP intersects the line x=1 at 2^(1/3) I read your first post as asking for the slope of the line tangent to the cissoid at point P. Your second post made it clear that is not what you meant. Just now, I was about to post essentially the same thing you have above. Pleased to see you beat me to it! 8-) === Subject: Another coin betting game... posting-account=CYtergoAAAAOD8k-T_NdzknPJOBXmt8x CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Currently, you have $100. You go to play a game. Each time, flip a coin, which appears head with probability 0.75, appears tail with probability 0.25. Each time, you can bet x dollars, if it's head, you get x , if it's tail, you lose x. You can play 10 times. What's the maximum expected payoff after 10 times? === Subject: Re: Another coin betting game... posting-account=G4HhgwkAAAB0h1X9lFcN2_L63S7FUgWk 1.0.3705; .NET CLR 2.0.50727; .NET CLR 1.1.4322; Media Center PC 3.1; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Currently, you have $100. You go to play a game. Each time, flip a coin, which appears head with probability 0.75, > appears tail with probability > 0.25. Each time, you can bet x dollars, if it's head, you get x , if it's > tail, you lose x. You can play 10 times. What's the maximum expected payoff after 10 > times? You've got a winning proposition, so bet the farm! If you bet everything you have on each of the ten rolls, you'll lose everything most of the time. But in (3 / 4) ^ 10, or 5.631% of the games, you'll have $102,400.00 in front of you at the end. Play often enough and you'll walk away with an average of $5672 per game! Of course, Las Vegas lives for mathematicians like me :) === Subject: Re: Another coin betting game... > You've got a winning proposition, so bet the farm! If you bet > everything you have on each of the ten rolls, you'll lose everything > most of the time. But in (3 / 4) ^ 10, or 5.631% of the games, you'll > have $102,400.00 in front of you at the end. > Play often enough and you'll walk away with an average of $5672 per > game! The above two paragraphs do not match. The first paragraph says bet *everything*, but the second paragraph says *play more than once*, which isn't possible if you already bet everything. Two analysis in this thread suggested a win rate right around 1/20, which means that if you played 20 times you would win on average just once. But that's an average, and not to be counted on. A better plan would be to apply a level of certainty to winning. For instance, if I'm going to bet everything I own, with nothing left to ever play again, then I'll need a very high confidence of winning. Start at maybe 99%. The question then becomes how many times do I need to make this gamble, in order to be assured of at least one victory?, then split up your assets into exactly that many tries, and bet one portion on each run. === Subject: Re: Another coin betting game... > Currently, you have $100. You go to play a game. Each time, flip a coin, which appears head with probability 0.75, > appears tail with probability > 0.25. Each time, you can bet x dollars, if it's head, you get x , if it's > tail, you lose x. You can play 10 times. What's the maximum expected payoff after 10 > times? If x is not a variable, there is no way to do a maximization in your scheme, *unless* you mean that on any round the player has the option to play that round or not. If you allow that, your question is: when should the player quite (on or before round 10) in order to maximize the terminal winnings? It is intuitive that if a player skips a round for reasons of optimality, then that player should also skip all following rounds, because nothing about the player's wealth has changed and there are even fewer opportunities to catch up before the 10th and final round. So, it seems the optimal strategy is to play rounds 1..n for some n, 1 <= n <= 10, and to then stop playing. (The alternative of not playing at all is easy to assess.) You should be able to compute the expected net amount won in n plays, and then choose n to maximize that. R.G. Vickson === Subject: Re: Another coin betting game... posting-account=CYtergoAAAAOD8k-T_NdzknPJOBXmt8x CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Currently, you have $100. You go to play a game. Each time, flip a coin, which appears head with probability 0.75, > appears tail with probability > 0.25. Each time, you can bet x dollars, if it's head, you get x , if it's > tail, you lose x. You can play 10 times. What's the maximum expected payoff after 10 > times? If x is not a variable, there is no way to do a maximization in your > scheme, *unless* you mean that on any round the player has the option > to play that round or not. If you allow that, your question is: when > should the player quite (on or before round 10) in order to maximize > the terminal winnings? It is intuitive that if a player skips a round > for reasons of optimality, then that player should also skip all > following rounds, because nothing about the player's wealth has > changed and there are even fewer opportunities to catch up before > the 10th and final round. So, it seems the optimal strategy is to play > rounds 1..n for some n, 1 <= n <= 10, and to then stop playing. (The > alternative of not playing at all is easy to assess.) You should be > able to compute the expected net amount won in n plays, and then > choose n to maximize that. R.G. Vickson of course you are free to choose your x for each run... === Subject: Re: Another coin betting game... Luna Moon Toss the coin ten times in a row and bet everything you have each time. If tails ever comes up, you loose everything. The odds that tails never comes up is (3/4)^10 = .0563135147 If tails never comes up you win 100*(2^9) dollars. Expected value is .0563135147*100*512 = $2800 a twentieth of the time you win $51,200 . === Subject: Re: Another coin betting game... Toss the coin ten times in a row and bet everything you have each time. If tails ever comes up, you loose everything. The odds that tails never comes up is (3/4)^10 = .0563135147 If tails never comes up you win 100*(2^9) dollars. Expected value is > .0563135147*100*512 = $2800 a twentieth of the time you win $51,200 . You guys don't do any risk control? === Subject: Re: Another coin betting game... posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) Luna Moon Toss the coin ten times in a row and bet everything you have each time. If tails ever comes up, you loose everything. The odds that tails never comes up is (3/4)^10 = .0563135147 If tails never comes up you win 100*(2^9) dollars. Expected value is > .0563135147*100*512 = $2800 a twentieth of the time you win $51,200 . You guys don't do any risk control? YOU asked for maximum *expected* payoff. If you wanted a risk criterion you should have said so. R.G. Vickson === Subject: card game What's the expected numer of cards that need to be turned ovr in a regular 52-card deck in order to see the first ace? My friend answered: 52/4=13. And he was told his answer was wrong... Any thoughts? === Subject: Re: card game > What's the expected numer of cards that need to be turned ovr in a > regular 52-card deck in order to see the first ace? My friend answered: 52/4=13. And he was told his answer was wrong... Any thoughts? Let P(N,K) = N!/(N-K)! = number of permutations of N objects taken K at a time. To find the probability that the first ace is in position n, look at all the permutations of the 52 cards with that property. There are P(48,n-1) permutations for the non-aces in positions 1.. (n-1), there are 4 ways to pick the ace in position n, and that leaves 48 - (n-1) = 49-n non-aces plus three aces (= 52-n cards) to be distributed among the 52-n remaining spaces. There are (52-n)! such permutations. Altogether, the number of permutations with the desired property is 4*P(48,n-1)*(52-n)!, so the probability that the first ace is card n is P[n] = 4*P(48,n-1)*(52-n)!/52! = c*(52-n)/(49-n)!, where c = 1/1624350. Using Maple we get sum(p[n],n=1..49) = 1, as it should be, and the mean value = sum(n*p[n],n=1..49) = 53/5 = 10.6. R.G. Vickson === Subject: Re: card game posting-account=K5WE3woAAAAXArsybjkbN6LjMxWdHtbX Gecko/20080829 Firefox/2.0.0.17,gzip(gfe),gzip(gfe) What's the expected numer of cards that need to be turned ovr in a > regular 52-card deck in order to see the first ace? My friend answered: 52/4=13. And he was told his answer was wrong... Any thoughts? Let P(N,K) = N!/(N-K)! = number of permutations of N objects taken K > at a time. To find the probability that the first ace is in position > n, look at all the permutations of the 52 cards with that property. > There are P(48,n-1) permutations for the non-aces in positions 1.. > (n-1), there are 4 ways to pick the ace in position n, and that leaves > 48 - (n-1) = 49-n non-aces plus three aces (= 52-n cards) to be > distributed among the 52-n remaining spaces. There are (52-n)! such > permutations. Altogether, the number of permutations with the desired > property is 4*P(48,n-1)*(52-n)!, so the probability that the first ace > is card n is P[n] = 4*P(48,n-1)*(52-n)!/52! = c*(52-n)/(49-n)!, Sorry: that should be c*(52-n)!/(49-n)!. RGV > where c = 1/1624350. Using Maple we get sum(p[n],n=1..49) = 1, as it > should be, and the mean value = sum(n*p[n],n=1..49) = 53/5 = 10.6. === Subject: Re: card game > What's the expected numer of cards that need to be turned ovr in a > regular 52-card deck in order to see the first ace? My friend answered: 52/4=13. And he was told his answer was wrong... Any thoughts? The probability of hitting the first ace on card number n is: P(all the first n - 1 cards are not ace) * P(nth card is ace | the first n - 1 cards are not ace) = ((50 - n) * (51 - n) * (52 - n) * (53 - n) / (52 * 51 * 50 * 49)) * 4 / (53 - n) = (50 - n) * (51 - n) * (52 - n) / 1624350 Adding together n * this expression over n = 1 to 49, I get exactly 53 / 5 = 10.6. --- J K Haugland http://home.no.net/zamunda === Subject: Re: card game posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) What's the expected numer of cards that need to be turned ovr in a > regular 52-card deck in order to see the first ace? My friend answered: 52/4=13. And he was told his answer was wrong... Any thoughts? The probability of hitting the first ace on card number n is: P(all the first n - 1 cards are not ace) * P(nth card is ace | the > first n - 1 cards are not ace) = ((50 - n) * (51 - n) * (52 - n) * (53 - n) / (52 * 51 * 50 * 49)) * > 4 / (53 - n) = (50 - n) * (51 - n) * (52 - n) / 1624350 Adding together n * this expression over n = 1 to 49, I get exactly > 53 / 5 = 10.6. --- > J K Hauglandhttp://home.no.net/zamunda ... which is just (number of total cards + 1)/(number of good cards +1) Why? With total = n cards and good = k cards, let E(n,k) denote the expected index of the first good card. E(n,n) is clearly 1. In general, E(n,k) = k/n * 1 + (n-k)/n * (1+E(n-1,k)) if n>k. Now we can prove E(n,k)=(n+1)/(k+1) by induction on n-k: True for n=k since E(n,n)=1. Otherwise, E(n,k) = k/n * 1 + (n-k)/n * (1+E(n-1,k)) = k/n + (n-k)/n*(1+n/(k+1)) = ... = (n+1)/(k+1). It feels like there should be an elegant direct way to show E(n,k) = (n+1)/(k+1), but how? === Subject: Re: card game What's the expected numer of cards that need to be turned ovr in a > regular 52-card deck in order to see the first ace? My friend answered: 52/4=13. And he was told his answer was wrong... Any thoughts? The probability of hitting the first ace on card number n is: P(all the first n - 1 cards are not ace) * P(nth card is ace | the > first n - 1 cards are not ace) = ((50 - n) * (51 - n) * (52 - n) * (53 - n) / (52 * 51 * 50 * 49)) * > 4 / (53 - n) = (50 - n) * (51 - n) * (52 - n) / 1624350 Adding together n * this expression over n = 1 to 49, I get exactly > 53 / 5 = 10.6. --- > J K Hauglandhttp://home.no.net/zamunda ... which is just (number of total cards + 1)/(number of good cards > +1) > Why? > With total = n cards and good = k cards, let E(n,k) denote the > expected > index of the first good card. > E(n,n) is clearly 1. > In general, E(n,k) = k/n * 1 + (n-k)/n * (1+E(n-1,k)) if n>k. > Now we can prove E(n,k)=(n+1)/(k+1) by induction on n-k: > True for n=k since E(n,n)=1. > Otherwise, > E(n,k) = k/n * 1 + (n-k)/n * (1+E(n-1,k)) > = k/n + (n-k)/n*(1+n/(k+1)) = ... = (n+1)/(k+1). It feels like there should be an elegant direct way to show E(n,k) = > (n+1)/(k+1), > but how? Balls in bins. The good cards are the bin markers. The other cards are the balls. k+1 bins; n-k balls. Expect (n-k)/(k+1) balls in the first bin. -- Michael Press === Subject: Re: card game posting-account=ogMREwkAAAC5xUr8sg7heGtsvzzF18LA InfoPath.2; .NET CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > What's the expected numer of cards that need to be turned ovr in a > regular 52-card deck in order to see the first ace? My friend answered: 52/4=13. And he was told his answer was wrong... Any thoughts? Here is my hand-waving argument- but it should explain why 13 is wrong. In the first 13 cards, you expect to find one card of each rank, since there are 13 ranks, and there are the same number of each in the deck. (obviously you may not, but by symmetry there is no bias for any specific rank to appear more than another) However, if you assume there is one card of each rank, the ace is just as likely to be the first card as the 13th, so in general it WILL appear before the 13th card. In this case, you expect the ace to appear by the 6.5th card, which I round up to 7, so the probability is over 50%. Thus, without doing the math behind it, I would guess that the expected position for the 1st ace should be the 7th card. (the other three aces being expected at the 20th, 33rd, and 46th cards). === Subject: Number with mod 100. Hello teacher~ 2^1000 = x (mod 100) Find the x. ------------------------------------------------------- I can find with many(?) computation. Namely, 2^2 = 4, 2^3 = 8, 2^4 = 16, ... 2^20 = 6 (mod 100). 2^1000 = (2^20)^50 = 6^50 6^2 = 36, 6^3 = 16, 6^4 = 96, 6^5 = 76 6^6 = 56, 6^7 = 36 (mod 100) so, 6^(2+5k) = 36 (mod 100). 2^1000 = 6^50 = (6^47)*(6^3) = 36*(16) = 76 (mod 100) Do you have simpler method ? === Subject: Re: Number with mod 100. > 2^1000 = x (mod 100) Find the x. > [...] > Do you have simpler method ? 100 = 25.874 (decomposition into distinct powers of primes) We first compute 2^25 = x (mod 25). Under 25, there are 20 numbers coprime to 25, so a^20 = 1 (mod 25) for all a coprime to 25. Hence 2^1000 = 1 (mod 25). For 4, the computation is simpler: Obviously 2^b = 0 mod 4 for all b = 2, 3, ... According to the Chinese Remainder Theorem, there is one and only one x in 0..100 with x = 1 (mod 25) and x = 0 (mod 4). This one is 76. -- Helmut Richter === Subject: Re: Number with mod 100. > Hello teacher~ 2^1000 = x (mod 100) Find the x. ------------------------------------------------------- > I can find with many(?) computation. > Namely, > 2^2 = 4, 2^3 = 8, 2^4 = 16, ... > 2^20 = 6 (mod 100). 2^1000 = (2^20)^50 = 6^50 6^2 = 36, 6^3 = 16, 6^4 = 96, 6^5 = 76 > 6^6 = 56, 6^7 = 36 (mod 100) > so, 6^(2+5k) = 36 (mod 100). 2^1000 = 6^50 = (6^47)*(6^3) = 36*(16) = 76 (mod 100) Do you have simpler method ? Hm... maybe, Euler.. 2^1000 = x (mod 100) <==> 2^1000 = x (mod 5^2) and 2^1000 = x (mod 2^2) (1) 2^1000 = x (mod 25) 2^20 = 1 (mod 25) by Euler. 2^1000 = (2^20)^50 = 1 (mod 25) (2) 2^1000 = x (mod 2^2) 2^1000 = 0 (mod 2^2) by trivial. x = 1 (mod 25) and x = 0 (mod 4) x = 4t ==> 4t = 1 (mod 25) ==> t = 19 (mod 25) so, x = 4(19 + 25s) = 76 + 100.s = 76 (mod 100). === Subject: intuition behind min and max of IID random variables... Suppose we have N IID uniform (0, 1) random variables, What should be the expected value of Y=min(X1, ..., XN), Z=max(X1, ..., XN), and Var(Y), Var(Z), Cov(Y, Z) and Corr(Y, Z)? Any non calculation and intuitive way of getting the results? === Subject: Re: intuition behind min and max of IID random variables... > Suppose we have N IID uniform (0, 1) random variables, What should be the expected value of Y=min(X1, ..., XN), > Z=max(X1, ..., XN), and Var(Y), Var(Z), Cov(Y, Z) and Corr(Y, Z)? Any non calculation and intuitive way of getting the results? > What? Another homework like question? I propose that you write down the work you have done in addition to the bare question, if only to avoid the appearance of dishonesty. First I would write down a relationship between E(Y) and E(Z). Then I would observe a similarity between this question and another you asked for help on. What would you do? -- Michael Press === Subject: Re: question on enriched (bi)functors Dnia 21-10-2008 o 08:09:39 napisaü(a): > Please let me give you an example of my problem. If take a category C, > then one gets a functor hom(X,-) from C to sets for every object X of > C and a functor hom(-,Y) from Cop to sets for every Y. Then one says > that this defines a bifunctor hom(-,-) from (Cop x C) to sets.What > does one have to check *exactly* in this case? As Mariano has pointed out - in any case - you have to check functoriality (i.e. whether compositions and identities are preserved) in respect to the product category. Just think of your bifunctor (BTW: I don't like this terminology) hom(-,-) as of a functor from C^op times C to Set. -- Michaü Przybyüek === Subject: Number with prime p^4-5p^2+9. Hello teacher~ p is a prime. p^4 - 5.p^2 + 9 is a prime. Find the number of p. Answer : 2 ----------------------------------------------------------- If p = 2, then 2^4 - 5.(2^2) + 9 = 5. Yes. If p = 3, then 3^4 - 5.(3^2) + 9 = 45. No. If p = 5, then 5^4 - 5.(5^2) + 9 = 509. Yes. Since answer is 2, maybe it's only two possible cases. I want to know your normal solution. === Subject: Re: Number with prime p^4-5p^2+9. > Hello teacher~ p is a prime. p^4 - 5.p^2 + 9 is a prime. Find the number of p. Answer : 2 ----------------------------------------------------------- > If p = 2, then 2^4 - 5.(2^2) + 9 = 5. Yes. > If p = 3, then 3^4 - 5.(3^2) + 9 = 45. No. > If p = 5, then 5^4 - 5.(5^2) + 9 = 509. Yes. Since answer is 2, maybe it's only two possible cases. > I want to know your normal solution. Look at the possible values modulo 5. Phil -- The fact that a believer is happier than a sceptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality. -- George Bernard Shaw (1856-1950), Preface to Androcles and the Lion === Subject: Re: Number with prime p^4-5p^2+9. posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/2008072820 Firefox/3.0.1,gzip(gfe),gzip(gfe) > Hello teacher~ p is a prime. p^4 - 5.p^2 + 9 is a prime. Find the number of p. Answer : 2 ----------------------------------------------------------- > If p = 2, then 2^4 - 5.(2^2) + 9 = 5. Yes. > If p = 3, then 3^4 - 5.(3^2) + 9 = 45. No. > If p = 5, then 5^4 - 5.(5^2) + 9 = 509. Yes. Since answer is 2, maybe it's only two possible cases. > I want to know your normal solution. You should have tried a few more exmaples (p=7, p=11, p=13) and would have stumbled over a striking pattern (just look at the last digit). Use modular arithmetic with a suitable modulus and little Fermat to show that p=5 and p^4-5p^2+9=5 are the only possible cases. hagman === Subject: Re: Number with prime p^4-5p^2+9. > Hello teacher~ p is a prime. p^4 - 5.p^2 + 9 is a prime. Find the number of p. Answer : 2 ----------------------------------------------------------- > If p = 2, then 2^4 - 5.(2^2) + 9 = 5. Yes. > If p = 3, then 3^4 - 5.(3^2) + 9 = 45. No. > If p = 5, then 5^4 - 5.(5^2) + 9 = 509. Yes. Since answer is 2, maybe it's only two possible cases. > I want to know your normal solution. Hm...maybe, this is also Euler. I can show that p^4 - 5.p^2 + 9 = 0 (mod 5) for p = 7, 11, 13, 17, ... Since p^4 = 1 (mod 5) for p = 7, 11, 13, 17, ... p^4 - 5.p^2 + 9 = 10 - 5.p^2. so, 5 | (10 - 5.p^2). so, answer is p = 2, 5. === Subject: Re: Number with prime p^4-5p^2+9. @localhost.localdomain: > Hello teacher~ p is a prime. p^4 - 5.p^2 + 9 is a prime. Find the number of p. Answer : 2 ----------------------------------------------------------- > If p = 2, then 2^4 - 5.(2^2) + 9 = 5. Yes. > If p = 3, then 3^4 - 5.(3^2) + 9 = 45. No. > If p = 5, then 5^4 - 5.(5^2) + 9 = 509. Yes. Since answer is 2, maybe it's only two possible cases. > I want to know your normal solution. I think you've misread the problem. If the polynomial is p^4 + 5p^2 +9, then the only solution is p=2. The polynomial you give has lots of solutions. B. -- Cheerfully resisting change since 1959. === Subject: Selling solution manual for Introduction to Digital Systems (Ch 2-15) / Ercegovac posting-account=aAiiggoAAACwFXWVrvlIhLGgMWLAsV2N MathPlayer 2.10a; .NET CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.2; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Price is one dollar per chapter, payed via paypal. Email me at solutionaire77@yahoo.com is you are interested. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Yes, my name here is Salviati. You may call me Dr. Blumschein. I may call you anything I like, but I choose to call you something > relating to like ECKHARD OR BLUMSCHEIN. Quite right, let's not forget he is ECKARD BLUMSCHIEN. Unlike Salviati his real name makes many amusing anagrams. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > Dave Seaman schrieb im Newsbeitrag Proofs of what? That each natural number is finite? What do you mean by > finite? The definition I know is that a set is called finite if it > can be mapped bijectively to some natural number, viewed as a set. With > that definition, your question reduces to a triviality. You must have > some different meaning of finite in mind, and that's why I ask the > question. Let me criticise this unfortunately very common idea. Let us criticize this very uncommon criticism, which avoids anything like a direct response to 'What do you mean by finite?'. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? David R Tribble schrieb im Newsbeitrag >> Cantor and Dedeking agreed on that there must be more real numbers as >> compared to the rational ones. This is seemingly correct if one considers >> the reals as the sum of rationals and irrationals. Using the same logic: > There must be more rationals as compared to the naturals. > This is seemingly correct if one considers the rationals as the > sum of naturals and non-integral rationals. Or perhaps: > There must be more naturals as compared to the odd naturals. > This is seemingly correct if one considers the naturals as the > sum of odd naturals and even naturals. Except, of course, these arguments are incorrect. The solution is simple: Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > David R Tribble schrieb im Newsbeitrag >> Cantor and Dedeking agreed on that there must be more real numbers as >> compared to the rational ones. This is seemingly correct if one considers >> the reals as the sum of rationals and irrationals. Using the same logic: > There must be more rationals as compared to the naturals. > This is seemingly correct if one considers the rationals as the > sum of naturals and non-integral rationals. Or perhaps: > There must be more naturals as compared to the odd naturals. > This is seemingly correct if one considers the naturals as the > sum of odd naturals and even naturals. Except, of course, these arguments are incorrect. The solution is simple: There are different measures of set size, subset versus injection. By the subset measure {1,2,3} and {4,5,6} cannot be compared for size, but y injectively, they are of equal size. Similarly with the set of odd naturals and the set of even naturals. For {1,2} and {3,4,5,6}, they are subset-uncomparable but injectively {1,2} is strictly smaller === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Virgil schrieb im Newsbeitrag > Virgil schrieb im Newsbeitrag > Of course, it is difficult for mathematicians to imagine real numbers. >> It seems to be much more difficult for non-mathematicians, like you. >> No. What I consider real numbers fits to the good old foundations of >> mathematics. The good old foundations for real analysis are based on Dedekind cuts > or Cauchy sequences, both of which lead to uncountable sets. No no. I meant the older one. Dedekind cuts are already elusive. Cauchy hesitated to make his pragmatism a rigorous theory. >> He (Galilei) did not depend on circumstances but, using the same method, >> he found a convincing soluting which was ignored by Cantor and >> Dedekind. Since Galileo's solution did not take into account Cantor's insights, > you have no grounds for that claim. What proven insight do you attribute to Cantor? Do you have any literature that reveals how Cantor refuted Galilei? >> A continuum cannot be a set of points except for the fiction >> of infinitely much of fictitious elements. Then lets round-file all references to continua as irrelevant to set > theory. Yes. The question is whether we should abandon > the nice idea of continuum or the naive set theory. >> Just separate them. Do not try to make sets and contunua work together. >> You are correct in that here meet two different realms. We have no >> alternative but to calculate with rational numbers as if they were real. They are real, or at least there is a subfield of the reals which is > isomorphic to the ordered field of rationals, which is all that > mathematics needs. Mathematics needs pragmatism in this case. Sometimes applications need a more exact theory, so far missing. >> However, when it comes to interpretation, we should know what we are >> actually doing. Then it is a shame you don't. I can just point to some mistakes and resulting consequences. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > Virgil schrieb im Newsbeitrag > Virgil schrieb im Newsbeitrag > Of course, it is difficult for mathematicians to imagine real numbers. >> It seems to be much more difficult for non-mathematicians, like you. >> No. What I consider real numbers fits to the good old foundations of >> mathematics. The good old foundations for real analysis are based on Dedekind cuts > or Cauchy sequences, both of which lead to uncountable sets. No no. I meant the older one. Prior to Dedekind cuts and equivalence classes of Cauchy seqeunces, there was no coherent logical foundation for the reals, just handwaving. > Dedekind cuts are already elusive. Not as elusive as what went before. Those cuts lead to an at least logically cohesive system, which did not previously exist. > Cauchy hesitated to make his pragmatism a rigorous theory. But it got done, nevertheless. >> He (Galilei) did not depend on circumstances but, using the same method, >> he found a convincing soluting which was ignored by Cantor and >> Dedekind. Since Galileo's solution did not take into account Cantor's insights, > you have no grounds for that claim. What proven insight do you attribute to Cantor? Do you have any literature > that reveals how Cantor refuted Galilei? Cantors first proof as well as the diagonal construction validate a notion of different sizes of infiniteness. > Just separate them. Do not try to make sets and contunua work together. >> You are correct in that here meet two different realms. We have no >> alternative but to calculate with rational numbers as if they were real. They are real, or at least there is a subfield of the reals which is > isomorphic to the ordered field of rationals, which is all that > mathematics needs. Mathematics needs pragmatism in this case. Regarding isomorphic objects as being for all practical purposes, the way that mathematics habitually does is eminently pragmatic., > Sometimes applications need a > more exact theory, so far missing. You are asking for opposites simultaneously, pragmaticism and greater exactness. >> However, when it comes to interpretation, we should know what we are >> actually doing. Then it is a shame you don't. I can just point to some mistakes and resulting consequences. They are your mistakes and their consequences affect only you. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? >> When Cantor tried to show that the reals are uncountable, >> he characterized reals, not just irrationals, as having an infinite >> decimal representation. Sometimes you surprise me. Amidst of Stalin, Cantor in the madhouse and > all I am not aware of any link between Stalin and mathematics. > the other crap, you start out writing something sensible. Yes, rationals > e.g. like 0.5 are represented 0.499999... Hopefully you got aware that one cannot immediately handle any real number except via its rational representation or like an unresolved problem, e.g. sqrt(2) > Can we imagine to exactly subtract >> a number b from a number a if they possibly differ from each other >> at a decimal that we cannot even hope to know? You only can't, when you keep insisting to do it the old fashioned way > digit by digit, as you can with finite decimals. But since reals are > completely different than rationals alone, you cannot expect, that > arithmetic operations work the same way. But using lim(x_n - y_n) = x - y, where x = lim x_n and y = lim y_n, > you'll > easily see, that 0.999... equals 1,000... The construciton of reals using > equivalence classes of cauchy sequences reflects that, too. So, instead of > tediously performing operations on digits, we use theorems, that do it > quite easily. Isn't this way more efficient? Not really. Pragmatism is valuable. Of course. However one should know what one actually does. Is it really justified to assume equivalence classes across the abyss between rationals and reals? So, just widen your horizon a little, I consider my horizon wider as compared to yours. And I did my homework. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > Yes, rationals e.g. like 0.5 are represented 0.499999... Hopefully you got aware that one cannot immediately handle > any real number except via its rational representation > or like an unresolved problem, e.g. sqrt(2) You on the other hand must be aware, that I think it's bull. And even more, it's the very core of your mental blockade towards modern math. As long as you don't quit this thinking, your mathematical understanding will remain retarded. > Not really. Pragmatism is valuable. Of course. > However one should know what one actually does. You don't need to know, what you do. You only need to know, that the proofs are valid. > Is it really justified to assume equivalence classes > across the abyss between rationals and reals? Yes, equivalence classes are useful for solving ambiguations. And: no abyss any more, because the old rationals have been thrown away, and there is no abyss within the reals. > I consider my horizon wider as compared to yours. Possibly, but why do you keep sticking to an unsufficient notion? I think that's narrow-minded. > And I did my homework. Actually I didn't realize the connontation, because I reject that crappy notion of unresolved problems. It has nothing to do with math. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > Klaus Cammin schrieb im Newsbeitrag >> When Cantor tried to show that the reals are uncountable, >> he characterized reals, not just irrationals, as having an infinite >> decimal representation. Sometimes you surprise me. Amidst of Stalin, Cantor in the madhouse and > all I am not aware of any link between Stalin and mathematics. the other crap, you start out writing something sensible. Yes, rationals > e.g. like 0.5 are represented 0.499999... Hopefully you got aware that one cannot immediately handle > any real number except via its rational representation > or like an unresolved problem, e.g. sqrt(2) 0 and 1 and 2, etc., are all real numbers, as well as being rationals and integers and naturals within the set of reals. Not really. Pragmatism is valuable. Of course. > However one should know what one actually does. That eckard.blumschein does not know what he does, does not mean no one else does. > Is it really justified to assume equivalence classes > across the abyss between rationals and reals? The ordered field of rationals embeds isomorphically in the complete ordered field of reals, so there is, mathematically, no essentail difference between the original and its image, except that the rational as image also a real. > So, just widen your horizon a little, I consider my horizon wider as compared to yours. One always does, even when there is strong evidence against it being so. > And I did my homework. Not the math part osf ti. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Virgil schrieb im Newsbeitrag >> There IS no complete list of natural numbers. >> Salviati There is no such physical list, but it is easy enough to imagine one. Sorry, to me it is obviously even in a dream impossible to imagine something nonsensical and self-contradictory. > And as numbers exist only in imaginations, such a complete list exists > as certainly as the numbers in it exist. Such style of thinking let Buridan's donkey rise again as a cat. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > Virgil schrieb im Newsbeitrag > There IS no complete list of natural numbers. >> Salviati There is no such physical list, but it is easy enough to imagine one. Sorry, to me it is obviously even in a dream impossible > to imagine something nonsensical and self-contradictory. > And as numbers exist only in imaginations, such a complete list exists > as certainly as the numbers in it exist. Such style of thinking let Buridan's donkey rise again as a cat. The issue is whether imagining such a complete list introduces any logical contradictions into the system one imagines such a complete list arises from. E.g., ZFC or NBG. And, so far, there are no logical contradictions known in either ZFC or NBG. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Virgil schrieb im Newsbeitrag >> You may believe me that he actually believed that infinity is a firm >> quantity. Do you mean to claim that Cantor did NOT believe that that infiniteness > of sets was a quality that some, but not all, sets possessed? If I recall correctly, he never used the expression infiniteness. >> As soon as bijectability between sets is shown to be an equivalence >> relation, all of them. >> As far as I know, Cantor's proofs merely dealt with the distinction >> between >> countables and uncountables. How many classes are that? More than two? Cantor's results have been extended since his time. theory. Hilbert admitted that axiomatic theory merely provided a new possibility for maintaining the belief in certain Zusammenhaenge. Students still have to learn naive set theory. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > Virgil schrieb im Newsbeitrag > You may believe me that he actually believed that infinity is a firm >> quantity. Do you mean to claim that Cantor did NOT believe that that infiniteness > of sets was a quality that some, but not all, sets possessed? If I recall correctly, he never used the expression infiniteness. The issue is whether he used the idea of infiniteness. Which he did. >> As soon as bijectability between sets is shown to be an equivalence >> relation, all of them. >> As far as I know, Cantor's proofs merely dealt with the distinction >> between >> countables and uncountables. How many classes are that? More than two? Cantor's results have been extended since his time. Not really. Yes Really! === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Klaus Cammin schrieb im Newsbeitrag >> However, I am convinced his own >> answer was unable to resolve the challenge by Buridan's donkey. I've told you this before: there's is an unbroken chain of theorems and > their proofs, that lead to real numbers. Assuming that real numbers are an archimedian ordered field, I am not sure if this assumption is correct. > 1. you can prove, that Dedekind cuts and the LUB property are equivalent. I do not doubt that because even serious mathematicans object against the LUB. > 2. then you prove, that every bounded sequence converges to its LUB. I am just cutious: Is pi a LUB? > 3. next, you can prove, that interval nesting works, i.e. that you get a > real number from it. With nested intervals I am catched within the rationals because I will never arrive at a difference equal to zero for as many steps as you like. > 4. and then you can prove, that every decimal expansion is a real number, It undoubterly is, provided it is represented by an infinite amount of elements. > and vice versa, that every real number has a decimal expansion. No objection. > 5. further you can prove, that IR is complete, i.e. that the limit of > every > convergent cauchy sequence actually is a real number. I do not doubt that with the tiny addition that the limit does not belong to the rational numbers. Didn't Meray correctly write fictitious limit? > However, his sentence >> The natural numbers were made by God shows that he could not offer an >> alternative to Cantor Until today nobody has ever offered something better for IR than the > above. Really? Were not Euclid, Archimedes, Spinoza, Galilei, Gauss, and many others correct while Cantor's rejection of all rationality was at least highly questionable? The question I see is not whether or not there is a real number zero but does it share the same properties with the rational number zero? I am arguing, real numbers are something quite different even if it is mostly impossible and therefore of course not at all necessary to use it differently as compared to rationals. > And you and the other cranks This is an insult because there are actually people who are even more naive than Cantor who discredit serious criticism. > with their ridiculous philosphical pretexts > won't either - where the latter is, I believe, a clear consequence of the > former. A pretext is a reason which is pretended to have caused you to do something. If you insist that your sentence contibuted something worth to understand than tell me please what you mean by latter and former. > And: if I were Buridan's donkey, > I'd flip a coin, eat one haystack, and the other the next day ... With the sympathy of a senior I realize that you didn't realize why this metaphor is extremely valuable as a hint to still lacking understanding of a peculiarity of real numbers. Did you ever take into account the possibility that Schroedingers cat is just a modern version of Buridan's donkey, and quantum computing originates from it? Tell me please, who has a quantum computer that works as promised. ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? >> Assuming that real numbers are an archimedian ordered field, > I am not sure if this assumption is correct. See Michael, this can be proved, too. Actually the chain starts with first order logic. So we do have a pretty good answer to the question, what numbers are. >> 1. you can prove, that Dedekind cuts and >> the LUB property are equivalent. > I do not doubt that because even serious mathematicans object > against the LUB. What's his name? I doubt, it's a mathematician. > I am just cutious: Is pi a LUB? For the sequence pi-1/n forall n, pi is a LUB. But see Virgil, too. And: distinguish between LUB (least upper bound) and the LUB axiom: for every S c IR: if S has an upper bound, S has a LUB. > With nested intervals I am catched within the rationals because I will > never arrive at a difference equal to zero for as many steps as you > like. Since the prove is true (it's based on 2.), your thinking is wrong. When you quit this thinking, you won't have this problem any longer. (Hell, put your arm down!) >> 4. and then you can prove, that every decimal expansion is a real >> number, and vice versa, that every real number has a decimal expansion. No objection. That's a pretty charming property of your reasoning: when you assert 4., you must assert 1.-3., too, because it's a direct consequence. When you deny only 3., then 4. isn't true, and you're no longer entitled to say No objection! Albrecht deserves credit for denying the existence of pi. Hence, when I had to decide, who of the two of you is silly and who is insane, I proudly award you the medal of unsilliness. > I do not doubt that with the tiny addition that the limit does not > belong to the rational numbers. Right, since we're talking about infinite decimal expansions (reals), no limit belongs to the rational numbers. > Didn't Meray correctly write fictitious limit? You may do that, but since all reals are fictionous in our new sense, the term is superfluous. It's advantageous to do so, because emotional qualifications might be attached to this term, i.e. outer-mathematical reasoning. >> Until today nobody has ever offered something better for IR than the >> above. > Really? Were not Euclid, Archimedes, Spinoza, Galilei, Gauss, and many > others correct while Cantor's rejection of all rationality was at least > highly questionable? Do you think math is progressing while going back in time? Until now I thought it's the other way round ... Yes, the modern theory of numbers is better than theirs. (Yes, they were great guys, but that's irrelevent here. And yes, they also produced some crap.) > I am arguing, real numbers are something quite different Agreed, hence my advice: throw away the old rationals. > [...] therefore of course not at all necessary to use it > differently as compared to rationals. You'll see later, that you can save some wanted properties of the former rationals, but first you MUST throw away the old rationals. Come on. Give yourself a revolutionary push! > A pretext is a reason which is pretended to have caused you to do > something. If you insist that your sentence contibuted something worth > to understand than tell me please what you mean by latter and former. Sorry, I got it wrong with pretext. I meant to say bias or prejudice. And I meant, what you expressed responding to 3. and what I keep seeing in my tutorials: the whole crappy babbling about a sequence approaching its limit without ever reaching it. This thinking puts a completely false focus on the matter. In modern math, the sequences (1,1,1,...) and (1,0.5,0.333...,0.25,...) satisfy the same definition of a limit. One reaches its limit and one doesn't. So, it was a very sensible thing to drag away our ancestors' beady eyes from a completely sterile notion, and instead say for instance, that only finite many members of a sequence are outside an arbitrary epsilon- environment. Much more clearer, simplier, more capable and omitting outer-mathematical babbling. When someone talks about Einstein's relativity theory and doesn't know the definition of a tensor, I know that his babbling has nothing to do with physics. When someone talks about limits and doesn't understand the definition of a limit, .... > With the sympathy of a senior I realize that you didn't realize why this > metaphor is extremely valuable as a hint to still lacking understanding > of a peculiarity of real numbers. Do you refer to your reasoning about unfullfillable tasks like sqrt(2)? If so, you might easily conclude, that I think it's bull. Viele Gr.9f¤e Klaus === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > Klaus Cammin schrieb im Newsbeitrag >> However, I am convinced his own >> answer was unable to resolve the challenge by Buridan's donkey. I've told you this before: there's is an unbroken chain of theorems and > their proofs, that lead to real numbers. Assuming that real numbers are an archimedian ordered field, I am not sure if this assumption is correct. Then you also cannot be sure it is incorrect. 1. you can prove, that Dedekind cuts > and the LUB property are equivalent. I do not doubt that because even serious mathematicans object > against the LUB. Your doubts do not impress. 2. then you prove, that every bounded sequence converges to its LUB. I am just cutious: Is pi a LUB? Ever real is both a LUB of the set of smaller rationals and a GLB of the set of larger rationals. 3. next, you can prove, that interval nesting works, i.e. that you get a > real number from it. With nested intervals I am catched within the rationals because I will never > arrive at a difference equal to zero for as many steps as you like. But there will always be at least one real between the LUB of the left endpoints of those intervals, and the GLB of their right endpoints. And if the lengths of those intervals converge to zero, then that LUB equals that GLB. 4. and then you can prove, that every decimal expansion is a real number, It undoubterly is, provided it is represented by an infinite amount of > elements. and vice versa, that every real number has a decimal expansion. No objection. 5. further you can prove, that IR is complete, i.e. that the limit of > every > convergent cauchy sequence actually is a real number. I do not doubt that with the tiny addition that the limit does not belong to > the rational numbers. Since, in the usual sense, every rational is also a real(one can inject the rationals into the reals by a field isomorphism) that addition is unnecessary. >> However, his sentence >> The natural numbers were made by God shows that he could not offer an >> alternative to Cantor Until today nobody has ever offered something better for IR than the > above. The question I see is not whether or not there is a real number zero > but does it share the same properties with the rational number zero? In mathematics, one only needs to specify things up to isomorphism, so that one need not distinguish the rational subset of the reals from the rational field from which the reals are constructed, at least as far as their properties as an ordered field are concerned, as the two are isomorphic as ordered fields. > I am arguing, real numbers are something quite different even if it is > mostly impossible and therefore of course not at all necessary to use it > differently as compared to rationals. It would help if you knew more about mathematics. And you and the other cranks This is an insult because there are actually people who are even more naive > than Cantor who discredit serious criticism. Speaking of yourself? === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Virgil schrieb im Newsbeitrag >> I am just curious: Is pi a LUB? Ever real is both a LUB of the set of smaller rationals and a GLB of > the set of larger rationals. If the reals include the rationals, is every rational a LUB and a GLB of the sets you mentioned? >> 3. next, you can prove, that interval nesting works, i.e. that you get >> a >> real number from it. >> With nested intervals I am catched within the rationals because I will >> never >> arrive at a difference equal to zero for as many steps as you like. But there will always be at least one real between the LUB of the left > endpoints of those intervals, and the GLB of their right endpoints. You seem to agree that the process will never end. How to distinguish between reals and rationals under such condition? Is 1,000... a rational? Is 0.999... a rational? Is there something between the two? Please abstain from offering brutal definitions. I prefer the beauty of mathematics. > And if the lengths of those intervals converge to zero, then that LUB > equals that GLB. I imagine that the rationals then lost their property of being countable. >> 5. further you can prove, that IR is complete, i.e. that the limit of >> every >> convergent cauchy sequence actually is a real number. >> I do not doubt that with the tiny addition that the limit does not belong >> to >> the rational numbers. Since, in the usual sense, every rational is also a real (one can inject > the rationals into the reals by a field isomorphism) that addition is > unnecessary. Does a white paper really exhibit a morphism? Field isomorphism sounds to me like something without sound basis. > However, his sentence > The natural numbers were made by God shows that he could not offer > an > alternative to Cantor >> Until today nobody has ever offered something better for IR than the >> above. > The question I see is not whether or not there is a real number zero >> but does it share the same properties with the rational number zero? In mathematics, one only needs to specify things up to isomorphism, so > that one need not distinguish the rational subset of the reals from the > rational field from which the reals are constructed, at least as far as > their properties as an ordered field are concerned, as the two are > isomorphic as ordered fields. Sounds to me like largely sound but dishonest and unclean pragmatism. >> I am arguing, real numbers are something quite different even if it is >> mostly impossible and therefore of course not at all necessary to use it >> differently as compared to rationals. It would help if you knew more about mathematics. My person does not matter. Set theory led to endless debates. Someone even asked: Is there a school of cranks? Buridan's donkey and Heisenberg's cat are still to be understood .... Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > ... the limit of > every > convergent cauchy sequence actually is a real number. I do not doubt that with the tiny addition that the limit does not belong to > the rational numbers. Didn't Meray correctly write fictitious limit? The limit is an equivalence class of sequences of rationals. (Indeed the rationals are equivalence classes of pairs of integers). Whether you decide to call these equivalence classes real numbers or fictitious limits or cross grained kumquats does not matter. The rationals are (or are represented by, or are isomorphic to, or ...) a subset (or generalized subset, or lithuanian breadfruit, or ...) of these equivalence classes. but does it share the same properties with the rational number zero? Yes. In particular if x is a member of an equivalence class, and y is a member of an equivalence class, then x = y if and only if - William Hughes === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? William Hughes schrieb im Newsbeitrag >> I do not doubt that with the tiny addition that the limit does not belong >> to >> the rational numbers. Didn't Meray correctly write fictitious limit? > The limit is an equivalence class of sequences of rationals. (Indeed > the rationals are equivalence classes of pairs of integers). > Whether you decide to call these equivalence classes real numbers or > fictitious limits or cross grained kumquats does > not matter. The rationals are (or are represented by, or are > isomorphic to, or ...) a subset (or generalized subset, or lithuanian > breadfruit, or ...) of these equivalence classes. The relationship between reals and rationals can be compared with that between persons who are alife and dead. This is not my idea. Bishop Berkeley >> The question I see is not whether or not there is a real number zero >> but does it share the same properties with the rational number zero? > Yes. In particular if x is a member of an equivalence class, > and y is a member of an equivalence class, then x = y if and only if > x-y is a member of the equivalence class, zero. While I do not deny the possiblity to define such equivalence classes I would like to point out that these elements cannot be claimed to have the same properties as do the rationals. One may calculate with them as if they were rationals ir this is possible at all. However, equivalence classes that include irrationals primarily behave as if they were irrational. So the notion equivalence class is a red herring in this case. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Distribution: world > However, I am convinced his own > answer was unable to resolve the challenge by Buridan's donkey. >> I've told you this before: there's is an unbroken chain of theorems and >> their proofs, that lead to real numbers. >> Assuming that real numbers are an archimedian ordered field, I am not sure if this assumption is correct. The assumption was only for the sake of a brief post. It's possible to prove this, starting with a definition of reals as equivalence classes of Cauchy sequences of rationals. I've never done it myself, but _Real Analysis_ by Haaser and Sullivan has the whole development. (Inspired by this, I proved for my own edification the equivalent statement for rationals defined as equivalence classes of ordered pairs of integers. It's not difficult, just a *lot* of typing.) -- Michael F. Stemper #include If this is our corporate opinion, you will be billed for it. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Michael Stemper schrieb > The assumption was only for the sake of a brief post. It's possible to > prove this, starting with a definition of reals as equivalence classes > of Cauchy sequences of rationals. This is already the moot point. Can a real number be incorporated into the realm of the rational numbers? The proponents of set theory spoke of volles Buergerrecht (unrestricted citizenship). While 1/2 is of course the same as is 1000/2000 and 1.000/2.000 this does not hold for 1.000.../2.000... The difference is hard to understand for mathematicians. However we need it as to actually understand what makes real numbers different from the rational ones. We need it as to convincingly resolve Buridand's Donkey and Heisenberg's cat. And I found several further examples. > I've never done it myself, but _Real Analysis_ by Haaser and Sullivan > has the whole development. (Inspired by this, I proved for my own > edification the equivalent statement for rationals defined as equivalence > classes of ordered pairs of integers. It's not difficult, just a *lot* > of typing.) Whether or not a proof is conclusive, depends on its basic assumptions. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Virgil schrieb im Newsbeitrag > Virgil schrieb im Newsbeitrag Fraenkel 1923 explained >> Salviati seems unable to think for himself, but relies only on >> arguments >> to authority. In mathematics, arguments to authority do not prove >> anything. >> I just refuse to fulfill unnecessary demandings. It is your duty to read >> original literature. On the contrary, my only obligation is to accept valid proofs when they > are presented to me. And so far you have no presented any. It is only historians who have any duty to read the literature. If something seems to be wrong, then one should not disdain the possibility to search for reasons. > You are hung up on the false notion that there can only be one sort > of > size, your own. Size is still a pretty common and reasonable notion. >> How many sizes do you have? A neck size, a waist size, a leg length a >> sleeve length, a coat size, a shoe size, etc., etc. >> So there is room for more that one size for sets. >> One sort is by membership. >> Another sort by subset, >> And a third by injection/bijectiom/surjection. >> Size is always relative to a basic unity. How many independent basic unities do you have? One could quite > probably rack up 20 or so, just for you quite easily. If we agree on just one system of numbers, we have to agree on one unity 1. > Nor may we conclude that there are more than countably many real > numbers. >> You may not be able to, but your limitations are not my concern. >> Even Hausdorff did not write ueberabzaehlbar (= more than countable) but >> correctly unabzaehlbar (uncountable). But Cantor did define a perfectly legitimate ordering of sets sizes in > which injectable-to means less-than-or-equal-to. 0.999...9 < 1.000...0 (countable) but 0.999... = 1.000... (uncountable) Cantor naively veiled the distinction between countable elements and uncountable elements in a self-deceptive manner. Ebbinghaus (by Lessing) called this an obvious error and simultaneously a valuable truth. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > 0.999...9 < 1.000...0 (countable) but 0.999... = 1.000... (uncountable) > Cantor naively veiled the distinction between countable elements and > uncountable elements in a self-deceptive manner. Ebbinghaus (by Lessing) > called this an obvious error and simultaneously a valuable truth. Ascribing your stupid nonsense to innocent third parties is a case of libel. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Ralf Bader schrieb im Newsbeitrag >> 0.999...9 < 1.000...0 (countable) but 0.999... = 1.000... (uncountable) >> Cantor naively veiled the distinction between countable elements and >> uncountable elements in a self-deceptive manner. Ebbinghaus (by >> Lessing) >> called this an obvious error and simultaneously a valuable truth. Ascribing your stupid nonsense to innocent third parties is a case of > libel. Libel is something written which wrongly accuses someone of something, and which is therefore against the law. Everybody may look into the book Numbers by Ebbinghaus et al. The chapter on set theory was written by Ebbinghaus himself. It begins with a quotation from Gotthold Ephraim Lessing. I do not have the book at hand. However, I am sure that Lessing spoke of an obvious error that led to a valuable truth. I am also sure that the following text did nowhere explain what was meant by the obvious error. The author refused to give an explanation. Further comments are perhaps unnecessary. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > Virgil schrieb im Newsbeitrag > Virgil schrieb im Newsbeitrag Fraenkel 1923 explained >> Salviati seems unable to think for himself, but relies only on >> arguments >> to authority. In mathematics, arguments to authority do not prove >> anything. >> I just refuse to fulfill unnecessary demandings. It is your duty to read >> original literature. On the contrary, my only obligation is to accept valid proofs when they > are presented to me. And so far you have no presented any. It is only historians who have any duty to read the literature. If something seems to be wrong, then one should not disdain the possibility > to search for reasons. In this case, I would first ask a psychiatrist to examine eckard.blumschein in any attempt to locate what is wrong. > You are hung up on the false notion that there can only be one sort > of > size, your own. Size is still a pretty common and reasonable notion. >> How many sizes do you have? A neck size, a waist size, a leg length a >> sleeve length, a coat size, a shoe size, etc., etc. >> So there is room for more that one size for sets. >> One sort is by membership. >> Another sort by subset, >> And a third by injection/bijectiom/surjection. >> Size is always relative to a basic unity. How many independent basic unities do you have? One could quite > probably rack up 20 or so, just for you quite easily. If we agree on just one system of numbers, we have to agree on one unity 1. We are talking of sets which need not be sets of numbers, and there are several ways of comparing such sets, each with its own equivalent of your basic unity. > But Cantor did define a perfectly legitimate ordering of sets sizes in > which injectable-to means less-than-or-equal-to. 0.999...9 < 1.000...0 (countable) but 0.999... = 1.000... (uncountable) What is allegedly uncountable about either 0.999... or 1.000...? > Cantor naively veiled the distinction between countable elements and > uncountable elements in a self-deceptive manner. On the contrary, Cantor never applied those adjectives to elements, only to sets. So the only self-deception is eckard.blumschein's. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Virgil schrieb > S> Size is always relative to a basic unity. >> V>> How many independent basic unities do you have? One could quite >> probably rack up 20 or so, just for you quite easily. >> S>> If we agree on just one system of numbers, we have to agree on one unity 1. > V> We are talking of sets which need not be sets of numbers, and there are > several ways of comparing such sets, each with its own equivalent of > your basic unity. Ok, I recall that sets were explained to me like areas within a closed loop. In this case, it is pretty obvious that one cannot compare the size of an unlimited area A with the size of an also unlimited area B: xxxxxxxxxxxxx x oooooooooooooooooo x o A o x B o x o x xxxxxxxxxxxxx o oooooooooooooooooo Neither A nor B alone has a finite size. In order to measure the common area we need some unit. S>> 0.999...9 < 1.000...0 (countable) but 0.999... = 1.000... (uncountable) > V> What is allegedly uncountable about either 0.999... or 1.000...? The infinite and therefore uncountable amount of nines or zeroes, respectively. >> Cantor naively veiled the distinction between countable elements and >> uncountable elements in a self-deceptive manner. On the contrary, Cantor never applied those adjectives to elements, > only to sets. He intended to count the reals because he naively believed that there must be more reals as compared to the rationals. Initially he called a Menge (set) an Inbegriff and Maechtigkeit what he later renamed into cardinality. Cantor's naivity is obvious from his letters to Dedekind who also imagined that there are infinitely MANY more reals as compared wit the rationals. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R . === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > venkat.6123@gmail.com a .8ecrit : > I'm an amateur in this topic. When my friend told me about Cantor's >> cardinalities, the first thing that seemed odd to me was the 1-to-1 >> mapping used for comparing the sizes of two infinite sets. I'm not >> sure if it is proved that 1-to-1 correspondence is valid for comparing >> two sets of infinite size. Not because of the practical difficulties >> in comparing or the time it takes for the comparing process to >> complete. I think, atleast it should be proved logically that such >> mapping is valid between two infinite sets. >> For example, the mappings such as y=2x showing that there are as many >> even numbers as integers seem dubious to me because there is no proof >> that such function is meaningful when values of x and y reach >> infinite. > But x and y never do reach infinite. The issue is that in comparing N, the set of naturals and , say E, for > the set of evens, is there a correspondence which does not use twice nor > omit any member of either set. That is certainly a valid comparison for finite sets. Can you think of a better way of comparing sets for relative sizes that > works for finite sets? If not why object to it for infinite sets? >> The 1-to-1 correspondence works well as long as n is different from n >> +1, which is true for a natural number since it is defined to be >> finite. May be the roots of my perceived problem is in the definition >> of natural numbers as an infinite set while claiming that none of its >> elements is infinite. >> I read that the natural number n is defined as {0,1,2,...,n-1} where n >> can't be called infinite and n is also the count of all numbers in the >> finite set. Then we extended this to set containing all natural >> numbers but we now find none of its members can represent its size? >> First of all, there doesn't seem to be any such thing as all natural >> numbers. The standard mathematical language saying For all x in the >> set I, x+1 also belongs to I, and zero belongs to I does seem to do >> justice in the usage of the word all. What do you mean by all (or >> every)? By using this word one is assuming that all is already >> understood and defined in this context, which, in fact, is what we are >> trying to define. >> At this stage, you are obviously trolling. I take the opportunity to note > you use the same devices that known trolls. Is there a troll school > somewhere ? [plonk] Was there a school for heretics in the middle ages? Didn't most respectable persons like Kronecker, Poincar.8f, Brouwer, and Weyl reject Cantor's claims at least in part? I tend to attribute some responsibility to Gauss and his strive for rigorosity. Cantor wrongly claimed having refuted the opinion of Gauss: so protestiere ich zuvoerders gegen den Gebrauch einer unendlichen Groesse als einer Vollendeten welcher in der Mathematik niemals erlaubt ist. Das Unendliche ist nur eine facon de parler, indem man eigentlich von Grenzen spricht denen gewisse Verh.8altnisse so nahe kommen als man will, waehrend anderen ohne Einschraenkung zu wachsen verstattet ist. Gauss's pupil Dedekind frankly uttered his reason for his belief that it is possible to attribute either > or < or = to the entity of all rationals. My comment is the same as for the naturals: While every single rational is countable, the entity of all rationals is irrational and uncountable. Every single real number is fundamentally different from a rational in that it contains an uncountable amount of elements. Dedekind was correct in that the matter is difficult and pretty irrelevant. However, meanwhile Cantor's paradise led to serious mistakes in physics. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > venkat.6123@gmail.com a .8ecrit : > I'm an amateur in this topic. When my friend told me about Cantor's >> cardinalities, the first thing that seemed odd to me was the 1-to-1 >> mapping used for comparing the sizes of two infinite sets. I'm not >> sure if it is proved that 1-to-1 correspondence is valid for comparing >> two sets of infinite size. Not because of the practical difficulties >> in comparing or the time it takes for the comparing process to >> complete. I think, atleast it should be proved logically that such >> mapping is valid between two infinite sets. >> For example, the mappings such as y=2x showing that there are as many >> even numbers as integers seem dubious to me because there is no proof >> that such function is meaningful when values of x and y reach >> infinite. > But x and y never do reach infinite. The issue is that in comparing N, the set of naturals and , say E, for > the set of evens, is there a correspondence which does not use twice nor > omit any member of either set. That is certainly a valid comparison for finite sets. Can you think of a better way of comparing sets for relative sizes that > works for finite sets? If not why object to it for infinite sets? >> The 1-to-1 correspondence works well as long as n is different from n >> +1, which is true for a natural number since it is defined to be >> finite. May be the roots of my perceived problem is in the definition >> of natural numbers as an infinite set while claiming that none of its >> elements is infinite. >> I read that the natural number n is defined as {0,1,2,...,n-1} where n >> can't be called infinite and n is also the count of all numbers in the >> finite set. Then we extended this to set containing all natural >> numbers but we now find none of its members can represent its size? >> First of all, there doesn't seem to be any such thing as all natural >> numbers. The standard mathematical language saying For all x in the >> set I, x+1 also belongs to I, and zero belongs to I does seem to do >> justice in the usage of the word all. What do you mean by all (or >> every)? By using this word one is assuming that all is already >> understood and defined in this context, which, in fact, is what we are >> trying to define. >> At this stage, you are obviously trolling. I take the opportunity to note > you use the same devices that known trolls. Is there a troll school > somewhere ? [plonk] Was there a school for heretics in the middle ages? > Didn't most respectable persons like Kronecker, Poincar.8f, Brouwer, and > Weyl reject Cantor's claims at least in part? The list of respectable persons who accepted Cantor's claims is more impressive, and includes almost all of those mathematicians you did not cite. > While every single rational is countable, the entity of > all rationals is irrational and uncountable. Since Cantor's definition of countability applies only to such entities, and not at all to the numbers of which they are formed, eckard.blumschein is essentially saying countable = uncountable. > Every single real number is fundamentally different from a rational > in that it contains an uncountable amount of elements. If a real number is some form of a Dedekind cut, then it is either a set of rationals, which is countable, or a pair of sets of rationals, which is both countable and finite. In either case, eckard.blumschein does not know what he is talking about. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > ... it is true that the reals cannot > be written within a list, The reals cannot be written within a list. The rationals can. The rationals are a subset of the reals. Why do you think that putting different words around these facts will change them? - William Hughes === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Virgil schrieb im Newsbeitrag > You asked for an evidence. >> Cantor was a charismatic figure who had good friends supporting him when >> he >> came up with his so called second diagonal argument DA2 as to show that >> there are more reals than rationals. While it is true that the reals >> cannot >> be written within a list, those who tried in vain to refute DA2 did so >> with >> mathematical means without sufficient understanding of the fact that >> reals >> are something quite different from the rationals. Cantor assumed that any >> real number can be written as a rational with infinitely many >> decimals/digits. Cantor did NOT assume that every real can be written as a rational. > A rational expressed in decimal form is either a terminating or an > eventually repeating decimal, which no irrational can be, and Cantor was > aware of this difference. Nonetheless, he did not have any change but to express his list of all reals by rationals p/q of infinite p and infinite q. The problem is: As soon as one freezes p and q, as did Cantor for DA2, p and q are no longer infinite, and p/q is a rational number. >> The perhaps decisive reason for international acceptance of Cantor's >> naive >> schizophrenic ideas was the imprudent und unnecessary desire to have a >> rigorous theory that unites rational and irrational numbers, cf. >> Dedekind's >> Stetigkeit und Irrationale Zahlen. And perhaps it was the prudent and necessary desire to follow where > logic lead. Cantors logic was just tempting naive illusion. He ignored the relation | |. >> Cantor interpreted his DA2 as evidence for his claim that there must be >> more real numbers as compared to the rationals. Cantor was not the one who applied the diagonal proof to numbers. His > proof only applied to sets of binary sequences. I recommend to you reading his original work in German. It is available via Digitalisierungszentrum Goettingen. Presently I have at hand the very nice book Grundlagen der Mathematik in geschichtlicher Entwicklung by Oskar Becker. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > Virgil schrieb im Newsbeitrag > You asked for an evidence. >> Cantor was a charismatic figure who had good friends supporting him when >> he >> came up with his so called second diagonal argument DA2 as to show that >> there are more reals than rationals. While it is true that the reals >> cannot >> be written within a list, those who tried in vain to refute DA2 did so >> with >> mathematical means without sufficient understanding of the fact that >> reals >> are something quite different from the rationals. Cantor assumed that any >> real number can be written as a rational with infinitely many >> decimals/digits. Cantor did NOT assume that every real can be written as a rational. > A rational expressed in decimal form is either a terminating or an > eventually repeating decimal, which no irrational can be, and Cantor was > aware of this difference. Nonetheless, he did not have any change but to express his list of all reals > by rationals p/q of infinite p and infinite q. Non-sense. Where did Cantor ever claim infinite integers? Cantors logic was just tempting naive illusion. He ignored the relation | > |. What relation is || supposed to be? > Cantor interpreted his DA2 as evidence for his claim that there must be >> more real numbers as compared to the rationals. Cantor was not the one who applied the diagonal proof to numbers. His > original proof only applied to sets of binary sequences. I recommend to you reading his original work in German. It is available via > Digitalisierungszentrum Goettingen. Presently I have at hand the very nice > book Grundlagen der Mathematik in geschichtlicher Entwicklung by Oskar > Becker. Since reading it in German seems only to have confused you about what he said, I'll pass. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Nntp-Posting-Host: hera.cwi.nl > Virgil schrieb im Newsbeitrag ... >> Cantor interpreted his DA2 as evidence for his claim that there must be >> more real numbers as compared to the rationals. > > Cantor was not the one who applied the diagonal proof to numbers. His > proof only applied to sets of binary sequences. > > I recommend to you reading his original work in German. You should also do it. DA2 was *not* about numbers at all. It was about sequences consisting of the letters 'm' and 'w'. -- 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: Is one-to-one mapping valid for comparing infinite-sized sets? William Hughes schrieb im Newsbeitrag > Let me criticise this unfortunately very common idea. Is there really an >> a >> priori existing a set of natural numbers? >> Why? It makes sense to talk about an infinite entity and its > properties. Why should you think it matters what you call it? > It does not matter if you say it exists, actually exists, does > not exist > but is a quasi-imaginary concept ... It makes sense to talk about > an infinite entity and its properties. > I do not merely intend talking about existence. My final concern are consequences. What about the notion of mathematically existent, let me remind of x=sqrt (-1). There is no solution to this equation, unless we expand the numbers to include the imaginary ones. Perhaps I am the first one who argues that the entity of all natural numbers is likewise an expansion to all natural numbers. Natural numbers can be reached by counting, their entity cannot. The entity of all natural numbers is definitely something quite different from any natural numbers. I put it into the drawer of the uncountables. So natural numbers are countable while the entity of them is uncountable. Well, I agree that it makes sense to not just talk about but actually operate with the notion infinite entity. Engineers like me used to very often operate with oo as if it was a quantity. We also do not hesitate to interprete 1/0 as oo. However, at least WM and I and perhaps Vencat, Chrisopher Henrich, and many many others who capitulated when facing arrogant rejection are not aware of any tangible advantage of how Dedekind and Cantor naively did the forbidden split by putting uncountable reals on the same level as the countable rationals. Salviati: ... in ultima conclusione, gli attributi di eguale maggiore e minore non aver luogo ne gl'infiniti, ma solo nelle quantit.88 terminate. IR >|> IR+ | | R === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? > William Hughes schrieb im Newsbeitrag > Let me criticise this unfortunately very common idea. Is there really an >> a >> priori existing a set of natural numbers? >> Why? It makes sense to talk about an infinite entity and its > properties. Why should you think it matters what you call it? > It does not matter if you say it exists, actually exists, does > not exist > but is a quasi-imaginary concept ... It makes sense to talk about > an infinite entity and its properties. > I do not merely intend talking about existence. > My final concern are consequences. What about the notion of mathematically existent, > let me remind of x=sqrt (-1). > There is no solution to this equation, > unless we expand the numbers to include the imaginary ones. Which has been done successfully for some time. So that it appears that the set of numbers that exist depends on context. And if one chooses a context in which reals exist, then they exist in that context, just as do complex numbers in their own contest.. Perhaps I am the first one who argues that the entity of > all natural numbers is likewise an expansion to all natural numbers. > Natural numbers can be reached by counting, their entity cannot. > The entity of all natural numbers is definitely something > quite different from any natural numbers. That entity is the context in which it makes sense to talk of natural numbers existing. > I put it into the drawer of the uncountables. Which is its own context in which those uncountably many reals exist. > So natural numbers are countable while the entity of them is > uncountable. The entity of them is one object, which I have just counted. But if one allows countable to mean what Cantor defined it to mean, then the set (entity if you prefer) of naturals is by that definition countable. Well, I agree that it makes sense to not just talk about > but actually operate with the notion infinite entity. > Engineers like me used to very often operate with oo > as if it was a quantity. I'll bet that such use was mathematically, at least in essence, a one or two point compactification of the real topology, and never as a cardinality. We also do not hesitate to interprete 1/0 as oo. Mathematicians do, at least until they have a context in which to make it sensible. > However, at least WM and I and perhaps Vencat, Chrisopher Henrich, > and many many others who capitulated when facing arrogant rejection > are not aware of any tangible advantage of how Dedekind and Cantor > naively did the forbidden split by putting uncountable reals > on the same level as the countable rationals. What is a cardinal sin in the religion of eckard.blumschein's and his ilk is their problem, and not a problem to those lacking that creed. === Subject: Re: Is one-to-one mapping valid for comparing infinite-sized sets? Nntp-Posting-Host: hera.cwi.nl ... > Perhaps I am the first one who argues that the entity of > all natural numbers is likewise an expansion to all natural numbers. You the *first* one? Why do you think mathematicians talk about ordinal and cardinal numbers? > Natural numbers can be reached by counting, their entity cannot. > The entity of all natural numbers is definitely something > quite different from any natural numbers. Nobody has ever argued anything else. (Or perhaps you are meaning Cantors statements about Ganze Zahlen? But he also talks about different classes of Ganze Zahlen.) > I put it into the drawer of the uncountables. > So natural numbers are countable while the entity of them is > uncountable. Thereby using a word commonly used in mathematics in a completely different sense, making discussion more difficult. > However, at least WM and I and perhaps Vencat, Chrisopher Henrich, > and many many others who capitulated when facing arrogant rejection > are not aware of any tangible advantage of how Dedekind and Cantor > naively did the forbidden split by putting uncountable reals > on the same level as the countable rationals. Who did the forbidding? -- 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: Is one-to-one mapping valid for comparing infinite-sized sets? <48fec809$0$17381$9b4e6d93@newsspool1.arcor-online.net> <48ff2d12$0$16177$9b4e6d93@newsspool2.arcor-online.net> posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I >> Let me criticise this unfortunately very common idea. Is there really an >> a >> priori existing a set of natural numbers? Why? It makes sense to talk about an infinite entity and its > properties. Why should you think it matters what you call it? > It does not matter if you say it exists, actually exists, does > not exist > but is a quasi-imaginary concept ... It makes sense to talk about > an infinite entity and its properties. I do not merely intend talking about existence. > My final concern are consequences. Good. Then stop briging up the sterile discussion about whether an infinite entity should be called a set. The entity of all natural numbers is definitely something > quite different from any natural numbers. > I put it into the drawer of the uncountables. > So natural numbers are countable while the entity of them is > uncountable. And now you are nattering about the meaning of uncountable. We can meaninfully talk about entities X that we cannot pair up one-one with entity of all the integers. You cannot make this fact go away by a discussion (however psuedo-erudite) about what words one should use to describe it. - William Hughes === Subject: Re: JSH: Explaining for the angry idiots > I will admit that I spent a lot of time chasing FLT as, well, a > crackpot, and had this proof that went up in flames after I'd argued > with people over it for months, and was pondering the stupidity of > what I was doing. And then I realized I just LIKED it, a lot. Whether I was right or > not, endlessly scribbling out equations, was just for some bizarre > reasons an extremely fun activity for me, and I told myself to no > longer care so much if I'd figure out anything great or not, but to > just enjoy it. That was back sometime in 1999. By December 1999 I found myself > babbling about these things I eventually named tautological spaces. Who are you, and what have you done to the real James Harris? -- T. Wasell === Subject: Graph cycle question posting-account=zzTJjAoAAAA2omwN6AdEiKAFbfSBZBmO Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) Let G = (V;E) be an undirected, weigthed graph with weight function w. Let w(e) > 0 for all e 2 E and let all weights be di¬erent. Prove or disprove the following: Consider an arbitrary simple cycle C in G. What am I supposed to understand by arbitrary simple cycle? My English fails me here, and I don't understand what the dictionary says (I can make no sense of it). Maybe someone can explain in other words? === Subject: Re: Graph cycle question days. My association with the Department is that of an alumnus. >Let G =3D (V;E) be an undirected, weigthed graph with weight function >w. Let w(e) > 0 for all e 2 E and let all weights be di=AEerent. >Prove or disprove the following: Consider an arbitrary simple cycle C >in G. Surely the problem continues? It does not end there. >What am I supposed to understand by arbitrary simple cycle? It means: let C be any simple cycle in G. It means, C is a simple cycle in G, and you have no further hypothesis on C. Any simple cycle; whatever the statement that follows must apply to all simple cycles in G. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Graph cycle question posting-account=zzTJjAoAAAA2omwN6AdEiKAFbfSBZBmO Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > Surely the problem continues? It does not end there. Nope, it goes on: Let e0 be the edge on C with minimum weight. Then e0 is in every minimal spanning tree for G. But I didn't post this because I don't want the excercise solved for me. So C is just a simple cycle? It was the word arbitrary that confused me. === Subject: Re: Graph cycle question days. My association with the Department is that of an alumnus. >> Surely the problem continues? It does not end there. Nope, it goes on: >Let e0 be the edge on C with minimum weight. Then e0 is in every >minimal spanning tree for G. But I didn't post this because I don't want the excercise solved for >me. You might have wanted to indicate the problem statement continued anyway. >So C is just a simple cycle? It was the word arbitrary that confused >me. The word arbitrary, in this context, means that there is to be no assumption made on C other than that it is a simple cycle. Another way to phrase it would be: For every C, if C is a simple cycle of G, then the edge with minimum weight on C will be in every minimal spanning tree for G. That is, the condition will hold for each and every simple cycle in the graph. So the assertion is that, no matter which simple cycle you take, the edge with minimum weight in that cycle will be in every minimal spanning tree for G. So if C_1 and C_2 are two distinct simple cycles, and e0 and f0 are the corresponding edges of minimum weight, then e0 will be in every minimal spanning tree for G, and f0 will be in every minimal spanning tree for G. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: prototyping GPS <00be4635$0$10791$c3e8da3@news.astraweb.com> posting-account=ML9J7goAAAAS6N3k57S2qk4WxczSttuL 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; InfoPath.1; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) [...] > They also check the prevailing attitude on sci.physics > to see whether or not to apply the relativistic corrections. > Love it!. Just my thing. First time today I've laughed out loud. > ... ahahaha... Uncle Henry has truly class. He once insisted > that TV factories hire relativity scientists... ... ahaha... And to uncle Henry's further credit it must be stated that > he once posted that he is an Almost-victim... up in his > pomerado backyard in the San Jacinto Range... > Maybe it's from his apple cider or maybe he slipped > on an apple peel that turned him into an Almost-victim... AHAHAHAHA... ahhahaha... ahahahanson: I was truly a victim a year ago this week. The 2007 fire burned down time, however, the power was out for four days so I lost everything in the freezer that I didn't cook the first day. === Subject: Re: prototyping GPS This raises still another potentially interesting question. How and >from where do GPS receiver manufacturers obtain their software >algorithms? >>The idea is simple. You decode phase-shifted timing data from three >>to four geosynchronized satellites (known locations). Any good second >>year engineering student can do it. The difficult part is building >>low power and low cost components. >Is it a product of US government funded R&D, and >because of this in the public domain? >>Yes. > I think you are trying to make this much harder than it is. I do radar imaging and the process is the same. Actually, I doubt if one of my second year engineering students could > even absorb the problem parameters as you appear to do. We have had undergraduates working with the location algorithms. Realize that the GPS receive has absolutely no knowledge of the > satellites orbital positions nor do the satellites themselves. The receivers have to have the position of the satellites or it will not work. Part of the data stream is the orbital parameters. That is why some GPSs will give you a sky map of the satellites. As soon as you know the position and time from the satellites, it is a simple geometry problem to determine your position. If you know the distances from one satellite, you are located to the surface of a sphere. If you know the distance from two satellites, the intersection of the two spheres makes a circle. If you add a third satellite, the intersection of this circle and the new sphere gives you two points. This is why you can get a location with three satellites if you assume you are on the surface of the earth. With a fourth satellite, you remove the ambiguity of the two points and have a 3d solution. The atmospheric corrections are details that have to be added. GPS > receivers, on the other hand have not idea of the directional > information on the signals that they are receiving, and typically must > receive signal from three satellites simultaneously. Since to you The idea is simple so explain the basics of the > algorithm that you would use to decode phase-shifted timing data from > three > to four geosynchronized satellites (known locations) Slight problem > genius, neither the satellites nor the GPS receiver knows its current > location. So, I receive (my microwave reciever guys are very gifted) > a number of timing signals from a source that may or my not identify > itself, and has not concept of its current orbital location. So genius, what is the first, second and third step to identify the > GPS receiver's position on earth? See above. It's a remarkably interesting question, for which you haven't the hint > of a clue, Dork. Next... === Subject: Roots of AX^2 + BX + C Recently a student and I talked about the probability of the roots a quadratic equation (AX^2 + B*X + C) to be complex. As we all know, the roots will be complex if (B^2-4*A*C) is less than zero. The student and I both thought that for any three randomly chosen POSITIVE numbers, there is a 50% chance of (B^2-4*A*C) to be negative. However, when I ran a small simulation using MATLAB, I found that the roots are positive only 25% of the time. I ran the simulation several times and the results are quite consistent. Can someone please explain why three (uniformly distributed) randomly chosen numbers, A, B, C would behave such that B^2-4*A*C is positive for only 25% of the time ? (faculty member in Mechanical Engineering) MATLAB code follows: % an m-file to check b^2-4ac clear all; close all; rand('twister',sum(100*clock)); N = 1000; % number of sample points randarray = rand(3*N,1); A = randarray(1:N); B = randarray(N+1:2*N); C = randarray(2*N+1:3*N); R = B.^2-4*A.*C; LA1 = R>0; %Logical array in MATLAB. apositive = sum(LA1); anegative = sum(~LA1); apos_percent = (apositive/N)*100; aneg_percent = (anegative/N)*100; bar([apos_percent aneg_percent]) ylabel('pecentages'); [apos_percent aneg_percent] === Subject: Re: Roots of AX^2 + BX + C posting-account=sFP0HgkAAADJMwhdrXAaC5VX7Tc3BtzY .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Recently a student and I talked about the probability of the roots a > quadratic equation (AX^2 + B*X + C) to be complex. As we all know, the > roots will be complex if (B^2-4*A*C) is less than zero. The student and I both thought that for any three randomly chosen > POSITIVE numbers, there is a 50% chance of (B^2-4*A*C) to be negative. I can't imagaine why you would think that. > However, when I ran a small simulation using MATLAB, I found that the > roots are positive only 25% of the time. I ran the simulation several > times and the results are quite consistent. Can someone please explain why three (uniformly distributed) randomly > chosen numbers, A, B, C would behave such that B^2-4*A*C is positive > for only 25% of the time ? I assume you mean uniformly distributed on some interval [0, r]. We can put r = 1 wlog. Given A and C, the probability that B is large enough that B^2 - 4AC is positive is 1 - sqrt(4AC) if AC < 1/4, and 0 if AC > 1/4. If A < 1/4, then AC < 1/4 for all C in [0, 1]. If A > 1/4, then AC < 1/4 for 0 < C < 1/(4A) while AC > 1/4 for 1/(4A) < C < 1. So I get the following for the probability. It's a bit late so I won't guarantee that it's correct, but the numerical value I end up with is close to what you said your simulations gave. int{0, 1/4} int{0, 1} (1 - sqrt(4AC)) dC dA + int{1/4, 1} int{0, 1/(4A)} (1 - sqrt(4AC)) dC dA = int{0, 1/4} ([{0, 1} (C - 2/3 sqrt(4AC^3))) dA + int{1/4, 1} ([{0, 1/(4A)} (C - 2/3 sqrt(4AC^3))) dA = int{0, 1/4} (1 - 2/3 sqrt(4A)) dA + int{1/4, 1} (1/(12A)) dA = ([{0, 1/4} (A - 8/9 A^(3/2))) + 1/12 ([{1/4, 1} ln A) = 1/4 - 1/9 + 1/12 ln 4 = 5/36 + (ln 2)/6 = 0.2544... --- J K Haugland http://home.no.net/zamunda === Subject: Re: Roots of AX^2 + BX + C >Recently a student and I talked about the probability of the roots a >quadratic equation (AX^2 + B*X + C) to be complex. As we all know, the >roots will be complex if (B^2-4*A*C) is less than zero. The student and I both thought that for any three randomly chosen >POSITIVE numbers, there is a 50% chance of (B^2-4*A*C) to be negative. >However, when I ran a small simulation using MATLAB, I found that the >roots are positive only 25% of the time. I ran the simulation several >times and the results are quite consistent. Can someone please explain why three (uniformly distributed) randomly >chosen numbers, A, B, C would behave such that B^2-4*A*C is positive >for only 25% of the time ? >(faculty member in Mechanical Engineering) If you constrain each of a, b, and c to [N,-N], the probability of getting a positive discriminant is 41/72 + 1/12 log(2) which is approximately 62.72%. See Rob Johnson take out the trash before replying === Subject: Re: Roots of AX^2 + BX + C >>Recently a student and I talked about the probability of the roots a >>quadratic equation (AX^2 + B*X + C) to be complex. As we all know, the >>roots will be complex if (B^2-4*A*C) is less than zero. >>The student and I both thought that for any three randomly chosen >>POSITIVE numbers, there is a 50% chance of (B^2-4*A*C) to be negative. >>However, when I ran a small simulation using MATLAB, I found that the >>roots are positive only 25% of the time. I ran the simulation several >>times and the results are quite consistent. >>Can someone please explain why three (uniformly distributed) randomly >>chosen numbers, A, B, C would behave such that B^2-4*A*C is positive >>for only 25% of the time ? >>(faculty member in Mechanical Engineering) If you constrain each of a, b, and c to [N,-N], the probability of >getting a positive discriminant is 41/72 + 1/12 log(2) which is >approximately 62.72%. See types ac < 0 where all a, b, and c yield positive discriminants ac > 0 we can subtract 1/2 and double to get the probability of getting a positive discriminant in an octant with ac > 0: 5/36 + 1/6 log(2) which is approximately 25.44% as J K Haugland has said. Rob Johnson take out the trash before replying === Subject: Re: Roots of AX^2 + BX + C > Recently a student and I talked about the probability of the roots a > quadratic equation (AX^2 + B*X + C) to be complex. As we all know, the > roots will be complex if (B^2-4*A*C) is less than zero. The student and I both thought that for any three randomly chosen > POSITIVE numbers, there is a 50% chance of (B^2-4*A*C) to be negative. > However, when I ran a small simulation using MATLAB, I found that the > roots are positive only 25% of the time. I ran the simulation several > times and the results are quite consistent. Can someone please explain why three (uniformly distributed) randomly > chosen numbers, A, B, C would behave such that B^2-4*A*C is positive > for only 25% of the time ? I suspect that your random number generator was not producing numbers ->uniformly<- distributed on the open interval from 0 to oo, which is what you imply by randomly chosen positive numbers. In any case, it is the ratios A/B and C/B which will determine the sign of (B^2 - 4*A*C), so you can set B = 1 and only worry about A and C. But I am afraid that even MATLAB will have difficulty with generating random members of a uniform distribution on (0,oo). > (faculty member in Mechanical Engineering) > MATLAB code follows: > % an m-file to check b^2-4ac > clear all; > close all; rand('twister',sum(100*clock)); N = 1000; % number of sample points > randarray = rand(3*N,1); A = randarray(1:N); > B = randarray(N+1:2*N); > C = randarray(2*N+1:3*N); R = B.^2-4*A.*C; > LA1 = R>0; %Logical array in MATLAB. apositive = sum(LA1); > anegative = sum(~LA1); apos_percent = (apositive/N)*100; > aneg_percent = (anegative/N)*100; > bar([apos_percent aneg_percent]) > ylabel('pecentages'); [apos_percent aneg_percent] === Subject: Re: proving not(c = aleph_aleph_null) diagonally without choice <7572070.1223937676002.JavaMail.jakarta@nitrogen.mathforum.org> <87wsgc9iom.fsf@phiwumbda.org> <48f43154$0$7935$7a628cd7@news.club-internet.fr> <48feddfa$0$23646$7a628cd7@news.club-internet.fr> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.3) Gecko/2008092414 Firefox/3.0.3,gzip(gfe),gzip(gfe) > This is really ridiculous : you want to prove c cannot be aleph_omega. > If c has a well-ordering, it is an aleph and you can use cofinality > arguments. If it has not, it certainly cannot be aleph_omega, not any > other aleph, for that matter... hey denis! glad to see you back in the conversation! was wondering what your thoughts were on my very first point initially that c^w = c requires at least countable choice because well c =/= 2^w (to be clear: 2^w = w) c is 2^(aleph_0) and the math being done is cardinal math not ordinal and everything i can find says the step i mentioned very early on is the place where choice comes in so i was wondering what your thoughts were particularly since everything i can find says gitik proved that _all_ uncountable cardinals are singular levy has shown (work that gitik was building off of) that in ZF without C the continuum could be modelled as a countable union of countable sets gitik has apparently developed a strongly compact version of prikry forcing (which i am still trying to understand) that can force _all_ cardinals between w and some given strongly compact cardinal to have cofinality aleph_0 so please tell me denis what are your thoughts? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: proving not(c = aleph_aleph_null) diagonally without choice > This is really ridiculous : you want to prove c cannot be aleph omega. > If c has a well-ordering, it is an aleph and you can use cofinality > arguments. If it has not, it certainly cannot be aleph omega, not any > other aleph, for that matter... hey denis! glad to see you back in the conversation! was wondering what your thoughts were > on my very first point initially > that c^w = c requires at least countable choice because > well c =/= 2^w (to be clear: 2^w = w) c is 2^(aleph 0) and the math being done is cardinal math > not ordinal and everything i can find > says the step i mentioned very early on > is the place where choice comes in so i was wondering what your thoughts were particularly since everything i can find > says gitik proved that all uncountable cardinals are singular levy has shown actually it appears this was a joint result of feferman and levy i like feferman and even once went through a bunch of his stuff (never saw this one though...) i think this result is sufficient for the conversation (but who knows) > (work that gitik was building off of) > that in ZF without C > the continuum could be modelled > as a countable union of countable sets gitik has apparently developed > a strongly compact version of prikry forcing > (which i am still trying to understand) > that can force all cardinals > between w and some given strongly compact cardinal > to have cofinality aleph 0 so please tell me > denis what are your thoughts? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: proving not(c = aleph_aleph_null) diagonally without choice <87wsgc9iom.fsf@phiwumbda.org> <48f43154$0$7935$7a628cd7@news.club-internet.fr> <48f5d1df$0$7931$7a628cd7@news.club-internet.fr> <8dsEhkDyxd$IFwV0@212648.invalid> <48feddfa$0$23646$7a628cd7@news.club-internet.fr> In message >> hey denis! >> glad to see you back in the conversation! >> was wondering what your thoughts were >> on my very first point initially >> that c^w = c requires at least countable choice >> because >> well >> c =/= 2^w >> (to be clear: 2^w = w) >> c is 2^(aleph_0) >> and the math being done is cardinal math >> not ordinal ??!! Denis started his proof by defining w = aleph_0 and c = 2^w. 2^k =/= k for any cardinal. -- David Hartley === Subject: Re: proving not(c = aleph_aleph_null) diagonally without choice <48feddfa$0$23646$7a628cd7@news.club-internet.fr> <$HXkXeAiMHAJFwpF@212648.invalid> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.3) Gecko/2008092414 Firefox/3.0.3,gzip(gfe),gzip(gfe) > In message > hey denis! > glad to see you back in the conversation! > was wondering what your thoughts were >> on my very first point initially >> that c^w = c requires at least countable choice > because >> well > c =/= 2^w > (to be clear: 2^w = w) > c is 2^(aleph 0) > and the math being done is cardinal math >> not ordinal ??!! Denis started his proof by defining w = aleph 0 and c = 2^w. 2^k =/= k for any cardinal. yes and yet he thought there was a canonical map that gives aleph 0 x aleph 0 = aleph 0 without a choice principle this was the first thing i questioned and although i was fairly confident of it both denis and yourself seemed to ignore the point i accepted the possibility i could be wrong and have now found several models in which that assumption fails without choice there have been attempts to claim denis' result still holds with the models i have found but the levy / feferman result plainly shows those wrong and focuses directly back upon aleph 0 x aleph 0 = aleph 0 (which doesn't hold without some choice) i think it is now time that both you and denis accept the possibility that both of you are wrong i've done some legwork i've studied up on prikry forcing and am starting to understand the relation between collapsing cardinals in an initial model and the existence of certain large cardinals there comes a time when after my point has been ignored or skipped over and no details appear forthcoming and i've done the study work necessary that i have to start assuming that i have learned this properly and both of you are wrong it's an assumption maybe i'm just an ass but there are so many readily available sources that say aleph 0 x aleph 0 = aleph 0 requires (countable) choice that i've come to think it's a pretty safe assumption -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: proving not(c = aleph_aleph_null) diagonally without choice <87wsgc9iom.fsf@phiwumbda.org> <48f43154$0$7935$7a628cd7@news.club-internet.fr> <48feddfa$0$23646$7a628cd7@news.club-internet.fr> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.3) Gecko/2008092414 Firefox/3.0.3,gzip(gfe),gzip(gfe) > This is really ridiculous : you want to prove c cannot be aleph omega. > If c has a well-ordering, it is an aleph and you can use cofinality > arguments. If it has not, it certainly cannot be aleph omega, not any > other aleph, for that matter... hey denis! glad to see you back in the conversation! was wondering what your thoughts were > on my very first point initially > that c^w = c requires at least countable choice because > well c =/= 2^w (to be clear: 2^w = w) c is 2^(aleph 0) and the math being done is cardinal math > not ordinal and everything i can find > says the step i mentioned very early on > is the place where choice comes in so i was wondering what your thoughts were particularly since everything i can find > says gitik proved that all uncountable cardinals are singular ^ could be sorry for the buggy line the rest below says it right > levy has shown > (work that gitik was building off of) > that in ZF without C > the continuum could be modelled > as a countable union of countable sets gitik has apparently developed > a strongly compact version of prikry forcing > (which i am still trying to understand) > that can force all cardinals > between w and some given strongly compact cardinal > to have cofinality aleph 0 so please tell me > denis what are your thoughts? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: The year of Cyber Tester posting-account=UEdI2woAAABTdf0LcYPp9Pe_P88Ck_vM Gecko/20011019 Netscape6/6.2,gzip(gfe),gzip(gfe) A selected bug detected once by our 24/365 VM machine: (* Mathematica 6.0.3 *) N[Integrate[Re[z], {z, 0, Sqrt[I]}]] > NIntegrate[Re[z], {z, 0, Sqrt[I]}] 0.+ 0.5 I > 0.25+ 0.25 I N[Integrate[Im[z], {z, 0, Sqrt[I]}]] > NIntegrate[Im[z], {z, 0, Sqrt[I]}] 0. > 0.25+ 0.25 I > Does this also happen with Integrate[Re[z], {z, 0, Sqrt[I]}] and Integrate[Im[z], {z, 0, Sqrt[I]}] i.e. without prefixing these by N[]? Martin. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=6xUtvgkAAAD_jypmLa2oo2HnrV0e8X9q rv:1.7.13) Gecko/20060414,gzip(gfe),gzip(gfe) You are unable to understand (generally as usual) that unit itself > is a unit. The unit of counting. If you count apples, apple is the > unit. If you count material things, material thing in general is the > unit. If you count people, person is the unit. And so on. If you count, one is the unit! Again that eerie impression, that you and WM believe, that the modern > notion of uncountability violates basic laws of nature - human nature in > your case What are you talking from? I have no idea how the human nature might > violate the notion of uncountability - or in reverse. Albrecht has so often claimed knowledge and certainty on these very > issues, that his objection rings false. The only problem lays in the fact that an infinite totality is an > inconsitent object. It may contradict Albrecht's weltanschauung, as it seems to contradict > WM's, but that does not constitute anything like an inconsistency in > mathematics, which is quite fortunately totally independent of the > worldviews of either. Mathematics is more as you think. It is a searching after truth, not only after consistent systems. This aspect creates the freedom of the mathematics. In fact, you are the one who restrict mathematics (e.g. to a dogma). True mathematics can incorporate the concept of actual infinity without the lost of it's consistency and without the lost of it's power of prediction. In fact, it is a kind of worldview wherein the focus of mathematics lays. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor You are unable to understand (generally as usual) that unit itself > is a unit. The unit of counting. If you count apples, apple is the > unit. If you count material things, material thing in general is the > unit. If you count people, person is the unit. And so on. If you count, one is the unit! Again that eerie impression, that you and WM believe, that the modern > notion of uncountability violates basic laws of nature - human nature in > your case What are you talking from? I have no idea how the human nature might > violate the notion of uncountability - or in reverse. Albrecht has so often claimed knowledge and certainty on these very > issues, that his objection rings false. The only problem lays in the fact that an infinite totality is an > inconsitent object. It may contradict Albrecht's weltanschauung, as it seems to contradict > WM's, but that does not constitute anything like an inconsistency in > mathematics, which is quite fortunately totally independent of the > worldviews of either. Mathematics is more as you think. It is a searching after truth, not > only after consistent systems. This aspect creates the freedom of the > mathematics. In fact, you are the one who restrict mathematics (e.g. > to a dogma). On the contrary. I am one of those who would free it from your dogma. I only refuse to reject any system until it is shown to be internally inconsistent, and no one has shown standard math, or any of Cantor's results, to be internally inconsistent. Constructionism may be, by the above standard, a perfectly legitimate system, but until it is demonstrably the ONLY legitimate system, which it has not yet been shown to be, I reject your, or anyone else's, attempts to force it on me. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor <6m5dp1Feqc45U1@mid.uni-berlin.de> posting-account=6xUtvgkAAAD_jypmLa2oo2HnrV0e8X9q rv:1.7.13) Gecko/20060414,gzip(gfe),gzip(gfe) FYI: http://www.albrecht-storz.homepage.t-online.de/ > What are you talking from? When you claim, that numbers exist in reality/nature independant of man, > i.e. they're at least as objective as other things(*), isn't it too far > fetched, if basic ways of doing math (e.g. counting) were a matter of that > reality/nature, too? (The question remains, whether there is only one > possibility that nature leaves us.) (*) I admit, that the distinction between material and immaterial objects > might be misleading or not to the point. The only problem lays in the fact that an infinite totality is an > inconsitent object. Not mathematically, it's only a problem, when you apply improper concepts > to it. The notion of potential infinity has been abandoned for good > reasons, namely e.g. inability to provide for complete real numbers. As far as I know, constructive methods have turned out to be not too > fruitful. What convinces me least, is the high burden on the existence of > mathematical objects. And there seem to be no justification other than > taste or philosophical reservations to draw the line of existence just > there, where constructivists usually do. The objects they would accept are as existent or non existent like the > objects they deny their existence, since everything in math is about > concepts. Thus, the question is rather about the quality of a theory, and > there's no principal reason to forbid any type of concept, **provided the > theory benefits from it!** Face it: you'r waging a lost war, trying to preserve a way of thinking, > which at least in math has been sorted out. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > What are you talking from? When you claim, that numbers exist in reality/nature independant of man, > i.e. they're at least as objective as other things(*), isn't it too far > fetched, if basic ways of doing math (e.g. counting) were a matter of that > reality/nature, too? (The question remains, whether there is only one > possibility that nature leaves us.) Maybe the fact, that e.g. the the material helium exists is due to the gathering of two neutrons and two protons is a hint that counting is matter of the reality of counting. (*) I admit, that the distinction between material and immaterial objects > might be misleading or not to the point. Sure, I think. The only problem lays in the fact that an infinite totality is an > inconsitent object. Not mathematically, it's only a problem, when you apply improper concepts > to it. Maybe an aspect of the definition of mathematics. Or of the sense of mathematics, or of the intentions we project in the mathematics. > The notion of potential infinity has been abandoned for good > reasons, namely e.g. inability to provide for complete real numbers. This notion never has been abandoned as long as branches like constructivism are part of the mathematics. The Limes is exact the notion of potential infinity. As far as I know, constructive methods have turned out to be not too > fruitful. What convinces me least, is the high burden on the existence of > mathematical objects. And there seem to be no justification other than > taste or philosophical reservations to draw the line of existence just > there, where constructivists usually do. Inside the sure part of the mathematics constructive methods may be not as fruitful as methods including the idea of actual infinity. But the fact, that the concept of actual infinity is incoherent remains. Incoherent not inside the mathematics, as it seems, but incoherent in our reality. The question is: should mathematics reflect something (reality) or should it only be coherent intrinsically. The objects they would accept are as existent or non existent like the > objects they deny their existence, since everything in math is about > concepts. But in any mathematics, the basic and indispensable part of the mathematics is the 1 and the law: 1+1=2 (or, in my notation: I + I = II). Concepts can vary. But counting is unquestionable fixed. There is no mathematics in which holds: I + I = III (without an underlay of the basic truth I + I = II). The mathematical 'concept' of the natural numbers is nothing more than the attempts of fomulating a natural law. An elementary natural law. The natural numbers are here. And we try to catch them. How do you think about the fact, that seemingly we can't konw if there is a last pair of primes? > Thus, the question is rather about the quality of a theory, and > there's no principal reason to forbid any type of concept, **provided the > theory benefits from it!** The question is: how do you prove benefits? Is this a subjective qualifying or are there proveable criteria to identify benefits? Face it: you'r waging a lost war, trying to preserve a way of thinking, > which at least in math has been sorted out. > Viele Gr??e > Klaus Gruss Albrecht === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > Inside the sure part of the mathematics constructive methods may be > not as fruitful as methods including the idea of actual infinity. But > the fact, that the concept of actual infinity is incoherent remains. Constructive methods have resulted in: Computer Programming. Which makes them a successful form of mathematics in practice. There also does exist a purely theoretical (proper) construcivism, which is to be considered IMO as far less relevant, as well as unsuccessful, indeed. > Incoherent not inside the mathematics, as it seems, but incoherent in > our reality. Incoherent even _in_ mathematics. All kind of artificialities are needed to prevent the building from collapsing. Examples are: -- Wrong limits do not commute -- Natural Densities are Probabilities -- Calculus XOR Probability > The question is: should mathematics reflect something (reality) or > should it only be coherent intrinsically. Only a tiny pinch of salt is needed to make mathematics consistent with the rest of the world and that is: the acceptance of potential infinity as the _only_ form of infinity. Hence accept that there is an incredible small error in any of the real numbers. And accept that the naturals are not a set (i.e. an unfinished set at best) but, instead, a proper class. Han de Bruijn === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > What are you talking from? When you claim, that numbers exist in reality/nature independant of man, > i.e. they're at least as objective as other things(*), isn't it too far > fetched, if basic ways of doing math (e.g. counting) were a matter of that > reality/nature, too? (The question remains, whether there is only one > possibility that nature leaves us.) Maybe the fact, that e.g. the the material helium exists is due to the > gathering of two neutrons and two protons is a hint that counting is > matter of the reality of counting. Or maybe it is our confused interpretation of a reality in which 1 + 1 is not equal to 2. > (*) I admit, that the distinction between material and immaterial objects > might be misleading or not to the point. Sure, I think. > The only problem lays in the fact that an infinite totality is an > inconsitent object. Not mathematically, it's only a problem, when you apply improper concepts > to it. Maybe an aspect of the definition of mathematics. Or of the sense of > mathematics, or of the intentions we project in the mathematics. The objects of mathematics, being non-physical, need not be constrained to behave as if they were. > The notion of potential infinity has been abandoned for good > reasons, namely e.g. inability to provide for complete real numbers. > This notion never has been abandoned as long as branches like > constructivism are part of the mathematics. The Limes is exact the > notion of potential infinity. > As far as I know, constructive methods have turned out to be not too > fruitful. What convinces me least, is the high burden on the existence of > mathematical objects. And there seem to be no justification other than > taste or philosophical reservations to draw the line of existence just > there, where constructivists usually do. > Inside the sure part of the mathematics constructive methods may be > not as fruitful as methods including the idea of actual infinity. But > the fact, that the concept of actual infinity is incoherent remains. It is only incoherent to constructionalists, but not to the rest of humankind. > Incoherent not inside the mathematics, as it seems, but incoherent in > our reality. Perhaps in physicality, but that is not where mathematics is done anyway. The question is: should mathematics reflect something (reality) or > should it only be coherent intrinsically. The standard mathematics of real analysis 'reflects' reality well enough to support miracles of engineering. Forcing constructionism on mathematics would only make mathematics harder to use with no compensating advantages, other than to the egos of those very constructionists. The objects they would accept are as existent or non existent like the > objects they deny their existence, since everything in math is about > concepts. > But in any mathematics, the basic and indispensable part of the > mathematics is the 1 and the law: 1+1=2 (or, in my notation: I + I = > II). Concepts can vary. But counting is unquestionable fixed. There is no > mathematics in which holds: I + I = III (without an underlay of the > basic truth I + I = II). The mathematical 'concept' of the natural numbers is nothing more than > the attempts of fomulating a natural law. An elementary natural law. > The natural numbers are here. And we try to catch them. How do you think about the fact, that seemingly we can't konw if there > is a last pair of primes? > Thus, the question is rather about the quality of a theory, and > there's no principal reason to forbid any type of concept, **provided the > theory benefits from it!** > The question is: how do you prove benefits? Is this a subjective > qualifying or are there proveable criteria to identify benefits? Face it: you'r waging a lost war, trying to preserve a way of thinking, > which at least in math has been sorted out. > Viele Gr??e > Klaus Gruss > Albrecht === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=6xUtvgkAAAD_jypmLa2oo2HnrV0e8X9q rv:1.7.13) Gecko/20060414,gzip(gfe),gzip(gfe) > You are unable to understand (generally as usual) that unit itself > is a unit. The unit of counting. If you count apples, apple is the > unit. If you count material things, material thing in general is the > unit. If you count people, person is the unit. And so on. In any measurement, all the units of measure must either be identical, > or, when it does not come out even in the original unit, some fraction > of the original unit may be used at the end, as in yards, feet, inches > and simple fractions of an inch. In counting, the objects being counted need not be all of the same size > or shape or anything alike. If one is counting people, a baby and a > basketball star count the same, but they do not measure the same. You don't have any clue what counting is? If you count, at first you have to identify a category of objects which are to count. You may count sheeps in a yard, hairs on a head, peoples in a room, elements in a set, whatever you want. There is no need that objects, which are to count to be of the same size or anything similar. The objects have only to be of the same category, and it has to be clear which objects are to count and which are not. The abstract object of the category is the unit. It is the same if you count sheeps, kilometers, protons, stars, ... . In one aspect you are right: There is a difference between objects which could only be count as a whole, like persons, ideas, tomatoes, etc., and 'objects' which could be count also as parts of the unit, like kilometers, pounds, Kelvins, meters per seconds, etc. > In measuring, one can, and usually will, end up having fractional parts > of the unit of measure. In counting people one does not, or at least > should not, ever end up with a fraction of a person. So that naturals are natural for counting, but one needs at least the > possibility of fractions of units for measuring. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > You are unable to understand (generally as usual) that unit itself > is a unit. The unit of counting. If you count apples, apple is the > unit. If you count material things, material thing in general is the > unit. If you count people, person is the unit. And so on. In any measurement, all the units of measure must either be identical, > or, when it does not come out even in the original unit, some fraction > of the original unit may be used at the end, as in yards, feet, inches > and simple fractions of an inch. In counting, the objects being counted need not be all of the same size > or shape or anything alike. If one is counting people, a baby and a > basketball star count the same, but they do not measure the same. You don't have any clue what counting is? If you count, at first you have to identify a category of objects > which are to count. You may count sheeps in a yard, hairs on a head, > peoples in a room, elements in a set, whatever you want. There is no > need that objects, which are to count to be of the same size or > anything similar. Precisely what I said! So that Albrecht, while having no clue about what I said, still manages to object to it, and then says the same thing himself. In measuring, one can, and usually will, end up having fractional parts > of the unit of measure. In counting people one does not, or at least > should not, ever end up with a fraction of a person. So that naturals are natural for counting, but one needs at least the > possibility of fractions of units for measuring. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > > [Blah, blah, blah....] > Virgil, in base two or three, or base one or base infinity, the interpretation of the expansion list as matrix in the construction of the anti-diagonal is different than in other bases (not base-EEs, bas-es) and radices. It is well known that you can't construct an antidiagonal that isn't necessarily on the list, because of dual representation of the binary and trinary forms, and other structural features of the unary and single code natural rpresentation. There are other ways to construct rules, and those rules get as unwieldy as the complexity of the rules, you see. This happens for example in the exponential with the asymptotic and the concrete in the mathematics. Constructing the anti-diagonal rule is simple, easy, where the construction is even a primitive through the coordinate basis. That is where, there is much in the matrix mathematics. There is more ignored in the matrix mathematics than dreamt of in the current usage of the general matrix products, recomposably. Consider for example the orthogonants and other functional topological operators and so on, torsion groups, in particularly the general products and general inverses of matrices. The diagonal just has the same coordinate in each coordinate matrix block address with the recomposable blocks. To place all the rules in these more complicated systems, maintaining the addressing and offset in the space-filling of the enumerative algorithm, happens in matrices in the finite. Then, it's all very clear and simple along classically founded axiomatic lines the generality of utility in the, how-you-say, simple mathematical algorithm, the applied rule (composition of cause and effect, compositions). In that way, as described above resource-efficient machine representations of these numerical structures easily follow even natural lines. Ross F. === Subject: Re: A consideration concerning the diagonal argument of G. Cantor posting-account=EL3hgwoAAABtyRFrR2z7EBO1tnJeMiO7 Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) On Oct 22, 6:57pm, Ross A. Finlayson Imagine Sarah Palin posting to sci.logic: > Then, it's all very clear and simple along classically founded axiomatic > lines the generality of utility in the, how-you-say, simple mathematical > algorithm, the applied rule (composition of cause and effect, > compositions). In that way, as described above resource-efficient > machine representations of these numerical structures easily follow even > natural lines. MoeBlee === Subject: Re: A consideration concerning the diagonal argument of G. Cantor > On Oct 22, 6:57pm, Ross A. Finlayson Imagine Sarah Palin posting to sci.logic: Then, it's all very clear and simple along classically founded axiomatic > lines the generality of utility in the, how-you-say, simple mathematical > algorithm, the applied rule (composition of cause and effect, > compositions). In that way, as described above resource-efficient > machine representations of these numerical structures easily follow even > natural lines. MoeBlee Does sound a bit like her. === Subject: Residue theorem for beginners posting-account=f7Q9vAoAAAA9NrtfQEoVAs9UXkEuc4Rm Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) Good evening, I'm a newbie here and currently taking a course in complex analysis. I would appreciate if someone could answer my stupid questions. Let f(z) be a meromorphic function (i.e., f is holomorphic with singularities). Now, let's consider the integral of f over the closed curve C, so int_C f(z) dz. To evaluate this integral we can simply use the residue theorem: http://en.wikipedia.org/wiki/Residue_theorem My problem is that I am not quite sure on how to evaluate a complex integral for the following case: let us fix a real number b and b - i*oo up to b + i*oo the b-line denoted by (b). So (b) is the line through the point (b, 0) in the complex plane and parallel to the imaginary axis. Let us consider a concrete example: let f(z) = Gamma(z) be the Gamma function. Then f(z) has poles at the negative integers. Let b > 0 be positive. How do we concretely evaluate the integral of f(z) over the (b)-line, i.e., how do you evaluate int_(b) f(z) dz? Can we consider the (b)-line as closed curve and apply the residue theorem? If the (b)-line is a closed curve would it contain the singularities of f(z)? I mean, the curve could contain the half-plane with Re(z) < 0 or the other half-Plane with Re(z) > 0. In the first case all singularities are contained and in the latter one no singularities are contained... Does it make a difference if we choose b < 0? Elodie === Subject: Re: Residue theorem for beginners > Good evening, I'm a newbie here and currently taking a course in complex analysis. > I would appreciate if someone could answer my stupid questions. Let f(z) be a meromorphic function (i.e., f is holomorphic with > singularities). No, holomorphic with poles, I think. Not every singularity is a pole. > Now, let's consider the integral of f over the closed curve C, so > int_C f(z) dz. To evaluate this integral we can simply use the residue theorem: > http://en.wikipedia.org/wiki/Residue_theorem My problem is that I am not quite sure on how to evaluate a complex > integral for the following case: let us fix a real number b and b - > i*oo up to b + i*oo the b-line denoted by (b). So (b) is the line > through the point (b, 0) in the complex plane and parallel to the > imaginary axis. Let us consider a concrete example: let f(z) = Gamma(z) be the Gamma > function. Then f(z) has poles at the negative integers. Let b > 0 be > positive. How do we concretely evaluate the integral of f(z) over the > (b)-line, i.e., how do you evaluate int_(b) f(z) dz? In general, if you're going to use The Residue Theorem to evaluate an integral over (b), what you do is make a closed contour that includes part of (b), e.g., a rectangle one of whose sides goes from b - i Q to b + i Q for some large Q, integrate around it, then take a limit as Q goes to infinity. If the gods are willing, you'll be able to show that the integrals along the other three sides go to zero as Q goes one goes to zero), leaving just the integral along (b) and the residues of the poles. Whether this works for Gamma, I wouldn't know without trying. === Subject: Re: Residue theorem for beginners ) I have posted an answer before - but it seems it didn't come through. Okay that is my second attempt then. In general, if you're going to use The Residue Theorem to evaluate > an integral over (b), what you do is make a closed contour that includes > part of (b), e.g., a rectangle one of whose sides goes from b - i Q > to b + i Q for some large Q, integrate around it, then take a limit > as Q goes to infinity. If the gods are willing, you'll be able to show > that the integrals along the other three sides go to zero as Q goes > one goes to zero), leaving just the integral along (b) and the residues > of the poles. Wow, thank-you very much for your answer! Things are much clearer to me now! I don't know why I didn't understand the stuff during the lecture. I have a couple of more questions to ensure that I've completely understood on what is going on (I hope that is okay): Let b and c be different real numbers and f be a meromorphic function with poles. We consider a rectangle containing the (b) line as the right line and assume that the integral along the top and the bottom consider a rectangle containing (c) line with the same property. Also, we assume that all poles of f(z) are the negative integers with z_1 = -1, z_2 = -2, ... Case 1: We assume that all poles of f are to the left of b *and* c. Then the integral over (b) is the same as the integral over (c), isn't it? So int_(b) f(z)dz = int_(c) f(z)dz? Case 2: We assume that c < b and that c is between the first z_1 and second pole z_2 of f(z). Then, is the following statement correct int_(b) f(z)dz = int_(c) f(z)dz + Residue at z_1? > Whether this works for Gamma, I wouldn't know without > trying. Could you please provide me with a function g(z) where you know that the sum of the integrals along the three other paths of the rectangle are adding up to zero? So I can try to show it myself! === Subject: Re: Residue theorem for beginners posting-account=f7Q9vAoAAAA9NrtfQEoVAs9UXkEuc4Rm Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) On 22 Oct, 23:44, Gerry Myerson an integral over (b), what you do is make a closed contour that includes > part of (b), e.g., a rectangle one of whose sides goes from b - i Q > to b + i Q for some large Q, integrate around it, then take a limit > as Q goes to infinity. If the gods are willing, you'll be able to show > that the integrals along the other three sides go to zero as Q goes > one goes to zero), leaving just the integral along (b) and the residues > of the poles. Wow, thank-you very much for your answer! Things are much clearer to me now! I don't know why I didn't understand the stuff during the lecture. I have a couple of more questions to ensure that I've completely understood on what is going on (I hope that is okay): Let b and c be different real numbers and f be a meromorphic function with poles. We consider a rectangle containing the (b) line as the right line and assume that the integral along the top and the bottom consider a rectangle containing (c) line with the same property. Also, we assume that all poles of f(z) are the negative integers with z_1 = -1, z_2 = -2, ... Case 1: We assume that all poles of f are to the left of b *and* c. Then the integral over (b) is the same as the integral over (c), isn't it? So int_(b) f(z)dz = int_(c) f(z)dz? Case 2: We assume that c < b and that c is between the first z_1 and second pole z_2 of f(z). Then, is the following statement correct int_(b) f(z)dz = int_(c) f(z)dz + Residue at z_1? > Whether this works for Gamma, I wouldn't know without > trying. Could you please provide me with a function g(z) where you know that the three other paths of the integral tend to zero for large Q? So I can === Subject: Re: Residue theorem for beginners > On 22 Oct, 23:44, Gerry Myerson an integral over (b), what you do is make a closed contour that includes > part of (b), e.g., a rectangle one of whose sides goes from b - i Q > to b + i Q for some large Q, integrate around it, then take a limit > as Q goes to infinity. If the gods are willing, you'll be able to show > that the integrals along the other three sides go to zero as Q goes > one goes to zero), leaving just the integral along (b) and the residues > of the poles. Wow, thank-you very much for your answer! Things are much clearer to me now! I don't know why I didn't > understand the stuff during the lecture. I have a couple of more questions to ensure that I've completely > understood on what is going on (I hope that is okay): Let b and c be different real numbers and f be a meromorphic function > with poles. We consider a rectangle containing the (b) line as the > right line and assume that the integral along the top and the bottom > consider a rectangle containing (c) line with the same property. > Also, we assume that all poles of f(z) are the negative integers with > z_1 = -1, z_2 = -2, ... Case 1: We assume that all poles of f are to the left of b *and* c. > Then the integral over (b) is the same as the integral over (c), > isn't it? > So int_(b) f(z)dz = int_(c) f(z)dz? I think under your hypotheses this goes by considering a rectangle bounded by (parts of) the (b) and (c) lines, the (b) & (c) integrals ought to be equal. > Case 2: We assume that c < b and that c is between the first z_1 and > second pole z_2 of f(z). Then, is the following statement correct > int_(b) f(z)dz = int_(c) f(z)dz + Residue at z_1? Again, this should follow by considering the rectangle between the (b) and (c) lines. > Could you please provide me with a function g(z) where you know that > the three other paths of the integral tend to zero for large Q? So I Well, I guess it should work for 1 / z^3, although that won't help much since there aren't any residues. Maybe if you wanted to integrate 1 / (z - 1) (z - 2) (z - 3) along the (4)-line. But the best thing to do may be to crack open a complex variables text book (or the Schaum's Outline), which ought to have a few examples. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: please recommend books on intuition behind regression models... posting-account=CYtergoAAAAOD8k-T_NdzknPJOBXmt8x CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Hi all, I am looking for books or notes that talk about intuition behind regression models and ANOVA. For example, from the geometric perspective, how does the Sum of Squares correspond to the projection on a plane in the vector space. I am very interested in this sort of intuitive views and geometric perspectives in statistics, please recommend books and notes... === Subject: Re: please recommend books on intuition behind regression models... posting-account=XGudtAoAAAA0DMbOqV6ZmwLzRym7uNr7 Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Hi all, I am looking for books or notes that talk about intuition behind > regression models and ANOVA. For example, from the geometric > perspective, how does the Sum of Squares correspond to the projection > on a plane in the vector space. I am very interested in this sort of > intuitive views and geometric perspectives in statistics, please > recommend books and notes... > For simple linear regression, The Cartoon Guide to Statistics is pretty good. -- Bruce Weaver bweaver@lakeheadu.ca When all else fails, RTFM. === Subject: Re: please recommend books on intuition behind regression models... posting-account=CYtergoAAAAOD8k-T_NdzknPJOBXmt8x CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Hi all, I am looking for books or notes that talk about intuition behind > regression models and ANOVA. For example, from the geometric > perspective, how does the Sum of Squares correspond to the projection > on a plane in the vector space. I am very interested in this sort of > intuitive views and geometric perspectives in statistics, please > recommend books and notes... > For simple linear regression, The Cartoon Guide to Statistics is > pretty good. -- > Bruce Weaver > When all else fails, RTFM. It has the geometric and projection view about regression? === Subject: Re: please recommend books on intuition behind regression models... >> Hi all, >> I am looking for books or notes that talk about intuition behind >> regression models and ANOVA. For example, from the geometric >> perspective, how does the Sum of Squares correspond to the projection >> on a plane in the vector space. If you have any difficulty with this, you need to relearn the Pythagorean theorem. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Incredible set <1sjLk.5482$be.1050@nlpi061.nbdc.sbc.com> <48fdc772$0$11945$426a74cc@news.free.fr> <48fdda1e$0$10799$426a74cc@news.free.fr> posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN rv:1.9.0.3) Gecko/2008092414 Firefox/3.0.3,gzip(gfe),gzip(gfe) Here you are getting at something closer to what I was asking you for. > The local set theories which are equivalent to certain toposes are > quite axiomatic, and I suspect you could construct one in which your > set of all infinite sets exists; the fact that the axiom of > separation from ZF must necessarily fail in a local set theory in > which the set of all infinite sets exists could be accounted for by > having an unusual subobject classifier (part of the structure of a > topos), one with more than two points. This might be an interesting > thing to look at. > [ignoring the furthur posts caution] > yes > this has been hashed to death around here (sci.math) > a while back > the poster tommy1729 described a set theory > where x is an element of {x} > the recent proving not(c = aleph_aleph_null) diagonally without > choice > is one recent outgrowth of this discussion on x e {x} I assume galathaea means xex, not xe{x}, since xe{x} is > actually a theorem of ZFC -- but I know that it was just a > typo on her part. it was one of those read it in my head problems the e was actually me saying equals but got switched is an element after some bizarre brain machinations in the particular model discussed both x c x (subset) and x = {x} (equals) obtain i need to stop and breathe sometimes... > I'm still following the thread which she mentions, BTW. I > actually have something more to say about the theory, but > I want to wait to see the matter of whether ~c=aleph_omega > is a theorem of ZF settled before I say anything more in > that thread. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: sine and modulus ? <19263745.1224706743059.JavaMail.jakarta@nitrogen.mathforum.org>, > but p-adics are not finite sets ... so , im not saying denis is wrong but how does it relate since its not about finite sets such as mod p ? The integers are not a finite set. Does that mean that no facts about the integers have any implications mod p? There are ways to learn about finite sets by studying infinite ones. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: prime formula 30 <30147766.1224708284815.JavaMail.jakarta@nitrogen.mathforum.org>, > i was looking at formula 30 : http://mathworld.wolfram.com/PrimeProducts.html > which gives 4/pi. what about the analogue with primes mod 6 ? where character q(1) gives 1 and character q(5) gives -1 how to compute the product then ?? tommy1729 id say this is clear enough ! when you ask : which question , look for the ?? When I ask, which question? I expect that the person who wants an answer to the question will tell me which question he wants answered. Anyway. The formula in question is product over all odd primes of 1 - (chi(p) / p) is pi / 4, where chi(p) is 1 if p is 1 mod 4, -1 if p is 3 mod 4. The formulas before that evaluate the product of 1 + (chi(p) / p). They do it by a simple manipulation that relates the product to zeta(2) and to L(chi, 1). I assume formula 30 is established the same way. And if you want the analogue for primes mod 6, it is clear that you should try to do a similar sequence of manipulations to bring in L(chi, 1) for the analogous L-function. And, since you've written papers on L-functions, this should all be pretty routine for you. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Opportunity Cost I'm not really sure how to go about this question. You earn $30/hr while working. You are trying to decide between buying an album in iTunes and downloading the songs, vs. finding the album using a new Peer-2-Peer (P2P) application that you have discovered. ´ There is no time cost to download the application. ´ The price of the iTunes application is $0. In iTunes, the price of the album is $10. It will take you 1 minute to find the album in iTunes, and 5 minutes to download the album. ´ The price of the P2P application is $5. The album price is $0. It will take you 20 minutes to find the album, and 10 minutes to download the album. ´ You do not have to stop working while downloading the album. a. What is the opportunity cost of finding an album for each application? For iTunes: $5 For P2P: 20 mins b. What is the total cost of obtaining an album for each application? c. If you already own the P2P application, then what is the total cost of obtaining an album? === Subject: Re: Space is the (round) surface of the 4th dimemsion posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) > BURT mused: > BURT mused: > BURT mused: > 3D Space is the surface of the 4D hypersphere. Mitch Raemsch > God put the universe somewhere. There is no outside of the 4th dimension. The hypersphere > expands into nothing. > So God didn't put all that encompasses the 4th dimension >> anywhere? That is no-thing. > It's a yes or no question. Did God put all that encompases the >> 4th dimension anywhere?- Hide quoted text - > - Show quoted text - No its a question about nothing. It doesn't exist. Neither does your concept of the 4th dimension, unless you have some > evidence to support that claim.- Hide quoted text - - Show quoted text - The universe is closed in the 4th dimension theoretically. It can never be proven or disproven. It is not subjected to proof. Mitch Raemsch === Subject: Re: Space is the (round) surface of the 4th dimemsion In BURT mused: >> BURT mused: >> BURT mused: >> BURT mused: >3D Space is the surface of the 4D hypersphere. >Mitch Raemsch >> God put the universe somewhere. >There is no outside of the 4th dimension. The hypersphere > expands into nothing. >> So God didn't put all that encompasses the 4th dimension >> anywhere? >That is no-thing. >> It's a yes or no question. Did God put all that encompases the >> 4th dimension anywhere?- Hide quoted text - >> - Show quoted text - >No its a question about nothing. It doesn't exist. >> Neither does your concept of the 4th dimension, unless you have >> some evidence to support that claim.- Hide quoted text - >> - Show quoted text - The universe is closed in the 4th dimension theoretically. It can > never be proven or disproven. It is not subjected to proof. Do you have any *evidence* at all to support that theory? === Subject: Re: Is cyclotomic polynomial irreducible? I have read the proof of irreducibility here, and have one question about it. Robin Chapman states that the fact that the nth cyclotomic polynomial is a factor of x^n-1 implies it cannot have a repeated factor (mod p). I do not see the conclusion, but maybe I am missing something trivial or simply misunderstanding something. I am tired, so please forgive me if I am having a round shoulders and flat forehead moment. It is clear x^n-1 has no repeated factors over the complex plane (or the splitting field containing the nth roots of unity), but why cannot a repeated factor (mod p) exist if, for example, there are two integer monic factors over Z whose corresponding coefficients differ by multiples of p? In another resource, it was stated that the irreducible factors are all distinct (mod p) since x^n - 1 and its derivative are relatively prime (mod p) - again, why is this condition sufficient? I guess my question is why either of these clearly valid properties for functions in the complex plane should be preserved by the (mod p) ring homomorphism. My interest in this is related to a similar problem, computing all the periodic points of the polynomial x^2 - 2, which can be found using elementary trigonometry. They are seen to form cycles within the root sets of the minimal polynomials for 2 cos(2*pi/n) for odd n; the irreducibility of which can be inferred from that of the cyclotomic polynomials. Completing the proof here is the last part of the argument that needs to be understood for complete gratification. === Subject: Re: Is cyclotomic polynomial irreducible? <6554350.1224724784593.JavaMail.jakarta@nitrogen.mathforum.org>, > I have read the proof of irreducibility here, and have > one question about it. Robin Chapman states that the fact that the nth > cyclotomic polynomial is a factor of x^n-1 implies it > cannot have a repeated factor (mod p). I do not see the > conclusion, but maybe I am missing something trivial or > simply misunderstanding something. I am tired, so > please forgive me if I am having a round shoulders and > flat forehead moment. It is clear x^n-1 has no repeated factors over the > complex plane (or the splitting field containing the nth > roots of unity), but why cannot a repeated factor > (mod p) exist if, for example, there are two integer > monic factors over Z whose corresponding coefficients > differ by multiples of p? In another resource, it was stated that the irreducible > factors are all distinct (mod p) since x^n - 1 and its > derivative are relatively prime (mod p) - again, why is > this condition sufficient? I guess my question is why either of these clearly valid > properties for functions in the complex plane should be > preserved by the (mod p) ring homomorphism. You can develop the theory of polynomials over an arbitrary field K, introduce a formal derivative by (x^k)' = k x^(k-1) and linearity, and then prove repeated factor if and only if common factor between polynomial and derivative. A good textbook that does field theory should show you this in some detail. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Is cyclotomic polynomial irreducible? forgotten since my algebra course in 1967. I just realized the chain and product rules (mod p) would produce repeated factors to one lower power in the derivative...just like in elementary calculus (whether or not they were images of repeated factors over Z)! Darn!! If anybody is interested in the overall results (related to the Sarkovskii theorems) and some intersting but archaic constructions, please email me at wheierman@corunduminium.com or at williamh@wcjc.edu, and I can send reply with a preliminary draught. Give me a day or two to insert this final piece of the puzzle and check for errata... Eventually, a paper on the consolidated Sarkovskii theorems with a visual, intuitive proof will be available - interested? (Email if so.) === Subject: Re: JSH: So why no legal DVD copying? posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > What's the point of making a legal backup if it doesn't work if you > lose or damage the original? (The following is not a comment on whether or not JSH's scheme is > correct or feasible). You wouldn't use the backup as a backup. You'd use the backup as your > active copy, and keep the original as a backup. E.g., you buy a DVD > for your kids. You make a 'backup', and give that to the kids. You put > the original in a safe place. Yup. If (when) the kids destroy their copy, you take the original out of the > safe place, and make a new copy for the kids. If, by some miracle, the kids go 30 days without destroying their copy, > you take the original out of the safe place, and re-authorize the kid's > copy for another 30 days. Because the original only comes out when you need to make a new copy for > the kids, or every 30 days (whichever comes from first), and is only > used by you and then put away, you should be able to keep the original > undamaged. -- > --Tim Smith And that at least seemed to stop the arguing back and forth. I find it REMARKABLE how some of you put a label on a person of > crackpot and then refuse to acknowledge anything they say as > valuable so you will argue day and night with that person until > someone else comes in and then you quiet down. The behavior is telling. You wish to permanently label certain people > in a negative way and hold that label against any new evidence. This story shows how dramatic that is as well, where the fight is > against a simple way to allow legal copying of bought DVD's, where > that is concrete enough for people to understand just how rigid is the > hostility against supposed crackpots. I've joked in the past that many of you would reject a cure for cancer > because of the source. Short of it is that your opinions against people are often meaningless > as to what they state or the facts but more often than not reflect > rigid personal biases which you quickly form against CERTAIN people > and then refuse to shift. Oh, and of course there is also an experimental side to my posting on > this subject! DMESE is a fairly simple idea which because of the nature of the > problem with copying digital media actually closes the problem space, > as in, it's either it, or nothing, and so far industry has sided > towards nothing with many declaring DRM to be dead, which has a > potential cost in the billions of dollars over time, if artists, for > instance, are incapable of making as much from music sales because > widespread piracy is considered unstoppable. Past experience indicates I take over web search results on topics > fairly quickly, which I've demonstrated often with definition of > mathematical proof or DMESE itself, as well as many others including > disparate things like: business plan for Internet radio or prime number compression (Google them, as most of my search domination is with Google though it > is often on Yahoo! as well.) But is that being done through a proper process? I've been concerned > enough about it that I've contacted my U.S. senator and representative > about the issue. (Senator Feinstein gave me a form letter reply. She > may need to re-visit that later though as this situation develops > while I'm deferring a bit, because of the financial crisis, in > pursuing the issue.) If things stay to form though I should start showing up in search > results on RealNetworks and their DVD copying software. If I don't > then there may be more to the story of how Google and Yahoo! determine > those rankings. Already I have atypical behavior on my blogs where hits have gone DOWN > dramatically lately, since I started pushing this issue. Weird seemingly atypical behavior is where science starts... So we all lose. Yup, we are all losers. That's the real secret. So Uncle Al and the rest of the crackpot haters, so what? You're losers along with us. You'll never get recognition for your either. Waste your lives. Put it out there like anybody cares when it's a world run by fools. Knowledge isn't power. Crap is power. And idiots rule the world... JSH === Subject: Re: JSH: So why no legal DVD copying? posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) What's the point of making a legal backup if it doesn't work if you > lose or damage the original? (The following is not a comment on whether or not JSH's scheme is > correct or feasible). You wouldn't use the backup as a backup. You'd use the backup as your > active copy, and keep the original as a backup. E.g., you buy a DVD > for your kids. You make a 'backup', and give that to the kids. You put > the original in a safe place. Yup. If (when) the kids destroy their copy, you take the original out of the > safe place, and make a new copy for the kids. If, by some miracle, the kids go 30 days without destroying their copy, > you take the original out of the safe place, and re-authorize the kid's > copy for another 30 days. Because the original only comes out when you need to make a new copy for > the kids, or every 30 days (whichever comes from first), and is only > used by you and then put away, you should be able to keep the original > undamaged. -- > --Tim Smith And that at least seemed to stop the arguing back and forth. I find it REMARKABLE how some of you put a label on a person of > crackpot and then refuse to acknowledge anything they say as > valuable so you will argue day and night with that person until > someone else comes in and then you quiet down. The behavior is telling. You wish to permanently label certain people > in a negative way and hold that label against any new evidence. This story shows how dramatic that is as well, where the fight is > against a simple way to allow legal copying of bought DVD's, where > that is concrete enough for people to understand just how rigid is the > hostility against supposed crackpots. I've joked in the past that many of you would reject a cure for cancer > because of the source. Short of it is that your opinions against people are often meaningless > as to what they state or the facts but more often than not reflect > rigid personal biases which you quickly form against CERTAIN people > and then refuse to shift. Oh, and of course there is also an experimental side to my posting on > this subject! DMESE is a fairly simple idea which because of the nature of the > problem with copying digital media actually closes the problem space, > as in, it's either it, or nothing, and so far industry has sided > towards nothing with many declaring DRM to be dead, which has a > potential cost in the billions of dollars over time, if artists, for > instance, are incapable of making as much from music sales because > widespread piracy is considered unstoppable. Past experience indicates I take over web search results on topics > fairly quickly, which I've demonstrated often with definition of > mathematical proof or DMESE itself, as well as many others including > disparate things like: business plan for Internet radio or prime number compression (Google them, as most of my search domination is with Google though it > is often on Yahoo! as well.) But is that being done through a proper process? I've been concerned > enough about it that I've contacted my U.S. senator and representative > about the issue. (Senator Feinstein gave me a form letter reply. She > may need to re-visit that later though as this situation develops > while I'm deferring a bit, because of the financial crisis, in > pursuing the issue.) If things stay to form though I should start showing up in search > results on RealNetworks and their DVD copying software. If I don't > then there may be more to the story of how Google and Yahoo! determine > those rankings. Already I have atypical behavior on my blogs where hits have gone DOWN > dramatically lately, since I started pushing this issue. Weird seemingly atypical behavior is where science starts... So we all lose. Yup, we are all losers. That's the real secret. So Uncle Al and > the rest of the crackpot haters, so what? You're losers along with us. You'll never get recognition for your either. Waste your lives. Put it out there like anybody cares when it's a > world run by fools. Knowledge isn't power. Crap is power. And idiots rule the world... JSH- Hide quoted text - - Show quoted text - Welcome! === Subject: JSH: Moving to real fun posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) Back when I was naive and stupid, and idealistic, I dreamed of making great discoveries to advance the state of humanity. But then I made some discoveries, got years of abuse from mathematicians, and now all I want to do is make as many academics look like fools, destroy endowments, end departments and generally wreak havoc across the academic world, as I think I kind of went bonkers as a result. And so I no longer care, can no longer be convinced, but I'm still brilliant, can keep making discoveries indefinitely, and fantasize about shutting down, Princeton! Yup, I want to shut down Princeton. That sounds like a better goal than stupid nonsense like furthering and advancing the state of the human species as its full of ingrates, and people who don't value knowledge, but they care about cable! Yeah, give them their cable. Give the hoi polloi their television with its nincompoops and actors who are worth nothing but get so much from a world of fools. This ship of fools. Well at least I can make your lives miserable. Take away your funds and force you to do what I can do. Make you live up to my standard. Public funding for research is stupid. All scientists should learn to live in the free market. Do their research as best they can without welfare. It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! Like me. James Harris === Subject: Re: Moving to real fun > Back when I was naive and stupid, and idealistic, I dreamed of making > great discoveries to advance the state of humanity. you are still naive and stupid, and idealistic. > But then I made some discoveries, got years of abuse from > mathematicians, and now all I want to do is make as many academics > look like fools, destroy endowments, end departments and generally > wreak havoc across the academic world, as I think I kind of went > bonkers as a result. you are still bonkers. And so I no longer care, can no longer be convinced, but I'm still > brilliant, can keep making discoveries indefinitely, and fantasize > about shutting down, Princeton! nope, you were never brilliant. Yup, I want to shut down Princeton. Only to satisfy your ego, troll boy. That sounds like a better goal than stupid nonsense like furthering > and advancing the state of the human species as its full of ingrates, > and people who don't value knowledge, but they care about cable! Yeah, give them their cable. Give the hoi polloi their television > with its nincompoops and actors who are worth nothing but get so much > from a world of fools. This ship of fools. Well at least I can make your lives miserable. Take away your funds > and force you to do what I can do. dream on troll-boy. Make you live up to my standard. how low can a hacker go? really low. Public funding for research is stupid. JSH research is stupid. All scientists should learn to live in the free market. Do their > research as best they can without welfare. It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! Like me. is it time for you to start working for a living ? James Harris === Subject: Re: JSH: Moving to real fun > All scientists should learn to live in the free market. Do their > research as best they can without welfare. It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! I think it is actually true that there's a kind of laxity in US academic math jobs which is abused by many people--there are plenty of American schools where you can complete a math PhD and go on to get a comfortably-paid permanent teaching job at some school without much research funding, all while working well under 40 hours a week. At the same time there are those of us who work very, very hard and work very long hours--I personally have given up many things which meant a great deal to me so that I could have some kind of decent career in mathematics research, and right now I spend about 60 hours/week in my office, plus time I spend working at home. I am afraid eliminating government funding for research would not help this situation--based on what I have seen of free markets, the situation would certainly become worse. === Subject: Re: JSH: Moving to real fun > Yeah, give them their cable. Give the hoi polloi their television > with its nincompoops and actors who are worth nothing but get so much > from a world of fools. This ship of fools. > Written by a man who bases his life on an actor from a cable TV show. God, you're stupid. James Harris M === Subject: Re: JSH: Moving to real fun posting-account=d-ESTAkAAAAG0l03yI1WJgsTVXx4ebeJ Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > It's time for you all to start working for a living! Like me. James Harris DOOD! You are either seriously ed or scored some REALLY good ! All I can say is that based on this rant, I assume your other job is playing lead guitar in a HM band. === Subject: Re: JSH: Moving to real fun posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) It's time for you all to start working for a living! Like me. James Harris DOOD! You are either seriously ed or scored some REALLY good > ! > All I can say is that based on this rant, I assume your other job is > playing lead guitar in a HM band. Gullible. === Subject: Re: JSH: Moving to real fun >It's time for you all to start working for a living! > Like me. > James Harris >> DOOD! You are either seriously ed or scored some REALLY good >> ! >> All I can say is that based on this rant, I assume your other job is >> playing lead guitar in a HM band. Gullible. Where do you play? === Subject: Re: JSH: Moving to real fun > And so I no longer care, can no longer be convinced, but I'm still > brilliant, can keep making discoveries indefinitely, and fantasize > about shutting down, Princeton! And I fantasize about: ´ Fixing ACPI on my computer ´ Fixing in Thunderbird ´ Actually being good at the piano ´ Accidentally hitting sudo rm -rf / on a computer ´ Finishing any one of my four aborted programming projects ´ Finishing any one of my countless incomplete video games ´ Actually being good at, say, drawing ´ Actually writing down one of my stories ´ Winning the Fields Medal/Turing Award/Nobel Prize ´ Becoming president ´ Flying a plane ´ A world where people are actually polite to each other in conversation ´ A world where Linux is the primary desktop OS But these are all just fantasies, about as likely to occur as the Earth spontaneously ceasing to exist. > That sounds like a better goal than stupid nonsense like furthering > and advancing the state of the human species as its full of ingrates, > and people who don't value knowledge, but they care about cable! You've really detached yourself from society if you think people care about cable. It's all about the broadband internet. Give most people I know a 50 Mb/s downlink and they're reduced to salivating wrecks. === Subject: Re: JSH: Moving to real fun posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > And so I no longer care, can no longer be convinced, but I'm still > brilliant, can keep making discoveries indefinitely, and fantasize > about shutting down, Princeton! And I fantasize about: > ´ Fixing ACPI on my computer > ´ Fixing in Thunderbird > ´ Actually being good at the piano > ´ Accidentally hitting sudo rm -rf / on a computer > ´ Finishing any one of my four aborted programming projects > ´ Finishing any one of my countless incomplete video games > ´ Actually being good at, say, drawing > ´ Actually writing down one of my stories > ´ Winning the Fields Medal/Turing Award/Nobel Prize > ´ Becoming president > ´ Flying a plane > ´ A world where people are actually polite to each other in conversation > ´ A world where Linux is the primary desktop OS But these are all just fantasies, about as likely to occur as the Earth > spontaneously ceasing to exist. > not at all, I think you have a very nice list there. The polite conduct is actually a self fulfilling prophecy. Thinking the Earth cant spontaneously cease to exist is another one. The non US media is very upset about the global drop in oxygen levels. They discarded all the global warming data claiming it is no longer correct, things are now moving a hundred times as fast as planed. So, I'm sorry to tell you it might just happen very soon sir. Go tell your officials to grow hemp everywhere. Tossing around seeds doesn't take a lot of effort or budget. We need 2 big trees per person or 20 m2 worth of green in the sun. In the shadow it will be more like 50 m2. If we don't we will all tip over and cease to exist I'm afraid. Sorry to bring the bad news but do spread the word please. > That sounds like a better goal than stupid nonsense like furthering > and advancing the state of the human species as its full of ingrates, > and people who don't value knowledge, but they care about cable! You've really detached yourself from society if you think people care > about cable. It's all about the broadband internet. Give most people I > know a 50 Mb/s downlink and they're reduced to salivating wrecks. Pacman and internets are not really important. We need air, water, vegetables and a place to sleep. When you have those you should entertain yourself by helping others getting the same. Hollywood is not real. It's to distract you from the economic activities so that you don't bother the money makers. You'd really want to keep an eye on those in stead. Dial Up is good enough for such. If you know something about music perhaps you can take a look at this. === Subject: Re: JSH: Moving to real fun posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) > Back when I was naive and stupid, and idealistic, James Harris I think you could have stopped there - at least these statements are still true! Get a life - you are an angry idiot. Please seek professional help before you hurt another human being! Go drink beer, chase women and destroy the last remaining brain cell that may still exist in your delusional mind. possibly can. Maybe that will knock some sense into you. Don't be a bitter fool - get a life and do something to help as opposed to moaning, crying and peeing all over yourself! === Subject: Re: JSH: Moving to real fun posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) Back when I was naive and stupid, and idealistic, James Harris I think you could have stopped there.... Yes, but you are still naive, stupid and idealistic. :-) === Subject: Re: JSH: Moving to real fun posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Back when I was naive and stupid, and idealistic, I dreamed of making > great discoveries to advance the state of humanity. But then I made some discoveries, got years of abuse from > mathematicians, and now all I want to do is make as many academics > look like fools, destroy endowments, end departments and generally > wreak havoc across the academic world, as I think I kind of went > bonkers as a result. And so I no longer care, can no longer be convinced, but I'm still > brilliant, can keep making discoveries indefinitely, and fantasize > about shutting down, Princeton! Yup, I want to shut down Princeton. That sounds like a better goal than stupid nonsense like furthering > and advancing the state of the human species as its full of ingrates, > and people who don't value knowledge, but they care about cable! Yeah, give them their cable. Give the hoi polloi their television > with its nincompoops and actors who are worth nothing but get so much > from a world of fools. This ship of fools. Well at least I can make your lives miserable. Take away your funds > and force you to do what I can do. Make you live up to my standard. Public funding for research is stupid. All scientists should learn to live in the free market. Do their > research as best they can without welfare. It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! Like me. James Harris What's yer occupation? === Subject: Re: JSH: Moving to real fun It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! Like me. I'd like to see some actual evidence that you get paid for posting kookery on Usenet. There's just too much amateur talent for that to be a viable occupation. -- Bill Snyder [This space unintentionally left blank] === Subject: Re: JSH: Moving to real fun posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I'd like to see some actual evidence that you get paid for posting Pay attention billy. You just tolled us you get paid to post lies here? no? Here is some work for you. Is survival not a good pay? === Subject: Re: JSH: Moving to real fun posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I'd like to see some actual evidence that you get paid for posting Pay attention billy. You just tolled us you get paid to post lies here? no? > Toll: 1. To exact as a toll. 2. To charge a fee for using (a structure, such as a bridge). Told: 1. Past tense and past participle of tell It is easy to excuse misspellings - but at least use the correct words. === Subject: Re: JSH: Moving to real fun posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) It is easy to excuse misspellings - but at least use the correct > words. you claim it is easy but not for you. Yo gramma is so dumb you cant even address the topic. === Subject: Re: JSH: Moving to real fun >> I'd like to see some actual evidence that you get paid for posting Pay attention billy. You just tolled us you get paid to post lies here? no? You should take something for those hallucinations. >Here is some work for you. >Is survival not a good pay? Stupidity and insanity are available cheap at outlets everywhere. -- Bill Snyder [This space unintentionally left blank] === Subject: Re: JSH: Moving to real fun >> I'd like to see some actual evidence that you get paid for posting Pay attention billy. You just tolled us you get paid to post lies here? no? You should take something for those hallucinations. Here is some work for you. >Is survival not a good pay? Stupidity and insanity are available cheap at outlets everywhere. -- > Bill Snyder [This space unintentionally left blank] Oh, what a typical Hill Billy thing to say. :-) === Subject: Re: JSH: Moving to real fun posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! Like me. I'd like to see some actual evidence that you get paid for posting > kookery on Usenet. There's just too much amateur talent for that > to be a viable occupation. It is fun though. I did put fun in the title so yeah, the post was meant to be at least somewhat humorous. I especially liked the part about shutting down Princeton. But seriously, the message from me to academia is, you do your thing, and I'll do mine. We don't work together. Not today and not tomorrow. I'm still reeling from my latest research results and that weirdness with RealNetworks and Hollywood blocking all legal copying of bought DVD's, so I'm puzzling and wondering and reaching the point where I'm telling myself, yeah, I don't give a damn. I'm brilliant. I accept it. I like myself so it's ok. I see your academic world as primitive. Yeah, I know. You don't like being called primitive and think some of you are brilliant too, well that's ok too. You can feel what you like. It doesn't matter if it's not true, right? Wow. What a weird reality. When I was a kid I so looked up to all these people and wondered, what if? What if it were me and now I'm like, thank God you people are primitive. Yeah. It's a good world. I like it. I really do. Wow. James Harris === Subject: Re: JSH: Moving to real fun posting-account=_k7REQoAAACDJL2M6OKCfSBj5_wTcvrO Gecko/20080702 Firefox/2.0.0.16 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > But seriously, the message from me to academia is, you do your thing, > and I'll do mine. we can now just get on with our thing. This whole James Harris saga has been occupying everyone's mind for the last 10 years. === Subject: Bruce Sturman Bruce Sturman === Subject: Re: Modern Aether Properties I'm not sure what to make of this rambling. As for modern Aether theory, it based on and founded in Maxwell/Helmholtz/Kelvin's atomic vortex hypothesis, a.k.a. the vortex sponge supersolid. A state I've come to call the supercrystal. If the universe is a supercrystal spectral splitting such as the Lamb shift should come as no surprise. Here are some related papers, http://www.springerlink.com/content/hxt258554588753w/ http://arxiv.org/ftp/physics/papers/9906/9906019.pdf http://aps.arxiv.org/PS_cache/physics/pdf/0612/0612210v3.pdf http://www.znaturforsch.com/aa/v58a/s58a0231.pdf a good general overview of vortex dynamic, http://www.dtu.dk/upload/fluid.dtu/fys/hassan%20aref/aam_chap3_final.pdf worth a mention... http://www.astrosciences.info/Ring.html === Subject: Re: inverse AGM > let AGM(1,x) = y AGM is the aritmetic geometric mean. how to inverse the AGM with similar iterations of means ? AGM^-1(y) = (1,x) ? is it possible with rational means ? I don't know the answers to your questions. Nonetheless, you or others may find the following approximations for the inverse AGM to be of interest. Let AGM1(x) denote the arithmetic-geometric mean of 1 and x (with x >= 0) and let invAGM1 denote its inverse function. We approximate invAGM1(y) in three cases, depending on whether y is near 0, is near 1, or is large. --------------------------- Case 1: y is near 0 invAGM1(y) ~ 4 exp( -pi/(2 y) ) Numerical example: 4 exp( -pi/(2 AGM1(0.001)) ) = 0.000999998... --------------------------- Case 2: y is near 1 (by which I really mean that y is neither near 0 nor large) invAGM1(y) = 1 + 2 (y-1) + 1/2 (y-1)^2 - 1/4 (y-1)^3 + 3/16 (y-1)^4 - 5/32 (y-1)^5 + 33/256 (y-1)^6 - 25/256 (y-1)^7 + ... More terms of the series could be provided easily, if required. Numerical examples using just those terms explicitly stated above: Substituting AGM1(0.6) for y, we obtain 0.5999997... Substituting AGM1(2) for y, we obtain 1.99990... --------------------------- Case 3: y is large invAGM1(y) ~ -2/pi y W_{-1}( -pi/(8 y) ) where W_{-1} denotes the -1 branch of the Lambert W function. Numerical example: Substituting AGM1(1000) for y in the approximation, we obtain 999.9997... For those who prefer not to use the Lambert W function, we can give an alternative approximation which avoids it: Letting u = log(pi/(8 y)), invAGM1(y) ~ -2/pi y (u - log(-u) + log(-u)/u) Numerical example: Substituting AGM1(1000) for y in the alternative approximation, we obtain 1000.03... David W. Cantrell === Subject: Re: charge and mass excert forces on each other posting-account=4zkFhgoAAABRkhghJDznKV6E74qpdaXo 5.1),gzip(gfe),gzip(gfe) > F proo nuke=(+/-)0.7749N(m/kg)(m/C) M1/R Q2/R > F anti nuke=(-/+)0.7749N(m/kg)(m/C) M2/R Q1/R Check my math but doesn't that mean F proo nuke = F anti nuke? === Subject: Re: charge and mass excert forces on each other in sci.physics.relativity: >> F_proo_nuke=(+/-)0.7749N(m/kg)(m/C) M1/R Q2/R >> F_anti_nuke=(-/+)0.7749N(m/kg)(m/C) M2/R Q1/R Check my math but doesn't that mean F_proo_nuke = F_anti_nuke? Actually F_proo_nuke = - F_anti_nuke. However, I don't understand what he is trying to accomplish with those formulas. The terms aren't properly defined. -- // The TimeLord says: // Pogo 2.0 = We have met the aliens, and they are us! === Subject: Interesting linear inequality question Hi all, I have the following problem. Given a matrix A[mxn] (m>n), I wanna find 1 vector w[nx1] such that A*w>=0. Which is the best (fast) method to solve this? Nhat Vo. === Subject: Re: Interesting linear inequality question > Hi all, I have the following problem. Given a matrix A[mxn] (m>n), I wanna find 1 > vector w[nx1] such that A*w>=0. Which is the best (fast) method to solve > this? Write down the 0 vector. OK, that's probably not what you really want. You could try linear programming. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Asking for friend: how to analyse two time series and how to make forecast? posting-account=CYtergoAAAAOD8k-T_NdzknPJOBXmt8x CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) Real business forecasting examples wanted, hopefully using statistical package such as R. Hi all, I am playing with some companies' balance sheets and income statements and want to apply what I've just learned from Stats class to see if I can forecast the companies earnings, sales and gross margin in the short term (3rd and 4th Quarter), mid-term (2009) and long term (2011, etc. ) I pulled up some data from companies' financial statements over the past a few years. The companies are techs so don't have long history. And given regime switching and big changes in the macro condition, I think long historical data have no use any way... so I have quarterly and annual data... and I probably need some other time series, for example the GDP, interest rate, consumer spending, etc. How do I forecast the performance of the firms using R? And how do I study several time series, such as historical performance vs. US GDP, US real consumer expenditure, and find their relation, etc? I don't have much experience yet, for example, I probably have to de-trend the timeseries first, and then do some seasonality adjustment, etc. but I think it's very fun. Are there good hands-on tutorials about how to do these things in R, with realistic examples? === Subject: Re: Asking for friend: how to analyse two time series and how to make forecast? > Real business forecasting examples wanted, hopefully using statistical > package such as R. Go on with ensuring that the financial system be inherently unstable. Business forecasting is making up your own self-fulfilling prophecy. Han de Bruijn === posting-account=5ApcPgoAAABKcgEyKsQmJVb3Rz63IGGL .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; WWTClient2),gzip(gfe),gzip(gfe) The wheel of a car rotates at the same speed as it rolls down the street. These two equal motions slow down the wheel's time. The wheel of a car is double slow since it has two motions to be taken into account both of which are equal in effect. The rest of the car only slows down from one motion or its motion down the street. Mitch Raemsch === Subject: Re: when push comes to shove, the only way Special Relativity can be proven is by first assuming it is correct posting-account=vma-PgoAAABrctSmMdefNKZ-c5S8buvP > The fact that phi(v) CAN have any value at all proves that this group > is numerically unbounded, thus has an infinite number of > members. :-) = (-: glird Dabbling in group theory, Lebau? You didn't even get past 9-th grade (operations with polynomials), imbecile. === Subject: Re: when push comes to shove, the only way Special Relativity can be proven is by first assuming it is correct <48FCF3F4.9050301@guesswhatuia.no> posting-account=ASfdVQoAAADpvgPOQMrGb2BSzAnk0jYp Gecko/2008091620 Firefox/3.0.2,gzip(gfe),gzip(gfe) On Oct 20, 5:11 pm, Paul B. Andersen > glird skrev: > So is your point that the Lorentz transform doesn't follow from >> the postulates of SR? There are three such postulates. > 1. the equations of Physics should hod good regardless of the chosen > frame of reference. > 2. The speed of light in empty space is a constant, equal to c. > 3. If we set clocks of all systems to MEASURE that speed as c, they > will do so regardless of the state of inertial motion of any such system. > The Lorentz Transformation group is only ONE of the infinite number > of groups that obey and allow those postulates to hold good. > If you want to see them all, look at Einstein's group: > tau = phi(v)beta(t - vx/c^2, > xi = phi(v)beta(x - vt) > eta = phi(v)y, > zeta = phi(v)z; > where beta = 1/(sqrt(1 - v^2/c^2) and phi(v) can have any value at all. > The fact that phi(v) CAN have any value at all proves that this group > is numerically unbounded, thus has an infinite number of members. So let me rephrase the question: > If we define two coordinate systems like Einstein did in 'electrodynamics', > then can this coordinate transform between these two systems: > tau = beta(t - vx/c^2, > xi = beta(x - vt) > eta = y, > zeta = z; > where beta = 1/(sqrt(1 - v^2/c^2) > be shown to follow from the postulates of SR? No! Unless an entirely different method is used, at least one more postulate is needed; that phi(+/- v) = 1 as plotted by all systems regardless of their relative velocities wrt each other. (As proved in A Flower; whether by Einstein or several others, reasons of symmetry is NOT a proof!) glird === Subject: Re: when push comes to shove, the only way Special Relativity can be proven is by first assuming it is correct > On Oct 20, 5:11 pm, Paul B. Andersen >> glird skrev: >> So is your point that the Lorentz transform doesn't follow from >> the postulates of SR? > There are three such postulates. > 1. the equations of Physics should hod good regardless of the chosen > frame of reference. > 2. The speed of light in empty space is a constant, equal to c. > 3. If we set clocks of all systems to MEASURE that speed as c, they > will do so regardless of the state of inertial motion of any such system. > The Lorentz Transformation group is only ONE of the infinite number > of groups that obey and allow those postulates to hold good. > If you want to see them all, look at Einstein's group: > tau = phi(v)beta(t - vx/c^2, > xi = phi(v)beta(x - vt) > eta = phi(v)y, > zeta = phi(v)z; > where beta = 1/(sqrt(1 - v^2/c^2) and phi(v) can have any value at all. > The fact that phi(v) CAN have any value at all proves that this group > is numerically unbounded, thus has an infinite number of members. >> So let me rephrase the question: >> If we define two coordinate systems like Einstein did in 'electrodynamics', >> then can this coordinate transform between these two systems: >> tau = beta(t - vx/c^2, >> xi = beta(x - vt) >> eta = y, >> zeta = z; >> where beta = 1/(sqrt(1 - v^2/c^2) >> be shown to follow from the postulates of SR? No! Your opinion noted. > Unless an entirely different method is used, at least one more > postulate is needed; that phi(+/- v) = 1 as plotted by all systems > regardless of their relative velocities wrt each other. > (As proved in A Flower; whether by Einstein or several others, > reasons of symmetry is NOT a proof!) glird I am not going to waste my time arguing. -- Paul http://home.c2i.net/pb_andersen/ === Subject: ? no superposition for NL system then why NL e-vectors posting-account=H-IscAoAAABkDNrURGSxo9jPN3MJ3a8A 1.0.3705; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) Hi: The superposition principle applies only for a linear system, and this is why we can decomposit one such system with its eigenvector. A nonlinear system, on the other hand, does not satisfy this principle of superposition. But we do know that there are nonlinear eigenvectors. This is confusing to me then. Does it mean that a NL system can be superimposed by its NL eigenvectors? If not, what are the uses of the NL eigenvectors? === Subject: Re: a support exercise > Let X be an infinite set. Consider a real function f : X -> R and a >> subset A of X : We say f has *support* A iff f is nonzero exactly on A. >> Let W be the (rational) linear span of finitely many real functions on >> X. Prove or disprove : Among the (possibly) infinite set W of >> functions, there are only finitely many different supports. Correct! Next exercise: >Let X = [0,1]. The support of a *continuous* real function f is >closed, so it has a minimum, call it the starting point of $f$. >Let W be the (rational) linear span of finitely many continuous >real functions on [0,1]. Prove or disprove: Among the set W of >functions, there are only finitely many different starting points. > There are at most dim(W) starting points: if f_1,... f_n have distinct starting points then they are linearly independent. (For suppose f_i has starting point s_i, and arrange things so that s_1 < ... < s_n. If the f_i's are linearly dependent then for some k we have f_k in the linear span of {f_{k+1}, ..., f_n}, which implies that s_k >= s_{k+1}, a contradiction.) Do you have a third exercise up your sleeve? RS === Subject: Re: a support exercise Originator: grubb@lola >Let X be an infinite set. Consider a real function f : X -> R and a >subset A of X : We say f has *support* A iff f is nonzero exactly on A. >Let W be the (rational) linear span of finitely many real functions on >X. Prove or disprove : Among the (possibly) infinite set W of >functions, there are only finitely many different supports. Let X=[0,1], f(x)=1, and g(x)=x. Linear combinations are simply linear functions. The supports, as defined, are the complements of singletons of rational points. --Dan Grubb === Subject: JSH: State of the art of mathematics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I've now successfully used tautological spaces against only two problems. Where the latest which lead me to an alternate solution for binary quadratic Diophantine equations is rather spectacular for quite a few reasons because of the history around the problem. Obviously, I can contemplate other areas of mathematics on which to use tautological spaces. But I also know about their limitations as the second problem I tried to use tautological spaces on, and failed with them, was the factoring problem, so they are not infallible, just unimaginably powerful. I've been watching the television show House a lot lately. I've kind of become addicted to it which has impacted how I post, as in, I'm more insulting, and it feels good. But like Dr. House, my insults are for a purpose, or I think they are, though they may simply mean I'm juvenile? In any event, the show has given me a different perspective on the problem space. So what do we know? This notion of using identities in such a clever way clobbered the hell out of binary quadratic Diophantines in a way so devastating it mostly shut up even the angry idiots. But not a lot has changed. There's no reason to suspect much will change, so we can move on. And when I say we, I mean me, and my team of those people who bizarrely for reasons that puzzle many, and probably themselves, help me. So yes, I admit it. There are people who help me evaluate ideas and I mostly insult them. Like Dr. House. So I have a paradigm around which to fit my behavior and I am watching more and more episodes of the TV show to refine my behavior, and improve my insults. Hilbert came up with some problems. I can clobber quite a few of them but wonder if I care. Russell thought to formalize a few things, put some logic into it, but came up short, but I think a few simple ideas bridge things in a way he didn't imagine. But at the end of the day, does anybody really care? I know I find it hard to care. So what other problems for those who are silly enough to care might be amenable to tautological spaces in their new and raw form, which is new and raw as I just invented the concept, or discovered it, whatever, a few years ago? (No one say Collatz. It's verboten.) Of course I HAVE solved problems without using tautological spaces! But that's too advanced for this initial post. I'm thinking about cleaning out some crap left over from the old geezers who used to do math before they keeled over and died, like Russell and Hilbert. Still I'm kind of picky, so the idiots among you need not apply. Ok, anyone can of course as usual reply as if I try to put that kind of criteria on then no one is qualified but me to talk to myself and I do enough of that already. My suspicion is that you will fail me. That is good. It just means I'm more brilliant than I previously thought. So what is the state of the art of mathematics? In flux. The old ways have failed. I've collapsed 2000 years of old ideas about some simple binary whatevers into a page or two of rather potent theory. Mathematics can be revolutionized like the sciences. You CAN push out the old professors hogging up the works by outdoing them, re-inventing, discovering, starting from scratch, or not, if you're not smart enough. Get started. Or not. I'll insult you either way you go. Or not. James Harris === Subject: JSH Re: State of the art of mathematics > I've now successfully used tautological spaces you mean tautological flatulences kind of become addicted to it which has impacted how I post, as in, > I'm more insulting, and it feels good. But like Dr. House, my insults > are for a purpose, or I think they are, though they may simply mean > I'm juvenile? yes, it simply mean you are a juvenile. In any event, the show has given me a different perspective on the > problem space. So what do we know? > mostly insult them. Like Dr. House. Go for being like Mr. Ed. So I have a paradigm around which to fit my behavior and I am watching > more and more episodes of the TV show to refine my behavior, and > improve my insults. Good, moron is as moron does. Get started. Or not. I'll insult you either way you go. Or not. because you are only a troll crackpot, and NOT anything like a real mathematician. James Harris === Subject: Re: JSH: State of the art of mathematics ) > I've now successfully used tautological spaces against only two > problems. Where the latest which lead me to an alternate solution for > binary quadratic Diophantine equations is rather spectacular for quite > a few reasons because of the history around the problem. Obviously, I can contemplate other areas of mathematics on which to > use tautological spaces. But I also know about their limitations as the second problem I tried > to use tautological spaces on, and failed with them, was the factoring > problem, so they are not infallible, just unimaginably powerful. > ... > So what other problems for those who are silly enough to care might be > amenable to tautological spaces in their new and raw form, which is > new and raw as I just invented the concept, or discovered it, > whatever, a few years ago? I looked at some old posts mentioning tautological spaces. By tautological space do you mean a space of equations which hold true modulo some particular integer? So to use tautological spaces is to solve an equation modulo a prime n for some sufficiently large set of choices for n, and then lift the mod n solutions to a solution in the integers? Let me know if I understand this correctly-- if this is really what's going on then I definitely have a problem for you to look at. === Subject: Re: JSH: State of the art of mathematics > I've now successfully used tautological spaces against only two > problems. Where the latest which lead me to an alternate solution for > binary quadratic Diophantine equations is rather spectacular for quite > a few reasons because of the history around the problem. You will not have any trouble picking up women if you simply explain to them how your tautological space works. They will find it irresistible. === Subject: Re: What numbers do I enter? > So I'm using this new environmental database program > to graph some soil boring analyses. After I set my > query criteria, I hit the button to lauch the Map > module to get the following dialog box. Map > Ranges of values will be displayed as colored circles. > Enter values for the cutoffs between the colors. Min Value: 28 > Max Value: 28 So I try to run the map and I get an error: Error! > Number must be smaller than the Max Value. Hmm...I seem to have a conundrum. What number would be > both larger and smaller than 28? > 20 <= 20.000 <= 20 of course. > Nice to see that the slackers who flunked high school math > can still be productive members of society by getting jobs > as programmers. > Why bother with color graphs? Use bar graphs and be done with it. Quit wasting your time with art work and get on with a scientific report instead of a slick popular publication for slackers. While you're at it, return the software for being defective. === Subject: Re: What numbers do I enter? > So I'm using this new environmental database program > to graph some soil boring analyses. After I set my > query criteria, I hit the button to lauch the Map > module to get the following dialog box. Map > Ranges of values will be displayed as colored circles. > Enter values for the cutoffs between the colors. Selected Parameter: Arsenic (As) (Max for Station) Min Value: 28 > Green > ----------[ 50] > Blue > ----------[100] > Yellow > ----------[200] > Magenta > ----------[400] > Red > Max Value: 28 (The cutoff values shown were from the previous query). So I try to run the map and I get an error: Error! > Number must be smaller than the Max Value. and highlights the 400. Ok, so I change it to 27 only > to get a new error: Error! > Number must be larger than value above. Since I can't make it any larger, I make the number > above 26. This, of course, gives an error saying the > 26 must be larger than the value above. Ok, I'll > make them all [ 24] > [ 25] > [ 26] > [ 27] only to get yet another error: Error! > Number must be larger than the Min Value. with the 24 highlighted. Hmm...I seem to have a conundrum. What number would be > both larger and smaller than 28? My guess is that you're being asked to lower the Min Value so that all of your numbers fit between your Max and your Min. [alt.math deleted since my newsserver won't go there] -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: What numbers do I enter? posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) > On Oct 22, 5:34pm, Gerry Myerson to graph some soil boring analyses. After I set my > query criteria, I hit the button to lauch the Map > module to get the following dialog box. Map > Ranges of values will be displayed as colored circles. > Enter values for the cutoffs between the colors. Selected Parameter: Arsenic (As) (Max for Station) Min Value: 28 > Green > ----------[ 50] > Blue > ----------[100] > Yellow > ----------[200] > Magenta > ----------[400] > Red > Max Value: 28 (The cutoff values shown were from the previous query). So I try to run the map and I get an error: Error! > Number must be smaller than the Max Value. and highlights the 400. Ok, so I change it to 27 only > to get a new error: Error! > Number must be larger than value above. Since I can't make it any larger, I make the number > above 26. This, of course, gives an error saying the > 26 must be larger than the value above. Ok, I'll > make them all [ 24] > [ 25] > [ 26] > [ 27] only to get yet another error: Error! > Number must be larger than the Min Value. with the 24 highlighted. Hmm...I seem to have a conundrum. What number would be > both larger and smaller than 28? My guess is that you're being asked to lower the Min Value > so that all of your numbers fit between your Max and your Min. You can't set Min & Max, they are the range of values returned > by the query. There is, of course, only one value being returned > which is why both Min and Max are the same number. But it is > entirely legitimate that multiple records could all have the same > value (say, all were the Reporting Limit baecause all were > undetected), so they can't use that as an excuse. To be fair, I've made plenty of mistakes in my lifetime. But I never sold them to anyone. [alt.math deleted since my newsserver won't go there] -- > Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - Perhaps they want you to purchase a maintenance contract. Did the original contract stipulate that the product had to work? ;-) This is the sad state of getting quality work today. === Subject: K isomorphism C(t) C[X, Y, Z], I = {Z*X - Y^2, Z*Y - X^6, Z^2 - Y*X^5} (1)proof ; I is Prime ideal. (2)K =quotient field of C[X, Y, Z]/I proof ; K isomorphism C(t) === Subject: Re: Number with 0!=1. >> ... >> Try this one for example: >> n! = n * (n-1)! for n>1 >> n! = 127.24 if n=0 >> ... >> n! = 1 if n=1 >> But that breaks the recursion, since one of the basic properties of the >> factorial function is that n! = n * (n-1)! for all n > 0. > It is about a new definition of n!, see the recursion rule in this new >> definition. It shows that the new definition is arbitrary in a way that the original > rule is not. > I am not saying my factorial function is better or worse than the standard definition; I am merely pointing out that we can define factorial and ! however we like, specifically including making 0! = 127.24 This isn't maths; this is a discussion about the meaning of the English word factorial and ASCII symbol !. Those are the work of man. === Subject: Re: Number with 0!=1. <48fe9218$0$31806$afc38c87@news.optusnet.com.au> <49002b21$0$31800$afc38c87@news.optusnet.com.au... > Try this one for example: > n! = n * (n-1)! for n>1 > n! = 127.24 if n=0 > ... > n! = 1 if n=1 But that breaks the recursion, since one of the basic properties of the > factorial function is that n! = n * (n-1)! for all n > 0. >It is about a new definition of n!, see the recursion rule in this new > definition. >> It shows that the new definition is arbitrary in a way that the original >> rule is not. >I am not saying my factorial function is better or worse than the standard > definition; I am merely pointing out that we can define factorial and ! > however we like, specifically including making 0! = 127.24 We can also define the sin function however we like, specifically including making sin(0) = 127.24. > This isn't maths; this is a discussion about the meaning of the English word > factorial and ASCII symbol !. Those are the work of man. The same is true of sin. However, symbols have definitions, and according to standard definitions and conventions, 0! = 1. For one thing, it's an empty product, and the empty product in a monoid yields the identity element. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. === Subject: Re: Number with 0!=1. posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) >> It is about a new definition of n!, see the recursion rule in this new >> definition. It shows that the new definition is arbitrary in a way that the original > rule is not. The factorial function also generalizes to the Gamma function rather nicely, which is problematic for the new function. === Subject: Re: Number with 0!=1. <33203485.1224492326339.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=b4rQkQoAAADqqftYxNCieauSttMs6_zU 1.1.4322; .NET CLR 2.0.50727; InfoPath.1; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) On Oct 20, 4:44am, riderofgiraffes That way when n < 1, the sum is a sum of >> no numbers, is zero, the additive identity. That's arbitrary. Although the way the poster explained it is not > as clear as it might be, it is emphatically *not* > arbitrary that the empty sum is 0, the empty > product is 1, the empty conjuction is True, the > empty disjunction is False, and so on. Each of > these limiting/extremal cases follows from a > pattern. Further, in effect they all follow from > the same pattern. Saying that it's arbitrary indicates a lack of > perception of that pattern. Well said. It is interesting to see how often in mathematics a situation arises where you have something which is formally true just by definition but where the definition is so nonarbitrary that it is virtually a theorem. Usually a theorem is lurking in the background in that you can prove that the definition in question is the only one which satisfies certain natural constraints. I wonder if any logician has studied this sort of phenomenon. === Subject: field extensions and schemes posting-account=NRTuygoAAADRX_yQjy_PckUFiZZ72o1z Gecko/2008092510 Ubuntu/8.04 (hardy) Firefox/3.0.3,gzip(gfe),gzip(gfe) Hi let X and Y be schemes and f from X to Y a morphism, then there is a field extension k(X):k(Y) of the residue fields, right? I am not able to determine the degree of such an extension, even if the schemes are affine. Suppose X is speck[x,y]/(y-x^2), Y is k[x] and f the projection. How can I calculate the degree? Yours Alex === Subject: Re: field extensions and schemes posting-account=XH7n1goAAAC_c_M3JtSDv3QFzlbwvvMO .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Hi let X and Y be schemes and f from X to Y a morphism, then there is a > field extension k(X):k(Y) of the residue fields, right? I am not able > to determine the degree of such an extension, even if the schemes are > affine. Suppose X is speck[x,y]/(y-x^2), Y is k[x] and f the > projection. How can I calculate the degree? Yours > Alex It only makes sense to speak of the extension of residue fields when you choose some particular geometric point x in X and look at the residues at x as a field extension of the residues at f(x). If you want a field extension coming from the entire schemes X and Y themselves, look at what happens at the generic points--for X and Y affine, you will be looking at the fraction field of the global sections of X as an extension of the fraction field of the global sections of Y. For the example you gave, though, this particular field extension has degree 1, since the map at the generic points is the identity map on k(x). === Subject: Re: solve > Please let me know. ArcSin[a - x] - ArcSin[b - x] == ArcSin[c] > x=__________________________________. > c = sin(arcsin(a - x) - arcsin(b - x)) = sin arcsin(a - x) * cos arcsin(b - x) - cos arcsin(a - x) * sin arcsin(b - x) = (a - x).sqr(1 - (b - x)^2) - (sqr(1 - (a - x)^2))(b - x) c^2 = (a - x)^2 (1 - (b - x)^2) + (b - x)^2 (1 - (a - x)^2) - 2(a - x)(b - x).sqr((1 - (b - x)^2)(1 - (a - x)^2) = (a - x)^2 + (b - x)^2 - 2(a - x)^2 (b - x)^2 - 2(a - x)(b - x).sqr(1 - (a - x)^2 - (b - x)^2 + (a - x)^2 (b - x)^2) = k - 2(a - x)(b - x).sqr(1 - k), where k = the long expression. 4(a - x)^2 (b - x)^2 (1 - k) = k^2 - 2kc^2 + c^4 Simplify and solve for x. === Subject: Commutative composition of functions posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) I 've got a process to build commutative functions from not commutative ones. aj o bi <> bi o aj and ak o bp <> bp o ak But (aj o bi ) o (ak o bp ) = (ak o bp ) o (aj o bi ) COMMUTE Functions a[](x) belong to A, functions b[](x) belong to B . Example :(a4 o b5 ) o (a3 o b2 ) =(a3 o b2 ) o (a4 o b5 ) (1) a3(x) = (15*x +7)/(1 +2*x)/8 a4(x) = (31*x +15)/(1 +2*x)/16 b5(x) = -(x +31)/(x +15)/2 b2(x) = -(x +3)/(x +1)/2 Partial compounds : a4(b5(x)) = (x +511)/512 a3(b2(x)) = (x +3)/ 32 Final result, verifies (1) (x + 16383)/16384 REMARK :we also have (a4 o b5 )o(a3 o b2 ) = (a4 ob2)o(a3 ob5) NOTA : we thus might build a functional equation like : F(aj o bi ) = F(ak o bp ) = F(ap o bq) = .... = F Functions a and b being independant, Alain === Subject: Re: Commutative composition of functions I 've got a process to build commutative functions from > not commutative ones. > aj o bi <> bi o aj > and ak o bp <> bp o ak > But (aj o bi ) o (ak o bp ) = (ak o bp ) o (aj o bi ) COMMUTE > Functions a[](x) belong to A, functions b[](x) belong to B . Example :(a4 o b5 ) o (a3 o b2 ) =(a3 o b2 ) o (a4 o b5 ) (1) > a3(x) = (15*x +7)/(1 +2*x)/8 > a4(x) = (31*x +15)/(1 +2*x)/16 > b5(x) = -(x +31)/(x +15)/2 > b2(x) = -(x +3)/(x +1)/2 Partial compounds : a4(b5(x)) = (x +511)/512 > a3(b2(x)) = (x +3)/ 32 > Final result, verifies (1) (x + 16383)/16384 Note that these Moebius transformations correspond to nonsingular matrices: f(x) = (a*x + b)/(c*x + d) corresponds to the matrix [ a b ] [ c d ] or really to the pencil of matrices [ ta tb ] [ tc td ], t <> 0. Thus the Moebius transformations corresponding to matrices A and B commute if and only if AB is a scalar multiple of BA (which for 2 x 2 matrices implies that A and B either commute or anticommute). -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Goldstein Solution Manual. posting-account=cdDUIAoAAADxOr92q7CG3WqOVCgNfLjg Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I am Doing PhD at IIT Kanpur. I really need the classsical > mechanics (Goldstein) Solution manual. Please help me. My email hello i am cynthia i am from mexico and i need the same book do you > have? if is really can you send me thmak you hi iam susobhan paul doing M.Sc @ university college of science & technology,calcutta university.i need the classical mechanics solution manual(goldstein) === Subject: 2nd order differential equation equates to constant, not F(x) posting-account=q_4miwoAAAAf9jjFCVIqQL4AUcZad_nf Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) In general, is it feasible to find solutions to 2.nd order differential equations that equate to a constant? Specifically, I have this equation in mind: y'' - a*y/ = -b For a the specific problem and a given set of initial conditions, the claim is that this equation is a valid solution to the above: y = (c1-c2)*exp( -x/m) + c2 I've tried to substitute this solution back into the original equation and then solving for m, but the result is rather ugly and does not seem to fit the bill. I was hoping that juggling around with algebra would somehow reduce this mess into the original equation, but so far my attempts have been futile. Also, for my specific application, there exist a relationship between the constants which is: a = b / c2 m should look like. This assumption is m ~ c2/sqtr( b) m ~ c1/sqrt(b) or a linear combination of these. So my questions are: 1) Is there a straightforward way to solve the original equation? 2) Do you see some trick about my specific problem that I am missing? === Subject: Re: 2nd order differential equation equates to constant, not F(x) > In general, is it feasible to find solutions to 2.nd order > differential equations that equate to a constant? Specifically, I have > this equation in mind: y'' - a*y/ = -b For a the specific problem and a given set of initial conditions, the > claim is that this equation is a valid solution to the above: y = (c1-c2)*exp( -x/m) + c2 I've tried to substitute this solution back into the original equation > and then solving for m, but the result is rather ugly and does not > seem to fit the bill. I was hoping that juggling around with algebra > would somehow reduce this mess into the original equation, but so far > my attempts have been futile. Also, for my specific application, there exist a relationship between > the constants which is: > a = b / c2 m should look like. This assumption is > m ~ c2/sqtr( b) > m ~ c1/sqrt(b) > or a linear combination of these. So my questions are: > 1) Is there a straightforward way to solve the original equation? > 2) Do you see some trick about my specific problem that I am > missing? Start with a simpler form for the trial solution: Let y = A*exp(B*x) + C, A,B,C are unknown constants. Then: y' = A*B*exp(B*x) y'' = A*B^2*exp(B*x) Plug y and y'' back into original equation and stir well: y'' - a*y + b = 0 A*B^2*exp(B*x) - a*(A*exp(B*x) + C) + b = 0 (A*B^2 - a*A)*exp(B*x) - a*C + b = 0 A*(B^2 - a)*exp(B*x) + (b - a*C) = 0 We can see immediately that for (b - a*C) to be zero that C = b/a. We can also see that either A = 0 or (B^2 - a) = 0. A = 0 is a trivial case that reduces the test solution to y = C, which is uninteresting. So we look at the other case: Then we have: y = A*exp(+/- sqrt(a)*x) + b/a Where only A remains to be determined (need an initial condition). You can determine your c1,c2,c3 by equating coefficients. === Subject: Re: 2nd order differential equation equates to constant, not F(x) posting-account=eFTX7goAAACJBRoHHOnDC9IuTZeZT1_H 1.1.4322),gzip(gfe),gzip(gfe) > In general, is it feasible to find solutions to 2.nd order > differential equations that equate to a constant? Specifically, I have > this equation in mind: y'' - a*y/ = -b For a the specific problem and a given set of initial conditions, the > claim is that this equation is a valid solution to the above: y = (c1-c2)*exp( -x/m) + c2 I've tried to substitute this solution back into the original equation > and then solving for m, but the result is rather ugly and does not > seem to fit the bill. I was hoping that juggling around with algebra > would somehow reduce this mess into the original equation, but so far > my attempts have been futile. Also, for my specific application, there exist a relationship between > the constants which is: > a = b / c2 m should look like. This assumption is > m ~ c2/sqtr( b) > m ~ c1/sqrt(b) > or a linear combination of these. So my questions are: > 1) Is there a straightforward way to solve the original equation? > 2) Do you see some trick about my specific problem that I am > missing? Or rather w = ay - b === Subject: Re: 2nd order differential equation equates to constant, not F(x) posting-account=eFTX7goAAACJBRoHHOnDC9IuTZeZT1_H 1.1.4322),gzip(gfe),gzip(gfe) > In general, is it feasible to find solutions to 2.nd order > differential equations that equate to a constant? Specifically, I have > this equation in mind: y'' - a*y/ = -b For a the specific problem and a given set of initial conditions, the > claim is that this equation is a valid solution to the above: y = (c1-c2)*exp( -x/m) + c2 I've tried to substitute this solution back into the original equation > and then solving for m, but the result is rather ugly and does not > seem to fit the bill. I was hoping that juggling around with algebra > would somehow reduce this mess into the original equation, but so far > my attempts have been futile. Also, for my specific application, there exist a relationship between > the constants which is: > a = b / c2 m should look like. This assumption is > m ~ c2/sqtr( b) > m ~ c1/sqrt(b) > or a linear combination of these. So my questions are: > 1) Is there a straightforward way to solve the original equation? > 2) Do you see some trick about my specific problem that I am > missing? Try letting w = ay + b === Subject: Re: 2nd order differential equation equates to constant, not F(x) > In general, is it feasible to find solutions to 2.nd order > differential equations that equate to a constant? Specifically, I have > this equation in mind: y'' - a*y/ = -b > Do I ignore the / or is there a missing divisor? > For a the specific problem and a given set of initial conditions, the > claim is that this equation is a valid solution to the above: y = (c1-c2)*exp( -x/m) + c2 I've tried to substitute this solution back into the original equation > and then solving for m, but the result is rather ugly and does not > seem to fit the bill. I was hoping that juggling around with algebra > would somehow reduce this mess into the original equation, but so far > my attempts have been futile. Also, for my specific application, there exist a relationship between > the constants which is: > a = b / c2 m should look like. This assumption is > m ~ c2/sqtr( b) > m ~ c1/sqrt(b) > or a linear combination of these. So my questions are: > 1) Is there a straightforward way to solve the original equation? > 2) Do you see some trick about my specific problem that I am > missing? > === Subject: Re: 2nd order differential equation equates to constant, not F(x) posting-account=q_4miwoAAAAf9jjFCVIqQL4AUcZad_nf Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) In general, is it feasible to find solutions to 2.nd order > differential equations that equate to a constant? Specifically, I have > this equation in mind: y'' - a*y/ = -b Do I ignore the / or is there a missing divisor? Yes, ignore the /, that's a typo. The correct equation is y'' - a*y = -b === Subject: Re: 2nd order differential equation equates to constant, not F(x) posting-account=q_4miwoAAAAf9jjFCVIqQL4AUcZad_nf Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) In general, is it feasible to find solutions to 2.nd order > differential equations that equate to a constant? Specifically, I have > this equation in mind: y'' - a*y/ = -b Do I ignore the / or is there a missing divisor? For a the specific problem and a given set of initial conditions, the > claim is that this equation is a valid solution to the above: y = (c1-c2)*exp( -x/m) + c2 I've tried to substitute this solution back into the original equation > and then solving for m, but the result is rather ugly and does not > seem to fit the bill. I was hoping that juggling around with algebra > would somehow reduce this mess into the original equation, but so far > my attempts have been futile. Also, for my specific application, there exist a relationship between > the constants which is: > a = b / c2 m should look like. This assumption is > m ~ c2/sqtr( b) > m ~ c1/sqrt(b) > or a linear combination of these. So my questions are: > 1) Is there a straightforward way to solve the original equation? > 2) Do you see some trick about my specific problem that I am > missing? Sorry, the / is a typo. The correct differential equation is: y'' - a*y = -b === Subject: Re: Making a List from a set of construced elements posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) ... the list is required that exists at the time presenting it. > Another list or a later one is out of interest. Clearly I did not understand you. > I will try to define the question exactly. > Is this form acceptable? Let H be the set of numbers constructed between > midnight and t 2. This set exists and is finite. Let L be a list containing all the elements of H. This list may exist, but it is not that one required. (Compare: A Cantor list exists, but it is not that one required, .e., containing all real numbers.) Let t 2 be the time at which L is constructed. Does L exist. Any list filled with real numbers may exist. What does not exis is a Cantor list containing all real numbers and a Hughes list containing all antidiagonals constructed until the list is presented. By the way you can avaoid the timing. Consider a list that contains all anti-diagonals ever constructed (by you or by mankind or by whom ever). === Subject: Re: Making a List from a set of construced elements posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > ... the list is required that exists at the time presenting it. > Another list or a later one is out of interest. Clearly I did not understand you. > I will try to define the question exactly. > Is this form acceptable? We cannot have a discussion until we agree on the question. Proposed question Let H be the set of numbers constructed between midnight and t_2. Let L be a list containing all the elements of H. Let t_2 be the time at which L is constructed. Does L exist? Is this question acceptable? Yes or No. - William Hughes === Subject: Re: Making a List from a set of construced elements posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > We cannot have a discussion until we agree on the question. > Proposed question Can you construct a list that contains all anti-diagonals ever constructed by a certain replacement rule? === Subject: Re: Making a List from a set of construced elements posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I We cannot have a discussion until we agree on the question. > Proposed question Can you construct a list that contains all anti-diagonals ever > constructed by a certain replacement rule? > Rejected. all anti-diagonals ever constructed is something that can change. We need to talk about something that does not change To do this explicitly mention the time period during which the anti-diagonals were constructed. Proposed Question Let H be the set of numbers constructed between midnight and t_2. Let L be a list containing all the elements of H. Let t_2 be the time at which L is constructed. Can you construct L? - William Hughes === Subject: Re: Arguing against negative quantities by themselves as being real posting-account=n26igQkAAACeF9xA2Ms8cKIdBH40qzwr Gecko/20070505 Iceape/1.0.9 (Debian-1.0.13~pre080323b-0etch3),gzip(gfe),gzip(gfe) > Or we could just remember that numbers aren't real in the first > place... The idea of assigning a number to a group of objects is an > abstraction,. What about 0? What about fractions? Decimals (which are > really just fractions)? Irrationals? Because we can find a > measurable object with those lengths, they're ok? Does 0 fit that > bill? Up and down is a relative and never negative. But if we call up positive, then down must be negative, right? > It doesn't matter anyway, because the concept of number is relative to > what unit of measure we're using. Also our tradition is to build the integers then build the continuum from the integers. This is a false mathematical construction. We never witness such discrete members of the continuum. So three oranges versus three feet ought not to use the same name of three. I've often tried to consider a geometry with an origin labelled unity, zero being far away. Hence all distances will be from unity to zero, which cannot be confused with the discrete numbers. This could normalize the universe too. Since modern physics accepts a fixed value of some fifteen billion light years this simply gets scaled and reciprocaled. As the universe ages this value remains fixed. Then the classical force equations can read F = m1 m2 r r rather than messing with the inverse square. For charge equations we'd be entering the sign domain and since charge does model discretely it can take a sign and a distance. So the classical equation could be whittled down to F = q1 q2 where these q are discrete charges whose magnitude is their distance in the new domain and whose sign is their charge. This would be a pretty simplification. There is also a fixup for the old division by zero since all math is in product form. There is even the suggestion of a Planck length since the transform is Y = 1 / ( 1 + X ) where X is traditional distance and Y is the transformed value. This unity addition is Planck-like. Unity distance is local. Things far away are meaningless; zero. Things local are important; not zero. So the origin as zero ought to be brought into question. No infinite force will exist in this domain locally. These maxima are still puzzling but they are more workable. - Tim === Subject: Re: Arguing against negative quantities by themselves as being real posting-account=orrqAQoAAACc_RA9N1zrNaxvUyLr_2M_ rv:1.9.0.3) Gecko/2008092414 Firefox/3.0.3,gzip(gfe),gzip(gfe) On Oct 23, 7:47am, Timothy Golden BandTechnology.com Or we could just remember that numbers aren't real in the first > place... The idea of assigning a number to a group of objects is an > abstraction,. What about 0? What about fractions? Decimals (which are > really just fractions)? Irrationals? Because we can find a > measurable object with those lengths, they're ok? Does 0 fit that > bill? Up and down is a relative and never negative. But if we call up positive, then down must be negative, right? > It doesn't matter anyway, because the concept of number is relative to > what unit of measure we're using. Also our tradition is to build the integers then build the continuum > from the integers. > This is a false mathematical construction. We never witness such > discrete members of the continuum. > So three oranges versus three feet ought not to use the same name of > three. I've often tried to consider a geometry with an origin labelled unity, > zero being far away. > Hence all distances will be from unity to zero, which cannot be > confused with the discrete numbers. > This could normalize the universe too. Since modern physics accepts a > fixed value of some fifteen billion light years this simply gets > scaled and reciprocaled. As the universe ages this value remains > fixed. Then the classical force equations can read > F = m1 m2 r r > rather than messing with the inverse square. For charge equations we'd > be entering the sign domain and since charge does model discretely it > can take a sign and a distance. So the classical equation could be > whittled down to > F = q1 q2 > where these q are discrete charges whose magnitude is their distance > in the new domain and whose sign is their charge. This would be a > pretty simplification. There is also a fixup for the old division by > zero since all math is in product form. There is even the suggestion > of a Planck length since the transform is > Y = 1 / ( 1 + X ) > where X is traditional distance and Y is the transformed value. This > unity addition is Planck-like. > Unity distance is local. Things far away are meaningless; zero. Things > local are important; not zero. So the origin as zero ought to be > brought into question. No infinite force will exist in this domain > locally. These maxima are still puzzling but they are more workable. - Tim What do you mean false mathematical construction? Start with {} and build numbers... there are only so many ways to make everything from nothing... Anyway, this is a fun notion, but local measurements are much easier to make than universal ones... Also, you would have rationals budging their way in there... rationals which have a fun habit of confusing themselves with integers which we understand so much better. Cory My name has 1548/172 letters Knapp === Subject: Re: Arguing against negative quantities by themselves as being real > On Oct 23, 7:47 am, Timothy Golden BandTechnology.com Or we could just remember that numbers aren't real in the first >place... The idea of assigning a number to a group of objects is an >abstraction,. What about 0? What about fractions? Decimals (which are >really just fractions)? Irrationals? Because we can find a >measurable object with those lengths, they're ok? Does 0 fit that >bill? >>Up and down is a relative and never negative. >But if we call up positive, then down must be negative, right? >It doesn't matter anyway, because the concept of number is relative to >what unit of measure we're using. >>Also our tradition is to build the integers then build the continuum >>from the integers. >>This is a false mathematical construction. We never witness such >>discrete members of the continuum. >>So three oranges versus three feet ought not to use the same name of >>three. >>I've often tried to consider a geometry with an origin labelled unity, >>zero being far away. >>Hence all distances will be from unity to zero, which cannot be >>confused with the discrete numbers. >>This could normalize the universe too. Since modern physics accepts a >>fixed value of some fifteen billion light years this simply gets >>scaled and reciprocaled. As the universe ages this value remains >>fixed. Then the classical force equations can read >> F = m1 m2 r r >>rather than messing with the inverse square. For charge equations we'd >>be entering the sign domain and since charge does model discretely it >>can take a sign and a distance. So the classical equation could be >>whittled down to >> F = q1 q2 >>where these q are discrete charges whose magnitude is their distance >>in the new domain and whose sign is their charge. This would be a >>pretty simplification. There is also a fixup for the old division by >>zero since all math is in product form. There is even the suggestion >>of a Planck length since the transform is >> Y = 1 / ( 1 + X ) >>where X is traditional distance and Y is the transformed value. This >>unity addition is Planck-like. >>Unity distance is local. Things far away are meaningless; zero. Things >>local are important; not zero. So the origin as zero ought to be >>brought into question. No infinite force will exist in this domain >>locally. These maxima are still puzzling but they are more workable. >> - Tim > What do you mean false mathematical construction? Start with {} and > build numbers... there are only so many ways to make everything from > nothing... Anyway, this is a fun notion, but local measurements are much easier > to make than universal ones... OTOH universal equations will generate the local numbers you might seek. > Also, you would have rationals budging their way in there... rationals > which have a fun habit of confusing themselves with integers which we > understand so much better. Cory My name has 1548/172 letters Knapp === Subject: Re: Arguing against negative quantities by themselves as being real posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) >> Assuming that your income dropped to zero but you still >> owed money to bill collectors, would you say your net worth >> was a positive value? Of course. My self-worth would be unaffected by my economic > misfortunres. What a silly question! What a silly answer. You may not believe in negative numbers, but accountants and bankers surely do. === Subject: Re: Arguing against negative quantities by themselves as being real >> Assuming that your income dropped to zero but you still >> owed money to bill collectors, would you say your net worth >> was a positive value? > Of course. My self-worth would be unaffected by my economic > misfortunres. What a silly question! What a silly answer. You may not believe in negative numbers, but accountants > and bankers surely do. I owe no one in the negative. === Subject: Re: The Reality of Numbers (was: Arguing against negative quantities by themselves as being real) > You enter an elevator, see a board of buttons from Basement to Floor > 29. Will the circuit board get you up to Floor 29. A good test would > be to unplug it and take it with you. Now, whenever you need to get to > Floor 29, you press button 29! See? Obvious! Numbers don't exist or appear by themselves. They are part of a set of agreements between people that is used as a tool to reach our goals. It might be possible that we could achieve the same results without numbers at all. -- Alex. === Subject: Re: The Reality of Numbers (was: Arguing against negative quantities by themselves as being real) <48faba1e$0$14063$9b4e6d93@newsspool3.arcor-online.net> posting-account=orrqAQoAAACc_RA9N1zrNaxvUyLr_2M_ rv:1.9.0.3) Gecko/2008092414 Firefox/3.0.3,gzip(gfe),gzip(gfe) > They didn't have an LHC, though. And neither do we, at the moment. ;) Cory Always laughing at physicists Knapp. === Subject: Re: Arguing against negative quantities by themselves as being real posting-account=Z3AipgkAAABkoMfyNwddSxsYhXHi5CDt CLR 1.1.4322; InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > On 22 Oct, 06:32, Achava Nakhash, the Loving Snake On 22 Oct, 03:14, Achava Nakhash, the Loving Snake Negative quantities don't exist by themselves. They exist as > subtraction of their absolute value from some equal or greater > positive value. > They are a negative operator attached to a postive quantity. Their > only reality is in the operation of subtraction. > Mitch Raemsch > Of course they don't exist. I am sure you have gotten a lot of > replies saying that they do and that you are an idiot. They are > trying to pull your leg. Nobody believes in negative numbers. If > weight could be negative we would all fly. That would be nice, but I > haven't seen it happening. I always tell my classes that if they get > a negative answer to a problem involving physical quantities they are > wrong and should check their work. > I'll remember that the next time I'm calculating acceleration when I hit > my brakes. Many people owe their lives to the fact that my truck > (occasionally) experiences negative acceleration. > I used to be a mathematician, but I had to quit because the anxiety > over losing my income and social standing when we were all found was > ruining my health. I am glad to have found an ally. > Let me guess. You work on Wall Street now. ;-) > -- > Paul Hovnanian mailto:P...@Hovnanian.com > ------------------------------------------------------------------ > Hey, Paul. No I don't work on Wall Street now. But if I did, I could > probablyl accuractely say, Before I worked on Wall Street, I was > broke, but now I am broker. > Achava. > That will be $2, please. http://clarkevans.com/writings/pacioli.html Numbers and the world! THINK! Double enrty is not only for burglars! > -- > foolsrushin. > ---------------------------------------------------------------------------[ CapitalEth][CapitalEth][CapitalEth]- Minister: > How did all this happen? Senior Secretary: > Dodgy, worthless products ... . Minister: > Why did anybody buy them, then? Senior Secretary: > Commission, Minister, commision ..., and no controls. Minister: > And I suppose I and my colleagues take the blame? Senior Secretary: > I fear so, and there is something I must discuss with you ... . Minister: > An offer? What? Senior Secretary: > Well, I have had an offer from the City. I know this is not an easy > time for the Government, yet ... . > -- > foolsrushin. This is a joke, right? >http://clarkevans.com/writings/pacioli.html Do you check you bank statements, mate? You can't take 12 from 10. Put out 10 pennies > and ask anyone, even a fairly young child, to take 12 of them. They > will just look at you. You will find it excructiatingly embarrasing. Achava. You are a fine burly young fellow and needed in your village and in > alt.knitting. Go there at once: they are missing you. Eating sheeps' > heads was never a good idea! Nothing to be done about it: one more > demand on the public purse - when that habit of your tribe hits! A bizarre flow of imagination unconnected with much of anything. > Sometimes even Alan Truelove makes sense: we never really fell out, > anyway! We just stopped talking to each other! I would never doubt. I tried googling Alan Truelove. He appears to be a middle-aged fan of Bill O'Reilly who lives in or near Northern Virginia. Is there some reason we should care about him? > You are obviously not an accountant. How do you think the present mess > arose? Prat! Deficit=deceit. Idiot! I am not now nor have I even been an accountant. Is that a bad thing? Is that a good thing? I have heard a great many things about how the present mess (I assume you mean the economic problems in the United States. You never say. This is an international forum.) variety of reasons. I am also not an economist. You now know more than most about what I am not. > You probably think that because you have a stoopid foreigin name you > will gain sympathy. Not from me, mate. Learn to stand on your own > feet! Your imagination is flowing freely again. You actually know effectively nothing about me. However, your comment about my stoopid foreigin (sic) name is beyond the limit of civilized discourse and into the pathetic and disgusting area of irrational prejudice You should be ashamed of yourself. I am certainly ashamed of you. Achava === Subject: Re: Arguing against negative quantities by themselves as being real > Your imagination is flowing freely again. You actually know > effectively nothing about me. However, your comment about my stoopid > foreigin (sic) name is beyond the limit of civilized discourse and > into the pathetic and disgusting area of irrational prejudice You > should be ashamed of yourself. I am certainly ashamed of you. > Achava Exactly! Irony, Achava, irony! We know nothing about each other! Perhaps I was too subtle or too unsubtle. The name Achva means 'camaraderie'. Usenet is so unintimate: all you have is a stream of text: you could be dealing with anyone, and any agenda! Who or what is talking to who or what? Now, consider the following from a certain David Greene. Greene has decided that I am bogus, fictive, a sham. He has posted to wreck.org.mensa for ever, tho I have never ffound out what he does or who he is! He has no obvious way of deciding which I am, don't you see! And here goes: >Scum: how do you add 2 and two? Joke, twit! Fool. Slimebag, snotrocket, retard: it is pretty easy to figure it out. >> Anyway, on a lighter note: I still have my original copy >> of The ISMM :^) It has been worth its weight in gold, >> especially the DOT technique at my place of employment. >> Dave Greene. >DOT technique? Wot's that, mate? Go buy the book. >4-warmed is forewarned, Green. Thick is not the word! You accused me >of having no students. Anonymous posers such as yourself make all kinds of unsubstantiated claims. [ Note: we know everything about him!] > Still your stoopid contention? Going > through your history, creepy stuff: what do you mean by 'the potus', creep? What is creepy is you searching through the archives. If you could have done even a half-assed job you would have known from the context what the POTUS meant. http://en.wikipedia.org/wiki/POTUS >http://en.wikipedia.org/wiki/Toady http://en.wikipedia.org/wiki/Martinet >Green, the NSA and the CIA ain't going to lock me up. >Why? I'm not interested in your psychotic phantasies. Dave Greene foolsrushin View profile More options 22 Oct, 00:16 rec.org.mensa Local: Wed 22 Oct 2008 00:16 === Subject: Re: Faraday was a Charlatan and a Fraud Reply | Reply to author | Forward | Print | Individual message | Show original | Report this message | Find messages by this author [Various snips.] > Slimebag, snotrocket, retard: it is pretty easy to figure it out. >Green, the NSA and the CIA ain't going to lock me up. >Why? > I'm not interested in your psychotic phantasies. A pity: the DVD is now only $5. > Dave Greene. Greene is the color of your kind - or green is the colour of your kind. All your posts show a tendency to flatter establishment values and to be accepted by the Establishment. One thing in your favor is that, when not being silly, you make an attempt to disuss issues. Interesting! Why? All you want, Greene, is to set yourelf up amongst your fellows! The monkeys and typewriters stuff stung a bit, didn't it? -- foolsrushin. --------------------------------- See? Obvious! -- foolsrushin. === Subject: Re: The Reality of Numbers (was: Arguing against negative quantities by themselves as being real) <48faba1e$0$14063$9b4e6d93@newsspool3.arcor-online.net> posting-account=orrqAQoAAACc_RA9N1zrNaxvUyLr_2M_ rv:1.9.0.3) Gecko/2008092414 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Numbers, adjectives, etc. most surely exist, but are not what they > refer to. Were this to be untrue, you could order the universe by > manipulating numbers, for example, Stock Exchange reports, thus > altering the value of shares. Neither Goethe or Newton managed to do > anything of the sort, as far as we know! You enter an elevator, see a board of buttons from Basement to Floor > 29. Will the circuit board get you up to Floor 29. A good test would > be to unplug it and take it with you. Now, whenever you need to get to > Floor 29, you press button 29! See? Obvious! > -- > foolsrushin. My point exactly. There are adjectives but only as abstract descriptors. They don't exist in nature. The 29 on the switchboard is a button (not a number) and it refers to a floor (not a number). A computer scientist explanation is that numbers are not data objects, but references to data objects. === Subject: Re: --- x^2 + y^2 = 1 (mod n) >>Found a bug in previous post. >>We should use all x in range 0<=x<=(p-1)/2. >>And y_0^2,y_1^2,...,y_{(p-1)/2}^2 is permutation >> of >> 0^2,1^2,...,((p-1)/ >>2)^2 (mod p) >> Yes, you need to include 0, since (y_x)^2 = 0 when >> x >> = 1, and to get >> (y_x)^2 = 1, you need x = 0. On Oct 20, 7:12am, quasi Prove or disprove: >> If n in N is such that, for all x in Z, >> there >> exists y in Z, such that >> x^2 + y^2 = 1 (mod n), then n|12. >> quasi We could prove that for prime p>=5, >p| 1^2+2^2+...+((p-1)/2)^2 >Let's assume for x when 1<=x<=(p-1)/2 >x^2+y_x^2=1 (mod p) >then >y_1^2,y_2^2,...,y_{(p-1)/2}^2 (mod p) > is an permutation of 1^2,2^2,...,{(p-1)/2}^2 >> (mod p) >so >p|1^2+y_1^2+2^2+y_2^2+...+((p-1)/2)^2+y_{(p-1)/2}^2=( >> p-1)/2 So we have proved that all prime factors of n >> must be 2 or 3. >Next we only need check that the result is not >> true for n=8 and 9. Lovely proof. quasi >> only 5 posts for such an intresting thread ... >> from 19th till 22th >> is this really sci.math ? >> im a bit confused ... >> quasi are you claiming >> x^2 + y^2 = 1 only for (mod 12n) ? >> x = 0 y = 1 will do for all mod n ?? >> clearly x and y are cos^2 and sin^2 as mentioned in >> my own recent thread. oh ic for all x exists an y ... Right. so we must have mod 12 n The claim is n|12, not 12|n. >but what about 12 n + 18 ? Not applicable. >since prime factors had to be 2 or 3 and results requires mod m ; m >=12. No. >what is the smallest mod m such that for all a there exists a b such that a^2 + b^2 = 1 mod m ?? m = 1 >can we parametrize all possible values of m ? ( probably ) The only positive integer values of m for which the conditions hold are 1, 2, 3, 4, 6, 12. quasi === Subject: Re: --- x^2 + y^2 = 1 (mod n) >>Found a bug in previous post. >>We should use all x in range 0<=x<=(p-1)/2. >>And y_0^2,y_1^2,...,y_{(p-1)/2}^2 is permutation of >> 0^2,1^2,...,((p-1)/ >>2)^2 (mod p) >> Yes, you need to include 0, since (y_x)^2 = 0 when x >> = 1, and to get >> (y_x)^2 = 1, you need x = 0. > Prove or disprove: >> If n in N is such that, for all x in Z, there >> exists y in Z, such that >> x^2 + y^2 = 1 (mod n), then n|12. >> quasi We could prove that for prime p>=5, >p| 1^2+2^2+...+((p-1)/2)^2 >Let's assume for x when 1<=x<=(p-1)/2 >x^2+y_x^2=1 (mod p) >then >y_1^2,y_2^2,...,y_{(p-1)/2}^2 (mod p) > is an permutation of 1^2,2^2,...,{(p-1)/2}^2 >> (mod p) >so >p|1^2+y_1^2+2^2+y_2^2+...+((p-1)/2)^2+y_{(p-1)/2}^2=( >> p-1)/2 So we have proved that all prime factors of n >> must be 2 or 3. >Next we only need check that the result is not >> true for n=8 and 9. Lovely proof. quasi only 5 posts for such an intresting thread ... from 19th till 22th is this really sci.math ? >im a bit confused ... quasi are you claiming x^2 + y^2 = 1 only for (mod 12n) ? No. >x = 0 y = 1 will do for all mod n ?? Sure, but the claim was quantified. >clearly x and y are cos^2 and sin^2 as mentioned in my own recent thread. That's one interpretation. quasi === Subject: limits help posting-account=5cbwMAoAAAAZoYdmWfebkP1kkdh09sZi CLR 1.1.4322; .NET CLR 3.0.04506.30; InfoPath.1; .NET CLR 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) I've been trying to expand the following using series : y = x(1+ e^- x) / (1 - e^-x). I've tried just about everything, from expanding e^-x using taylor expansion, to expanding the denominator in the fraction using the binomial theorem, but nothing seems to work. Can anyone benjamin === Subject: Re: limits help I've been trying to expand the following using series : y = x(1+ e^- > x) / (1 - e^-x). I've tried just about everything, from expanding e^-x > using taylor expansion, to expanding the denominator in the fraction > using the binomial theorem, but nothing seems to work. Can anyone benjamin you might have to try harder! If you really need help, you should tell us more about the circumstances: Is it homework? What are you supposed to do? (e.g. at which point would you asked to perform the series expansion?) And what are you supposed to know already? I would guess, that this is some homework, after you learned Taylor series and did some similar examples in school. Have a look at these examples, try to solve them without help! Another hint: Multiply both the nominator and the denominator with e^(x/2), do you recognize the fraction? Alois -- Alois Steindl, Tel.: +43 (1) 58801 / 32558 Inst. for Mechanics and Mechatronics Fax.: +43 (1) 58801 / 32598 Vienna University of Technology, A-1040 Wiedner Hauptstr. 8-10 === Subject: Re: limits help posting-account=5cbwMAoAAAAZoYdmWfebkP1kkdh09sZi AppleWebKit/523.15.1 (KHTML, like Gecko) Version/3.0.4 Safari/523.15,gzip(gfe),gzip(gfe) I've been trying to expand the following using series : y = x(1+ e^- > x) / (1 - e^-x). I've tried just about everything, from expanding e^-x > using taylor expansion, to expanding the denominator in the fraction > using the binomial theorem, but nothing seems to work. Can anyone benjamin you might have to try harder! > If you really need help, you should tell us more about the > circumstances: Is it homework? What are you supposed to do? (e.g. at which > point would you asked to perform the series expansion?) And what > are you supposed to know already? > I would guess, that this is some homework, after you learned Taylor > series and did some similar examples in school. Have a look at these > examples, try to solve them without help! > Another hint: Multiply both the nominator and the denominator with > e^(x/2), do you recognize the fraction? Alois -- > Alois Steindl, Tel.: +43 (1) 58801 / 32558 > Inst. for Mechanics and Mechatronics Fax.: +43 (1) 58801 / 32598 > Vienna University of Technology, A-1040 Wiedner Hauptstr. 8-10 numerator and denominator by e^(x/2) the function y becomes the function hyperbolic cotangent x multiplied by x. After checking wikipedia on the series expansion of y = coth x and multiplying it by x, i now see why the terms containing the odd powers of x disappear. However, where did you get the idea of multiplying both the top and === Subject: Re: limits help posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I've been trying to expand the following using series : y = x(1+ e^- > x) / (1 - e^-x). I've tried just about everything, from expanding e^-x > using taylor expansion, to expanding the denominator in the fraction > using the binomial theorem, but nothing seems to work. Can anyone benjamin Are you trying to find the limit as x -> infinity? Hint: What happens to the e^(-x) terms as x is increasing? Can you deduce anything from that? === Subject: Re: limits help >> I've been trying to expand the following using series : y = x(1+ e^- >> x) / (1 - e^-x). I've tried just about everything, from expanding e^-x >> using taylor expansion, to expanding the denominator in the fraction >> using the binomial theorem, but nothing seems to work. Can anyone >> benjamin Are you trying to find the limit as x -> infinity? I'd bet he wants the limit as x ->0. If so, I'd note that x/(1-e^{-x}) -> 1 by L'hospital. B. -- Cheerfully resisting change since 1959. === Subject: Re: limits help posting-account=5cbwMAoAAAAZoYdmWfebkP1kkdh09sZi AppleWebKit/523.15.1 (KHTML, like Gecko) Version/3.0.4 Safari/523.15,gzip(gfe),gzip(gfe) What the question asks really is if y be expanded in a series of ascending powers of x, show that the coefficients of all the odd powers of x are zero, and determine the expansion as far as the term === Subject: Re: limits help >What the question asks really is if y be expanded in a series of >ascending powers of x, show that the coefficients of all the odd >powers of x are zero, and determine the expansion as far as the term Hint: consider how x (1+ e^-x)/(1 - e^-x) is related to x coth(x/2) === Subject: Re: limits help > What the question asks really is if y be expanded in a series of > ascending powers of x, show that the coefficients of all the odd > powers of x are zero, and determine the expansion as far as the term Then I think Alois' idea is the right tack. Rewrite as x*coth(x). B. -- Cheerfully resisting change since 1959. === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > For all nodes means all nodes. It does not > mean that you have to pick something and then look. In order to be sure that a finite initial segment of naturals contains a > particular natural, one does have to look. The finite initial segment ending in 3 does not contain 4. > But as there is a finite initial segment for every natural, it is not necessary to pick anything other than that segment that contains every natural and all smaller natuarls. Iff all exist, then no choice is required. You simply take all. Therefore not *all* exist. === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > For every real number you know, at least in principle, all the digits > of the decimal expansion, namely for every path in the binary tree. > Therefore all real numbers are constructed. Conclusion: The real numbers form a countable set. Great! WRONG AGAIN! The set-of-real-numbers-YOU-KNOW may be countable, but that only says > that the set of all reals is AT LEAST countably large, unless you can > prove that you know ALL reals. All elements of a set must be distinct. {a, b, c, c, c, } = {a, b, c}. > Therefore there cannot be any set with uncounably many elements. The naturals are all distinct. Unless you consider physical constraints, yes. But if you do so, then you see that it is quite impossible to pick most of the naturals, because you cannot name them. > Cauchy seqeunces) all distinct. No, one cannot (if there are uncountably many, and if uncountable means 2^aleph 0). There are at most aleph 0 names to call up a real number x in order to compare it with another real number y. How would you do that? You can call up pi or sqrt(2) or Liouville's number no.1 or no.100 (if you have enumerated them). You can call up a number as the limit of a series or sequence or by its digits sequence. But the latter requires a finite sequence, because you do not live forever. That restricts your call to rational numbers. Hence most real numbers, namely those that were not lucky enough to get baptized, cannot be used for your purposes (or for any other purpose). They have an infinite amount of information stored within their body, alas no one can use it. They cannot exist in the Platonic heaven of numbers [CapitalEth] at most they exist in the hell of ill logic. So there remain only aleph 0 numbers x to be compared with y. === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > P *is* the union. Nobody said that it was different from the union. > What people have said is that it is different from each finite path > p_n in the union. The situation is completely described by the two statements: (1) P is not equal to any of the paths p_0, p_1, p_2, etc. > (2) P is equal to the union of all those paths. These two statements are both true. They are wrong for every finite set of linear finite paths. Here is an example with a linear set of 3 paths: 1 11 111 111 There are finite sets of finite nonlinear paths which satisfy your claim: 1 01 101 111 Why do you think that an infinite set of finite linear paths has the properties of a finite set of nonlinear paths? The result of linearity demonstrated above for three paths does not get lost for any number of finite linear paths. Why do you think that your logic is logic? === Subject: Re: Why meta diagonals are irrelevant WM says... >> P *is* the union. Nobody said that it was different from the union. >> What people have said is that it is different from each finite path >> p_n in the union. >> The situation is completely described by the two statements: >> (1) P is not equal to any of the paths p_0, p_1, p_2, etc. >> (2) P is equal to the union of all those paths. >> These two statements are both true. They are wrong for every finite set of linear finite paths. That's true. That's the difference between finite and infinite sets. Let U = p_0, p_1, ... be an increasing sequence of sets. Then we have two cases: Case A: U has a largest element. In that case, U is finite, and union(U) = the largest element of U. Case B: U does not have a largest element. In that case, U is infinite, and union(U) is *not* an element of U. -- Daryl McCullough Ithaca, NY === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > WM says... >> P *is* the union. Nobody said that it was different from the union. >> What people have said is that it is different from each finite path >> p_n in the union. > The situation is completely described by the two statements: > (1) P is not equal to any of the paths p_0, p_1, p_2, etc. >> (2) P is equal to the union of all those paths. > These two statements are both true. They are wrong for every finite set of linear finite paths. That's true. That's the difference between finite and infinite > sets. Let U = p_0, p_1, ... be an increasing sequence of sets. > Then we have two cases: Case A: U has a largest element. In that case, U is finite, > and union(U) = the largest element of U. Case B: U does not have a largest element. In that case, > U is infinite, and union(U) is *not* an element of U. because union(U) does not exist. But that is not the meaning of set theory. === Subject: Re: Why meta diagonals are irrelevant WM says... > That's true. That's the difference between finite and infinite >> sets. Let U = p_0, p_1, ... be an increasing sequence of sets. >> Then we have two cases: >> Case A: U has a largest element. In that case, U is finite, >> and union(U) = the largest element of U. >> Case B: U does not have a largest element. In that case, >> U is infinite, and union(U) is *not* an element of U. because union(U) does not exist. Yes, if you want to adopt the axiom there are no infinite sets, then it follows that Case B never arises. -- Daryl McCullough Ithaca, NY === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > WM says... > That's true. That's the difference between finite and infinite >> sets. Let U = p_0, p_1, ... be an increasing sequence of sets. >> Then we have two cases: > Case A: U has a largest element. In that case, U is finite, >> and union(U) = the largest element of U. > Case B: U does not have a largest element. In that case, >> U is infinite, and union(U) is *not* an element of U. because union(U) does not exist. Yes, if you want to adopt the axiom there are no infinite sets, > then it follows that Case B never arises. Then it follows that you invent a contradiction. It is clear for linear sets that case B does never arise and cannot be avided to arise. === Subject: Re: Why meta diagonals are irrelevant WM says... > WM says... >That's true. That's the difference between finite and infinite > sets. Let U = p_0, p_1, ... be an increasing sequence of sets. > Then we have two cases: >Case A: U has a largest element. In that case, U is finite, > and union(U) = the largest element of U. >Case B: U does not have a largest element. In that case, > U is infinite, and union(U) is *not* an element of U. >>because union(U) does not exist. >> Yes, if you want to adopt the axiom there are no infinite sets, >> then it follows that Case B never arises. Then it follows that you invent a contradiction. No, because I *don't* adopt the axiom there are no infinite sets. -- Daryl McCullough Ithaca, NY === Subject: Re: Why meta diagonals are irrelevant <1m295r4a4ma3h.1102mtunsb6vc$.dlg@40tude.net> posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > Translating into facts about natural numbers gets at the heart > of WM's problem. He believes the following two statements are > contradictory, as well: 1. Every natural number is in some finite initial segment of the naturals. > 2. No finite initial segment of the naturals contains every natural. They are contradictory, *if* the set of all finite initial segments exits. But that is not my problem. A. One proof is that you must understand not all when you say every and all. This is clear from the statement that you agreed to: Every node in the infinite path Q is in a finite path u together with all its preceding nodes. I every node exists, then there is no chance to avoid that consequence that 2. is false. Therefore your rules of logic are not logical. B. Here s another proof. Let the union of all finite paths exist and let all paths be linear (each one contains one 1 more than its predecessor) as given here: 0.1 0.11 0.111 ... Then the union is Q = 0.111.... And no 1 of Q is missing in the union of us. Then there must be a path 0.111... in the set of paths u that have been united. Another possibility to achieve this result would be using paths with gaps like these: 0.1 0.11 0.1-1 0.-111 and so on. Yes, here we can have a union Q that contains more 1s than every path u of the union. The same is also possible for a potentially infinite set of linear paths, because the union is not fixed, neither is the number of paths. But it is impossible for an actually infinite set. The result is of course not that an infinite finite path exists. The result is simply, generalized to N, that there is no actually infinite set of finite numbers, because it would yield the contradiction that an infinite finite number must exist. If your logic leads you astray, yielding every and all is not all, then you should correct it. If your sets are time depend or depending on some choice or pick, then you should confess that your sets and all their elements do not exist in the static way you always claim. === Subject: Re: Why meta diagonals are irrelevant WM says... >> Translating into facts about natural numbers gets at the heart >> of WM's problem. He believes the following two statements are >> contradictory, as well: >> 1. Every natural number is in some finite initial segment of the naturals. >> 2. No finite initial segment of the naturals contains every natural. They are contradictory, *if* the set of all finite initial segments >exits. But that is not my problem. That is *exactly* your problem. 1&2 are both *true*. They are not contradictory. You need to consider the following two statements: 1. Every man has a woman who is his mother. 2. No woman is every man's mother. Do you understand that both of these are true? -- Daryl McCullough Ithaca, NY === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > WM says... >> Translating into facts about natural numbers gets at the heart >> of WM's problem. He believes the following two statements are >> contradictory, as well: > 1. Every natural number is in some finite initial segment of the naturals. >> 2. No finite initial segment of the naturals contains every natural. They are contradictory, *if* the set of all finite initial segments >exits. But that is not my problem. That is *exactly* your problem. 1&2 are both *true*. They > are not contradictory. You need to consider the following two statements: 1. Every man has a woman who is his mother. > 2. No woman is every man's mother. Do you understand that both of these are true? Of course. Do you want to imply that this would touch the path problem? === Subject: Re: Why meta diagonals are irrelevant WM says... 1. Every natural number is in some finite initial segment of the naturals. > 2. No finite initial segment of the naturals contains every natural. >> You need to consider the following two statements: >> 1. Every man has a woman who is his mother. >> 2. No woman is every man's mother. >> Do you understand that both of these are true? Of course. Do you want to imply that this would touch the path >problem? Yes. The fact about natural numbers is no more contradictory than the fact about mothers. Let's make the fact about naturals more explicit. 1. The natural number n is in the finite initial segment { 0, 1, 2, 3, ..., n } 2. The natural number n+1 is *not* in the finite segment { 0, 1, 2, 3, ..., n } Surely you agree with both 1 and 2? -- Daryl McCullough Ithaca, NY === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322),gzip(gfe),gzip(gfe) > WM says... 1. Every natural number is in some finite initial segment of the naturals. > 2. No finite initial segment of the naturals contains every natural. >> You need to consider the following two statements: > 1. Every man has a woman who is his mother. >> 2. No woman is every man's mother. > Do you understand that both of these are true? Of course. Do you want to imply that this would touch the path >problem? Yes. The fact about natural numbers is no more contradictory > than the fact about mothers. Let's make the fact about naturals more explicit. 1. The natural number n is in the finite initial segment > { 0, 1, 2, 3, ..., n } 2. The natural number n+1 is *not* in the finite segment > { 0, 1, 2, 3, ..., n } Surely you agree with both 1 and 2? Of course. This is always true and you get the statement: There is no finite segment containing all natural numbers or linear paths. But at the same time you get the statement there is no natural number that is outside of a sequence containng all natural numbers less than itself. The complement of none is all and not every one you may pick in order to look for another one. You try to forget that statements about all natural numbers concern all natural numbers and not only one picked as an example. === Subject: Re: Why meta diagonals are irrelevant WM says... > Let's make the fact about naturals more explicit. >> 1. The natural number n is in the finite initial segment >> { 0, 1, 2, 3, ..., n } >> 2. The natural number n+1 is *not* in the finite segment >> { 0, 1, 2, 3, ..., n } >> Surely you agree with both 1 and 2? Of course. Well, 1. implies the following: 1.1 Forall natural numbers n, the initial segment {0,1,2,...,n} contains n. which is a special case of the more general statement 1.2 Forall natural numbers n, there is an initial segment containing n. The only difference between 1.1 and 1.2 is that in 1.1 we say explicitly which initial segment contains n, while in 1.2, we are just saying that there is *some* initial segment containing n, we are just not saying which one. Do you agree with 1.2? If not, why not? On the other hand, 2. is a special case of the following: 2.1 Forall finite initial segments S, the natural number max(S)+1 is not contained in S. (where max(S) means the largest element of S; every finite set of naturals has a largest element). But 2.1 is a special case of 2.2 Forall finite initial segments S, there is a natural number that is not contained in S. The only difference between 2.2 and 2.1 is that in 2.1 we say explicitly which natural number is not contained by S, (namely, max(S)+1), while in 2.2 we are just saying that there is *some* natural number not contained in S, we are just not saying which one. So we have the following two statements: 1.2 Forall natural numbers n, there is an initial segment containing n. 2.2 Forall initial segments S, there is a natural number not contained in S. Do you still claim that 1.2 and 2.2 are contradictory? -- Daryl McCullough Ithaca, NY === Subject: Re: Why meta diagonals are irrelevant posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I Of course. This is always true and you get the statement: There is no > finite segment containing all natural numbers or linear paths. But at the same time you get the statement there is no natural number that is outside of _all_ sequences containing all natural numbers less than itself. Look! Over there! A pink elephant! But at the same time you get the statement there is no natural number > that is outside of a sequence containng all natural numbers less than > itself. > The complement of none is all and not a - William Hughes === Subject: Re: Why meta diagonals are irrelevant posting-account=X9VdBgoAAAA0ZF8HT8BN_JvL2DEZQ6_G CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Of course. This is always true and you get the statement: There is no > finite segment containing all natural numbers or linear paths. But at the same time you get the statement there is no natural number > that is outside of _all_ sequences containing all natural numbers less > than > itself. > A. One proof is that you must understand not all when you say every and all. This is clear from the statement: Every node in the infinite path Q is in a finite path u together with all its preceding nodes. I every node exists, then there is no chance to avoid that consequence that 2. is false. Therefore your rules of logic are not logical. B. Here s another proof. Let the union of all finite paths exist and let all paths be linear (each one contains one 1 more than its predecessor) as given here: 0.1 0.11 0.111 ... Then the union is Q = 0.111.... And no 1 of Q is missing in the union of us. Then there must be a path 0.111... in the set of paths u that have been united. === Subject: Re: Why meta diagonals are irrelevant posting-account=1lE9SQkAAADFrJsDv61dh1YXcJ_ahy5I > Of course. This is always true and you get the statement: There is no > finite segment containing all natural numbers or linear paths. But at the same time you get the statement there is no natural number > that is outside of _all_ sequences containing all natural numbers less > than > itself. > We agree that i: there is no natural number that is outside of _all_ sequences containing all natural numbers less than itself. is true. However, without Pink Elephants we cannot replace i with ii: there is no natural number that is outside of _a_ sequence containing all natural numbers less than itself. - William Hughes === Subject: Re: Why meta diagonals are irrelevant Am 22 Oct 2008 03:25:10 -0700 schrieb Daryl McCullough: Translating into facts about natural numbers gets at the heart > of WM's problem. > At least this is my strategy when arguing with a crank: try to simplify the issue as much as possible (to get at the heart of the problem). He believes the following two statements are contradictory, as well: 1. Every natural number is in some finite initial segment of the naturals. > 2. No finite initial segment of the naturals contains every natural. > Right. By the way, here lies the reason why Cantor's diagonal argument must fail [...]. Every number smaller than omega is a finite number, and it is surpassed by other finite numbers [Cantor, p. 406]. Hence, if the list contains every number smaller than omega, then there are /other/ numbers, not contained in the list. The list is not complete. (W. M.9fckenheim, A severe inconsistency of transfinite set theory) Herb === Subject: infinitely differentiable function with compact support posting-account=tRBaOAoAAAB7qCLaIYH-yq0oSzOukouW Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) Hello! I found in one book such an exercise: Construct an infinitely differentiable function g with compact support (let call it idf with cs) such that: g(x) = 0, if |x|>=1 and 1 if |x| <= 0.5 I know that function defined as: h(x) = exp(-1/(1-x^2)), |x|<1 and 0 otherwise is idf with cs but here we dont have this 1 inside for interval [-0.5,+0.5]. Has someone idea how to define such a function with condtion above? === Subject: Re: infinitely differentiable function with compact support > Hello! I found in one book such an exercise: Construct an infinitely differentiable function g with compact support > (let call it idf with cs) such that: g(x) = 0, if |x|>=1 and 1 if |x| <= 0.5 I know that function defined as: h(x) = exp(-1/(1-x^2)), |x|<1 and 0 otherwise is idf with cs but here we dont have this 1 inside for interval [-0.5,+0.5]. Has someone idea how to define such a function with condtion above? I'll give you the basic idea, but you have to figure out the details yourself. Start with the function: f(x) = exp(-1/x), which is C^{oo} construct a g which has compact support [-1,1]: g(x) = f(1-x^2) adjust g's height to have g(0) = 1 (Can you see how?) g is C^{oo} @ 0, so you can split it up into two symmetric (about x=0) pieces, and interject the function h(x) = 1 in the middle. Doing this, would require you to also adjust g's support a bit. Good luck. -- I.N. Galidakis === Subject: Re: infinitely differentiable function with compact support > Start with the function: f(x) = exp(-1/x), which is C^{oo} I mean the function: f(x) = {exp(-1/x), x>0, 0 otherwise}, of course. > construct a g which has compact support [-1,1]: g(x) = f(1-x^2) adjust g's height to have g(0) = 1 (Can you see how?) g is C^{oo} @ 0, so you can split it up into two symmetric (about x=0) pieces, > and interject the function h(x) = 1 in the middle. Doing this, would require you to also adjust g's support a bit. Good luck. > -- I.N. Galidakis === Subject: help with solving Riccati equation posting-account=47d2bQoAAAAcDL51xmeqgNOnCj4rEC_I 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) here's the problem: show that y1(x) = 1/x is a solution of the riccati equation y'=1/(x^2) - y/x - y^2 and then find the general solution of y'=1/(x^2) - y/x - y^2 im learning this from the book and whatever other sources i can find. i started solving it by since y = 1/x => y = 1/x + 1/z => dy/dx = 1/(x^2) - 1/(z^2) dz/dx and which point i have no idea what to do next. im not even sure if the above is right. if you have any sample problems similiar to this i could use or how to solve this particular one, i will greatly appreciate it. === Subject: diff eq word problem HELP posting-account=twH9PwoAAABVFxHWRVsv2WYMPEStgjmE 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) problem is the following A man swims across a river 100ft wide, always heading for a tree directly across from his starting point. He can swim at the rate of 3 ft/sec Find the equation of his path if the current is carrying him downstream at the rate 2ft/sec, 3 ft/sec, 5 ft/sec all comments on how to solve this are welcome and are appreciated! === Subject: What numbers do I enter? posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) So I'm using this new environmental database program to graph some soil boring analyses. After I set my query criteria, I hit the button to lauch the Map module to get the following dialog box. Map Ranges of values will be displayed as colored circles. Enter values for the cutoffs between the colors. Selected Parameter: Arsenic (As) (Max for Station) Min Value: 28 Green ----------[ 50] Blue ----------[100] Yellow ----------[200] Magenta ----------[400] Red Max Value: 28 (The cutoff values shown were from the previous query). So I try to run the map and I get an error: Error! Number must be smaller than the Max Value. and highlights the 400. Ok, so I change it to 27 only to get a new error: Error! Number must be larger than value above. Since I can't make it any larger, I make the number above 26. This, of course, gives an error saying the 26 must be larger than the value above. Ok, I'll make them all [ 24] [ 25] [ 26] [ 27] only to get yet another error: Error! Number must be larger than the Min Value. with the 24 highlighted. Hmm...I seem to have a conundrum. What number would be both larger and smaller than 28? Nice to see that the slackers who flunked high school math can still be productive members of society by getting jobs as programmers. === Subject: Re: What numbers do I enter? > So I'm using this new environmental database program > to graph some soil boring analyses. After I set my > query criteria, I hit the button to lauch the Map > module to get the following dialog box. Map > Ranges of values will be displayed as colored circles. > Enter values for the cutoffs between the colors. Min Value: 28 > Max Value: 28 So I try to run the map and I get an error: Error! > Number must be smaller than the Max Value. Hmm...I seem to have a conundrum. What number would be > both larger and smaller than 28? > 20 <= 20.000 <= 20 of course. > Nice to see that the slackers who flunked high school math > can still be productive members of society by getting jobs > as programmers. > Why bother with color graphs? Use bar graphs and be done with it. Quit wasting your time with art work and get on with a scientific report instead of a slick popular publication for slackers. While you're at it, return the software for being defective. === Subject: Re: Fear of Numbers > it's only the kids in the US not the rest of the world. >> In most other parts of the world, people aren't called nerds just because >> they get good grades or like math and science. >> this is about the only place in the world you can get beat up for having >> good grades >> so this is the only place where in order to be cool you have to pretend >> to >> be stupid to fit in LOL, I assure you that at my schools in the '70s and '80s, being a brain > did > not make you cool! > exactly, and didn't that stink, that a person couldn't be proud of studying hard and getting good grades? (hey! there's a comb in your email address!!! what a cool name) mk5000 I used to work in the record store. I had everything before anyone. I was there in the Paradise Garage DJ booth with Larry Levan. I was there in Jamaica during the great sound clashes. --lcd soundsystem, i'm losing my edge === Subject: Re: Fear of Numbers reply-type=response Importance: Normal > exactly, and didn't that stink, that a person couldn't be proud of > studying hard and getting good grades? I think Phineas and Ferb are pretty cool. === Subject: Re: Question about E=1/2 mv^2 reply-type=response > Assume a 10m diameter tunnel and a 200 km wind blowing through the tunnel. > How do I calculate the energy of the system? What will the units of the > result be? > Depends on how long the tunnel is. I will firstly take your question at face value - that you want the energy expended in accelerating a cylinder of air, 10 m in diameter and of unspecified length that I will call L to 200 kms/hr. Their is a great thing about air; its mass is very close to 1 kg per cubic metre, which in SI units gives it a density of almost exactly 1 kg/m^2. The volume of air inside the cylinder is pi * r^2 * L m^3 Its mass is therefore pi * r^2 * L kg r=5, so m = 80*L kgs So we know m. v = 200 kms/hr = 2 * 10^5 m/hr = 2*10^5 m /3600 seconds = 55 m/s E = 1/2 m v^2 = 0.5 * 80*L (55)^2 Joules = L * 3 * 10^6 Joules (where L is the tunnel length in metres.) If, however, you meant power, then the length of the tunnel makes no difference, and the power can be calculated in a similar manner. But that's not using E = 0.5 mv^2 to directly find the answer, as your subject line implies. === Subject: Solutions manual to Engineering Mechanics - Dynamics (11th ) by R.C.HIBBELER posting-account=jBP0yQoAAADkjbQMT90jR5JPojXnsRCF Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) Solution manual solutions manual solution manual I am a solutions manual collector, I offer solutions manual services Note: all solutions manual in soft copy that mean in Adobe Acrobat Reader (PDF ) format. if you want any book not just solutions just contact with us.81B to get the solution manual you want .81Cplease send message to happyren2008@hotmail.com .81Chappyren2008(at)hotmail.com.81C replace (at) to @ ,please email to me . 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Johnston Solutions manua to Young & Freedman,University Physics, 12th Edition Basic Technical Mathematics with Calculus SI Version 8th (INSTRUCTORÍS SOLUTIONS MANUAL) By John R. Martin(chapter 1 to chapter15) solution manual for Probability and Statistical Inference ( 7th edition by Hogg & Tanis) solution manual for Fundamentals of Communication Systems by John G. Proakis ,Masoud Salehi solution manual for Materials and Processes in Manufacturing,9thBy Degarmo Advanced Industrial Economics 2nd by Martin Fundamentals of Corporate Finance 8th by Ross Principles of Corporate Finance 8th by Brealey Myers Alen Investments Student Solutions Manual 6e by Zvi Bodie Statistical Inference 2nd by George Casella Solution Manual of Econometrics of Financial Market options futures and other derivatives 3rd Solution by hull === Subject: Re: A cute little problem with logs > .... > Notation: log_p(q) means the base-p logarithm of q. Given: log_a(x) = c and log_b(x) = d. > Find: log_ab(x), the log of x in base (a*b), in > terms of c and d.... A useful elementary formula, perhaps almost forgotten, is log_y(x) = 1/log_x(y). That converts the problem entirely into logs to base x, making it easy to do in your head. Ken Pledger. === Subject: Re: JSH: So why no legal DVD copying? posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > What's the point of making a legal backup if it doesn't work if you > lose or damage the original? (The following is not a comment on whether or not JSH's scheme is > correct or feasible). You wouldn't use the backup as a backup. You'd use the backup as your > active copy, and keep the original as a backup. E.g., you buy a DVD > for your kids. You make a 'backup', and give that to the kids. You put > the original in a safe place. Yup. If (when) the kids destroy their copy, you take the original out of the > safe place, and make a new copy for the kids. If, by some miracle, the kids go 30 days without destroying their copy, > you take the original out of the safe place, and re-authorize the kid's > copy for another 30 days. Because the original only comes out when you need to make a new copy for > the kids, or every 30 days (whichever comes from first), and is only > used by you and then put away, you should be able to keep the original > undamaged. -- > --Tim Smith And that at least seemed to stop the arguing back and forth. I find it REMARKABLE how some of you put a label on a person of > crackpot and then refuse to acknowledge anything they say as > valuable so you will argue day and night with that person until > someone else comes in and then you quiet down. The behavior is telling. You wish to permanently label certain people > in a negative way and hold that label against any new evidence. This story shows how dramatic that is as well, where the fight is > against a simple way to allow legal copying of bought DVD's, where > that is concrete enough for people to understand just how rigid is the > hostility against supposed crackpots. I've joked in the past that many of you would reject a cure for cancer > because of the source. Short of it is that your opinions against people are often meaningless > as to what they state or the facts but more often than not reflect > rigid personal biases which you quickly form against CERTAIN people > and then refuse to shift. Oh, and of course there is also an experimental side to my posting on > this subject! DMESE is a fairly simple idea which because of the nature of the > problem with copying digital media actually closes the problem space, > as in, it's either it, or nothing, and so far industry has sided > towards nothing with many declaring DRM to be dead, which has a > potential cost in the billions of dollars over time, if artists, for > instance, are incapable of making as much from music sales because > widespread piracy is considered unstoppable. Past experience indicates I take over web search results on topics > fairly quickly, which I've demonstrated often with definition of > mathematical proof or DMESE itself, as well as many others including > disparate things like: business plan for Internet radio or prime number compression (Google them, as most of my search domination is with Google though it > is often on Yahoo! as well.) But is that being done through a proper process? I've been concerned > enough about it that I've contacted my U.S. senator and representative > about the issue. (Senator Feinstein gave me a form letter reply. She > may need to re-visit that later though as this situation develops > while I'm deferring a bit, because of the financial crisis, in > pursuing the issue.) If things stay to form though I should start showing up in search > results on RealNetworks and their DVD copying software. If I don't > then there may be more to the story of how Google and Yahoo! determine > those rankings. Already I have atypical behavior on my blogs where hits have gone DOWN > dramatically lately, since I started pushing this issue. Weird seemingly atypical behavior is where science starts... So we all lose. Yup, we are all losers. That's the real secret. So Uncle Al and the rest of the crackpot haters, so what? You're losers along with us. You'll never get recognition for your either. Waste your lives. Put it out there like anybody cares when it's a world run by fools. Knowledge isn't power. Crap is power. And idiots rule the world... JSH === Subject: Re: JSH: So why no legal DVD copying? posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) What's the point of making a legal backup if it doesn't work if you > lose or damage the original? (The following is not a comment on whether or not JSH's scheme is > correct or feasible). You wouldn't use the backup as a backup. You'd use the backup as your > active copy, and keep the original as a backup. E.g., you buy a DVD > for your kids. You make a 'backup', and give that to the kids. You put > the original in a safe place. Yup. If (when) the kids destroy their copy, you take the original out of the > safe place, and make a new copy for the kids. If, by some miracle, the kids go 30 days without destroying their copy, > you take the original out of the safe place, and re-authorize the kid's > copy for another 30 days. Because the original only comes out when you need to make a new copy for > the kids, or every 30 days (whichever comes from first), and is only > used by you and then put away, you should be able to keep the original > undamaged. -- > --Tim Smith And that at least seemed to stop the arguing back and forth. I find it REMARKABLE how some of you put a label on a person of > crackpot and then refuse to acknowledge anything they say as > valuable so you will argue day and night with that person until > someone else comes in and then you quiet down. The behavior is telling. You wish to permanently label certain people > in a negative way and hold that label against any new evidence. This story shows how dramatic that is as well, where the fight is > against a simple way to allow legal copying of bought DVD's, where > that is concrete enough for people to understand just how rigid is the > hostility against supposed crackpots. I've joked in the past that many of you would reject a cure for cancer > because of the source. Short of it is that your opinions against people are often meaningless > as to what they state or the facts but more often than not reflect > rigid personal biases which you quickly form against CERTAIN people > and then refuse to shift. Oh, and of course there is also an experimental side to my posting on > this subject! DMESE is a fairly simple idea which because of the nature of the > problem with copying digital media actually closes the problem space, > as in, it's either it, or nothing, and so far industry has sided > towards nothing with many declaring DRM to be dead, which has a > potential cost in the billions of dollars over time, if artists, for > instance, are incapable of making as much from music sales because > widespread piracy is considered unstoppable. Past experience indicates I take over web search results on topics > fairly quickly, which I've demonstrated often with definition of > mathematical proof or DMESE itself, as well as many others including > disparate things like: business plan for Internet radio or prime number compression (Google them, as most of my search domination is with Google though it > is often on Yahoo! as well.) But is that being done through a proper process? I've been concerned > enough about it that I've contacted my U.S. senator and representative > about the issue. (Senator Feinstein gave me a form letter reply. She > may need to re-visit that later though as this situation develops > while I'm deferring a bit, because of the financial crisis, in > pursuing the issue.) If things stay to form though I should start showing up in search > results on RealNetworks and their DVD copying software. If I don't > then there may be more to the story of how Google and Yahoo! determine > those rankings. Already I have atypical behavior on my blogs where hits have gone DOWN > dramatically lately, since I started pushing this issue. Weird seemingly atypical behavior is where science starts... So we all lose. Yup, we are all losers. That's the real secret. So Uncle Al and > the rest of the crackpot haters, so what? You're losers along with us. You'll never get recognition for your either. Waste your lives. Put it out there like anybody cares when it's a > world run by fools. Knowledge isn't power. Crap is power. And idiots rule the world... JSH- Hide quoted text - - Show quoted text - Welcome! === Subject: JSH: Moving to real fun posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) Back when I was naive and stupid, and idealistic, I dreamed of making great discoveries to advance the state of humanity. But then I made some discoveries, got years of abuse from mathematicians, and now all I want to do is make as many academics look like fools, destroy endowments, end departments and generally wreak havoc across the academic world, as I think I kind of went bonkers as a result. And so I no longer care, can no longer be convinced, but I'm still brilliant, can keep making discoveries indefinitely, and fantasize about shutting down, Princeton! Yup, I want to shut down Princeton. That sounds like a better goal than stupid nonsense like furthering and advancing the state of the human species as its full of ingrates, and people who don't value knowledge, but they care about cable! Yeah, give them their cable. Give the hoi polloi their television with its nincompoops and actors who are worth nothing but get so much from a world of fools. This ship of fools. Well at least I can make your lives miserable. Take away your funds and force you to do what I can do. Make you live up to my standard. Public funding for research is stupid. All scientists should learn to live in the free market. Do their research as best they can without welfare. It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! Like me. James Harris === Subject: Re: Moving to real fun > Back when I was naive and stupid, and idealistic, I dreamed of making > great discoveries to advance the state of humanity. you are still naive and stupid, and idealistic. > But then I made some discoveries, got years of abuse from > mathematicians, and now all I want to do is make as many academics > look like fools, destroy endowments, end departments and generally > wreak havoc across the academic world, as I think I kind of went > bonkers as a result. you are still bonkers. And so I no longer care, can no longer be convinced, but I'm still > brilliant, can keep making discoveries indefinitely, and fantasize > about shutting down, Princeton! nope, you were never brilliant. Yup, I want to shut down Princeton. Only to satisfy your ego, troll boy. That sounds like a better goal than stupid nonsense like furthering > and advancing the state of the human species as its full of ingrates, > and people who don't value knowledge, but they care about cable! Yeah, give them their cable. Give the hoi polloi their television > with its nincompoops and actors who are worth nothing but get so much > from a world of fools. This ship of fools. Well at least I can make your lives miserable. Take away your funds > and force you to do what I can do. dream on troll-boy. Make you live up to my standard. how low can a hacker go? really low. Public funding for research is stupid. JSH research is stupid. All scientists should learn to live in the free market. Do their > research as best they can without welfare. It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! Like me. is it time for you to start working for a living ? James Harris === Subject: Re: JSH: Moving to real fun posting-account=XH7n1goAAAC_c_M3JtSDv3QFzlbwvvMO .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > All scientists should learn to live in the free market. Do their > research as best they can without welfare. It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! I think it is actually true that there's a kind of laxity in US academic math jobs which is abused by many people--there are plenty of American schools where you can complete a math PhD and go on to get a comfortably-paid permanent teaching job at some school without much research funding, all while working well under 40 hours a week. At the same time there are those of us who work very, very hard and work very long hours--I personally have given up many things which meant a great deal to me so that I could have some kind of decent career in mathematics research, and right now I spend about 60 hours/week in my office, plus time I spend working at home. I am afraid eliminating government funding for research would not help this situation--based on what I have seen of free markets, the situation would certainly become worse. === Subject: Re: JSH: Moving to real fun posting-account=LlRppgoAAAD1KDQAbEw51E0RIOUzJ0up Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe) > Yeah, give them their cable. Give the hoi polloi their television > with its nincompoops and actors who are worth nothing but get so much > from a world of fools. This ship of fools. > Written by a man who bases his life on an actor from a cable TV show. God, you're stupid. James Harris M === Subject: Re: JSH: Moving to real fun posting-account=d-ESTAkAAAAG0l03yI1WJgsTVXx4ebeJ Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) > It's time for you all to start working for a living! Like me. James Harris DOOD! You are either seriously ed or scored some REALLY good ! All I can say is that based on this rant, I assume your other job is playing lead guitar in a HM band. === Subject: Re: JSH: Moving to real fun posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) It's time for you all to start working for a living! Like me. James Harris DOOD! You are either seriously ed or scored some REALLY good > ! > All I can say is that based on this rant, I assume your other job is > playing lead guitar in a HM band. Gullible. === Subject: Re: JSH: Moving to real fun >It's time for you all to start working for a living! > Like me. > James Harris >> DOOD! You are either seriously ed or scored some REALLY good >> ! >> All I can say is that based on this rant, I assume your other job is >> playing lead guitar in a HM band. Gullible. Where do you play? === Subject: Re: JSH: Moving to real fun > And so I no longer care, can no longer be convinced, but I'm still > brilliant, can keep making discoveries indefinitely, and fantasize > about shutting down, Princeton! And I fantasize about: ´ Fixing ACPI on my computer ´ Fixing in Thunderbird ´ Actually being good at the piano ´ Accidentally hitting sudo rm -rf / on a computer ´ Finishing any one of my four aborted programming projects ´ Finishing any one of my countless incomplete video games ´ Actually being good at, say, drawing ´ Actually writing down one of my stories ´ Winning the Fields Medal/Turing Award/Nobel Prize ´ Becoming president ´ Flying a plane ´ A world where people are actually polite to each other in conversation ´ A world where Linux is the primary desktop OS But these are all just fantasies, about as likely to occur as the Earth spontaneously ceasing to exist. > That sounds like a better goal than stupid nonsense like furthering > and advancing the state of the human species as its full of ingrates, > and people who don't value knowledge, but they care about cable! You've really detached yourself from society if you think people care about cable. It's all about the broadband internet. Give most people I know a 50 Mb/s downlink and they're reduced to salivating wrecks. === Subject: Re: JSH: Moving to real fun > And so I no longer care, can no longer be convinced, but I'm still > brilliant, can keep making discoveries indefinitely, and fantasize > about shutting down, Princeton! And I fantasize about: ... > ´ Actually being good at, say, drawing I gave up on that one when I realized that when I closed my eyes and tried to *imagine* drawing a good freehand circle, it came out lopsided, the same way it does when I try to actually draw one. -- --Tim Smith === Subject: Re: JSH: Moving to real fun posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > And so I no longer care, can no longer be convinced, but I'm still > brilliant, can keep making discoveries indefinitely, and fantasize > about shutting down, Princeton! And I fantasize about: > ´ Fixing ACPI on my computer > ´ Fixing in Thunderbird > ´ Actually being good at the piano > ´ Accidentally hitting sudo rm -rf / on a computer > ´ Finishing any one of my four aborted programming projects > ´ Finishing any one of my countless incomplete video games > ´ Actually being good at, say, drawing > ´ Actually writing down one of my stories > ´ Winning the Fields Medal/Turing Award/Nobel Prize > ´ Becoming president > ´ Flying a plane > ´ A world where people are actually polite to each other in conversation > ´ A world where Linux is the primary desktop OS But these are all just fantasies, about as likely to occur as the Earth > spontaneously ceasing to exist. > not at all, I think you have a very nice list there. The polite conduct is actually a self fulfilling prophecy. Thinking the Earth cant spontaneously cease to exist is another one. The non US media is very upset about the global drop in oxygen levels. They discarded all the global warming data claiming it is no longer correct, things are now moving a hundred times as fast as planed. So, I'm sorry to tell you it might just happen very soon sir. Go tell your officials to grow hemp everywhere. Tossing around seeds doesn't take a lot of effort or budget. We need 2 big trees per person or 20 m2 worth of green in the sun. In the shadow it will be more like 50 m2. If we don't we will all tip over and cease to exist I'm afraid. Sorry to bring the bad news but do spread the word please. > That sounds like a better goal than stupid nonsense like furthering > and advancing the state of the human species as its full of ingrates, > and people who don't value knowledge, but they care about cable! You've really detached yourself from society if you think people care > about cable. It's all about the broadband internet. Give most people I > know a 50 Mb/s downlink and they're reduced to salivating wrecks. Pacman and internets are not really important. We need air, water, vegetables and a place to sleep. When you have those you should entertain yourself by helping others getting the same. Hollywood is not real. It's to distract you from the economic activities so that you don't bother the money makers. You'd really want to keep an eye on those in stead. Dial Up is good enough for such. If you know something about music perhaps you can take a look at this. === Subject: Re: JSH: Moving to real fun posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) > Back when I was naive and stupid, and idealistic, James Harris I think you could have stopped there - at least these statements are still true! Get a life - you are an angry idiot. Please seek professional help before you hurt another human being! Go drink beer, chase women and destroy the last remaining brain cell that may still exist in your delusional mind. possibly can. Maybe that will knock some sense into you. Don't be a bitter fool - get a life and do something to help as opposed to moaning, crying and peeing all over yourself! === Subject: Re: JSH: Moving to real fun posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) Back when I was naive and stupid, and idealistic, James Harris I think you could have stopped there.... Yes, but you are still naive, stupid and idealistic. :-) === Subject: Re: JSH: Moving to real fun posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) > Back when I was naive and stupid, and idealistic, I dreamed of making > great discoveries to advance the state of humanity. But then I made some discoveries, got years of abuse from > mathematicians, and now all I want to do is make as many academics > look like fools, destroy endowments, end departments and generally > wreak havoc across the academic world, as I think I kind of went > bonkers as a result. And so I no longer care, can no longer be convinced, but I'm still > brilliant, can keep making discoveries indefinitely, and fantasize > about shutting down, Princeton! Yup, I want to shut down Princeton. That sounds like a better goal than stupid nonsense like furthering > and advancing the state of the human species as its full of ingrates, > and people who don't value knowledge, but they care about cable! Yeah, give them their cable. Give the hoi polloi their television > with its nincompoops and actors who are worth nothing but get so much > from a world of fools. This ship of fools. Well at least I can make your lives miserable. Take away your funds > and force you to do what I can do. Make you live up to my standard. Public funding for research is stupid. All scientists should learn to live in the free market. Do their > research as best they can without welfare. It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! Like me. James Harris What's yer occupation? === Subject: Re: JSH: Moving to real fun It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! Like me. I'd like to see some actual evidence that you get paid for posting kookery on Usenet. There's just too much amateur talent for that to be a viable occupation. -- Bill Snyder [This space unintentionally left blank] === Subject: Re: JSH: Moving to real fun posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I'd like to see some actual evidence that you get paid for posting Pay attention billy. You just tolled us you get paid to post lies here? no? Here is some work for you. Is survival not a good pay? === Subject: Re: JSH: Moving to real fun posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I'd like to see some actual evidence that you get paid for posting Pay attention billy. You just tolled us you get paid to post lies here? no? > Toll: 1. To exact as a toll. 2. To charge a fee for using (a structure, such as a bridge). Told: 1. Past tense and past participle of tell It is easy to excuse misspellings - but at least use the correct words. === Subject: Re: JSH: Moving to real fun posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) It is easy to excuse misspellings - but at least use the correct > words. you claim it is easy but not for you. Yo gramma is so dumb you cant even address the topic. === Subject: Re: JSH: Moving to real fun >> I'd like to see some actual evidence that you get paid for posting Pay attention billy. You just tolled us you get paid to post lies here? no? You should take something for those hallucinations. >Here is some work for you. >Is survival not a good pay? Stupidity and insanity are available cheap at outlets everywhere. -- Bill Snyder [This space unintentionally left blank] === Subject: Re: JSH: Moving to real fun posting-account=qHNgPQoAAAA40SLlrmjQxSJhnyKS76HY Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) >> I'd like to see some actual evidence that you get paid for posting Pay attention billy. You just tolled us you get paid to post lies here? no? You should take something for those hallucinations. Here is some work for you. >Is survival not a good pay? Stupidity and insanity are available cheap at outlets everywhere. -- > Bill Snyder [This space unintentionally left blank] Oh, what a typical Hill Billy thing to say. :-) === Subject: Re: JSH: Moving to real fun posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) It's time for you people to grow up! I'm tired of the world babying you, babies. It's time for you all to start working for a living! Like me. I'd like to see some actual evidence that you get paid for posting > kookery on Usenet. There's just too much amateur talent for that > to be a viable occupation. It is fun though. I did put fun in the title so yeah, the post was meant to be at least somewhat humorous. I especially liked the part about shutting down Princeton. But seriously, the message from me to academia is, you do your thing, and I'll do mine. We don't work together. Not today and not tomorrow. I'm still reeling from my latest research results and that weirdness with RealNetworks and Hollywood blocking all legal copying of bought DVD's, so I'm puzzling and wondering and reaching the point where I'm telling myself, yeah, I don't give a damn. I'm brilliant. I accept it. I like myself so it's ok. I see your academic world as primitive. Yeah, I know. You don't like being called primitive and think some of you are brilliant too, well that's ok too. You can feel what you like. It doesn't matter if it's not true, right? Wow. What a weird reality. When I was a kid I so looked up to all these people and wondered, what if? What if it were me and now I'm like, thank God you people are primitive. Yeah. It's a good world. I like it. I really do. Wow. James Harris === Subject: Re: JSH: Moving to real fun posting-account=_k7REQoAAACDJL2M6OKCfSBj5_wTcvrO Gecko/20080702 Firefox/2.0.0.16 (.NET CLR 3.5.30729),gzip(gfe),gzip(gfe) > But seriously, the message from me to academia is, you do your thing, > and I'll do mine. we can now just get on with our thing. This whole James Harris saga has been occupying everyone's mind for the last 10 years. === Subject: JSH: State of the art of mathematics posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/2008092417 Firefox/3.0.3,gzip(gfe),gzip(gfe) I've now successfully used tautological spaces against only two problems. Where the latest which lead me to an alternate solution for binary quadratic Diophantine equations is rather spectacular for quite a few reasons because of the history around the problem. Obviously, I can contemplate other areas of mathematics on which to use tautological spaces. But I also know about their limitations as the second problem I tried to use tautological spaces on, and failed with them, was the factoring problem, so they are not infallible, just unimaginably powerful. I've been watching the television show House a lot lately. I've kind of become addicted to it which has impacted how I post, as in, I'm more insulting, and it feels good. But like Dr. House, my insults are for a purpose, or I think they are, though they may simply mean I'm juvenile? In any event, the show has given me a different perspective on the problem space. So what do we know? This notion of using identities in such a clever way clobbered the hell out of binary quadratic Diophantines in a way so devastating it mostly shut up even the angry idiots. But not a lot has changed. There's no reason to suspect much will change, so we can move on. And when I say we, I mean me, and my team of those people who bizarrely for reasons that puzzle many, and probably themselves, help me. So yes, I admit it. There are people who help me evaluate ideas and I mostly insult them. Like Dr. House. So I have a paradigm around which to fit my behavior and I am watching more and more episodes of the TV show to refine my behavior, and improve my insults. Hilbert came up with some problems. I can clobber quite a few of them but wonder if I care. Russell thought to formalize a few things, put some logic into it, but came up short, but I think a few simple ideas bridge things in a way he didn't imagine. But at the end of the day, does anybody really care? I know I find it hard to care. So what other problems for those who are silly enough to care might be amenable to tautological spaces in their new and raw form, which is new and raw as I just invented the concept, or discovered it, whatever, a few years ago? (No one say Collatz. It's verboten.) Of course I HAVE solved problems without using tautological spaces! But that's too advanced for this initial post. I'm thinking about cleaning out some crap left over from the old geezers who used to do math before they keeled over and died, like Russell and Hilbert. Still I'm kind of picky, so the idiots among you need not apply. Ok, anyone can of course as usual reply as if I try to put that kind of criteria on then no one is qualified but me to talk to myself and I do enough of that already. My suspicion is that you will fail me. That is good. It just means I'm more brilliant than I previously thought. So what is the state of the art of mathematics? In flux. The old ways have failed. I've collapsed 2000 years of old ideas about some simple binary whatevers into a page or two of rather potent theory. Mathematics can be revolutionized like the sciences. You CAN push out the old professors hogging up the works by outdoing them, re-inventing, discovering, starting from scratch, or not, if you're not smart enough. Get started. Or not. I'll insult you either way you go. Or not. James Harris === Subject: JSH Re: State of the art of mathematics > I've now successfully used tautological spaces you mean tautological flatulences kind of become addicted to it which has impacted how I post, as in, > I'm more insulting, and it feels good. But like Dr. House, my insults > are for a purpose, or I think they are, though they may simply mean > I'm juvenile? yes, it simply mean you are a juvenile. In any event, the show has given me a different perspective on the > problem space. So what do we know? > mostly insult them. Like Dr. House. Go for being like Mr. Ed. So I have a paradigm around which to fit my behavior and I am watching > more and more episodes of the TV show to refine my behavior, and > improve my insults. Good, moron is as moron does. Get started. Or not. I'll insult you either way you go. Or not. because you are only a troll crackpot, and NOT anything like a real mathematician. James Harris === Subject: Re: JSH: State of the art of mathematics posting-account=XH7n1goAAAC_c_M3JtSDv3QFzlbwvvMO .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > I've now successfully used tautological spaces against only two > problems. Where the latest which lead me to an alternate solution for > binary quadratic Diophantine equations is rather spectacular for quite > a few reasons because of the history around the problem. Obviously, I can contemplate other areas of mathematics on which to > use tautological spaces. But I also know about their limitations as the second problem I tried > to use tautological spaces on, and failed with them, was the factoring > problem, so they are not infallible, just unimaginably powerful. > ... > So what other problems for those who are silly enough to care might be > amenable to tautological spaces in their new and raw form, which is > new and raw as I just invented the concept, or discovered it, > whatever, a few years ago? I looked at some old posts mentioning tautological spaces. By tautological space do you mean a space of equations which hold true modulo some particular integer? So to use tautological spaces is to solve an equation modulo a prime n for some sufficiently large set of choices for n, and then lift the mod n solutions to a solution in the integers? Let me know if I understand this correctly-- if this is really what's going on then I definitely have a problem for you to look at. === Subject: Re: JSH: State of the art of mathematics > I've now successfully used tautological spaces against only two > problems. Where the latest which lead me to an alternate solution for > binary quadratic Diophantine equations is rather spectacular for quite > a few reasons because of the history around the problem. Obviously, I can contemplate other areas of mathematics on which to > use tautological spaces. But I also know about their limitations as the second problem I tried > to use tautological spaces on, and failed with them, was the factoring > problem, so they are not infallible, just unimaginably powerful. I've been watching the television show House a lot lately. I've > kind of become addicted to it which has impacted how I post, as in, > I'm more insulting, and it feels good. But like Dr. House, my insults > are for a purpose, or I think they are, though they may simply mean > I'm juvenile? In any event, the show has given me a different perspective on the > problem space. So what do we know? This notion of using identities in such a clever way clobbered the > hell out of binary quadratic Diophantines in a way so devastating it > mostly shut up even the angry idiots. But not a lot has changed. > There's no reason to suspect much will change, so we can move on. And > when I say we, I mean me, and my team of those people who bizarrely > for reasons that puzzle many, and probably themselves, help me. So yes, I admit it. There are people who help me evaluate ideas and I > mostly insult them. Like Dr. House. OK, then here's a chance for you to play Dr. House. Let's consider your recent post where you started with x^2 - D y^2 = 1 and then apply the transformation x <= x + d y y <= x + y to get a sequence of equations of the form: x^2 - D y^2 = (1-D)^j Your conjecture is that if you start with x = 1, y = 0, and form this sequence of equations, you will often find one where x and y are both divisible by (1-D)^(j/2). You can then divide that out of x and y, and end up with a solution for the original equation. You tested this for D = 3, and it worked. You quickly got an equation where you could divide the right side off, and came up with a non-trivial solution to x^2 - 3 y^2 = 1. OK, that was cool. And now I've done some tests, checked out your conjecture (or in House terms, your diagnosis), and have proven that it is impossible to hit any equation in the chain where you can divide off that way unless D = 2^n + 1, for some integer n. D = 2, 3, and 5 work. D = 7, 10, 11, 13, 14, 15 cannot work. D = 17 might work. The next D that might work is 33, and so on. I also have checked numerically and found that D = 17 and D = 33 do not work for any reasonably small values of j. Examining *why* they fail in those cases gives me some strong hints that in fact 2, 3, and 5 are the only D for which your method works, but I cannot yet prove it. It looks like it depends on some details of the base 2 representation of some sums of the products of sums of binomial coefficients, and I need to do some serious relearning to boost my binomial coefficient Kung-Fu first. What House would do now, if he were a mathematician, rather than a doctor, is immediately see why I claim that D must be of the form 2^n+1, see why that has me now looking at these complicated binomial coefficient equations, and come up with a clever way to cut through all that, and either prove that it works for all 2^n+1, or that it is only 2, 3, 5. So, the ball's in your court now, House. It's time to stop flailing around with guesses and conjectures, and finish off this problem. Oh, BTW, here's something else related to this that might be of interest. At each level of the chain, x and y are polynomials in D. Let these, at even level j, be Xj(D) and Yj(D). Then what we have is: Xj(D)^2 - D Yj(D)^2 = (1-D)^j For example, at level 8, we have X8(D) = 1+ 28 D+ 70 D^2 + 28 D^3 + D^4 Y8(D) = 8 + 56 D + 56 D^2 + 8 D^3 and what we would like to do is divide both of these by (1-D)^4. Note that Y8(D) > 0, and the degree of Y8 is less than the degree of (1-D)^4, so for sufficiently large D, Y8(D) < (1-D)^4, and so can't be divisible by (1-D)^4. This pattern continues at all levels. Yj(D) has degree one less than the degree of (1-D)^(j/2), and there is some Dj for that level such that (1-D)^j cannot divide off for any D > Dj. Here are the values of Dj, for j = 2, 4, 6, ...: 3,6,10,15,20,26,33,40,47,55,63,72,81,90,100 That sequence is not in Sloane's. So, for example, if you were trying to search for examples where D=17 works, there is no need to check below level 10. === Subject: Re: JSH: State of the art of mathematics > I've now successfully used tautological spaces against only two > problems. Where the latest which lead me to an alternate solution for > binary quadratic Diophantine equations is rather spectacular for quite > a few reasons because of the history around the problem. You will not have any trouble picking up women if you simply explain to them how your tautological space works. They will find it irresistible. === Subject: post-docs in London Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The UK suffers from having no canonical centralised system for advertising maths post-docs jobs. Furthermore, post-doc jobs are typically rarer in the UK than in the US, because there is less money in the system. Still institutions typically demand different things from the applicant. My university, Imperial College London, has a few post-docs that are currently being advertised and I thought I'd mention them on s.m.r. Note that there are substantial tax breaks for e.g. US citizens who are employed in the UK for less than 2 years (they basically pay no tax at all, because of a tax treaty). Here are the jobs: http://www3.imperial.ac.uk/employment/research/ns2008176ju [that one is specific to mathematics] and http://www3.imperial.ac.uk/juniorresearchfellowships [that one is science-wide] Current exchange rates make the pay competitive with a decent US institution. Kevin Buzzard +++ Professor Kevin Buzzard buzzard@imperial.ac.uk Department of Mathematics, Tel: (+44) 20 7594 8523 Imperial College, 180 Queen's Gate, Fax: (+44) 20 7594 8517 London SW7 2AZ, ENGLAND http://www.ma.ic.ac.uk/~buzzard === Subject: Four papers published by AGT and five papers posted in preview by GT Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Four papers have been published by Algebraic & Geometric Topology in Volume 8; papers (1) and (2) complete issue 3 and papers (3) and (4) open issue 4: (1) Algebraic & Geometric Topology 8 (2008) 1811-1832 The homology of the stable nonorientable mapping class group by Oscar Randal-Williams URL: http://www.msp.warwick.ac.uk/agt/2008/08-03/p066.xhtml DOI: 10.2140/agt.2008.8.1811 (2) Algebraic & Geometric Topology 8 (2008) 1833-1853 Commensurability classes of (-2,3,n) pretzel knot complements by Melissa L Macasieb and Thomas W Mattman URL: http://www.msp.warwick.ac.uk/agt/2008/08-03/p067.xhtml DOI: 10.2140/agt.2008.8.1833 (3) Algebraic & Geometric Topology 8 (2008) 1855-1959 Model structures on the category of small double categories by Thomas M Fiore, Simona Paoli and Dorette Pronk URL: http://www.msp.warwick.ac.uk/agt/2008/08-04/p068.xhtml DOI: 10.2140/agt.2008.8.1855 (4) Algebraic & Geometric Topology 8 (2008) 1961-1987 The R(S^1)-graded equivariant homotopy of THH(F_p) by Teena Meredith Gerhardt URL: http://www.msp.warwick.ac.uk/agt/2008/08-04/p069.xhtml DOI: 10.2140/agt.2008.8.1961 Five papers have been posted on the preview page for Geometry & Topology Volume 13 (2009): (5) Geometry & Topology 13 (2009) 1-48 Gromov-Witten invariants of blow-ups along submanifolds with convex normal bundles by Hsin-Hong Lai URL: http://www.msp.warwick.ac.uk/gt/2009/13-01/p001.xhtml DOI: 10.2140/gt.2009.13.1 (6) Geometry & Topology 13 (2009) 49-86 K-duality for stratified pseudomanifolds by Claire Debord and Jean-Marie Lescure URL: http://www.msp.warwick.ac.uk/gt/2009/13-01/p002.xhtml DOI: 10.2140/gt.2009.13.49 (7) Geometry & Topology 13 (2009) 87-98 Global fixed points for centralizers and Morita's Theorem by John Franks and Michael Handel URL: http://www.msp.warwick.ac.uk/gt/2009/13-01/p003.xhtml DOI: 10.2140/gt.2009.13.87 (8) Geometry & Topology 13 (2009) 99-139 On the homology of the space of knots by Ryan Budney and Fred Cohen URL: http://www.msp.warwick.ac.uk/gt/2009/13-01/p004.xhtml DOI: 10.2140/gt.2009.13.99 (9) Geometry & Topology 13 (2009) 141-187 Snowflake groups, Perron-Frobenius eigenvalues and isoperimetric spectra by Noel Brady, Martin R Bridson, Max Forester and Krishnan Shankar URL: http://www.msp.warwick.ac.uk/gt/2009/13-01/p005.xhtml DOI: 10.2140/gt.2009.13.141 Abstracts follow (1) The homology of the stable nonorientable mapping class group by Oscar Randal-Williams Combining results of Wahl, Galatius--Madsen--Tillmann--Weiss and Korkmaz one can identify the homotopy type of the classifying space of the stable nonorientable mapping class group N_infty (after plus-construction). At odd primes p, the F_p-homology coincides with that of Q_0({HP}^infty_+), but at the prime 2 the result is less clear. We identify the F_2-homology as a Hopf algebra in terms of the homology of well-known spaces. As an application we tabulate the integral stable homology of N_infty in degrees up to six. As in the oriented case, not all of these cohomology classes have a geometric interpretation. We determine a polynomial subalgebra of H^*(N_infty;F_2) consisting of geometrically-defined characteristic classes. (2) Commensurability classes of (-2,3,n) pretzel knot complements by Melissa L Macasieb and Thomas W Mattman Let K be a hyperbolic (-2,3,n) pretzel knot and M its complement in the 3-sphere. For these knots, we verify a conjecture of Reid and Walsh: there are at most three knot complements in the commensurability class of M. Indeed, if n is not 7, we show that M is the unique knot complement in its class. We include examples to illustrate how our methods apply to a broad class of Montesinos knots. (3) Model structures on the category of small double categories by Thomas M Fiore, Simona Paoli and Dorette Pronk In this paper we obtain several model structures on DblCat, the category of small double categories. Our model structures have three sources. We first transfer across a categorification-nerve adjunction. Secondly, we view double categories as internal categories in Cat and take as our weak equivalences various internal equivalences defined via Grothendieck topologies. Thirdly, DblCat inherits a model structure as a category of algebras over a 2-monad. Some of these model structures coincide and the different points of view give us further results about cofibrant replacements and cofibrant objects. As part of this program we give explicit descriptions for and discuss properties of free double categories, quotient double categories, colimits of double categories, horizontal nerve and horizontal categorification. (4) The R(S^1)-graded equivariant homotopy of THH(F_p) by Teena Meredith Gerhardt The main result of this paper is the computation of TR^n_alpha(F_p;p) for alpha in R(S^1). These R(S^1)-graded TR-groups are the equivariant homotopy groups naturally associated to the S^1-spectrum THH (F_p), the topological Hochschild S^1-spectrum. This computation, which extends a partial result of Hesselholt and Madsen, provides the first example of the R(S^1)-graded TR-groups of a ring. These groups arise in algebraic K-theory computations and are particularly important to the understanding of the algebraic K-theory of non-regular schemes. (5) Gromov-Witten invariants of blow-ups along submanifolds with convex normal bundles by Hsin-Hong Lai When the normal bundle N_{Z/X} is convex with a minor assumption, we prove that genus-0 GW-invariants of the blow-up Bl_Z X of X along a submanifold Z, with cohomology insertions from X, are identical to GW-invariants of X. Under the same hypothesis, a vanishing theorem is also proved. An example to which these two theorems apply is when N_{Z/X} is generated by its global sections. These two main theorems do not hold for arbitrary blow-ups, and counterexamples are included. (6) K-duality for stratified pseudomanifolds by Claire Debord and Jean-Marie Lescure This paper continues our project started in [J. Funct. Anal. 219 (2005) 109-133] where Poincare duality in K-theory was studied for singular manifolds with isolated conical singularities. Here, we extend the study and the results to general stratified pseudomanifolds. We review the axiomatic definition of a smooth stratification S of a topological space $X$ and we define a groupoid T^{S}X, called the S-tangent space. This groupoid is made of different pieces encoding the tangent spaces of strata, and these pieces are glued into the smooth noncommutative groupoid T^{S}X using the familiar procedure introduced by Connes for the tangent groupoid of a manifold. The main result is that C^*(T^{S}X) is Poincare dual to C(X), in other words, the S-tangent space plays the role in K-theory of a tangent space for X. (7) Global fixed points for centralizers and Morita's Theorem by John Franks and Michael Handel We prove a global fixed point theorem for the centralizer of a homeomorphism of the two-dimensional disk D that has attractor-repeller dynamics on the boundary with at least two attractors and two repellers. As one application we give an elementary proof of Morita's Theorem, that the mapping class group of a closed surface S of genus g does not lift to the group of C^2 diffeomorphisms of S and we improve the lower bound for g from 5 to 3. (8) On the homology of the space of knots by Ryan Budney and Fred Cohen Consider the space of 'long knots' in R^n, K_{n,1}. This is the space of knots as studied by V Vassiliev. Based on previous work [Budney: Topology 46 (2007) 1-27], [Cohen, Lada and May: Springer Lecture Notes 533 (1976)], it follows that the rational homology of K_{3,1} is free Gerstenhaber-Poisson algebra. A partial description of a basis is given here. In addition, the mod-p homology of this space is a 'free, restricted Gerstenhaber-Poisson algebra'. Recursive application of this theorem allows us to deduce that there is p-torsion of all orders in the integral homology of K_{3,1}. This leads to some natural questions about the homotopy type of the space of long knots in R^n for n>3, as well as consequences for the space of smooth embeddings of S^1 in S^3 and embeddings of S^1 in R^3. (9) Snowflake groups, Perron-Frobenius eigenvalues and isoperimetric spectra by Noel Brady, Martin R Bridson, Max Forester and Krishnan Shankar The k-dimensional Dehn (or isoperimetric) function of a group bounds the volume of efficient ball-fillings of k-spheres mapped into k-connected spaces on which the group acts properly and cocompactly; the bound is given as a function of the volume of the sphere. We advance significantly the observed range of behavior for such functions. First, to each nonnegative integer matrix P and positive rational number r, we associate a finite, aspherical 2-complex X_r,P and determine the Dehn function of its fundamental group G_r,P in terms of r and the Perron-Frobenius eigenvalue of P. The range of functions obtained includes elta(x)= x^s, where s is arbitrary rational number at least 2. Next, special features of the groups G_r,P allow us to construct iterated multiple HNN extensions which exhibit similar isoperimetric behavior in higher dimensions. In particular, for each positive integer k and rational sgeq(k+1)/k, there exists a group with k-dimensional Dehn function x^s. Similar isoperimetric inequalities are obtained for fillings modeled on arbitrary manifold pairs (M,partial M) addition to (B^k+1,S^k). === Subject: Full Time Research Professor in Mathematics Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The Facultad de Ciencias of the University of Colima invites applications for a faculty position in Mathematics. Applicants must possess a Ph.D. and postdoctoral experience. The successful candidate will be expected to build a strong research program, supervise students and teach at the undergraduate level. More information can be found in the following website: http://www.ucol.mx/docencia/facultades/fciencias/plaza.php === Subject: Ripples in ponds and the wave equation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) When you drop a pebble in a pond and see a series of ripples, are those ripples a) part of the solution to the 2-dimensional wave equation on the surface of the water, or b) the result of vertical oscillations of the water emanating from points below where the pebble was dropped? The remainder of this post is a more elaborate version of the above question. In three dimensions, any spherically symmetric solution to the wave equation is of the form psi(r,t) = f(r-t)/r + g(r+t)/r. In two dimensions, this is false. In particular, consider initial conditions of the form psi(r,0) = 0 d psi(r,0)/dt = q(r) Then in three dimensions, the value of psi at a point (r,t) is determined completely by the values of q at points a such that (a,0) is on the past light cone of (r,t). By contrast, in two dimensions, the value of psi at a point (r,t) dependes on the values of q at points a such that (a,0) is either *on or inside* the past light cone of (r,t). (These facts follow from the Asgeirsson Mean Value Theorem, which I've quoted at http://www.landsburg.com/appsmr.txt . The 3-dimensional result generalizes to any odd dimension and the 2-dimensional result to any even dimension, but here I'll be concerned only with the 2 and 3 dimensional cases.) Many years ago, I had a conversation with an assistant professor here at Rochester, in which he observed that sounds---that is, disturbances in the three-dimensional air---propagate according to the 3-d wave equation, which is why, when you hum for one second, I hear the humming (after a slight delay) for exactly one second. By contrast, a pebble tossed into a pond creates a disturbance in the two-dimensional surface, which propagates according to the 2-d wave equation, which is why the initial wave crest is followed by a series of ripples. If sound waves were followed by analogous ripples, the world would be a very noisy place. That assistant professor has left academics and I can find no trace of him on the World Wide Web. However, in preparing this post, I've found two web sites and one paper (by Sigurdur Helgason) that make essentially the same assertion; I wouldn't be surprised if the Helgason paper inspired my assistant professor and both web sites. I've recently been engaged in an extended conversation with a physicist who doubts that the ripple story is correct. His guess ---based, I am sure, on far better intuition than mine---is this: 1) Although the 2-d wave solution to the wave equation allows for ripples (by which I mean effects of anything outside the past light cone), these ripples are likely to be small and insignificant at distances far from the source on the scale of a wavelength. 2) The ripples we actually see on ponds are therefore not part of the 2-d solution, but instead the result of nonlinear vertical oscillations of the water at the center of the disturbance initiated by the dropped pebble. In partial support of this view, one of my correspondent's colleagues makes the following observation: The reason you can hear the conversations of very distant fisherman so clearly when you're out fishing on a calm day, is that the sounds are carried by surface vibrations of the water (the intensity of which only diminishes as the inverse distance in 2 dimensions rather than the inverse square of the distance in 3 dimensions), which disagrees with your description of how sounds would get smeared out in 2D. (The conversations are not smeared out.) So my big question is: Question 1: Why does a dropped pebble cause ripples? Is it (primarily) because these ripples are part of the solution to the 2-d wave equation, or (primarily) because of vertical oscillations in the water, or something else? I can imagine two ways to tackle this question. Method A is to actually solve the 2-d wave equation (at least numerically) for an appropriate disturbance and see what it looks like. Method B is to drop pebbles into ponds of various depths; it seems to me that if vertical oscillations were the culprit, then the ripples would look very different in a shallow pond than in a deep pond. Question 2: Has anyone carried out either of these methods? What are the results? Question 3: Would either of these methods in fact be definitive or is there something I'm overlooking? Question 4: Is there some other way to decide this question? -- Steven E. Landsburg http://www.landsburg.com/about2.html === Subject: surveying scientists' use of computers Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Computers are as important to modern scientists as test tubes, but we know surprisingly little about how scientists develop and use software in their research. To find out, the University of Toronto, Simula Research Laboratory, and the National Research Council of Canada have launched an online survey in conjunction with American Scientist magazine. If you have 20 minutes to take part, please go to: http://softwareresearch.ca/seg/SCS/scientific-computing-survey.html We'd be grateful if you could circulate the link to your colleagues at other institutions as well. Dr. Jo Hannay (Simula Research Laboratory) Dr. Hans Petter Langtangen (Simula Research Laboratory) Dr. Dietmar Pfahl (Simula Research Laboratory) Dr. Janice Singer (National Research Council of Canada) Prof. Greg Wilson (University of Toronto) === Subject: max/min distance between points in the plane Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I imagine that the below is a classical problem but I'd not seen it before. Initially I thought it was circle-packing, but now I'm suspicious that in fact it's slightly different. Here's the background... The following came up in an undergraduate problem-solving group: given 6 distinct points in the plane we can consider the 6-choose-2 distances between pairs of distinct points. If M is the largest of these distances, and m is the smallest, then show that M/m>=sqrt(3). In fact sqrt(3) isn't optimal, but I don't know what the optimal number is. One might ask what the answer is with n>=3 points in the plane: a compactness argument shows that there will be some r(n)>=1 such that M/m is always at least r(n) and furthermore such that r(n) is attained by some configuration of n points. For n=3 an equilateral triangle gives M/m=1 which is clearly optimal, so r(3)=1, and I've convinced myself that r(4)=sqrt(2) [and that a square is the unique best idea]. The general case is no doubt called the XYZ problem for some XYZ, and either there's a general formula for the optimal ratio for all n>=3 or there has been work computing lots of values of r(n). Can anyone give me any leads? There seemed to be a general concensus amongst the people in the problem-solving group who had thought about the problem for n=6 that *perhaps* an optimal configuration in this case was a regular pentagon with a point in the middle, giving a ratio of 2.sin(2.Pi/5). At least this proves r(6)<=2.sin(2.Pi/5) but I'm sure that r(6) will be known exactly. Does anyone know what it is? Kevin Buzzard ===