mm-327 === Subject: Re: JSH: Mathematical clarity > Let r = GCD(A, p). Then A/r is an algebraic integer which is > relatively prime to p, and p/r is an algebraic integer which is > relatively prime to A No, both are false. A trivial example is r = GCD(4, 2). Another example is r = GCD(sqrt(2), 2). But they can not both be false, so the conclusion (p/r) is coprime to (A/r) is true in a gcd-domain, by the definition of gcd. I see no errors in the remainder. It basically uses that the algebraic integers are a Bezout domain. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Mathematical clarity Adjunct Assistant Professor at the University of Montana. > Let r = GCD(A, p). Then A/r is an algebraic integer which is > relatively prime to p, and p/r is an algebraic integer which is > relatively prime to A >No, both are false. A trivial example is r = GCD(4, 2). Another >example is r = GCD(sqrt(2), 2). But they can not both be false, A = 12, B = 3, p = 18, C = 2 (I know p is supposed to be a rational prime, but of course you can see the pattern). Then gcd(A,p) = 6, A/6 = 2, p/6 = 3, and 2 is not coprime to p AND 3 is not coprime to 12. So, if we do not assume gcd(C,p)=1, then they can both be false at the same time. > so >the conclusion > (p/r) is coprime to (A/r) >is true in a gcd-domain, by the definition of gcd. This one ->is<- of course true, but that was not the statement given... -- ============ === Subject: Re: JSH: Mathematical clarity > The central point that won't go away though is that with > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > while the factors *are* algebraic integers with an algebraic integer > x, something odd happens if you try to remove 7 and its factors from > both sides, as while the right remains in the ring of algebraic > integethe factors on the left are, in general, forced out. You've never proven this to be true nor have you established any significance if it is true. You have, however, asser it endlessly.By the way, what if you do *not* replace a_2 with b_2 - 1 ? Then you have your original equation: (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x*2 + 30x + 2) How does your Ôdivide by 7' exercise play? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com === Subject: Re: JSH: Mathematical clarity > general x) is that we have f_2 and 2/f_2 in there. And all that's > important and all you can prove is that the product (5a_1(x)/f_1 + f_2)(5b_2(x)/f_2 + 2/f_2) is an algebraic integer. That's just forced stupidity as my *point* is that you can't stay in > the ring of algebraic integers if 2/f_2 is not an algebraic integer, > while others have tried to argue that all that matters is that > (5b_2(x) + 2)/f_2 is an algebraic integer. James, pray pay attention to what a ring *is*. The ring axioms do not say anything about division. They say that multiplication is distributive over addition (i.e.: a * (b + c) = (a * b + a * c).) There is nothing said about a distributive property of division. When you write: (5b_2(x)/f_2(x) + 2/f_2(x)) rather than (5b_2(x) + 2)/f_2(x) you are abusing notation. It is the latter that matteand there is *no* theorem so that you can rewrite the latter to the former in general. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Mathematical clarity > (5a_1(x)/f_1 + f_2)(5b_2(x)/f_2 + 2/f_2) = 25x^2 + 30x + 2 > typo then the algebra tells you that the factors of 2 on the right are > f_2 > and 2/f_2, and if f_2 is a non-unit factor of 7, then you're not in > the ring of algebraic integers. James, it does not matter whether individual terms are or aren't > algebraic > integeonly that the whole expression is an algebraic integer. >> Do you deny then that the factors of 2 on the right are f_2 and 2/f_2 >> on the left? >> Can't you see that f_2 (2/f_2) = 2. >> Can you? >> Can you admit the truth here? > It's a simple thing we drum into our undergrads that the sum and > product of irrationals can be rational, an irrational number raised > to an > irrational power can be rational. >> How does that relate to the fact that the factors of two are f_2 and >> 2/f_2? >> You can't see the analogy? When operating in a ring, you have only the elements of that ring > available. > You admit this but fail to adhere to your own rules. It's not *my* rule! It's basic mathematics!!! I think your statement there tells a lot about you. Mathematics isn't about opinions or personalities. It's clear that you've lost sight of that fact. JSh has lost sight ot the fact that when a+b is in a ring it does not guarantee that either a or b need be in that ring. Consider that 3/7 + 4/7 is an integer but that neither 3/7 nor 4/7 is an integer, at least in standard mathematics. According to JSH, 3/7 + 4/7 cannot be an integer unless both 3/7 and 4/7 are already integers. This is not the arithmetic that most of us know. === Subject: Easy integration ? Sorry guys, I have another integration problem: The integral is: with p,q in |N. I tried integration by parts: Set: u(x) = (1-x)^q -> u'(x) = -q(1-x)^(q-1) v'(x) = x^p -> v(x) = 1/(p+1) * x^(p+1) I tried to calc: ((1-x)^q) * (1/(p+1) * x^(p+1)) - Integral from 0 to 1 over (-q(1-x)^(q-1)) * x^p dx The part before the minus gets zero, right, because filling in 1 and then 0 and subtracting these values from each other leads to 0 - 0 = 0. So there is still to calc the part after the minus: Integral from 0 to 1 over (-q(1-x)^(q-1)) * x^p dx But this does not look easier to me than the term before. Can anyone post the solution please ? I am fed up with this ! Thanks a lot. === Subject: Re: Easy integration ? > Sorry guys, I have another integration problem: The integral is: > Call the result I(p,q) with p,q in |N. I tried integration by parts: Set: u(x) = (1-x)^q -> u'(x) = -q(1-x)^(q-1) > v'(x) = x^p -> v(x) = 1/(p+1) * x^(p+1) I tried to calc: ((1-x)^q) * (1/(p+1) * x^(p+1)) - Integral from 0 to 1 over (-q(1-x)^(q-1)) > * x^p dx ^^^^^^ Error here. You want v*du (not v'*du) Well, you are almost done (evaluate from 0 to 1). Call the integral I(p,q). the first term is zero FOR p,q >0 and you have I(p,q) = (q/(p+1)) I(p+1,q-1) Now, do it again and again and again and again and again ... (just don't give up) again ... = q(q-1)/(p+1)(p+2) I(p+2,q-2) again ... = q(q-1)(q-2)/(p+1)(p+2)(p+3) I(p+3,q-3) etc. (until q=0) Then you only have to evaluate I(p+q,0) and put it all together. SPOILER BELOW: etc. =q(q-1)....1/(p+1)(p+2)...(p+q) I(p+q,0) Now you only have to do INT(x^(p+q)dx) = 1/(p+q+1) to get q!/(p+1)(p+2)...(p+q+1) = q!p!/(p+q+1)! === Subject: Re: Easy integration ? from integrating by parts once, you can see a pattern. the first part is 0 and the second part you should get int from 0 to 1 (-q(1-x)^(q-1) * x^(p+1) / (p+1) this integral is the same as the one yu star with except that p has increased by 1 and q decreased by 1. just factor q/p+1 and integrate by parts again, so you should get sth that cancels similarly and another integral with powers p+2 and q-2 this time. Inductively, integrating by parts q times, you should arrive eventually at a single integral. q!/(p+1)(p+2)....(p+q) times integral (x^(p+q) dx). which is straight forward to compute. I hope the idea is clear. > Sorry guys, I have another integration problem: > The integral is: > with p,q in |N. > I tried integration by parts: > Set: > u(x) = (1-x)^q -> u'(x) = -q(1-x)^(q-1) > v'(x) = x^p -> v(x) = 1/(p+1) * x^(p+1) > I tried to calc: > ((1-x)^q) * (1/(p+1) * x^(p+1)) - Integral from 0 to 1 over (-q(1-x)^(q-1)) > * x^p dx > The part before the minus gets zero, right, because filling in 1 and then 0 > and subtracting these values from each other leads to 0 - 0 = 0. > So there is still to calc the part after the minus: > Integral from 0 to 1 over (-q(1-x)^(q-1)) * x^p dx > But this does not look easier to me than the term before. > Can anyone post the solution please ? > I am fed up with this ! === Subject: Need help with a tricky proof (group theory?) I have been stuck on this for several days and can't find a solution (or a way to avoid it). Given: a+b = c+d e+b = c+f a+g = h+d Prove: e+g = h+f Here's tricky part: You can't use subtraction -- it's easy otherwise. You can use: 1. associativitiy of + 2. commutatity of + 3. left and right cancelability of + 4. the usual properties of = Any help would be GREATLY apprecia. Dan === Subject: Re: Need help with a tricky proof (group theory?) > Using only: + is associative, commutative, cancellative Given: a + b = c + d > b + e = c + f > a + g = d + h > ------------- > Prove: e + g = f + h Easy: add the 2nd and 3rd equations then use 1st to cancel a + b = c + d. QED Symmetry is revealed by renaming as below a + b = A + B OLD: c d e f g h b + c = B + C NEW: B A c C d D a + d = A + D ------------- c + d = C + D Now it is quite obvious. -Bill Dubuque === Subject: Re: Need help with a tricky proof (group theory?) |I have been stuck on this for several days and can't find a solution (or a |way to avoid it). | |Given: | |a+b = c+d |e+b = c+f |a+g = h+d | |Prove: | |e+g = h+f | |Here's tricky part: You can't use subtraction -- it's easy otherwise. You |can use: | |1. associativitiy of + |2. commutatity of + |3. left and right cancelability of + |4. the usual properties of = | |Any help would be GREATLY apprecia. well, doesn't it look as though what you really need to do here is to find some way of proving an equation of the form: e+g+(cancelable junk) = h+f+(the exact same cancelable junk) ? have you tried looking for a way to prove some equation of that form? how might you be able to prove some equation of this form? you need to get e and g on one side, and h and f on the other side... -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Need help with a tricky proof (group theory?) Adjunct Assistant Professor at the University of Montana. >I have been stuck on this for several days and can't find a solution (or a >way to avoid it). >Given: >a+b = c+d >e+b = c+f >a+g = h+d >Prove: >e+g = h+f >Here's tricky part: You can't use subtraction -- it's easy otherwise. You >can use: >1. associativitiy of + >2. commutatity of + >3. left and right cancelability of + >4. the usual properties of = >Any help would be GREATLY apprecia. Add all three equations as follows: a + b = c + d c + f = e + b h + d = a + g Then we have a+b+c+f+h+d = c+d+e+b+a+g. Cancelling a, b, c, and d from both sides we have h+f = e+g. -- ============ === Subject: Re: Need help with a tricky proof (group theory?) >I have been stuck on this for several days and can't find a solution (or a >way to avoid it). Given: a+b = c+d >e+b = c+f >a+g = h+d Prove: e+g = h+f Here's tricky part: You can't use subtraction -- it's easy otherwise. You >can use: 1. associativitiy of + >2. commutatity of + >3. left and right cancelability of + >4. the usual properties of = Any help would be GREATLY apprecia. Add all three equations as follows: a + b = c + d > c + f = e + b > h + d = a + g Then we have a+b+c+f+h+d = c+d+e+b+a+g. Cancelling a, b, c, and d from both sides we have h+f = e+g. ************************************************************* ** At the very best I find this an odd question: in the present context, cancelability is just another name for substraction... Tonio === Subject: Re: Need help with a tricky proof (group theory?) Adjunct Assistant Professor at the University of Montana. >>I have been stuck on this for several days and can't find a solution (or a >>way to avoid it). >>Given: >>a+b = c+d >>e+b = c+f >>a+g = h+d >>Prove: >>e+g = h+f >>Here's tricky part: You can't use subtraction -- it's easy otherwise. You >>can use: >>1. associativitiy of + >>2. commutatity of + >>3. left and right cancelability of + >>4. the usual properties of = >>Any help would be GREATLY apprecia. >> Add all three equations as follows: >> a + b = c + d >> c + f = e + b >> h + d = a + g >> Then we have >> a+b+c+f+h+d = c+d+e+b+a+g. >> Cancelling a, b, c, and d from both sides we have >> h+f = e+g. >************************************************************ *** >At the very best I find this an odd question: in the present context, >cancelability is just another name for substraction... Well, yes. The rules allowed mean that he is working in a cancellation monoid (or at least, a cancellation commutative semigroup); so the semigroup embedds in the enveloping group, from which you can get what to do. -- ============ === Subject: Re: Need help with a tricky proof (group theory?) >>I have been stuck on this for several days and can't find a solution (or a >>way to avoid it). >>Given: >>a+b = c+d >>e+b = c+f >>a+g = h+d >>Prove: >>e+g = h+f >>Here's tricky part: You can't use subtraction -- it's easy otherwise. You >>can use: >>1. associativitiy of + >>2. commutatity of + >>3. left and right cancelability of + >>4. the usual properties of = >>Any help would be GREATLY apprecia. >> Add all three equations as follows: >> a + b = c + d >> c + f = e + b >> h + d = a + g >> Then we have >> a+b+c+f+h+d = c+d+e+b+a+g. >> Cancelling a, b, c, and d from both sides we have >> h+f = e+g. >************************************************************ *** >At the very best I find this an odd question: in the present context, >cancelability is just another name for substraction... I am not sure exactly what you mean by that, but I think it is true that any commutative semigroup with cancellation embeds in a group. You can adjoin a 0 if there is not one already, and then do the standard trick of forming a group whose elements are equivalence classes of pairs (a,b) with equivalence relation (a,b) ~ (c,d) <=> a+d=b+c. To prove that works, you have to show transitivity of ~, which involves proving that a + d = b + c and c + f = d + e imply a + f = b + c. The problem above is similar but has one more step. It seems a reasonable question to me when viewed in that context. Note that it is not true that a non-commutative semigroup with left and right cancellation necessarily embeds in a group. An example was found by Mal'cev in 1937. Derek Holt. === Subject: Re: Need help with a tricky proof (group theory?) >>I have been stuck on this for several days and can't find a solution (or a >>way to avoid it). >>Given: >>a+b = c+d >>e+b = c+f >>a+g = h+d >>Prove: >>e+g = h+f >>Here's tricky part: You can't use subtraction -- it's easy otherwise. You >>can use: >>1. associativitiy of + >>2. commutatity of + >>3. left and right cancelability of + >>4. the usual properties of = >>Any help would be GREATLY apprecia. >Add all three equations as follows: >a + b = c + d >c + f = e + b >h + d = a + g >Then we have >a+b+c+f+h+d = c+d+e+b+a+g. >Cancelling a, b, c, and d from both sides we have >h+f = e+g. >-- In determining what to add, first give names to the three differences: eq1 = (a + b) - (c + d) eq2 = (e + b) - (c + f) eq3 = (h + d) - (a + g) Of course eq1, eq2, eq3 will all simplify to zero, but we give the equations symbolic names. You want a different formula for e + g. e and g appear only once each in the definitions of eq1, eq2, eq3. Solve for them e := c + f - b + eq2 g := h + d - a - eq3 So e + g - (h + f) = c + d - a - b + eq2 - eq3 = -eq1 + eq2 - eq3 So far, all of this work is on your scratch sheets, not in your submit solution. The minus signs mean you must reverse the two sides of the first and third equalities, but retain the second equality as is. c + d = a + b e + b = c + f a + g = h + d Now, since all three coefficients are 1, you multiply 1 by each equation and add the three pairs of products. Then you cancel as in the earlier solution. -- John Adams served two terms as Vice President and one as President, but lost reelection. Later his son became President despite losing the popular vote. That son lost his reelection attempt badly. Now history is repeating itself. pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: Re: Need help with a tricky proof (group theory?) Well done! That's exactly what I was looking for. Thanks! Dan >I have been stuck on this for several days and can't find a solution (or a >way to avoid it). Given: a+b = c+d >e+b = c+f >a+g = h+d Prove: e+g = h+f Here's tricky part: You can't use subtraction -- it's easy otherwise. You >can use: 1. associativitiy of + >2. commutatity of + >3. left and right cancelability of + >4. the usual properties of = Any help would be GREATLY apprecia. > Add all three equations as follows: > a + b = c + d > c + f = e + b > h + d = a + g > Then we have > a+b+c+f+h+d = c+d+e+b+a+g. > Cancelling a, b, c, and d from both sides we have > h+f = e+g. > -- > ============================================================= ========= > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin ( Calvin and Hobbes ) > ============================================================= ========= === Subject: Programmers Wan for Computer Graphics Startup Near adelphia My name is Stan Schwartz, and I'm a University of Pennsylvania Ph.D. and an independent inventor. I'm submitting two computer graphics patents to the USPTO during the next several weeks. Technologies derived from these patents have application in several different market segments, including still image photomanipulation, movie special effects and post-production, web animation, and video games. I'm developing and marketing one or more of the applications through a startup company. I've produced demo output that clearly demonstrates the viability of my approaches, which I've been showing to potential licensees and investors. This demo output is currently being genera through a loosely connec series of scripts. To be productized the scripts will need to be recoded into standalone industrial strength applications that are much faster, bullet-proof, and that have slick GUIs. I'm looking for a lead/senior programmer to participate in and organize the implementation side of the startup and probably one or more junior programmers. Successful candidates for the programming team should have some to many of the following skills/characteristics: Extensible knowledge of/comfort with math, as exemplified by one or more of: Computer vision training/experience - feature location/ extraction, motion estimation, object recognition; Image processing training/experience - you know what a convolution kernel is; you've implemen a morphing algorithm; Linear algebra training/experience - matrix manipulation, familiarity with numeric computation packages, such as Matlab, or equivalent; Art/math programming play, mathematical visualization; 3D modeling, especially for character animation, especially if you've hacked/manipula low-level 3D data representations in a variety of formats; GUI design/building, graphic design training a plus; Analysis of algorithms; C/C++, Linux/Unix a plus; Startup and/or major fielded product development experience required for the lead/senior programmer, a plus for juniors; Within commuting distance of adelphia - we may move and/or outsource, but this is where/how we'll start; Relevant educational background is a plus, but I'm more interes in what you can demonstrate to me that you know, can do, and have done, than what you've formally studied; Intellectual ßexibility, creativity, and ability to work and brainstorm collaboratively; Open-ended time commitment - After growing this, I have additional ideas in the queue that could be spun off into future products; and Demonstrable artistic skills and training in figurative painting, sculpture, and/or life-drawing are a plus. We have access to seed capital, and we'd prefer if participants accept some portion of salary as equity in our venture. If you're interes in participating, send me a cover letter and resume at cgstartup@earthlink.net. I'm willing to show you the demo and provide more details about the technology, although first you'll have to sign a non-disclosure agreement. The demo can also be pos temporarily to the web for online viewing, accompanied by an explanatory phone call. For Investors/Licensees: If you're a potential investor in the startup or a representative of a company that is potentially interes in licensing one of the applications and are interes in seeing the demo and hearing a description of the technologies, also under conditions of non-disclosure, please let me know through the means described above. Thanks for your time, Stan === Subject: Re: Modified Decker example, reference [snip] Posters trying to dispute the algebra when presen with the basic > algebra claim that it's the sum (5b_2(x) + 2) that mattebut that > is refu by noting that the 2 in that expression is a factor of 7(2) > in 7(25x^2 + 30x + 2) as the constants are NOT a product of the full sums, but are in fact > products of the constants 7 and 2 within those sums, so by trying to > include variable expressions they are relying on claims insupportable > mathematically. > Consider (4/3 + 2/3) (9/2 + 3/2) = (4/3)(9/2) + (4/3)(3/2) + (2/3)(9/2) + (2/3)(3/2) = 6 + 2 + 3 + 1 = 12 Here we have (a+b)(c+d) = ab + ac + bc + bd = g and (a+b), (c+d), ab, ac, bc, bd, and g are all integers. However none of a, b, c or d is an integer. So the fact that the partial sums as well as full sums are intege does not mean that a,b,c, and/or d must be integers. In particular, the fact that the product of the constant terms, b and d, is an integer does not mean that b and d must be integers. - William Hughes === Subject: Re: Modified Decker example, reference > [snip] > Posters trying to dispute the algebra when presen with the basic > algebra claim that it's the sum (5b_2(x) + 2) that mattebut that > is refu by noting that the 2 in that expression is a factor of 7(2) > in 7(25x^2 + 30x + 2) as the constants are NOT a product of the full sums, but are in fact > products of the constants 7 and 2 within those sums, so by trying to > include variable expressions they are relying on claims insupportable > mathematically. > Consider (4/3 + 2/3) (9/2 + 3/2) = (4/3)(9/2) + (4/3)(3/2) + (2/3)(9/2) + (2/3)(3/2) > = 6 + 2 + 3 + 1 = 12 > That's ok in the field of algebraic numbebut doesn't exist in the ring of algebraic integers. > Here we have (a+b)(c+d) = ab + ac + bc + bd = g The algebra doesn't care what the ring is. > and (a+b), (c+d), ab, ac, bc, bd, and g are all integers. > However none of a, b, c or d is an integer. > You sum first for your example, or you're out of the ring. But what about (x+2)/3? Now here's a pop quiz, for algebraic integer x, is (x+2)/3 in the ring of algebraic integers? Can you *add* first to say that it always is? > So the fact that the partial sums as well as full sums are intege > does not mean that a,b,c, and/or d must be integers. In particular, the fact that the product of the constant terms, > b and d, is an integer does not mean that b and d must be integers. - William Hughes Think about (x+2)/3. Subject: Re: Modified Decker example, reference === [non-math dele] > Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of > non-polynomial factors. Notice that despite not being polynomials they are algebraic integers > if x is an algebraic integer because a_1(x) and a_2(x) are the two > roots of a^2 - (x - 1)a + 7(x^2 + x). Using the substitution a_2(x) = b_2(x) - 1 with the factorization > gives (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) where a_1(0) = b_2(0) = 0. Then there must exist a_1(x) = 7b_1(x), and that substitution gives (5(7)b_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) and the 7 can now easily be removed from both sides giving (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2. But, it is rather easy to show that in general with algebraic integer > x, b_1(x) is not an algebraic integer. At this point it would seem obvious to conclude that the desire to replace a_1(x) with 7b_1(x) is not a valid replacement. Several posters have decided to argue against the correct conclusion > on several newsgroups. I have seen arguments over what the conclusion implies, but not its validity. However, their positions force you out of the ring of algebraic > integers as well, as for instance, if each of the factors (5a_1(x) + > 7) and (5b_2(x) + 2) have sqrt(7) as a factor, then from basic > algebra, you end up with sqrt(7) and 2/sqrt(7) as the factors of 2 which follows easily enough by simply dividing each factor by sqrt(2). In the algebraic numbeyes, in the algebraic integeno. Posters trying to dispute the algebra when presen with the basic > algebra claim that it's the sum (5b_2(x) + 2) that mattebut that > is refu by noting that the 2 in that expression is a factor of 7(2) > in 7(25x^2 + 30x + 2) as the constants are NOT a product of the full sums, but are in fact > products of the constants 7 and 2 within those sums, so by trying to > include variable expressions they are relying on claims insupportable > mathematically. Whether it's the sum that matters depends entirely on what you're trying to prove. If you want to prove that (5b_2(x)+2) has a particular factor, then worry only about the sum. If you want to find a factor of 2, then the sum is irrelevant. Which are you concerned about? [more non-math dele] Note: perhaps if you restric your discussions to the mathematics instead of the mathematicians, more responses would be limi to the math as well. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Modified Decker example, reference [non-math dele] Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of > non-polynomial factors. Notice that despite not being polynomials they are algebraic integers > if x is an algebraic integer because a_1(x) and a_2(x) are the two > roots of a^2 - (x - 1)a + 7(x^2 + x). Using the substitution a_2(x) = b_2(x) - 1 with the factorization > gives (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) where a_1(0) = b_2(0) = 0. Then there must exist a_1(x) = 7b_1(x), and that substitution gives (5(7)b_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) and the 7 can now easily be removed from both sides giving (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2. But, it is rather easy to show that in general with algebraic integer > x, b_1(x) is not an algebraic integer. At this point it would seem obvious to conclude that the desire to > replace a_1(x) with 7b_1(x) is not a valid replacement. > No it's not obvious. After all constants like 7 normally multiply through in a rather basic way. So it makes sense that it does here as well. It actually takes an illogical leap to conclude that somehow, someway the mathematics decides to behave in an inconsistent way. === Subject: Re: Modified Decker example, reference > Recently Rick Decker, a professor at Hamilton College, apparently > trying to refute my research *Successfully* refuting your research - and why don't you at least quote Decker's conclusion? Is this a propaganda tactic - not stating the opponent's argument because other people might actually see that it makes sense ? came up with a quadratic example, which I > like because it's a quadratic, and easier to manipulate than the > cubics I've used before. If you wish to see his original post here are some headers which also > show that he posts from Hamilton College: Subject: Re: Mathematical consistency, courage > Why is it so important to keep noting that he posts from Hamilton College? Might it not be more important to quote the *substance* of what he said, rather than where he posts from ??? > Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of > non-polynomial factors. > Just to clarify: these *are* polynomials in the number 5 . The coefficients a_1(x) and a_2(x) are not polynomial functions of the variable x. A few months ago, these expressions were given terms of two variables, m and x. The number 5 here plays the role of what was formerly x; the variable x corresponds to what was formerly called m . In order to conclude that the factors a_1(x) and a_2(x) are roots of a^2 - (x - 1)*a + 7*(x^2 + x) you must regard 5 as if it were a polynomial *variable* (which is of course rather bizarre, since 5 is a fixed number). We have accep this weird formulation, but it is probably a good idea to keep in mind where it had its origins. > Notice that despite not being polynomials they are algebraic integers > if x is an algebraic integer because a_1(x) and a_2(x) are the two > roots of a^2 - (x - 1)a + 7(x^2 + x). > Yes. > Using the substitution a_2(x) = b_2(x) - 1 with the factorization > gives (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) where a_1(0) = b_2(0) = 0. Then there must exist a_1(x) = 7b_1(x), and that substitution gives (5(7)b_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) and the 7 can now easily be removed from both sides giving (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2. But, it is rather easy to show that in general with algebraic integer > x, b_1(x) is not an algebraic integer. Which SHOULD tell you that you have chosen the wrong way to factor 7 out of the polynomial factors. But you're refusing to hear the message. > Several posters have decided to argue against the correct conclusion > on several newsgroups. > It's not correct. When you factor out 7, the result on the left side SHOULD be two algebraic integers. There is a theorem that says that should happen, *if* 7 is factored out in the correct way. You think there is only ONE way to factor out 7, and you note that when you do the factoring in that way, you find that not both factors can be algebraic integers. You are thus contradicting the theorem. There are only three possibilities: 1. The theorem is false, or 2. Your way of factoring out 7 is not the right way to do it. 3. Mathematics is inconsistent. I think we agree that mathematics is not inconsistent, or at least, if it is, it will not be shown demonstra at such a simple level. Is the theorem false? Let's see. Here it is: Theorem. Suppose A, B, C, and p are algebraic integeand p is coprime to C, and A*B = p*C. Then there exist algebraic integers f1 and f2 such that 1. f1*f2 = p 2. A/f1 and B/f2 are both algebraic integeand 3. (A/f1)*(B/f2) = C. Proof. By a theorem of Dedekind, GCD's exist in the ring of algebraic integers. Let f1 = GCD(A, p) and f2 = GCD(B, p). Clearly f1 and f2 have the desired properties 1., 2., and 3., because A/f1 and B/f2 are both coprime to p. QED. Here is how you apply the theorem to Rick Decker's quadratic: Let p = 7, C = 25*x^2 + 30*x + 2, A = (5 a_1(x) + 7), and B = (5 a_1(x) + 7) = (5 b_2(x) = 2). For *most* values of x (including x = 1, 2, 3, 4, 6, ...), C and 7 are coprime. So there is only one possible remaining way to escape the conclusion that your factorization is the wrong one. That is, if Dedekind's theorem about the existence of GCDs in the ring of algebraic integers is wrong. It is a deep and difficult theorem, requiring finiteness of the class number. Do you want to take on trying to find an error in Dedekind's proof? [That's really unnecessary, since we already know where you have made your error.] > However, their positions force you out of the ring of algebraic > integers as well, as for instance, if each of the factors (5a_1(x) + > 7) and (5b_2(x) + 2) have sqrt(7) as a factor, then from basic > algebra, you end up with sqrt(7) and 2/sqrt(7) as the factors of 2 > NOT TRUE. It is perfectly possible that (5 b_2(x) + 2)/sqrt(7) is an algebraic integer, even if 2/sqrt(7) is not. In fact, Decker showed explicitly that, for x = 1, (5 b_2(1) + 2)/sqrt(7) = (-5 *sqrt(-2) + sqrt(7)), and the latter expression - the thing that you keep not quoting - CLEARLY IS an algebraic integer. So here you have it. You have just said that (5 b_2(x) + 2)/sqrt(7) cannot be an algebraic integer. But obviously, from the above, it CAN be, and it is when x = 1. Let me quote you again, just so you can see the problem clearly. Just above you said: ...if each of the factors (5a_1(x) + 7) and (5b_2(x) + 2) have sqrt(7) as a factor, then from basic algebra, you end up with sqrt(7) and 2/sqrt(7) as the factors of 2 meaning, one must assume, that 2/sqrt(7) would have to be an algebraic integer. But Decker's calculation clearly shows that (5 b_2(1) + 2) has sqrt(7) as a factor, and this in NO WAY implies that 2 is divisible by sqrt(7) in the algebraic integers. > which follows easily enough by simply dividing each factor by sqrt(2). > You mean sqrt(7) here, not sqrt(2). > Posters trying to dispute the algebra when presen with the basic > algebra claim that it's the sum (5b_2(x) + 2) that mattebut that > is refu by noting that the 2 in that expression is a factor of 7(2) > in 7(25x^2 + 30x + 2) as the constants are NOT a product of the full sums, but are in fact > products of the constants 7 and 2 within those sums, so by trying to > include variable expressions they are relying on claims insupportable > mathematically. Continued arguments from posters indicate that at least these people > in math society have some agenda with a strong emotional component, > which must exist for them to push so desperately against basic > mathematics. It remains to be seen if their desperation is shared by mainstream > mathematicians. > It's not desperation. It's not even algebra. It's just plain arithmetic that proves you wrong. And STILL you refuse to quote exactly what that arithmetic says. Better start a new thread on this, eh? > For more on non-polynomial factorizations, see my blog archives: === Subject: Re: Modified Decker example, reference > Recently Rick Decker, a professor at Hamilton College, apparently > trying to refute my research > *Successfully* refuting your research - and why don't you at > least quote Decker's conclusion? Is this a propaganda tactic - > not stating the opponent's argument because other people might > actually see that it makes sense ? > No. came up with a quadratic example, which I > like because it's a quadratic, and easier to manipulate than the > cubics I've used before. If you wish to see his original post here are some headers which also > show that he posts from Hamilton College: Subject: Re: Mathematical consistency, courage > Why is it so important to keep noting that he posts from > Hamilton College? Might it not be more important to quote > the *substance* of what he said, rather than where he posts > from ??? > Why ask why? > Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of > non-polynomial factors. Just to clarify: these *are* polynomials in the number 5 . Irrelevant. What's important for the discussion is that the everything works. Here if you take the roots of a^2 - (x - 1)a + 7(x^2 + x) and substitute one in for a_1(x), and the other for a_2(x), then you have (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) as asser. Oh yeah, and why don't you off as I strongly sugges before? I'm sort of tired of you chasing after my posts like a lost puppy. === Subject: Re: Modified Decker example, reference > Recently Rick Decker, a professor at Hamilton College, apparently > trying to refute my research > *Successfully* refuting your research - and why don't you at > least quote Decker's conclusion? Is this a propaganda tactic - > not stating the opponent's argument because other people might > actually see that it makes sense ? No. came up with a quadratic example, which I > like because it's a quadratic, and easier to manipulate than the > cubics I've used before. If you wish to see his original post here are some headers which also > show that he posts from Hamilton College: Subject: Re: Mathematical consistency, courage > Why is it so important to keep noting that he posts from > Hamilton College? Might it not be more important to quote > the *substance* of what he said, rather than where he posts > from ??? Why ask why? > Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of > non-polynomial factors. Just to clarify: these *are* polynomials in the number 5 . What's important for the discussion is that the everything works. > Here if you take the roots of > a^2 - (x - 1)a + 7(x^2 + x) > and substitute one in for a_1(x), and the other for a_2(x), then you have > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > as asser. > Oh yeah, and why don't you off as I strongly sugges before? > I'm sort of tired of you chasing after my posts like a lost puppy. Really mature, James. Isn't there any way to refrain from F bombs? As for Nora, she shouldn't quit posting because she has a right to post. James just doesn't like her because she's proven him wrong over and over and over and...... well you get the picture. If you don't like it James, DON'T POST. Grow up and act like a man instead of a 2 year old. > === Subject: Re: Modified Decker example, reference where a_1(0) = b_2(0) = 0. >> Then there must exist a_1(x) = 7b_1(x), and that substitution gives Proof? Why must 7 divide a_1(x) for all x? As you've admit, Rick has provided a counter example at x=1. That is sufficient to demonstrate your assertion is wrong. Sticking with Z, 2 divides x+2 at x=0, but doesn't divide x+2 for all x. (5(7)b_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) and the 7 can now easily be removed from both sides giving (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2. But, it is rather easy to show that in general with algebraic integer > x, b_1(x) is not an algebraic integer. > And so 7 didn't divide a_1 as you claimed. You have some contradiction so what's the conclusion? That 7 doesn't factor as 7*1 in general or... > Several posters have decided to argue against the correct conclusion > on several newsgroups. However, their positions force you out of the ring of algebraic > integers as well, as for instance, if each of the factors (5a_1(x) + > 7) and (5b_2(x) + 2) have sqrt(7) as a factor, then from basic > algebra, you end up with sqrt(7) and 2/sqrt(7) as the factors of 2 > no we don't. This is only justifiable if we claimed that since (5b_2(x)+2)/sqrt(7) is integral that both 5b/sqrt(7) and 2/sqrt(7) are integral. We've given you examples showing that is not true, but you accuse us of peddling bull . > which follows easily enough by simply dividing each factor by sqrt(2). > typo? > Posters trying to dispute the algebra when presen with the basic > algebra claim that it's the sum (5b_2(x) + 2) that mattebut that > is refu by noting that the 2 in that expression is a factor of 7(2) > in 7(25x^2 + 30x + 2) > no it's not refu like this. > as the constants are NOT a product of the full sums, but are in fact > products of the constants 7 and 2 within those sums, so by trying to > include variable expressions they are relying on claims insupportable > mathematically. oh crap, not the variable/constant thing again... as you're not claiming anything about division in the ring of polynomials that's a little misleading. In fact your claim to greatness is that you're *not* looking at polynomial factorizations! Continued arguments from posters indicate that at least these people > in math society have some agenda with a strong emotional component, > which must exist for them to push so desperately against basic > mathematics. It remains to be seen if their desperation is shared by mainstream > mathematicians. For more on non-polynomial factorizations, see my blog archives: is this your conclusion? Wow, that's a big proof by contradiction. I don't recall some of these things being in your hypothesis though. === Subject: Re: Modified Decker example, reference where a_1(0) = b_2(0) = 0. >> Then there must exist a_1(x) = 7b_1(x), and that substitution gives > Proof? Why must 7 divide a_1(x) for all x? > Are you assuming a ring? If so, what ring? === Subject: Re: Modified Decker example, reference has star another new thread here only covering the same ground covered by a host of previous threads, but not yet filled with proofs of his many errors in covering that ground that his other threads have all attrac. It seems to be a standard JSH tactic: when faced with undeniable proof of errors in one thread, JSH merely restates them in a new thread. The material in this thread has been worked over in other threads. The only new stuff is JSH's wording of his astonishment that the mathematical world has so little regard for his genius. For those who read him for amusement, go ahead, enjoy. For any other purposes, this thread is of no use. === Subject: Re: Modified Decker example, reference > has star another new thread here only covering the same > ground covered by a host of previous threads, but not yet filled with > proofs of his many errors in covering that ground that his other threads > have all attrac. > It seems to be a standard JSH tactic: when faced with undeniable proof > of errors in one thread, JSH merely restates them in a new thread. > The material in this thread has been worked over in other threads. > The only new stuff is JSH's wording of his astonishment that the > mathematical world has so little regard for his genius. > For those who read him for amusement, go ahead, enjoy. For any other > purposes, this thread is of no use. This would probably be a great disclaimer to post after any JSH post (along with maybe some additional comments like: may cause nausea, diarrhea, ...) Jack === Subject: Re: Modified Decker example, reference > has star another new thread here only covering the same > ground covered by a host of previous threads, but not yet filled with > proofs of his many errors in covering that ground that his other threads > have all attrac. It seems to be a standard JSH tactic: when faced with undeniable proof > of errors in one thread, JSH merely restates them in a new thread. The material in this thread has been worked over in other threads. The only new stuff is JSH's wording of his astonishment that the > mathematical world has so little regard for his genius. For those who read him for amusement, go ahead, enjoy. For any other > purposes, this thread is of no use. This would probably be a great disclaimer to post after any JSH post > (along with maybe some additional comments like: may cause nausea, > diarrhea, ...) Jack Why editorialize at all? What are you afraid of? Why can't I just post and you let the post sit there without feeling a need to communicate something from yourself along with it? Can you drop the emotion long enough to answer those basic questions? I'm curious if you have any idea yourself what your motivation is. === Subject: Re: Modified Decker example, reference > has star another new thread here only covering the same > ground covered by a host of previous threads, but not yet filled with > proofs of his many errors in covering that ground that his other threads > have all attrac. It seems to be a standard JSH tactic: when faced with undeniable proof > of errors in one thread, JSH merely restates them in a new thread. The material in this thread has been worked over in other threads. The only new stuff is JSH's wording of his astonishment that the > mathematical world has so little regard for his genius. For those who read him for amusement, go ahead, enjoy. For any other > purposes, this thread is of no use. This would probably be a great disclaimer to post after any JSH post > (along with maybe some additional comments like: may cause nausea, > diarrhea, ...) Jack Why editorialize at all? What are you afraid of? If JSH is free to ,ake an ass of himself here on any and all occasions. I should have that same right on a few rare occasions, like JSH's postinig of a new thread with nothing but old content. But, now that I think of it, those occasions are neither few nor rare. Why can't I just post and you let the post sit there without feeling a > need to communicate something from yourself along with it? Maybe, James, I am infec with the same evil daemon that forces you to post all those new threads with old, and disproved, content. Can you drop the emotion long enough to answer those basic questions? I'm curious if you have any idea yourself what your motivation is. I am constitutionally unable to suffer fools gladly. Trying makes me itchy, and puncturing their conceits is the way I scratch. Curiosity satisfied? === Subject: Re: Modified Decker example, reference >> has star another new thread here only covering the same >> ground covered by a host of previous threads, but not yet filled with >> proofs of his many errors in covering that ground that his other threads >> have all attrac. >> It seems to be a standard JSH tactic: when faced with undeniable proof >> of errors in one thread, JSH merely restates them in a new thread. >> The material in this thread has been worked over in other threads. >> The only new stuff is JSH's wording of his astonishment that the >> mathematical world has so little regard for his genius. >> For those who read him for amusement, go ahead, enjoy. For any other >> purposes, this thread is of no use. >> This would probably be a great disclaimer to post after any JSH post >> (along with maybe some additional comments like: may cause nausea, >> diarrhea, ...) >> Jack >Why editorialize at all? What are you afraid of? >Why can't I just post and you let the post sit there without feeling a >need to communicate something from yourself along with it? Uh, this is usenet. People make posts, and other people comment on them. When the post is really stupid people say it's stupid. When the post comes from the sort of megalomaniac who says out loud that he's the misunderstood genius and the rest of us are people find it hilarious, and post mocking replies. What are _you_ afraid of? If you're right about the math the Truth will come out eventually, and the rest of us will all look silly. >Can you drop the emotion long enough to answer those basic questions? the following? I don't know how long it'll last, could be a few weeks. In the meantime, I suggest that people who actually like to read my posts take a break. It's likely to get uglier for a while. I'm doing what I can, but I can feel a tremendous need to unload the worst of it. It's all kind of like some kind of emotional dump. Hint: it was you. >I'm curious if you have any idea yourself what your motivation is. > ************************ === Subject: Re: Modified Decker example, reference > has star another new thread here only covering the same > ground covered by a host of previous threads, but not yet filled with > proofs of his many errors in covering that ground that his other threads > have all attrac. It seems to be a standard JSH tactic: when faced with undeniable proof > of errors in one thread, JSH merely restates them in a new thread. The material in this thread has been worked over in other threads. The only new stuff is JSH's wording of his astonishment that the > mathematical world has so little regard for his genius. For those who read him for amusement, go ahead, enjoy. For any other > purposes, this thread is of no use. > This would probably be a great disclaimer to post after any JSH post > (along with maybe some additional comments like: may cause nausea, > diarrhea, ...) > Jack Don't forget chest pain from laughing so hard.........on second thought, you shouldn't laugh at dimwits. David === Subject: Horror with integrals Guys, I am really fed up with some integrals. I just can't solve them... I am absolutely unable to show the convergence for the integral sin(x^2) von 0 bis +oo I know that it is called Fresnel but knowing the name does not help... Can anyone explain the convergence proof step by step for a dummy, please ? For cos(x^2) the proof will work similar, right ? And then I am also trying to calc the integral from 0 to 1 over x^p * (1-x)^q dx I am sure it works out with integration by parts, but I do not get the solution... I make a mistake at some point... Can anyone explain this step by step, too - so that I can see where I make the mistake ? Thank you sooo much ! Regards === Subject: Re: Horror with integrals >Guys, >I am really fed up with some integrals. >I just can't solve them... >I am absolutely unable to show the convergence for the integral sin(x^2) von >0 bis +oo >I know that it is called Fresnel but knowing the name does not help... >Can anyone explain the convergence proof step by step for a dummy, please ? You got detailed explanations in reply to your first post on this question. >For cos(x^2) the proof will work similar, right ? >And then I am also trying to calc the integral from 0 to 1 over x^p * >(1-x)^q dx >I am sure it works out with integration by parts, but I do not get the >solution... >I make a mistake at some point... >Can anyone explain this step by step, too - so that I can see where I make >the mistake ? >Thank you sooo much ! >Regards ************************ === Subject: Re: Horror with integrals Guys, I am really fed up with some integrals. >I just can't solve them... I am absolutely unable to show the convergence for the integral sin(x^2) von >0 bis +oo >I know that it is called Fresnel but knowing the name does not help... >Can anyone explain the convergence proof step by step for a dummy, please ? You got detailed explanations in reply to your first post on this > question. For cos(x^2) the proof will work similar, right ? >And then I am also trying to calc the integral from 0 to 1 over x^p * >(1-x)^q dx >I am sure it works out with integration by parts, but I do not get the >solution... >I make a mistake at some point... >Can anyone explain this step by step, too - so that I can see where I make >the mistake ? >Thank you sooo much ! >Regards ************************ Prof. Ullrich remarks that your question was already answered. Perhaps you did not read the answer when it was pos. Go to http://groups.google.com and do a Google Search on sci.math. Then do a compound search on Alexander Schmidt Fresnel and you will see that your question has appeared both in German- and English-language newsgroups. David Ames === Subject: Metric Engineering Super Cosmos: Men Like Gods Lecture 3 is at the end. I have also modified Lectures 1 and 2. Please delete previous versions. Also the page references to Rovelli's text book for this Star Fleet Academy course have been fixed to match Rovelli's and not Adobe's labeling. Rovelli makes a mistake it seems in confusing the tangent spaces of LIFs with the base space of LNIFs in the sense of MTW's Gravitation . Perhaps it is only a linguistic problem with Rovelli's English or perhaps it is something deep I have missed. So I give this Caveat. Sarfatti Lecture 1 on the Zero Point Dark Energy Metric Engineering of UFOs http://www.livingreviews.org/Articles/Volume4/2001-1carroll/ . Carroll, S.M., ñThe Cosmological Constant , Living Rev. Relativity, 4, (2001), ñGeneral relativity is a paradigmatic example of a scientific theory of impressive power and simplicity. The cosmological constant, meanwhile, is a paradigmatic example of a modification, originally introduced [81] to help fit the data, which appears at least on the surface to be superßuous and unattractive. Its original role, to allow static homogeneous solutions to Einstein's equations in the presence of matter, turned out to be unnecessary when the expansion of the universe was discovered [131], and there have been a number of subsequent episodes in which a nonzero cosmological constant was put forward as an explanation for a set of observations and later withdrawn when the observational case evapora. constant can be interpre as a measure of the energy density of the vacuum. This energy density is the sum of a number of apparently unrela contributions, each of magnitude much larger than the upper limits on the cosmological constant today; the question of why the observed vacuum energy is so celebra puzzle, although it is usually thought to be easier to imagine an unknown mechanism which would set it precisely to zero than one which would suppress it by just the right amount to yield an observationally accessible cosmological constant. This checkered history has led to a certain reluctance to consider further invocations of a nonzero cosmological constant; however, recent years have provided the best evidence yet that this elusive quantity does play an important dynamical role in the universe.î Note in the above excerpt: interpre as a measure of the energy density of the vacuum. This energy density is the sum of a number of apparently unrela contributions, each of magnitude much larger than the upper limits on the cosmological constant today; the question of why the observed vacuum become a celebra puzzleî I allege I have solved this problem using the idea of ñvacuum coherenceî missing from the orthodox theory in the precise way I use it. The idea of ñvacuum condensateî is in orthodox theory. It is a rela idea, but not exactly the way I mean it. EinsteinÍs GR local geometrodynamical field equation with the cosmological term / guv is Ruv [CapitalEth] (1/2)Rguv + / guv = 8pi(G/c^4)Tuv (1) Impose the large-scale coarse-grained isotropic homogeneous Friedman-Robert-Walker solution ds^2 = -(cdt)^2 + a^2(t)Ro^2[(dr)^2/(1 [CapitalEth] kr^2) + r^2dO^2] (2) r is dimensionless and the Her MajestyÍs Royal NavyÍs navigational ñcelestial sphereî spherical angular line element is dO^2 = (dtheta)^2 + (sintheta)^2(dphi)^2 (3) in usual polar coordinates for latitude theta and longitude phi on the 2D sphere of unit radius. a(t) is dimensionless = R(t)/Ro. Subscript o means ñnowî. k is +1 (closed universe in 3D like a sphere) or 0 (spatially ßat like an infinite plane in EuclidÍs geometry) or [CapitalEth] 1 like a hyperboloid. Both k = 0 and k = -1 are open universes of infinite spatial extent. The cosmological redshift z of retarded radiation from a co-moving source in the ñHubble ßowî where the Cosmic Black Body Radiation (CBR) is maximally isotropic from the past till now obeys the equation a(past) = [1 + z(now)]^-1 (4) The symmetric stress-energy density tensor Tuv for ñstuffî on the RHS of eq (1) is of the form Tuv = (energy density + pressure)UuUv + pressure guv = (energy density)[(1 + w)UuUv + wguv] (5) Uv is the dimensionless 4-velocity dx^u/ds of this ñcosmic ßuidî stuff. w = 0 for cold matter (6a) w = 1/3 for electromagnetic radiation (far field) (6b) w = -1 for any kind of zero point vacuum ßuctuation (ZPF) of any quantum field including string and brane fields (6c) An ñexotic vacuumî is any kind of virtual stuff with w = -1 and a non-vanishing pressure. has ñdarkî gravity and anti-gravity properties bending EM radiation signals for example. Dark matter exotic vacuum in a clump will act like a convex converging gravity lens. Dark energy in a clump will act like a diverging gravity lens. Dark matter makes a gravity red shift and dark energy makes a gravity blue shift. The global cosmic time t (~ 1/(Absolute temperature of CBR) as a convenient measure) dependent Hubble parameter is H(t) = a^-1(da/dt) (7) EinsteinÍs tensor field equation (1) forced into the highly symmetric ñKilling vector fieldî mold of FRW eq (2) simplifies to TWO ordinary differential equations: H^2 = 8pi(G/c^2)(energy density) + c^2/ /3 [CapitalEth]kc^2/a^2Ro^2 (8a) Note that H^2 is analogous to NewtonÍs Laplacian of the Gravity Potential Energy of ANY Source Stuff (real or virtual in sense of quantum field theoryÍs ñonî or ñoffî ñmass shellî) per unit test analogy to NewtonÍs Poisson equation in this spatially homogeneous strictly large-scale approximation is the second equation a^-1d^2a/dt^2 = -(4pi/3)(G/c^2)[(energy density) + 3(pressure)] + c^2/ /3 = -(4pi/3)(G/c^2)(energy density)[(1 + 3w) + c^2/ /3 (8b) to be continued PROBLEM SET #1 1.1 Term Paper Project # 1 What does Hal Puthoff write about zero point energy and gravity and its relation to ñinterstellar ßightî and UFOs? Go to http://www.earthtech.org/publications/ Compare what Puthoff and Davis profess with the physics in these lectures. Sarfatti Lecture 2 on Metric Engineering the Zero Point Dark Energy RovelliÍs formalism in http://www.cpt.univ-mrs.fr/~rovelli/book.pdf Note indices I,J,K,L.83 are in the LIF representation and indices u,v,w, l are in the LNIF representation. The two are connec by EinsteinÍs local version of the equivalence principle. ñPhysics is simple when it is local.î (John A. Wheeler). There is also a global version of the equivalence principle that Hal Puthoff prefers that I shall not use unless explicitly no. Problems of interpretation arise as mentioned by Paul Zielinski because the local tidal curvature tensor is an absolute distinction between a timelike geodesic LIF near a large source mass warping space-time and a timelike geodesic LIF far away from that same large source mass. Similarly with LNIFs near and far from the same large source mass. e^I(x) = eu^I(x)dx^u = ñgravitational fieldî Cartan 1-form in the LIF representation i.e. his eq. 2.1 p. 23 I in Minkowski ñLIFîspace, but Rovelli says u in Tangent vector space TxM not in base space M. This is odd since the local tangent space has the Minkowski metric and the curved base space has guv metric. *Note that the locally variable tetrad components eu^I(x) and its inverse eI^u(x) at space-time event x express EinsteinÍs Equivalence Principle (EEP) of the passive transformation between the Locally Inertial Frame (LIF) that is non-rotating on a timelike geodesic and a momentarily concident Locally Non-Inertial Frame (LNIF) that can be rotating and whose center is on a timelike non-geodesic that intersects the timelike geodesic at x. This is not the same as the active ñgauge-equivalentî diffeomorphisms where x -> xÍ =/= x. The globally ßat solution is eu^I(x) = (Kronecker Delta)u^I i.e. eq 2.44 p. 27 With a globally vanishing spin-connection WIJ = 0 in a region of space-time not only at a point or on timelike geodesic world lines. My elastic-plastic Kleinert ñworld crystalî distortion field must be in the LNIF representation on the LHS L^u(x) = [e^uI(x) - (Kronecker Delta)^uI]dx^I = (Loop Gravity Quantum of Area)arg(Vacuum Coherence),u (S1a) ,u is ordinary partial derivative in the LNIF representation L^I(x) = eu^I L^u(x) (S1b) in the LIF representation, Next introduce the antisymmetric Lorentz group algebra ñspin connectionî that will apply to spinning tops for example. The spin connection is also a Cartan 1-form W^IJ(x) = Wu^Ijdx^u i.e. eq. (2.2) p. 23 There are 6 of these generators of the Lie algebra so(3,1)of the 6 parameter Lorentz group of 3 space-time rotations (boosts) and 3 space-space rotations (total angular momentum) WIJ = - WJI (S2) in the LIF representation. Note that L^u(x) is the local gauge force compensating field restoring the broken 4-parameter translation group symmetry T4. However, while the T4 and so(3,1) are intertwined as a ñsemi-direct productî and as seen in the Thomas precession and the Sagnac effect, there is no local gauge force compensating field for Lie algebra so(3,1) when we use a torsion-free spin connection to define the tidal force tensor curvature that intimately uses so(3,1). Thus, the generally covariant partial derivative of General Relativity in the Cartan formalism starts from DuV^I = V^I,u + Wu^IJV^J i.e. (2.3) p. 23 in a mixed LNIF/LIF representation. More generally the generally covariant Cartan exterior derivative on a 1-form v^I is the 2-form DV^I = dV^I + W^IJ/ V^J i.e. (2.4) p. 24 in the LIF representation Where d is the Cartan exterior derivative and / is the wedge product. Do not confuse this ñ/ [CapitalOAcute] the exterior wedge product with ñ/ zpfî the ñmetric engineeringî random micro-quantum zero point exotic vacuum dark energy/matter induced curvature scale. What is the relation of / to Clifford algebra? Note that the Lorentz group Lie algebra so(3,1) is the fundamental connection field from the breaking of the global symmetry of T4 whose Lie algebra is total energy-momentum. This total ñmom-energyî (Wheeler) is not a natural concept in General Relativity because locally variable curvature breaks global translational symmetry even though curvatureÍs definition involves the rotations of so(3,1) since local curvature is the anholonomy disclination ñBerry phaseî in the orientation of a vector parallel transpor around a closed space-time loop as the area of the loop shrinks to zero. Torsion T^I means that the loop in tangent fiber space is broken or disloca for a closed loop in base space or vice versa. That is, with torsion, closed loops do not map to closed loops in parallel transport. The condition of zero torsion gauge force field is the vanishing Cartan torsion field 2-form T^I = De^I = de^I + W^IJ/ e^J = 0 i.e. (2.6) p. 24 Again note that it is the spin connection from so(3,1) that plays the vital role! Finally the tidal relative acceleration or timelike geodesic deviation is also a Cartan 2-form from the zero torsion so(3,1) spin connection (which allows fermions as well as bosons) R^IJ = dW^IJ + W^IK/ W^KJ i.e. (2.8) p. 24 i.e. tidal curvature is the nonlinear self-organizing spin connection covariant exterior derivative on itself! This is why there are curved vacuums independent of real mass-energy sources! Gravity acts on itself in a non-Abelian way. The Einstein local field equation with the zero point energy cosmological / zpf term is in Cartan formalism is the Euler-Lagrange equation [IJKL]e^I/ R^J^K + / zpf e^I/ e^J/ e^K = 0 i.e. (2.9) p. 24 where [IJKL] is the anti-symmetric 4-symbol Note the cubic nonlinearity in the zero point energy / ñmetric engineeringî term. The dynamical action density is ~ e^I/ e^J/ R^K^L + / zpf e^I/ e^J/ e^K/ e^L contrac with [IJKL] to make an invariant scalar i.e. eq. (2.10) p. 24 Sarfatti Lecture 3 on Metric Engineering Super Cosmos Rovelli in Footnote 3 p. 42 in http://www.cpt.univ-mrs.fr/~rovelli/book.pdf makes an important remark for metric engineering that the space-time stiffness factor is only important in the coupling of ordinary mass-energy to the warped spacetime geometrodynamic field. The space-time stiffness factor is G/c^4 with the dimensions of length/energy = (string tension)^-1 = WittenÍs alphaÍ/hc ~ (Loop Quantum of Area)/hc. Coupling of gravity warped space-time to ordinary mass-energy sources. Here one can think of the Kaluza-Klein extra space dimensions or, alternatively, internal dimensions, which is not as ñgeometrodynamicî lacking a common super-metric. 1. Electromagnetism spin 1 compensating gauge force field from breaking global U(1) phase symmetry: here the so(3,1) spin connection WIJ is replaced by the U(1) ñMaxwell potentialî connection Cartan 1-form in the extra dimensions A(x) = Au(x)dx^u i.e. eq. (2.25) p. 26 This 1x1 connection field for parallel transport in the extra dimensions restores the broken global U(1) symmetry genera by electric charge just as the so(3,1) spin connection restores the broken global T4 symmetry genera by mom-energy. A(x)Ís globally ßat exterior U(1) gauge-invariant derivative is the EM field tensor internal (fiber)curvature 2-form: F = dA = Fuvdx^u/ dx^v (S3a) Fuv = Au,v [CapitalEth] Av,u (S3b) The dynamical action density is Lem = (1/4)F*/ F i.e. Where F* is the Hodge star transform, in ßat spacetime F*IJ = [IJKL]F^K^L i.e. eq. (3) p. xiv * Always with summation over repea lower and upper pairs of tensor indices. And in curved spacetime Fuv* = (-detg)^1/2[uvwl]F^w&l = |dete|[uvwl]F^w&l i.e. eq. (4) p. xiv 2. Yang-Mills non-Abelian e.g. G -> SU(2) weak beta decay radioactive spin 1 W+0- bosons ñßavorî gauge force, or G -> SU(3) gluon ñcolorî strong sub-nuclear spin 1 gauge force, have the connection field 1-form A(YM) that is a matrix representation of the Lie algebra of whatever the Lie group G is e.g. (2.27) p. 26. 3. See also the boson scalar spin 1 fields and the fermion O(3,1) Lorentz group spinor fields on p. 26. The complete local Einstein field equation with the ordinary mass-energy source term and the exotic vacuum random micro-quantum zero point dark energy/matter term in Cartan form notation is in the locally quasi-ßat non-rotating timelike geodesic ñLIFî (Local Inertial Frame) is the Cartan 3-form equation [IJKL](e^I/ R^J^K + / zpf e^I/ e^J/ e^K) = 8pi(G/c^4)TI i.e. eq. (2.34) p. 27 Where the source stress-energy density Cartan 3-form is TI = T^uI[uvwl]dx^v/ dx^w/ dx^w i.e. eq. (2.35) p. 27 This stress-energy density 3-form comes from the functional derivative of the dynamical action of the ordinary matter-energy S(matter) with respect to the tetrad gravity field eu^I(x), i.e. in mixed LIF/LNIF representation T^uI(x) = &S(matter)/&eu^I(x) (2.36) p. 27 The stress-energy density tensor of ordinary vacuum is zero. The Yilmaz theory is plain wrong IMHO. There is a micro-quantum stress-energy tensor for exotic vacuum of course that is in pure LNIF representation Tuv(Exotic Vacuum Zero Point Energy) ~ (String Tension)/ zpf guv (S4) What is the anti-symmetric spin-connection.? Let o(3,1)IJ be the 6 independent generator elements of the Lie Algebra of the Lorentz group O(3,1). I propose for consideration WIJ(x) = [IJKL]eK^u(x)eL^u(argVacuum Coherence)[,u,v] (S5) ???? Where [ , ] is the anti-symmetrizer or commutator of the partial derivatives ,u and ,v. This is a ñGoldstoneî phase anholonomy where the mixed partial derivatives of the MACRO-QUANTUM vacuum coherent phase are not equal. This only happens in string superßuid ñvortex coresî that are topological defects where the virtual off-mass-shell superßuid condensate inßation field ñHiggsî amplitude |Vacuum Coherence| -> 0 inside a ñcoherence lengthî. This picture of string defects for the complex numbered Vacuum Coherence inßation field fits Hagen KleinertÍs world crystal lattice long-wave Infra Red (IR) curvature and torsion as string disclination and dislocation lattice defect densities. WIJ(x) = WIJ(x)disclination + WIJ(x)dislocation (S6) This idea gets us to the key notion of nonlocal string holonomies and ñWilson loopsîof lattice gauge theory and the resolution scale dependent ñwavelet transformî renormalization group ßows to ñfixed pointsî. Rovelli in 1.2.1 on pp. 10-11 traces the origin of this idea back to Michael FaradayÍs idea of the ñfieldî as ñlines of forceî in the early 19th Century. Given any connection field Cartan 1-form, the ñholonomyî is the parallel transport of the ñforce vectorî around a closed string loop. Now one can, no doubt, probably generalize this ñholonomyî to higher-dimensional ñbranesî like in the Clifford Algebra ñExtended Relativityî of Castro and Pavsic, for example, and in other probably equivalent algebraic matrix formalisms. In Loop Quantum Gravity (LQG) the nonlocal holonomy becomes a quantum matrix operator and an elementary closed string loop quantum state |l> is where the ñforce vectorî vanishes everywhere except on the closed string loop. An example is shown in Fig 1.1 with equation (1.6) on p. 11. Now our example in eq. (S6) is an inver image of what Rovelli is talking about. We have a phase singularity or anholonomy in which the Higgs amplitude |Vacuum Coherence| vanishes on the closed string loop so that the mixed second order partial derivatives of the Goldstone phase of the Vacuum Coherence field are not equal. 3D ñpre-space-likeî SU(2) group quantum bit (qubit) computer spintronic networks (I use ñspintronicî rather than ñspinî to call attention to the technology applications pursued with DARPA funding) are coherent superpositions of these elementary loop states. These become interesting when the loops intertwine and intersect and form ñknotsî. One can form a finite-norm orthogonal basis of a separable Hilbert space of these 3D spintronic network quantum computer states in the pre-geometry prior to the emergence of classical c-number warped space-time provided that there is a world crystal lattice structure whose unit cells do not vanish in the continuum limit. That is there is an indivisible ñLoop Gravity Quantum of Areaî. One can do continuous active diffeomorphisms on these discrete lattice qubit spintronic network states. Infinitesimal active diffeomorphisms on a spintronic net make ñgauge group equivalent representationî of the same nonlocal network state. These equivalent representations that are connec by infinitesimal active diffeomorphisms equivalence relation ~ must be factored out in a ñquotient set = set/~î of ñcosetsî algebraic process. See RovelliÍs argument on p. 12 how these ideas of gauge equivalence of infinitesimal continuous active diffeomorphisms with a quantum of area in a discrete combinatoric qubit computing network pre-space-time lead to a separable kosher quantum Hilbert space of spintronic networks that permit a dynamical ñbackground independentî quantum gravity theory and indeed a background independent quantum field string theory for all fields in a purely relational frame work. Rovelli pictures pre-space-time as the Hobbes-Melville ñGreat Leviathanî the ñGreat White Whaleî ñBig Fishî with the other fermion and boson fields like smaller fish clinging to it and sliding relative to it and each other. The condition that the loops cannot shrink in area to zero also leads to the Witten M-theory string duality generalized ñconformal inversionî w = z + z^-1 uncertainty principle in which Uncertainty in position ~ h/(uncertainty in momentum) + (cWitten alphaÍ)(uncertainty in momentum) (S7a) And its dual: Uncertainty in momentum = h/(uncertainty in position) + (1/cWitten alphaÍ)(uncertainty in position) (S7b) Recall that: Witten alphaÍ = (String Tension)^-1 = (Loop Gravity Quantum of Area)/hc (S7c) cWitten alphaÍ = Loop Gravity Quantum of Area/h (S7d) === Subject: Re: Metric Engineering Super Cosmos: Men Like Gods Sarfatti decoded a Sumarian clay tablet and transla: > .8c.8bGeneral relativity is a paradigmatic example of a scientific theory of > impressive > power and simplicity. No ? Is it subject to W32/Mydoom.a@MM virus? You talke like Microsoft came up with GR. Seems to me GR dropped it's Ôtheory' aspect long ago when it appeared for all intents to take on a Ôfact' aspect. My money is on the dice thrower. === Subject: definition of asymptote. which of these definitions is true of an asymptote and why? given f(x) and g(x), g(x) is an asymptote of f(x) if lim g(x) / f(x) = 1 as x -> infinity OR lim g(x) - f(x) = 0 as x -> infinity (as an example, does x^2 + x converge to x^2 even though they get further apart in absolute terms, their relative distance gets smaller) ...c8to === Subject: Re: definition of asymptote. > which of these definitions is true of an asymptote and why? given f(x) and g(x), g(x) is an asymptote of f(x) if lim g(x) / f(x) = 1 as x -> infinity OR lim g(x) - f(x) = 0 as x -> infinity The latter is true of all but vertical asymptotes. Subject: re:definition of asymptote. === A definition is what you want it to be, so there is no answer to why. As far as the accep definition of asymptote, it is the second (difference goes to 0). ----== Pos via Newsfeed.Com - Unlimi-Uncensored-Secure Usenet News==---- http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: Mahlo Cardinals >>I read a definition of Mahlo Cardinals in Rudy Rucker's book ( Infinity >>and the mind ). It's probably not the best place for such definitions, >>and I didn't look at more serious books yet. >>Rudy gives what he calls a formal definition. He says that a Mahlo >>Cardinal is an inaccessibe cardinal which for every set of ordinals >>smaller than itself with it's cardinality, there is smaller inaccessible >>cardinal kappa, for which the set of all ordinals in that set less than >>kappa has a cardinality of kappa. >>I'd like to know if that's the way they usually define Mahlo Cardinals. >For the converse, suppose rho is a weakly Mahlo cardinal, and let A be a >set of ordinals less than rho with cardinality rho. Define f : rho -> rho >as follows. f(0) is the smallest element of A. f(a+1) is the smallest >element of A-{f(b) : b < a or b = a} for all ordinals a < rho. If b < rho >is a limit ordinal, then f(b) = lim{f(a) : a < b}. Then f is a normal >function on rho, so that f has a regular fixed point, alpha, since rho is >a weakly Mahlo cardinal. Thr bit that went here should be replaced by: For any ordinal c < rho, define g_c : rho -> rho by g_c(a) = f(c+a) for all a < rho, then g_c is a normal function on rho, and so g_c has a regular fixed point alpha. Since alpha <= c+alpha <= f(c+alpha) = g_c(alpha) = alpha, where a <= b denotes that a is less than or equal to b, then c < alpha, c+alpha = alpha (i.e. alpha >= c omega), and f(alpha) = alpha, i.e. alpha is a fixed point of f. So for any ordinal less than rho, there is a regular fixed point of f which exceeds it. Therefore the set of regular fixed points of f is unbounded, and so the cardinality of the set B of regular fixed points of f is rho. >Define h : rho -> rho as follows. h(0) is the >smallest element of B. h(a+1) is the smallest element of B-{h(b) : b < a >or b = a} for all ordinals a < rho. If b < rho is a limit ordinal, then >h(b) = lim{h(a) : a < b}. Then h is a normal function on rho, so that h >has a regular fixed point, kappa, since rho is a weakly Mahlo cardinal. >Since kappa is a limit ordinal, then h(kappa) is a limit cardinal, and so >kappa is a limit cardinal since h(kappa) = kappa. Since kappa is a >regular limit cardinal, then kappa is inaccessible, so that kappa is an >inaccessible fixed point of h. Since kappa is a limit of fixed points of >f, then kappa is a fixed point of f, since f is continuous. So kappa is >an inaccessible fixed point of f. Since f(kappa) = kappa, and the set of >successor ordinals less than kappa has cardinality kappa, then the >intersection of A and {f(a) : a < kappa} has cardinality kappa. Since >f(a) < kappa if a < kappa (since f is increasing), then the cardinality of >the intersection of A and kappa is kappa, and so there exists an >inaccessible cardinal kappa < rho such that the set of ordinals in A less >than kappa has cardinality kappa. This demonstrates that Rudy's >condition correctly characterizes the weakly Mahlo cardinals. David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. === Subject: Re: solutions of equations Consider the following two equations in four unknowns: w + x + y + z = 7.11 (1) wxyz = 7.11 (2) w, x, y, z represent the prices of four items. Therefore, each can > assume values only upto 2 decimal places. Question: Is it possible to determine the values of w,x,y,z exactly > upto two > decimal places? By trial I find values as w = 3.06, x = 1.54, y = 1.00, z = 1.51 > Here w + x + y + z = 7.11 but wxyz = 7.1157. Maybe there is a better solution. > Any useful response will be apprecia. There is an exact solution. You can make this a problem in integers by convderting dollars to cents. w + x + y + z = 711 > amd w*x*y*z = 711000000 (multiplying each variable by 100). > The prime factorization of 711000000 is 2^6*3^2*5^6*79 So, what are the final values of w,x.y,z? === Subject: Re: solutions of equations Consider the following two equations in four unknowns: w + x + y + z = 7.11 (1) wxyz = 7.11 (2) w, x, y, z represent the prices of four items. Therefore, each can > assume values only upto 2 decimal places. Question: Is it possible to determine the values of w,x,y,z exactly > upto two > decimal places? By trial I find values as w = 3.06, x = 1.54, y = 1.00, z = 1.51 > Here w + x + y + z = 7.11 but wxyz = 7.1157. So, what are the final values of w,x.y,z? 1.25, 1.20, 3.16, 1.50 === Subject: elementary analysis I want to show lim_x->1 (x^3)=1 For any e>0, let d(>0) satisfy d^3+3d^2+3d|x^3-1|=|x-1||x^2+x+1| I want to show lim_x->1 (x^3)=1 > For any e>0, > let d(>0) satisfy d^3+3d^2+3d for all x on 0<|x-1| =>|x^3-1|=|x-1||x^2+x+1| I wonder wheather this is correct solution or not. > Maybe, there will be more simple way to seize d of part (i). > But I don't know the way. > if someone can help me, please post reply. > What you have is correct. (Although you do have to know that you CAN choose d small enough so that d^3+3d^2+3d < e, which is equivalent to knowing that lim_d->0 (d^3+3d^2+3d) = 0, which is a problem just as hard as the original problem.) In a way, you found the best possible d for any given e. But you can be a little cruder with the estimates to avoid all the algebra. For example, choose d to be the smaller of e/7 and 1. If 0<|x-1|Sorry guys, I have another integration problem: >The integral is: >with p,q in |N. >I tried integration by parts: >Set: >u(x) = (1-x)^q -> u'(x) = -q(1-x)^(q-1) >v'(x) = x^p -> v(x) = 1/(p+1) * x^(p+1) >I tried to calc: >((1-x)^q) * (1/(p+1) * x^(p+1)) - Integral from 0 to 1 over (-q(1-x)^(q-1)) >* x^p dx >The part before the minus gets zero, right, because filling in 1 and then 0 >and subtracting these values from each other leads to 0 - 0 = 0. >So there is still to calc the part after the minus: >Integral from 0 to 1 over (-q(1-x)^(q-1)) * x^p dx >But this does not look easier to me than the term before. >Can anyone post the solution please ? >I am fed up with this ! >Thanks a lot. If you are going to use integration by parts, it would make a lot more sense to set u= x^p so that du= p x^(p-1). Repea integration by parts will reduct this to a power of x-1. Actually, I would be inclined to use the binomial theorem to expand (x-1)^q as Sum(i=0 to q){qCi x^i} so that x^p(1-x)^q becomes Sum(i=0 to q){qCi x^p+i} and the integral is Sum(i=0 to q){(qCi/(p+i+1))x^(p+i+1)} === Subject: Re: When was zero inven? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0U1mBt15772; Just last semester I did a project on this very topic for one of my undergrad classes. What I found was very intriguing, albeit hard to digest by what we have become comfortable with to this day. Indeed, there are two fundamental uses for the number zero: as a placeholder, and as a magnitude. The first account (proved, etc.) of its use came from India, where it quickly spread to the Arabs and then to Europe. How did its need come about? It was for a need to the solutions of the roots of polynomials, only then was it significant in the true sense. One text (not referenced here) claimed that it was because of the unimportance (unlike today) of its meaning. It was just not a meaningful result (it was essentially Ônothing', to use the word crudely). Of course, it is hard to imagine how any society could have functioned without the concept; the value of it is certainly amazing in mathematics today. The same analog also happened with the cubic equation and imaginary numbers. Sometimes people discard important results that seem, at the time, unimportant or not worthwile to study (there is a moral to this story... ;-) MM. === Subject: Re: When was zero inven? > Indeed, > there are two fundamental uses for the number zero: as a placeholder, > and as a magnitude. The first account (proved, etc.) of its use came > from India, where it quickly spread to the Arabs and then to Europe. > How did its need come about? It was for a need to the solutions of the > roots of polynomials, only then was it significant in the true sense. > One text (not referenced here) claimed that it was because of the > unimportance (unlike today) of its meaning. It was just not a > meaningful result (it was essentially Ônothing', to use the word > crudely). Of course, it is hard to imagine how any society could have > functioned without the concept; the value of it is certainly amazing > in mathematics today. > If I understand you correctly, the Romans didn't know zero should be replaced with the Romans didn't know zero as a placeholder . But even this seems incorrect because abacus had the idea of zero as a placeholder. I also wonder, if a Roman merchant - or whoever - never transferred this idea from abacus to his book-keeping on paprys or whatever he used to scribble down the items he had for inventory or taxation purposes. It just sounds so much more practical to reduce 89 from 1 000 and come up with 911 instead of messing around with the confusing Roman numerals. I mean, how could the merchant even calculate with them if he didn't have an abacus at hand? Tomi === Subject: Re: When was zero inven? The ancients probably knew about zero but didn't consider it a number. If I remember right, the Greeks didn't consider one to be a number. One was special and in some cases so was two. Plain old numbers star with three. Russell - 2 many 2 count === Subject: Re: When was zero inven? The ancients probably knew about zero but didn't consider it a number. > If I remember right, the Greeks didn't consider one to be a number. > One was special and in some cases so was two. > Plain old numbers star with three. > Russell > - 2 many 2 count I didn't know you were that old. === Subject: Re: When was zero inven? dnYaG3N_miYfd4p2dnA@comcast.com: > The ancients probably knew about zero but didn't consider it a number. > If I remember right, the Greeks didn't consider one to be a number. > One was special and in some cases so was two. > Plain old numbers star with three. > It sounds reasonable. One was the Platonian god, the ulitimate idea . Two was a dialectical entity between the idea(s) and their representations or the representation itself in an abstract sense. Zero was vacuum, which didn't exist according to most osophers. But osohpical ponderings had little meaning to a merchant. He must have known zero both as a symbol for nothing and as a placeholder (fero of tens), me thinks. To him - if I'm right - zero was a number for all practical purposes. Tomi === Subject: Re: When was zero inven? > dnYaG3N_miYfd4p2dnA@comcast.com: The ancients probably knew about zero but didn't consider it a number. > If I remember right, the Greeks didn't consider one to be a number. > One was special and in some cases so was two. > Plain old numbers star with three. It sounds reasonable. One was the Platonian god, the ulitimate idea . > Two was a dialectical entity between the idea(s) and their > representations or the representation itself in an abstract sense. Zero was > vacuum, which didn't exist according to most osophers. But osohpical ponderings had little meaning to a merchant. He must have > known zero both as a symbol for nothing and as a placeholder (fero of > tens), me thinks. To him - if I'm right - zero was a number for all > practical purposes. One is a number for all practical purposes. Because with one, comes minus one. Zero is not even a number. It's a set of NOTHING, idiots. osophers are only well-aquain with vacuum cleane rather than vacuums. That trivially follows from the fact that dialectic is their preferred form of history, rather than intelligence. Tomi === Subject: Re: When was zero inven? Because with one, comes minus one. > Zero is not even a number. > It's a set of NOTHING, idiots. As has been mentioned in this thread already, zero IS a number and is very different from the empty set (that is, it is NOT a set of NOTHING, as you say.) An example you might be able to understand: When someone says it is zero degrees outside today it does NOT mean that there is no temperature today. J === Subject: Re: When was zero inven? One is a number for all practical purposes. > Because with one, comes minus one. > Zero is not even a number. > It's a set of NOTHING, idiots. As has been mentioned in this thread already, zero IS a number and is > very different from the empty set (that is, it is NOT a set of NOTHING, as > you say.) An example you might be able to understand: When someone says it is zero > degrees outside today it does NOT mean that there is no temperature > today. That follows trivially from your use of the words, degrees , outside , today idiot, not from your idiot suggestion that there is an absolute zero temperature. === Subject: Re: When was zero inven? One is a number for all practical purposes. > Because with one, comes minus one. > Zero is not even a number. > It's a set of NOTHING, idiots. As has been mentioned in this thread already, zero IS a number and is > very different from the empty set (that is, it is NOT a set of NOTHING, as > you say.) An example you might be able to understand: When someone says it is zero > degrees outside today it does NOT mean that there is no temperature > today. That follows trivially from your use of the > words, degrees , outside , today idiot, not from your idiot > suggestion that there is an absolute zero temperature. Unfortunately, zero exists on the number line; zero does not mean Ôabsolute zero,' but instead it means the integer between -1 and +1. J === Subject: Re: When was zero inven? Because with one, comes minus one. > Zero is not even a number. > It's a set of NOTHING, idiots. As has been mentioned in this thread already, zero IS a number and is > very different from the empty set (that is, it is NOT a set of NOTHING, as > you say.) In set theory, zero is usually defined as the empty set. -- G.C. === Subject: Re: When was zero inven? > As has been mentioned in this thread already, zero IS a number and is > very different from the empty set (that is, it is NOT a set of NOTHING, > as you say.) > In set theory, zero is usually defined as the empty set. Right. And so, in that light, the distinction between zero and nothing could be considered as precisely the same as that between the empty set and its content. David === Subject: Re: When was zero inven? In sci.math, Jim Nastos : > One is a number for all practical purposes. >> Because with one, comes minus one. >> Zero is not even a number. >> It's a set of NOTHING, idiots. As has been mentioned in this thread already, zero IS a number and is > very different from the empty set (that is, it is NOT a set of NOTHING, as > you say.) An example you might be able to understand: When someone says it is zero > degrees outside today it does NOT mean that there is no temperature > today. No, but it does mean there's not a lot of heat. If one says absolute zero there's no heat at all, in fact. Fortunately for those in Antarctica and on Earth, it never gets that cold, even down there. 0 degrees Celsius is freezing water. 0 degrees Fahrenheit is even colder -- but not cold enough to freeze air. As far as most temperatures are concerned it's a somewhat arbitrary (but useful) basis point; one could make the same case on the number line, for certain types of measurements and/or quantities. Sounds like ZZBunker should wake up and smell the coffee, which hopefully is nearer to 100 degrees than 0 degrees... (either scale, though 100 degrees F is a tad tepid) J > -- #191, ewill3@earthlink.net -- mmm, hot breakfast It's still legal to go .sigless. === Subject: Heisenberg Group, Lie algebra help? Dear all, I am reading along in a text and have found an example that I do not understand...maybe someone can help? Let H denote the Heisenberg group (the group of 3 x 3 upper triangular matrices) and h its lie algebra (the group of 3 x 3 matrices such that the diagonal and everything below is 0). Let G be a matrix lie group with lie algebra g and let f : h --> g be a lie algebra homomorphism. Then, there exists a uniue lie group homomorphism F : H ---> G such that F(e^X) = e^(f(X)) Now, to the proof: Define F : H ---> G by the formula F(A) = e^(f(log(A)). Show it's a lie group homomorphism. (I will make this shorter by skipping some easy to prove steps). If [X,Y] = XY-YX as usual (the commutator), then if X,Y are in the lie algebra of the heisenberg group, then [X,[X,Y]] = [Y,[X,Y]] = 0. Since f is a lie algebra hom, [f(X),[f(X),f(Y)]] = [f(Y),[f(X),f(Y)]] = 0. Want to show that F(AB) = F(A)F(B). Well, note that A can be written as e^X for a unique X in h (the lie algebra of the heisenberg group) and B can be written as e^Y for a unique Y in h. Thus, F(AB) = F(e^Xe^Y) = F(e^(X+Y+[X,Y]/2) (by a theorem that if X and Y commute with [X,Y], then e^Xe^Y = e^(X+Y+[X,Y]/2). Now, the following is the part that I can't understand. By how we define F and the fact that f is a lie algebra hom, F(AB) = exp (f(X) + f(Y) + [f(X),f(Y)]/2) WHY??? Well, follow the steps : F(AB) = F(e^(X+Y+[X,Y]/2) = exp (f(log(exp(X+Y+[X,Y]/2))) by our proposed definition F(A) = e^(f(log(A)). But then, I'm assuming that the author's next implicit calculation is log(exp(X+Y+[X,Y]/2)) = X + Y + [X,Y]/2. But isn't this only true if ||X + Y + [X,Y]/2|| < log(2) since only this is when log(exp(Q)) = Q for some n x n complex matrix Q? Where did we ever assume this? I don't think we did. I must be missing something....Can someone please help? Moshe Adrian === Subject: Re: Heisenberg Group, Lie algebra help? > Let H denote the Heisenberg group (the group of 3 x 3 upper triangular > matrices) Actually, it's the group of 3 x 3 upper triangular matrices *with nothing but 1's in the diagonal*. > and h its lie algebra (the group of 3 x 3 matrices such that the > diagonal and everything below is 0). Let G be a matrix lie group with lie > algebra g and let f : h --> g be a lie algebra homomorphism. Then, there > exists a uniue lie group homomorphism F : H ---> G such that F(e^X) = > e^(f(X)) Now, to the proof: Define F : H ---> G by the formula F(A) = e^(f(log(A)). Show it's a lie > group homomorphism. If [X,Y] = XY-YX as usual (the commutator), then if X,Y are in the lie > algebra of the heisenberg group, then [X,[X,Y]] = [Y,[X,Y]] = 0. Since f is > a lie algebra hom, [f(X),[f(X),f(Y)]] = [f(Y),[f(X),f(Y)]] = 0. Want to show that F(AB) = F(A)F(B). Well, note that A can be written as e^X > for a unique X in h (the lie algebra of the heisenberg group) and B can be > written as e^Y for a unique Y in h. Thus, F(AB) = F(e^Xe^Y) = > F(e^(X+Y+[X,Y]/2) (by a theorem that if X and Y commute with [X,Y], then > e^Xe^Y = e^(X+Y+[X,Y]/2). Now, the following is the part that I can't understand. By how we define F > and the fact that f is a lie algebra hom, F(AB) = exp (f(X) + f(Y) + [f(X),f(Y)]/2) WHY??? Well, follow the steps : F(AB) = F(e^(X+Y+[X,Y]/2) = exp > (f(log(exp(X+Y+[X,Y]/2))) by our proposed definition F(A) = e^(f(log(A)). But then, I'm assuming that the author's next implicit calculation is > log(exp(X+Y+[X,Y]/2)) = X + Y + [X,Y]/2. But isn't this only true if ||X + Y + [X,Y]/2|| < log(2) since only this is > when log(exp(Q)) = Q for some n x n complex matrix Q? Where did we ever > assume this? I don't think we did. Well, I do not know what's the definition of log that you are working with, but in the case of the Heisenberg group the exponential map is a diffeomorfism. Therefore, it would make sense to define log as the inverse of exp and so you do have log(exp(X+Y+[X,Y]/2)) = = X + Y + [X,Y]/2. Perhaps that you might want to know that the assertion whose proof you are trying to understand is valid for every simply connec Lie group. This is certainly the case of the Heisenberg group, since it is homeomorphic with R^3. I hope that this helps. Best === Subject: Re: Socrates' Nothing is Everything [SNIP] However, I take issue with the idea that a number needs to be > defined in terms which capture only the essense of counting numbers. > Suppose I define a number as any object which is an element of a set > for which there exists a binary operation and which satisfies some set > of axioms. (Vague, obviously, but I think you might get the idea). > Then our problem with 0 goes away. ÔNumber' usually means the kinds of things included in dictionary or > encyclopedic definitions eg integral, rational, real, complex, hypercomplex, > modular and other numbers that mathematicians have thought fit to invent. Your definition is too wide. For example, it would include sets of > propositions for which first order logic provides both axioms and binary > relations. Propositions are not numbers even if they have values which are > number-like in the some of their relations. I'm not so sure about this. In particular, this definition is a natural one in the context of model theory. The fact that propositions have number-like relations is just a consequence of the fact that the Boolean algebra used in sentential logic is isomorphic to the finite field with two elements. What is peculiar about numbers proper is that the systems they inhabit use > the operator Ô+'. Without this operator (and its ilk), the unique character > of number systems is lacking. The peculiarity is that this operator defies > closure and points the finger at infinity. This property of infinite > extension is abstrac in the Peano definition of number which deploys the > device of logical substitution into a string. ie Ô0S' is substitu into > Ô0X' indefinitely many times to produce, in principle, any possible integer. > In particular, the boolean algebra on which propositional logic is based on the operations + and * defined by: 1 + 1 = 1 0 + 0 = 0 1 + 0 = 1 0 + 1 = 1 and: 1*1 = 1 0*0 = 0 1*0 = 0 0*1 = 0 The semantics are simple: + denotes disjunction and * denotes conjunction. Negation is obvious. Either * with - or + with - forms a complete set of connectives for the propositional logic. > This isn't true of course, because the cosmos would be long dead or crunched > many times before > quite Ôsmall' numbers (eg 1,234!^9,999!) could be realised in this pedantic > way. Reminds me of the Law of Big Numbers. ( Most integers are very, very large. ) No matter how elaborate or generalised new kinds of numbers may be, the > property of infinite extension is a necessary one. Wittgenstein spent some time trying to define the meaning of the word > Ôgame', but gave up, concluding that the possible meanings were incapable of > definite reduction. The same could be said for the word Ônumber' or any term > which resists precise definition. > Ah! A Wittgenstein reader. Excellent. Perhaps you'll appreciate the following then. I believe we're using the word number in two different contexts. If by number you mean something along the lines of counting number, then infinity is obviously not a number. In this context, 0 may not be a number either, since one usually never counts zero objects. But I think there's a more appropriate interpretation of the term which allows for infinity to be considered a number with the same logical status as any other number. This perspective, I think, is more desireable, since it allows the term to apply to a larger class of objects which are often used in this (logical) context. By the way, I took this post off of sci.math and sci.physics, since this is only tangential in the first and irrelevant in the second. Ôcid Ôooh === Subject: Re: Socrates' Nothing is Everything Cancel-Lock: sha1:CgOICmD1GC4UKnsUTUhemEa7NMg= > Infinity is not a property, since it's not a set. > Sets are moronic, since Cantor didn't make sets, > morons made sets. > All of these statements are trivially proven, since some is a property > that only people with higher IQ's than physicists have. You're such a lame poser. This is simply not up to James Hunter's standards. Sets are moronic, since Cantor didn't make sets, morons made sets. Makes no sense. No bloody sense at all. Mind you, most things that aesthetically pleasing manner. Give it up. -- Jesse Hughes Well, you know as soon as you have a new number I will be happy to add it to the list. Don't try those childish tit-for-tat games with me. -- Finlayson on Cantor's theorem. === Subject: Re: Socrates' Nothing is Everything > Makes no sense. No bloody sense at all. Mind you, most things that > aesthetically pleasing manner. Are you talking about the same James Hunter that I remember? -- Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Socrates' Nothing is Everything Cancel-Lock: sha1:LOVnqPIWwE3tKnv6KeD0mZZifO0= >> Makes no sense. No bloody sense at all. Mind you, most things that >> aesthetically pleasing manner. > Are you talking about the same James Hunter that I remember? Of course not. I'm talking about the James Hunter that *I* remember. -- You lack the ability to reason, but instead get an idea in your head and hold on to it against all evidence. I don't find you credible, and reject your claims, as coming from a ßawed source. === Subject: Re: Socrates' Nothing is Everything Infinity is not a property, since it's not a set. > Sets are moronic, since Cantor didn't make sets, > morons made sets. All of these statements are trivially proven, since some is a property > that only people with higher IQ's than physicists have. You're such a lame poser. This is simply not up to James Hunter's > standards. Sets are moronic, since Cantor didn't make sets, morons made sets. Makes no sense. No bloody sense at all. Mind you, most things that > aesthetically pleasing manner. Cantor has *one* proof to his credit, concerning sets. Dedekind has *no* proofs to his credit concerning sets. Goedel has *two* proofs to his credit concerning sets. Russel not only doesn't have any proofs to credit concerning sets, his paradoxes aren't even interesting. Give it up. I like Einstein gave it up before it was inven. === Subject: Re: Socrates' Nothing is Everything Discussion, linux) Cancel-Lock: sha1:pFQ76NhMQQ+jON6F9hMuMK25DWo= > Cantor has *one* proof to his credit, concerning sets. > Dedekind has *no* proofs to his credit concerning sets. > Goedel has *two* proofs to his credit concerning sets. > Russel not only doesn't have any proofs to credit > concerning sets, his paradoxes aren't even interesting. See, this is far too specific for the *real* James Hunter. No use of the word moron . No reference to the superiority of engineering. Just silly. Of course, I won't ask for evidence for such silly claims. -- Jesse Hughes Casting [Demi] Moore as a woman who has come to the New World so that she can Ôworship without fear or persecution' in _The_Scarlet_Letter_ is like casting Bruce Willis as Young Rene Descartes. -Joe Queenan === Subject: Re: Socrates' Nothing is Everything Cantor has *one* proof to his credit, concerning sets. > Dedekind has *no* proofs to his credit concerning sets. > Goedel has *two* proofs to his credit concerning sets. > Russel not only doesn't have any proofs to credit > concerning sets, his paradoxes aren't even interesting. See, this is far too specific for the *real* James Hunter. No use of > the word moron . No reference to the superiority of engineering. > Just silly. I have never claimed the superiority of engineering. I have claimed the superiority of *U.S. Engineers*. If you math idiots got a problem with that, make an alliance with Iraq, France, Canada, India, or Australia. It's not like my Nuke Subs cares. Of course, I won't ask for evidence for such silly claims. === Subject: Re: Socrates' Nothing is Everything > That's nice. But nobody mentioned anything about > definitions of infinity. Somebody in this > discussion who wasn't me, said about infinity > being a *property*. A property of *WHAT* > is the question. A property of sets. For a given set ask if there exists a bijection from > that set to a proper subset of itself. There either does or there does not. Through the process of elimination, > we can elimanate: 1) Einstein: Since he didn't even give Hilbert > that much credit for being bright. Nonsense. I guess you never read much of Einstein. The only people in the entire universe he considered more stupid than the people who inven Logic, were the people who inven Probability Theory. Who he considered infinitely more stupid. So stupid in fact that he even called their Nobel Prizes -- stupid -- on more than one occasion. 2) Cantor: Since Cantor being an actual logician, > rather than an idiot physicist actually > knew something about logic. Cantor inven set theory version 1.0 Cantor was somebody who lived in Europe in the latter-half of the eighteen century who inven that were orthoganal to politics. So he inven brains, not set theory. Computer scientists inven 1.0, as if they actually had a clue what a 1, a point, or a 0 is. > 3) Socrates: Since Socrates thought > ALL scientists were idiots. Nonsense again. In Socrates' day there were no scientists. Since Plato was the preeminent scientist on all of Earth for about two thoushand yea you are a 21st Century extreme moron, rather than just a ordinary moron. > 4) Goedel: Since Goedel actually knew something about > proofs, rather than the stupid propositional > calculus, mathemalawyers call a langauge. Goedel proved his incompleteness theorems wrt to first order logic, with > posulates for arithmetic added on. Goedel proved that first order logic was inven > Bob Kolker === Subject: Re: Socrates' Nothing is Everything This *infinite* thingy is very funny. In equation c = 6.28*radius, > (1) if radius is infinte, circumference will be 6.28 times infinity!! > (2) if radius is zero, circumeference will be 6.28 times zero!! We we are talking about limits if k > 0 then lim kx = inf as x -> inf and lim kx = 0 as x -> 0 for any positive k. Bob Kolker === Subject: Re: More Accessible Algorithm-Maze > One error in my other reply to Teabag's post. Correc below in this > repost. > (But, in any case, the error did not involve the particular > algorithm-loop being critisized.) >This puzzle has no solution; it is defective. The 11th step directs >>you to skip the 11th step the second time you encounter it. You can't >>skip the step if the step itself directs you to skip it. >>Also I'm lazy, and it was easier to stop the first time I got to the >>11th step. >>I'll try it again tomorrow, reading the 11th step as it was probably >>intended. > Huh?... Step 11 is encountered AND NOT skipped once, the *first* time you get > to > it. It then, the FIRST time you encounter it, tells you to skip it the > SECOND time you encounter it. Analogously: (if my pseudo-coding is not erroneous) > f=0: g=0 ....(steps 1 through 4) 5) g=g+1: move to adjacent square with number: if f+g=3 goto 5; else > goto 6 ....(steps 6 through 10) 11) f=f+1: if f=1 goto 5; else goto 12 I hope this clears things up... [In other reply, in line-5 I had if f+g=2 , instead of if f+g=3 , > the latter being correct.] >5) Move to adjacent square with a number. >11) Repeat the fifth step (step-5 above) twice, then repeat >steps (6),(7),(8),(9) and (10) once each in-order. >(Skip this {11th} step the second time you encounter it.) 12) Which square are you now on, and which symbol is in that square?? >(This maze only has one solution, or does it??..) Leroy Quet Hi Leroy-- Since my Calculatrivia was mentioned in this thread, I'd like to comment. I, too, found there was one answer, and I didn't have any trouble with Step 11. I thought the level of difficulty was perfect for your sta objective of something kids can do. I thought of a few ways to increase the complexity. Obviously, if you allow diagonal moves, and then allow the path to c itself, that would be a start. And instead of right left up down, how about spelling North, South, East, and West. The pairs of similar endings would quickly multiply the possible paths that still spell a valid word. You can have East Cing West and which st ending do you take first? Maybe both will work for quite some distance. With some thought it could lead a long way on multiple paths before dead-ending on some. I thought the spelling of words, adding numbeand counting symbols to determine how far to move were all clever ideas. If you REALLY want to get tricky, it could run like a self altering computer program, with steps like change the symbol above to whatever symbol is below , or switching two symbols, etc. Lots of possibilities here. a long list of complica directions, and many of them told you to alter othechange words, add or subtract things, etc, and then you would loop around, or skip to out-of-order directions. It was a real challenge to do perfectly correctly and get through it successfully. This format could be much the same idea, only the changes are on the grid instead of the direction words. (Or both?!!) Kindest Bob Lodge === Subject: Re: More Accessible Algorithm-Maze > Hi Leroy-- > Since my Calculatrivia was mentioned in this thread, I'd like to > comment. I, too, found there was one answer, and I didn't have any > trouble with Step 11. I thought the level of difficulty was perfect for > your sta objective of something kids can do. > I thought of a few ways to increase the complexity. Obviously, if > you allow diagonal moves, and then allow the path to c itself, that > would be a start. And instead of right left up down, how about spelling > North, South, East, and West. Or NorthEast, or SouthWest, or... (or {pi*3*4^Fibonacci(10)+...}radians, or...) :) >The pairs of similar endings would > quickly multiply the possible paths that still spell a valid word. You > can have East Cing West and which st ending do you take first? > Maybe both will work for quite some distance. With some thought it > could lead a long way on multiple paths before dead-ending on some. > I thought the spelling of words, adding numbeand counting > symbols to determine how far to move were all clever ideas. If you > REALLY want to get tricky, it could run like a self altering computer > program, with steps like change the symbol above to whatever symbol is > below , or switching two symbols, etc. Lots of possibilities here. > a long list of complica directions, and many of them told you to > alter othechange words, add or subtract things, etc, and then you > would loop around, or skip to out-of-order directions. It was a real > challenge to do perfectly correctly and get through it successfully. > This format could be much the same idea, only the changes are on the > grid instead of the direction words. (Or both?!!) Kindest Bob Lodge I have pos a few puzzles along the same idea over the last year or so. None is especially difficult, nor without faults. But in each I was only trying to showcase the basic idea, which you are familiar with already...(just a little familiar, or so I've heard...) (My {relatively unimpressive and unnumerous} puzzles along this line varied from algorithm-mazes which are grid-mazes {involving math more than in this thread's puzzle}, polygon-based algorithm-mazes {move between vertexes}, purely abstract algorithm-mazes {like computer programs altering an integer}, to rule-based word-chains. I would like to make more some day, with fewer ßaws {hopefully}, but I am lazy.) === Subject: Partition of positive integers In the following, let us denote [.] = the integral part N = the set of positive integers. For p > 0 define the set A[p]= { [np] ; n in N } . Find all irrational numbers p , q ,p>1 ,q>1 , such that (1) A[p] U A[q] = N . If (1) is satisfied, it's true that 1/p + 1/q = 1 ? Perhaps, please give a counterexample. Thank you, Alex ============== === Subject: Re: Partition of positive integers |> In the following, let us denote |> |> [.] = the integral part |> |> N = the set of positive integers. |> |> For p > 0 define the set A[p]= { [np] ; n in N } . |> |> Find all irrational numbers p , q ,p>1 ,q>1 , such that |> |> (1) A[p] U A[q] = N . |> |> If (1) is satisfied, it's true that 1/p + 1/q = 1 ? |> |> Perhaps, please give a counterexample. Thank you, Alex |> ============== |> The Beatty theorem states that if 1/a + 1/b = 1 a,b irrational then the sequences [ßoor(na),n>=1] and [ßoor(nb),n>=1] make a partition of the integers. If 1/a +1/b is too small. then some integers will be missed. If 1/a + 1/b is too large, then some integers will be repea. Since you seem to be interes with only the unions, it is possible to have the unions equal to N: take for example a<2 and replace b (defined by 1/a +1/b =1) by c=b/2, that is also larger than 1. For example a=sqrt(2) and c=(2+sqrt(2))/2. Then A[c] contains A[2c]=A[b] obviously and N=A[a] union A[b] = A[a] union A[c]. -- Charles Delorme tous les m .8egalomanes LRI ont une signature cd@lri.fr .88 .8etages === Subject: Elegance is for tailors. Lecture #4 Lecture 4: Gauge Symmetries and Invariances as PlatoÍs Theory of Forms See 2.1.3 on p. 28 and study the complex list on p. 29 with the three kinds of gauge symmetries and how they act differently on the set of fundamental objects . (i) are the spin gauge force symmetries that can be viewed as internal symmetries of Lie group G, or as symmetries in extra dimensions like in the Kaluza-Klein hyperspace Geometrodynamics of electric charge from the radius R of a curled up (AKA ñcompactifiedî) fourth space dimension. This has been generalized to ñCalabi-Yauî spaces of ñstring theoryî in the sense of Brian GreeneÍs NOVA PBS TV show ñThe Elegant Universe.î The gravity field tetrad e from locally gauging the translation group T4 and the spin connection W for parallel transport in warped base space-time from the Lie algebra of the tangent vector fiber space Lorentz group of rotations in 4D space-time are invariants under these Yang-Mills electro-weak-strong spin 1 gauge force symmetry transformations. See eqs. (2.47) to (2.51) on p.29. Note (2.49) how the gauge force potentials, AKA the connection field for parallel transport along paths in the extra dimensions, do not transform homogeneously like a multi-linear tensor of the group G. (ii) LIF Lorentz transformation in the quasi-ßat tangent vector fiber space to the warped base space-time. I use the term ñquasi-ßatî because the local curvature tensor, if it does not vanish, can still be locally measured in a LIF whose approximate metric is that of globally ßat special relativity. See eqs. (2.52) to (2.56) on p. 29. Note (2.56) how the so(3,1) spin connection transforms inhomogeneously under O(3,1). This is what Tesla & ñadelphia Experimentî expertî Jim Corum called the ñanholonomic objectî in his Ph.D. electrical engineering thesis under John Kraus at Ohio State that caused a lot of confusion at ISSOÍs UFO Physics Project in 1999-2000. See my two books ñDestiny Matrixî and ñSpace-Time and Beyond IIî both published 2002 for that story. KrausÍs radio telescope also got the ñWOWî possible ET signal I think in the 1970Ís? Nick Herbert, author of ñQuantum Realtyî also worked on that telescope for Kraus. (iii) Finally the vexing ñactiveî (ñpullbackî) nonlocal x -> xÍ =/= x Diffeomorphisms and their relation to the local passive general coordinate transformations at fixed space-time event x that even seemed to confuse Einstein for awhile. Look at (2.57) to (2.61) on p. 29. Note (2.61) which is a tensor transformation, but only on the one LNIF index. The two other indices are LIF in class (ii) above. Go back to the mixed LIF/LNIF equation (2.56) in (ii) for the O(3,1) Group LIF -> LIFÍ transformations at fixed x. Note there is a typo error in the indices in RovelliÍs (2.56). Let o denote a O(3,1) LIF -> LIFÍ transformation. W^IvJ -> o^IKWv^KLo^LJ + o^Iko^KJ,v (2.56) The first term on the RHS is the homogeneous multi-linear O(3,1) tensor transformation. The second term on the RHS is the inhomogeneous part that spoils the O(3,1) tensor property for the spin connection field for these tangent space fiber transformations. How do we get EinsteinÍs 1915 Levi-Civita Christoffel connection field {^uvw} for parallel transport of tensors in the curved (warped) LNIF ñbaseî space-time? What is the relation of it to the spin connection? On p. 27 eq. (2.42) is {^uvw} = e^lJ e^J^u(elIe^I(,v w) + ewIe ^I[,vl] + evIe^I[,wl]) ( ) is symmetrized anti-commutator of the LNIF Diff(4)indices and [ ] is the anti-symmetrized commutator of the indices. The comma denotes ordinary partial differentiation. Note, for example that e^I(,v w) means (1/2)[e^Iw,v + e^Iv,w]. Applying eq. (2.60) to eq. (2.42) gives a non-tensor transformation under the Diff(4) gauge symmetry group that is similar in form to eq. (2.56) for a different group of course! Note that the active Diff(4) symmetry group of (iii) comes from locally gauging the global T4 translation group of the Poincare space-time symmetry group of special relativity. T4 is intertwined with the O(3,1) Lorentz group as a semi-direct product, which is why the spin connection valued in the Lorentz group comes into the notion of parallel transport even in the curved space-time. Look again at eq. (2.3) on p. 23 for the covariant derivative DuV^I = V^I,u + Wu^IJV^J i.e. (2.3) p. 23 in a mixed LNIF/LIF representation. V^I = e^IvV^v (S8a) Therefore V^I,u = (V^ve^Iv),u (S8b) Wu^IJ = Wu^vwe^IveJ^w (S8c) V^J = V^vÍe^JvÍ (S8d) Therefore (2.3) is equivalent to Du e^IvV^v = (V^ve^Iv),u + Wu^vwe^IveJ^w V^vÍe^JvÍ (S8e) Assume metricity, so that Due^Iv = 0 (S8f) Therefore, by the chain rule of elementary calculus Du e^IvV^v = e^IvDuV^v (S8g) Also by the product rule of elementary calculus (V^ve^Iv),u = e^IvV^v,u + e^Iv,uV^v (S8h) Therefore e^IvDuV^v = e^IvV^v,u + e^Iv,uV^v + e^IvWu^vweJ^w V^vÍe^JvÍ (S8i) Hence DuV^v = V^v,u + Wu^vweJ^w e^Jl V^l + eI^v^e^Il,uV^l (S8j) However, we know that DuV^v = V^v,u + {^vul}V^l (S8k) Therefore {^vul} = Wu^vweJ^w e^Jl + eI^v^e^Il,u (S8l) We now have deduced the relationship between the connection field in the curved base space-time and the spin connection in the tangent vector bundle as the above inhomogeneous non-tensor expression. Whew! Let me wipe the sweat off my brow. ñElegance is for tailors.î ;-) === Subject: Coefficients of Cyclotimic Polynomial How come, the coefficients of the (p-1)th cyclotomic polynomial (where p is a prime) is always equal to 0, 1 or -1? (or is this wrong?) === Subject: Re: Coefficients of Cyclotimic Polynomial > How come, the coefficients of the (p-1)th cyclotomic polynomial (where p is > a prime) is always equal to 0, 1 or -1? (or is this wrong?) As others have poin out, this is not true in general (although if you replace p-1 by p it certainly is, obviously). You might find this sequence useful, although it does not answer your question: Smallest order of cyclotomic polynomial containing n or -n as a coefficient. http://www.research.att.com/projects/OEIS?Anum=A013594 -Jim Ferry === Subject: Re: Coefficients of Cyclotimic Polynomial If not all the coefficients, what about just the first two(the first one not zero of course)? > How come, the coefficients of the (p-1)th cyclotomic polynomial (where p is > a prime) is always equal to 0, 1 or -1? (or is this wrong?) === Subject: Re: Coefficients of Cyclotimic Polynomial http://tinylink.com/?O4HnnZ5HE8 === Subject: Re: Coefficients of Cyclotimic Polynomial >How come, the coefficients of the (p-1)th cyclotomic polynomial (where p is >a prime) is always equal to 0, 1 or -1? (or is this wrong?) The coefficients of t^7 and t^41 in cyclotomic(210, t) are 2. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Coefficients of Cyclotimic Polynomial >>How come, the coefficients of the (p-1)th cyclotomic polynomial (where p is >>a prime) is always equal to 0, 1 or -1? (or is this wrong?) >The coefficients of t^7 and t^41 in cyclotomic(210, t) are 2. >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2 The 28210-th cyclotomic polynomial has coefficients as large as +-14. For a smaller example, the 2730-th cyclotomic polynomial has coefficients as large as +-4. | ^/| Maple 9 (SUN SPARC SOLARIS) MAPLE / All rights reserved. Maple is a trademark of <____ ____> Waterloo Maple Inc. | Type ? for help. > with(numtheory): Warning, the protec name order has been redefined and unprotec > sort(cyclotomic(2730, x)); 576 575 573 572 571 566 565 564 563 562 560 x + x - x - x - x + x + x + x - x - x + x 557 555 554 553 552 551 550 549 547 544 - x + x + x - x - x - x + x + x - x + x 542 540 539 537 536 533 532 530 529 527 - x + 2 x + x - x - x - x - x + x + 2 x - x 525 522 520 519 518 517 516 515 514 512 + x - x + x + x - x - x - x + x + x - x 511 510 509 508 507 506 505 504 503 501 + x + x + x - x - 2 x - 2 x + x + x + x + x 500 499 498 497 496 495 494 493 492 490 + x + x - 2 x - 2 x - x + x + x + x - x + x 489 488 487 486 485 484 482 481 480 + x - x - x - x + 2 x + 2 x - 2 x - x - x 479 477 475 474 473 472 471 470 469 + x - x + 3 x + 2 x + x - 2 x - 3 x - x + x 465 464 463 462 461 460 458 456 454 + x + 3 x + x - 2 x - 2 x - x + 2 x - x + x 451 449 448 445 442 441 440 439 438 - 2 x + 2 x + 2 x - 2 x - x - 2 x + x + x + 3 x 437 436 435 433 431 429 428 426 425 + x - x - 2 x - x - x + 2 x + 3 x - x - 2 x 424 423 422 421 419 418 416 415 413 - x + x + x - x + x + x - 2 x - 3 x + 2 x 412 411 408 407 406 405 403 402 401 + 2 x + x + x - x - 3 x - 2 x + 2 x + 3 x + x 400 397 396 395 393 392 390 389 387 - x - x - 2 x - 2 x + 3 x + 2 x - x - x + x 386 385 384 383 382 381 380 379 378 - x - x + x + 2 x + x - x - 4 x - 2 x + x 377 376 375 373 372 371 370 369 367 + 2 x + x + x + 2 x + x - 2 x - 4 x - 2 x + 3 x 366 363 361 360 359 358 357 356 354 + 2 x + x - x - 3 x - 3 x + x + 3 x + 2 x - x 353 352 350 349 348 347 346 345 344 - x + x - x - x + x + 2 x + 2 x - 2 x - 3 x 343 340 338 337 336 335 334 333 - 2 x + x + 2 x + 3 x + 2 x - 2 x - 3 x - 3 x 331 330 328 327 326 325 324 323 322 + x + x + x + x + x - x - 3 x - 2 x + x 321 320 318 315 314 312 311 310 309 + 2 x + x - x - x - 2 x + 2 x + 4 x + 2 x - x 308 307 306 305 304 302 301 300 - 3 x - 2 x - 2 x - x - x + 2 x + 4 x + 2 x 298 297 296 294 292 291 290 289 288 - 2 x - 2 x - x - x + x + 2 x + x - x - 3 x 287 286 285 284 282 280 279 278 276 - x + x + 2 x + x - x - x - 2 x - 2 x + 2 x 275 274 272 271 270 269 268 267 266 + 4 x + 2 x - x - x - 2 x - 2 x - 3 x - x + 2 x 265 264 262 261 258 256 255 254 253 + 4 x + 2 x - 2 x - x - x + x + 2 x + x - 2 x 252 251 250 249 248 246 245 243 242 - 3 x - x + x + x + x + x + x - 3 x - 3 x 241 240 239 238 236 233 232 231 - 2 x + 2 x + 3 x + 2 x + x - 2 x - 3 x - 2 x 230 229 228 227 226 224 223 222 220 + 2 x + 2 x + x - x - x + x - x - x + 2 x 219 218 217 216 215 213 210 209 207 + 3 x + x - 3 x - 3 x - x + x + 2 x + 3 x - 2 x 206 205 204 203 201 200 199 198 197 - 4 x - 2 x + x + 2 x + x + x + 2 x + x - 2 x 196 195 194 193 192 191 190 189 187 186 - 4 x - x + x + 2 x + x - x - x + x - x - x 184 183 181 180 179 176 175 174 173 + 2 x + 3 x - 2 x - 2 x - x - x + x + 3 x + 2 x 171 170 169 168 165 164 163 161 160 - 2 x - 3 x - x + x + x + 2 x + 2 x - 3 x - 2 x 158 157 155 154 153 152 151 150 148 + x + x - x + x + x - x - 2 x - x + 3 x 147 145 143 141 140 139 138 137 136 + 2 x - x - x - 2 x - x + x + 3 x + x + x 135 134 131 128 127 125 122 120 118 - 2 x - x - 2 x + 2 x + 2 x - 2 x + x - x + 2 x 116 115 114 113 112 111 107 106 105 - x - 2 x - 2 x + x + 3 x + x + x - x - 3 x 104 103 102 101 99 97 96 95 94 92 - 2 x + x + 2 x + 3 x - x + x - x - x - 2 x + 2 x 91 90 89 88 87 86 84 83 82 81 80 + 2 x - x - x - x + x + x - x + x + x + x - x 79 78 77 76 75 73 72 71 70 69 68 - 2 x - 2 x + x + x + x + x + x + x - 2 x - 2 x - x 67 66 65 64 62 61 60 59 58 57 56 54 + x + x + x - x + x + x - x - x - x + x + x - x 51 49 47 46 44 43 40 39 37 36 34 + x - x + 2 x + x - x - x - x - x + x + 2 x - x 32 29 27 26 25 24 23 22 21 19 16 14 + x - x + x + x - x - x - x + x + x - x + x - x 13 12 11 10 5 4 3 - x + x + x + x - x - x - x + x + 1 quit bytes used=914060, alloc=786288, time=0.20 -- John Adams served two terms as Vice President and one as President, but lost reelection. Later his son became President despite losing the popular vote. That son lost his reelection attempt badly. Now history is repeating itself. pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: Re: Coefficients of Cyclotimic Polynomial Peter L. Montgomery > The 28210-th cyclotomic polynomial has coefficients as large as +-14. > For a smaller example, the 2730-th cyclotomic polynomial > has coefficients as large as +-4. When I looked up the OEIS as Jim Ferry told, I got: ID Number: A013594 Sequence: 0,105,385,1365,1785,2805,3135,6545,6545,10465,10465,10465, 10465,10465,11305,11305,11305,11305,11305,11305,11305,15015, 11305,17255,17255,20615,20615 Name: Smallest order of cyclotomic polynomial containing n or -n as a coefficient. Offset: 1 So cyclotomic(1365) is the first with some +-4 as a coefficient. And you have +-14 at lower than a half of 28210, namely at 10465. The sequence is given up to +-27 and still is below the 28210-th cyclotomic polynomial. There's no need for you to give examples in lowest terms. But I just hope, I didn't get something wrong here. Best Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Coefficients of Cyclotimic Polynomial The original posting asked about the (p-1)-th cyclotomic polynomial where p is prime. >Peter L. Montgomery >> The 28210-th cyclotomic polynomial has coefficients as large as +-14. >> For a smaller example, the 2730-th cyclotomic polynomial >> has coefficients as large as +-4. >When I looked up the OEIS as Jim Ferry told, I got: >ID Number: A013594 >Sequence: 0,105,385,1365,1785,2805,3135,6545,6545,10465,10465,10465, > 10465,10465,11305,11305,11305,11305,11305,11305,11305,15015, > 11305,17255,17255,20615,20615 >Name: Smallest order of cyclotomic polynomial containing n or -n > as a coefficient. >Offset: 1 >So cyclotomic(1365) is the first with some +-4 as a coefficient. >And you have +-14 at lower than a half of 28210, namely at 10465. >The sequence is given up to +-27 and still is below the 28210-th >cyclotomic polynomial. >There's no need for you to give examples in lowest terms. But I >just hope, I didn't get something wrong here. >Best >Rainer Rosenthal >r.rosenthal@web.de -- John Adams served two terms as Vice President and one as President, but lost reelection. Later his son became President despite losing the popular vote. That son lost his reelection attempt badly. Now history is repeating itself. pmontgom@cwi.nl Microsoft Research and CWI Home: San Rafael, California === Subject: Re: Coefficients of Cyclotimic Polynomial Peter L. Montgomery > The original posting asked about the (p-1)-th cyclotomic > polynomial where p is prime. Aha! Thanks for the explanation: But I just hope, I didn't get something wrong here. > Now history is repeating itself. Hmmm... you are right: I faintly remember having made an error once or even twice. Rainer Rosenthal r.rosenthal@web.de === Subject: VonNeumann Gametheory Optimal Strategy on StockMarket; Cover 29JAN04 Portfolio of PAF as of 29JAN04: BCE 6,720 22.45 $150,864.00 BMY 50 28.82 $1,441.00 Q 14,200 4.00 $56,800.00 SBC 12,200 25.70 $313,540.00 realestate land 3APR03 of 3 lots $19,000. science-art of pictures,porcelain etc starting JAN03 for $14,143. realestate land 30JUL03 another lot $11,500. Today I saw a cover and took advantage of it, even though it was a mere 3 free shares of BCE. What is the old saying gain is better than nothing or worse a loss at 47.90 and with the proceeds bought 210 shares of BCE at 22.40 when cheaper BCE today than if I had bought those BCE back on the 8th of January. And even though I had a few dollars lost in selling MRK at 48.10 when I bought it at 47.90, the point of Cover is not paper-money gain or loss but a slow and steady rise in numbers of shares of stock. So by selling 100 MRK today and buying 210 BCE today I got 3 of those shares of BCE for free. I have it in my mind that 2 shares of BCE should be valued much more than Merck. The last time I looked, MRK had a total market capitalization of $110 billion and BCE $20 billion. That in my mind is skewed because MRK major patents expire and their pipeline is not as robust as in the past. And with the clamor of health insurance and prescription drugs, I do not see the drugs roaring ahead. On the other hand, I see evidence of a fight for wireless telephony and BCE not only controls most of Canada's fixed wire but also wireless as a bidding war for AT&T is taking place. So I think the price of BCE should be about $26 per share now and the price of Merck about $44 per share and should that come about, I would consider switching back into Merck, only with more than just a 100 shares. I contempla putting Merck and Medtronics into the portfolio and running a switching campaign with Cover on them. For it is conceivable that MRK may pull ahead of MDT soon, say $51 whereas MDT falls back to $48 and switch into MDT and then wait for MDT to surpass MRK and switch again. But the trouble with that campaign is that MDT pays such a minuscule dividend. As mothers so often tell their daughters it is just as easy to fall in love with a rich man as a poor man, I so often tell myself that it is just as easy to running switching campaigns with fat dividend payers as with tiny payers. Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: 1st-order wave equation We see frequently the wave equation of second order in various fields of engineering and physics. However the wave equation of first order is not seen so frequently as former one except in some specific fields. I tried to apply the wave equation of first order to Newtonian mechanics and quantum mechanics in the following site: http://139.134.5.123/tiddler2/wave/waveequation.htm === Subject: Re: Fractal problem >Of course, defined iteratively, this is a fractal. It's self-similar, but self-similarity isn't a necessary or sufficient condition for a set to be a fractal by a lot of definitions of fractal . The Hausdorff dimension of this figure is 1, which is the same as for any old smooth curve. The figure you describe has a finite length, in fact. === Subject: Advances in Complex Systems - Vol 6 No 4 Advances in Complex Systems View table-of-contents and abstracts at http://www.worldscinet.com/acs.html Contents: Modeling And Parametric Study Of A Fabric Drape Pascal Bruniaux, Adel Ghith And Christian Vasseur Analysis And Modeling Of Science Collaboration Networks Felix Putsch An Astrophysical Basis For A Universal Origin Of Life Stirling A. Colgate et al. Coupled Growing Networks Dafang Zheng And Guler Ergun The Birth And Death Processes Of Hypercycle Spirals Kazumasa Oida Complex Systems In Language Evolution: The Cultural Emergence Of Compositional Structure Kenny Smith, Henry Brighton And Simon Kirby Asymmetry in the hierarchy model of bonabeau et al. D. Stauffer and J. S. Sa Martins Community Structure In Jazz Pablo M. Gleiser and Leon Danon A System Identification Approach To Estimating Complex Impedance Gerardo Miramontes De Leon, David C. Farden and Lyle E. Mcbride Systemic Lupus Erythematosus In African-American Women: Cognitive Physiological Modules, Autoimmune Disease, And Structured Psychosocial Stress Rodrick Wallace Decision Spread In The Corporate Board Network Stefano Battiston, Gerard Weisbuch And Eric Bonabeau For more information, go to http://www.worldscinet.com/acs.html === Subject: Is this quotientan integer? for k, p, m, n in N*, let B(k,p,m,n)=((k+p+1)m)!((k+p+1)n)! / [(pm)!(pn)!(km+n)!(kn+m)!] Can somebody proove that it is an integer (if not, is it for some values of k)? Thanks for any help. Amically, Georges === Subject: groups and matrices Let V be a finite dim (say n>2) vector space over a field F. Suppose that G is a group of nxn invertible matrices with entries in F satisfying the following property: If S,T are subspaces of V, there is a g in G such that g(S)=T. Then G contains all the matrices of determinant one. Is this true? I thank any proof or counterexample. === Subject: Re: groups and matrices >Let V be a finite dim (say n>2) vector space over a field F. >Suppose that G is a group of nxn invertible matrices with entries in F >satisfying the following property: >If S,T are subspaces of V, there is a g in G such that g(S)=T. Do you want to assume that S,T have the same dimension? >Then G contains all the matrices of determinant one. >Is this true? I thank any proof or counterexample. Counterexamples are not hard to find. We could take G to be a subgroup of order 7 in GL(3,2). Since there are 7 subspaces of dimension 1 and 7 of dimension 2, then G is clearly transitive on both sets. Derek Holt. === Subject: Re: groups and matrices >Let V be a finite dim (say n>2) vector space over a field F. >Suppose that G is a group of nxn invertible matrices with entries in F >satisfying the following property: >If S,T are subspaces of V, there is a g in G such that g(S)=T. There cannot be such a G, unless V = {0}. >Then G contains all the matrices of determinant one. >Is this true? I thank any proof or counterexample. ************************ === Subject: Re: groups and matrices > Let V be a finite dim (say n>2) vector space over a field F. Suppose that G is a group of nxn invertible matrices with entries in F > satisfying the following property: If S,T are subspaces of V, there is a g in G such that g(S)=T. So there's a g in G with g({0}) = V? -- Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: The Lost Proof of Fermat > Using good ol' Fermat's equation: A^n + B^n = C^n only has integer solutions for n = 2 I think it also has a few integer solutions when n = 1. I believe I can find some for n = 0, too Although I'm not sure if n needs to be a natural number; I only remember seeing proofs for n>2. === Subject: Re: The Lost Proof of Fermat x^3 + (x+1)^3 = 2x^3 + 3x^2 + 3x + 1 is not a perfect cube, far all positive integex, ... no integer solutions. x^2 + (x+1)^2 = 2x^2 + 2x + 1 has perfect square solutions for specified values of integer, x ... Let x^3 + (x+1)^3 = 2x^3 + y y = 3x^2 + 3x + 1 3x^2 forms perfect cubes 3x forms perfect cubes 3x^2 + 3x forms perfect cubes 3x^2 + 1 does not 3x + 1 does not 3x^2 + 3x + 1 does not 2x^3 + y + K = z^3 Interesting... === Subject: Re: The Lost Proof of Fermat x^3 + (x+1)^3 = 2x^3 + y > y = 3x^2 + 3x + 1 > 3x^2 forms perfect cubes > 3x forms perfect cubes > 3x^2 + 3x forms perfect cubes > 3x^2 + 1 does not > 3x + 1 does not When x=0, 3x+1 = 1^3 When x=21, 3x+1 = 4^3 ... > 3x^2 + 3x + 1 does not > 2x^3 + y + K = z^3 > Interesting... Hmm? J === Subject: Re: The Lost Proof of Fermat by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QMT1d07965; >The limit of [A^x + B^x]^[1/x] , as x goes to infinity, is B. >For positive integers x,y,z,p >x^p + y^p = z^p >If >(x+y)^p = z^p + f(x,y) >(x-y)^p = z^p - f(x,y) >(x+y)^p + (x-y)^p = 2*z^p >[(x+y)^p + (x-y)^p ]/2 = x^p + y^p = z^p >then >p = 2 but only for p=2 it is true from your beginning: (x+y)^p = z^p + f(x,y) (x-y)^p = z^p - f(x,y) because for p=3 and all bigger p and even not primes (x+y)^3 = z^3 + f1(x,y); f1(x,y)= x^3 +3x^2 y +3xy^2 +y^3 - z^3 = 3xy(x+y) (x-y)^3 = z^3 - f2(x,y); f2(x,y)= -x^3 +3x^2 y -3xy^2 +y^3 +z^3 = 3xy(x-y) + 2y^3 also f1(x,y) # f2(x,y) ( You can check for some other p ) except of p=2 once f1(x,y)= f2(x,y)= 2xy Things are going rather rough, until You'll recover some Euler's findings and etc. till Andrew Wiles topics. But if You'll find some more time so You can judge some less perfect proofs: see topic Pure abstract demonstration of mine edi at 26.01.04 at sci.math Roman B. Binder love is the lonely hunter - similar with the extra number rules === Subject: Calculus - Continuity Can a function be continuous at an endpoint of an interval? For instance. f(x)=sec(x) The interval is [-pi/2, pi,2]. Is the function continuous on this interval? How about (-pi/2,pi/2)? I think this is obviously YES. Thanks === Subject: Re: Calculus - Continuity > Can a function be continuous at an endpoint of an interval? Yes > For instance. f(x)=sec(x) > The interval is [-pi/2, pi,2]. > Is the function continuous on this interval? The function *is not defined* at the endpoints and therefore the question is meaningless. > How about (-pi/2,pi/2)? I think this is obviously YES. Obviously. Best === Subject: Re: Calculus - Continuity Can a function be continuous at an endpoint of an interval? > Yes For instance. f(x)=sec(x) The interval is [-pi/2, pi,2]. Is the function continuous on this interval? > The function *is not defined* at the endpoints and therefore the > question is meaningless. Your answer, JCS, is good for Excel (who is probably a beginning calculus student). Nonetheless, I'll note that if we allow the codomain of the secant function to be the one-point extension of the reals, so that sec(-pi/2) and sec(pi/2) are defined, then sec(x) is continuous on [-pi/2, pi/2]. Best David Cantrell How about (-pi/2,pi/2)? I think this is obviously YES. > Obviously. > Best > === Subject: Learning maths I am a fan of Schaum's outline maths books, since they help to practice what you have read on other books (the theory). The offer a lot of exercises and show their solutions. Does anyone know if there is a Schaum's outline (or different ones), covering: - Algebra: Grouptheory (Groups, Rings, Fields) (hope these are the right english terms) - Discrete maths: Relations, Maps, modular arithmetics - Calculus: Sequences (finite, infinite), Series Which Schaum's cover these topics in detail ? Do you know similar literature (not for reading the theory, but for practicinng - with solutions !) ? Thanks sooo much ! === Subject: A particular cost function I have a sequence of observations (x(1), x(2), ...,) There is a set of functions f_i, such that y_i(t) = f_i(x(t-N),...,x(t-1)) I have the MSE cost function: C = sum_t sum_i (p_i(t) (y_i(t)-x(t)))^2 (1) which I try to minimise by gradient descent.. If p_i(t) is independent on the input x, or is directly dependent on the input (ala mixtures of experts) then things are very simple, but actually: p_i(t+1) = a p_i(t) + (1-a) frac { exp(e_i^2(t+1))} (2) { sum_j exp(e_j^2(t+1))} where e_i(t) = y_i(t) - x(t). In that case , what is the derivative of C with respect to y_i? Christos -- Christos Dimitrakakis === Subject: I need a routine to verify a BigInteger is a square by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0UE5Nv08635; I need this in my Java 2 code. Can anybody help? Kent Holing === Subject: Re: I need a routine to verify a BigInteger is a square >I need this in my Java 2 code. Can anybody help? Is there a builtin way to divide two BigIntegegetting the quotient rounded or trunca to a BigInteger? If so you could use Newton's method to search for sqrt(N). >Kent Holing ************************ === Subject: Re: Elementary Inequality for Fun by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0UE5ab08749; Actually, the complete inequality is: AB>3>CD, where C=(e/2)^(pi/2) and D=(pi/2)^(e/2). Let us observe that CD<3 is harder. As for AB>3 , it results easily from e^e>15 (since from this it follows that A>3/2, and it is not very difficult to see that B>2). Now, let us remark that e>19/7. Therefore, it suffices to prove 19^19>(15^7)*(7^19), which is indeed true, but I'm asking for an elegant manner to achieve this last inequality by hand. Best Ady. >Hi! >I'm asking for a classical (i.e., by hand, without using any calculators) solution of the following inequality: >Let A=(e/2)^(e/2) and B=(pi/2)^(pi/2). Then AB>3. >Best >Ady. === Subject: Re: Any news on odd perfect numbers? To Mark Griffith: Extremely hard to get your ideas. However, I tried my best. Comments follow. >Sorry about delay. >. >Since we know that the first odd perfect number [OP] >must be a product of a square [sq] and an odd power of >a single prime [pr] we start with the case where pr is >raised to the power of 1 and sq.pr = OP. >(1) Visualise OP as an array of pr-many squares, each >containing sq-many tiles, and as OP is perfect we >assign all factors as follows: >i] one square represents the factor sq itself; >ii] pr-2 squares contain factors of sq multiplied by >pr; >iii] one square contains factors of sq + an overßow >from [ii] which we call x. >We know that there is an overßow from the factors of >sq multiplied by pr, since [a] the factors of sq >cannot sum to more than sq [else the factors of OP >will sum to more than sq.pr], nor can they sum to >exactly sq since by hypothesis OP, not sq, is the >first odd perfect and also the factors of sq must have >an even sum [odd-many proper factors + 1]. >Therefore the factors of sq sum to sq-x where x is >odd. Further, x must be 1 or else divide into sq, >since [sq-x]pr = [pr-2]sq + x and so x.pr = sq + sq-x. >(2) But by the following argument the factors of an >odd square cannot sum to sq-x where x divides into sq. >Either x=1 or x>1. Suppose x=1. To complex. For x=1 proof takes 2 lines :) >We know that factors which are paired with >whole-number multiples of sqrt will occupy the same >number of tiles in one column as the factors which are >whole-number multiples of sqrt will occupy whole >columns elsewhere. >So if x=1 then sq-1 takes up (sqrt-1)(sqrt + 1) tiles. >We can either divide the factors into two groups, one >of which (larger) sums to (sqrt-1).sqrt and the other >(smaller) to (sqrt-1), or we cannot. >If not, then one larger factor either overlaps or >underlaps the divide between the (sqrt-1).sqrt area >and the (sqrt-1) area. In either case there is a >contradiction, since the larger-factor sum is either >too large to be sqrt times the smaller-factor sum >(overlapping), or too small to be sqrt times the >smaller-factor sum (underlapping). >So we must have a set of larger factors which sum to >an exact multiple of sqrt. But dividing the sum of >larger factors through by sqrt should then give a >whole number, sqrt-1. We note that with cases like sq = >abb.abb we get larger factors like bbb and bbbb and >with cases like sq = abc.abc larger factors like aabb, >aacc, bbcc which cannot sum to an integer total when >divided by sqrt. By induction on numbers being >multiplied up into products with more factowe note >that as xy can only multiply up to xyy or to xyz with >the addition of one factor then we will always have >incommensurable sums when dividing by sqrt. >Therefore the larger factors which are not already whole >multiples of sqrt will not sum to a whole multiple of >sqrt+1. Although sq-1 divides without remainder by >sqrt+1, we cannot divide the sum of factors into two >blocks measuring sqrt-1 and [sqrt-1].sqrt in size. >Therefore no odd squares have factors summing to >exactly sq-1. >(3) This leaves x>1 where x divides into sq. >Dividing sq-1 by sqrt+1 gives sqrt-1, so we know that, >since sqrt is odd, that there is an even multiple of >sqrt+1 in sq, with a final remainder of 1. >To test whether the factors of sq could sum to sq-x >where x divides into sq, we divide sqrt+1 both into >sq-x and into x and examine the remainder1 and r2. >Let [sq-x]/[sqrt+1] = p + r1, and x/[sqrt+1] = q + r2. I suppose you mean sq-x = p(sqrt+1)+r1, x=q(sqrt+1)+r2. (from your variant of equations r1 and r2 are not integers) >Since there is still the final remainder of 1 to >complete sq, if overßow x can divide into sq and >complete sq, then r1 + r2 must equal sqrt + 1 + 1. >As sqrt+1 cannot divide into sq, so not into sq-x or x >either, we know both r1 and r2 are non-zero. >But since sq-x is even, x is odd, and p + q odd [the >total number of sqrt+1 in sq-1 is even = sqrt-1 = p + >q + 1] we either have an odd p leaving an odd >remainder in sq-x and an even q leaving an odd >remainder in x, or an even p leaving an even remainder >in sq-x and an odd q leaving an even remainder in x. >Either way, r1 + r2 is even, but sqrt + 2 is odd. Here is my main question. sq-x = p(sqrt+1)+r1, x=q(sqrt+1)+r2 (previous comment). sq-x -- even, sqrt+1 -- even. Then r1 is even with any p( from 1st equation). x -- odd, sqrt+1 - even. Then r2 is odd with any q. >Therefore the factors of an odd square sq cannot sum >to sq-x if x>1 and divides into sq. >Since this obstacle holds equally for higher powers of pr, >we see there are no odd perfect numbers. >Mark Griffith >14th Jan 04 >. === Subject: Re: fundamental period Hello >What is the usual definition of the fundamental period of a function >f? I've see 2 different ones: (1) - The smallest number p such that >f(x+ p) = f(x) for every x (the same as minimum period) and (2) - the >smallest positive integer n such that f(x+n) = f(x) for every x. Have you really seen the second definition, _for_ functions > f : R -> R? This seems like a very strange definition - I know > I've never seen any such thing. (On the other hand, if we > were talking about f : Z -> R then it would make perfect > sense, being equivalent to the first one.) If we >adopt definition (2), then it's possible that a function is periodic >but has no fundamental period. This is possible even with definition 1, if we don't require that > f be continuous (and it can happen with a non-constant > function). For example, say f(x) = 0 if f is rational and > f(x) = 1 if x is irrational. Then any rational number is a > period of f, so there is no smallest period. If f has an irrational period, then is it possible tha f also has a >rational one? >Thank you >Amanda I think I made a mistake. The usual definition is actually the first one. If :R->R is continuous and periodic on R, then f is bounded and uniformly continuous. Let P be the set of periods of f. By definition, P is bounded below by 0 and, therefore, has a infimum w. I'm trying to prove that, if f is not constant, then w belongs to P, so that the fundamental period exists. But I got lost, could you help, please Tnak you. Amanda? === Subject: Re: fundamental period >>Hello >>What is the usual definition of the fundamental period of a function >>f? I've see 2 different ones: (1) - The smallest number p such that >>f(x+ p) = f(x) for every x (the same as minimum period) and (2) - the >>smallest positive integer n such that f(x+n) = f(x) for every x. >> Have you really seen the second definition, _for_ functions >> f : R -> R? This seems like a very strange definition - I know >> I've never seen any such thing. (On the other hand, if we >> were talking about f : Z -> R then it would make perfect >> sense, being equivalent to the first one.) >>If we >>adopt definition (2), then it's possible that a function is periodic >>but has no fundamental period. >> This is possible even with definition 1, if we don't require that >> f be continuous (and it can happen with a non-constant >> function). For example, say f(x) = 0 if f is rational and >> f(x) = 1 if x is irrational. Then any rational number is a >> period of f, so there is no smallest period. >>If f has an irrational period, then is it possible tha f also has a >>rational one? >>Thank you >>Amanda >I think I made a mistake. The usual definition is actually the first >one. >If :R->R is continuous and periodic on R, then f is bounded and >uniformly continuous. Let P be the set of periods of f. By definition, >P is bounded below by 0 and, therefore, has a infimum w. I'm trying to >prove that, if f is not constant, then w belongs to P, so that the >fundamental period exists. But I got lost, could you help, please >Tnak you. Amanda? Hint: If w is the inf of P then there exists a sequence p_n in P with p_n -> w. ************************ === Subject: Re: fundamental period I think I made a mistake. The usual definition is actually the first >one. If :R->R is continuous and periodic on R, then f is bounded and >uniformly continuous. Let P be the set of periods of f. By definition, >P is bounded below by 0 and, therefore, has a infimum w. I'm trying to >prove that, if f is not constant, then w belongs to P, so that the >fundamental period exists. But I got lost, could you help, please >Tnak you. Amanda? Hint: If w is the inf of P then there exists a sequence p_n in P > with p_n -> w. > Ok, lemme try. Fix an x in R. Then, x+ p_n -> x+w. Since f is continuous, f(x+p_n) -> f(x+w). But for every n, f(x+p_n) = f(x), which implies f(x+p_n) converges trivially to f(x). It follows f(x) = f(x+w) for every x, which shows w is a period and belongs to P. Therefore, if w>0 the fundamental limit exists. It remains to show that, if w=0, the f is constant. Fix x>0 in R (if x<0,the argiments are similar). Since w=0, for every natural n there's p_n in P such that 0< p_n < 1/n^2 and 0 < n*p_n < 1/n. Therefore, n*p_n -> 0. Given 0 0, we get f(x) = f(0). And since x is arbitrary, we conclude f is constant. By contraposition, it follows that, if f is not constant, then w>0 and the fundamental limit exists. I was told there's another proof based on the kernell and ideal of groups, but I don't know how it goes. Thank you, david Amanda === Subject: Re: fundamental period >>I think I made a mistake. The usual definition is actually the first >>one. >>If :R->R is continuous and periodic on R, then f is bounded and >>uniformly continuous. Let P be the set of periods of f. By definition, >>P is bounded below by 0 and, therefore, has a infimum w. I'm trying to >>prove that, if f is not constant, then w belongs to P, so that the >>fundamental period exists. But I got lost, could you help, please >>Tnak you. Amanda? >> Hint: If w is the inf of P then there exists a sequence p_n in P >> with p_n -> w. >Ok, lemme try. Fix an x in R. Then, x+ p_n -> x+w. Since f is >continuous, f(x+p_n) -> f(x+w). But for every n, f(x+p_n) = f(x), >which implies f(x+p_n) converges trivially to f(x). It follows f(x) = >f(x+w) for every x, which shows w is a period and belongs to P. >Therefore, if w>0 the fundamental limit exists. >It remains to show that, if w=0, the f is constant. Fix x>0 in R (if >x<0,the argiments are similar). Since w=0, for every natural n there's >p_n in P such that 0< p_n < 1/n^2 and 0 < n*p_n < 1/n. Therefore, >n*p_n -> 0. Given 0n*p_n < x (1). Was that exactly what you meant to say? I don't see why there exists n with x-h < n*p_n < x. Also don't see why you need that. All you need is p_n < epsilon and m*p_n in the interval (x-h, x). Which is maybe what you really meant. >Since p_n is a period of f, f(n*p_n) = f(0) for every Hmm. Here it seems _possible_ that you're misunderstand what the notation o(h) means. I don't see how you get o(h) and I don't see why you need it; what's clear is that you get o(1), and that's enough. (o(h) means some function such that it divided by h tends to 0 . All you need is some function that tends to 0 as h -> 0, which is also, it seems to me, all you have. That could be deno o(1) , for some function such that it divided by 1 tends to 0 , ie something that tends to 0 .) > In >virtue of inequalty (1), it follows that f(x) - f(n*p_n) = f(x) - f(0) >= o(n*p_n). Since n*p_n -> 0, we get f(x) = f(0). And since x is >arbitrary, we conclude f is constant. >By contraposition, it follows that, if f is not constant, then w>0 and >the fundamental limit exists. >I was told there's another proof based on the kernell and ideal of >groups, but I don't know how it goes. >Thank you, david >Amanda ************************ === Subject: Proof of poincare lemma on an open subset of R^n i want to know the proof of poincare lemma on differetial forms defined on open subset of R^n. could anyone suggest me a reference? i want a proof not involving the terminologies of manifolds. thank you in advance .. === Subject: Re: Is the non-existence of odd perfect number proved? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0UF7sE14064; >My 2 cents: The paper is poorly written. I can't say it's >right or wrong since I gave up trying to understand it. I have to agree with David Joyner on this one. I got lost in that first sentence, and either I'm not familiar with the notation, or it is not coming out right on my screen. I think the answer is no, there is still no proof or counterexample to the plausible conjecture that there are no odd perfect numbe unless this http://mathquest.com/discuss/sci.math/m/569936/570941 makes sense, in which case, yes there is. Mark G. . === Subject: Re: please find ßaws, anyone? Re: odd perfect numbers by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0UF7s014060; Extremely hard to get your ideas. However, I tried my best. Comments follow. >Sorry about delay. >. >Since we know that the first odd perfect number [OP] >must be a product of a square [sq] and an odd power of >a single prime [pr] we start with the case where pr is >raised to the power of 1 and sq.pr = OP. >(1) Visualise OP as an array of pr-many squares, each >containing sq-many tiles, and as OP is perfect we >assign all factors as follows: >i] one square represents the factor sq itself; >ii] pr-2 squares contain factors of sq multiplied by >pr; >iii] one square contains factors of sq + an overßow >from [ii] which we call x. >We know that there is an overßow from the factors of >sq multiplied by pr, since [a] the factors of sq >cannot sum to more than sq [else the factors of OP >will sum to more than sq.pr], nor can they sum to >exactly sq since by hypothesis OP, not sq, is the >first odd perfect and also the factors of sq must have >an even sum [odd-many proper factors + 1]. >Therefore the factors of sq sum to sq-x where x is >odd. Further, x must be 1 or else divide into sq, >since [sq-x]pr = [pr-2]sq + x and so x.pr = sq + sq-x. >(2) But by the following argument the factors of an >odd square cannot sum to sq-x where x divides into sq. >Either x=1 or x>1. Suppose x=1. To complex. For x=1 proof takes 2 lines :) >We know that factors which are paired with >whole-number multiples of sqrt will occupy the same >number of tiles in one column as the factors which are >whole-number multiples of sqrt will occupy whole >columns elsewhere. >So if x=1 then sq-1 takes up (sqrt-1)(sqrt + 1) tiles. >We can either divide the factors into two groups, one >of which (larger) sums to (sqrt-1).sqrt and the other >(smaller) to (sqrt-1), or we cannot. >If not, then one larger factor either overlaps or >underlaps the divide between the (sqrt-1).sqrt area >and the (sqrt-1) area. In either case there is a >contradiction, since the larger-factor sum is either >too large to be sqrt times the smaller-factor sum >(overlapping), or too small to be sqrt times the >smaller-factor sum (underlapping). >So we must have a set of larger factors which sum to >an exact multiple of sqrt. But dividing the sum of >larger factors through by sqrt should then give a >whole number, sqrt-1. We note that with cases like sq = >abb.abb we get larger factors like bbb and bbbb and >with cases like sq = abc.abc larger factors like aabb, >aacc, bbcc which cannot sum to an integer total when >divided by sqrt. By induction on numbers being >multiplied up into products with more factowe note >that as xy can only multiply up to xyy or to xyz with >the addition of one factor then we will always have >incommensurable sums when dividing by sqrt. >Therefore the larger factors which are not already whole >multiples of sqrt will not sum to a whole multiple of >sqrt+1. Although sq-1 divides without remainder by >sqrt+1, we cannot divide the sum of factors into two >blocks measuring sqrt-1 and [sqrt-1].sqrt in size. >Therefore no odd squares have factors summing to >exactly sq-1. >(3) This leaves x>1 where x divides into sq. >Dividing sq-1 by sqrt+1 gives sqrt-1, so we know that, >since sqrt is odd, that there is an even multiple of >sqrt+1 in sq, with a final remainder of 1. >To test whether the factors of sq could sum to sq-x >where x divides into sq, we divide sqrt+1 both into >sq-x and into x and examine the remainder1 and r2. >Let [sq-x]/[sqrt+1] = p + r1, and x/[sqrt+1] = q + r2. I suppose you mean sq-x = p(sqrt+1)+r1, x=q(sqrt+1)+r2. (from your variant of equations r1 and r2 are not integers) >Since there is still the final remainder of 1 to >complete sq, if overßow x can divide into sq and >complete sq, then r1 + r2 must equal sqrt + 1 + 1. >As sqrt+1 cannot divide into sq, so not into sq-x or x >either, we know both r1 and r2 are non-zero. >But since sq-x is even, x is odd, and p + q odd [the >total number of sqrt+1 in sq-1 is even = sqrt-1 = p + >q + 1] we either have an odd p leaving an odd >remainder in sq-x and an even q leaving an odd >remainder in x, or an even p leaving an even remainder >in sq-x and an odd q leaving an even remainder in x. >Either way, r1 + r2 is even, but sqrt + 2 is odd. Here is my main question. sq-x = p(sqrt+1)+r1, x=q(sqrt+1)+r2 (previous comment). sq-x -- even, sqrt+1 -- even. Then r1 is even with any p( from 1st equation). x -- odd, sqrt+1 - even. Then r2 is odd with any q. >Therefore the factors of an odd square sq cannot sum >to sq-x if x>1 and divides into sq. >Since this obstacle holds equally for higher powers of pr, >we see there are no odd perfect numbers. >Mark Griffith >14th Jan 04 >. > === Subject: Re: by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0UF7tm14076; . Good to see everyone has plenty of time on their hands over here... Mark >> Infinity is not a property, since it's not a set. >> Sets are moronic, since Cantor didn't make sets, >> morons made sets. >> All of these statements are trivially proven, since some is a property >> that only people with higher IQ's than physicists have. >> You're such a lame poser. This is simply not up to James Hunter's >> standards. >> Sets are moronic, since Cantor didn't make sets, morons made sets. >> Makes no sense. No bloody sense at all. Mind you, most things that >> aesthetically pleasing manner. > Cantor has *one* proof to his credit, concerning sets. > Dedekind has *no* proofs to his credit concerning sets. > Goedel has *two* proofs to his credit concerning sets. > Russel not only doesn't have any proofs to credit > concerning sets, his paradoxes aren't even interesting. > Give it up. > I like Einstein gave it up before it was inven. === Subject: Re: goldbach's and fermat's proofs >You may view my proofs at the Florida State web page: >www.math.fsu/Science/Specialized under Goldbach Proof and In >Defense of Mr. Fermat. Please mail me any questions. Kerry >Evans. I have tried to read both of your papers twice. But I could not > follow your notations. Could you rewrite your paper in MS Word, using > Object; Math Equation ? (Hopefully converting to PDF) I would be glad to read your paper. I have been a little depressed > with my Differential Equation class whose textbook is written for > machines that need to be programmed. erdos fan You might like to know that I am rewriting some segments of my > Goldbach proof. Meanwhile there is a pdf listing on > www.mathprints.com. Kerry > You repealy say things like Ôlet n and m be as defined' which is > really confusing. Either explain what holds in every instance, or make > a definition to define things. In the golbach one for Lemma I-b you seem to claim that if n-m=1, and > n+m is an odd prime then [(n+m-1)!]^4=(n+m+1)(mod 2*(n+m)) > Let n=2, m=3 > [(2+3-1)!]^4 (mod 2(2+3))=4^4 (mod 10)=4 (mod 10) > (2+3+1) (mod 2(2+3))=6 (mod 10) > Unfortunately 6 does not equal 4 in mod 10. > For the Fermat one, the definition of F(x,y), and the definition of {} > are really confusing. It could also would probably help make things > clearer if you explain upfront why they are defined that way. I also don't understand the following paragraph >Suppose now that (1) has been rewritten for some n, where n is equal > to or >greater than n' so that [r,a,b and n] is transformed to a reduced > form, >redefining [r,a, b and n] to be [u,v,w and n']. Thus n' is greater > than or >equal to 2 implies [v>u>w>0]. Consider the transformation (under > n'), T, >such that T:v to v, u to w, and w to u. Respecting (1). T(v,u,w) = > Tnot >(v,u,w) where Tnot signifies not T . It follows that [u>w] can be >transformed to [w>u] such that [u>w] is unaltered.! This can happen > only for >the case,u=w, which is moot. Conclusively, order must be assignable > when it >exists. ( ! denotes contradiction.). > I don't understand what exactly you mean by t, and I don't understand > what you mean by tnot. It almost looks like you are trying to say that > u^n+v^n=w^n implies u^n+w^n=v^n under some conditions, but I don't see > that justified. You also go on about ordering, although I'm not sure > what you mean. Obviously u^n+v^n=w^n is equivalent to v^n+u^n=w^n, > although u>v implies u^n>v^n for real n>0. Gershon Bialer Please. I will address your questions and email my answers. I should get caught up by mid-February. Thanks === Subject: Re: Transformation Geometry Problem > One Weird Dude > Define a collineation (sp?) to be a transformation that always maps > lines onto lines, and also, if necessary, define a transformation to > be a one-to-one mapping of the plane onto the plane. Now, exercise > 1.15 in my UTM Transformation Geometry text states Prove or Disprove: > A transformation that preserves betweeness is necessarily a > collineation. I have only the definitions, the Exterior Angle > Theorem, Pasch's Axiom, and general precalculus with which to work, > and maybe a few others on the same general level. Any HINTS would be > apprecia, please. > A transformation will be a collineation if it sends any three collinear > points to collinear points. And what can we say about any three collinear > points? > LH Thank you, but my skepticism is not fully quenched. A transformation that maps a half-open interval on to two half open intervals is *not* a collineation. This is the dilemma for me. Hopefully the terminology is not an obstacle here. -- Said Area Purposefully Kept Empty. [30 Jan] === Subject: SMSU Problem Corner The new problems (High School, Advanced, and Challenge) have been pos. Please visit us at http://math.smsu.edu/~les/POTW.html Thanks. === Subject: More graph-theory proof problems Learning from some of the replies to my previous posts, i tried to attack another graph-theory problem. But to no avail. The problem, as usual is i can not figure out a good strategy. Here is the problem: a) Suppose G = (X Union Y, E) is a bipartite graph with |X| = |Y| = n. Show that if k is the minimum degree of G then: |A| -|J(A)| <= n-k for all subsets A of X *My comments: earlier in this chapter J(A) was defined as {y in Y | xy in E for some x in A} where A is a subset of X. J(A) was also defined in one example like this: If a set of people A are collectively qualified for the set of jobs J(A) and |A| > J(A) then at least |A| -|J(A)| ppl will be dissapoin* b) Deduce that if |E| > (m-1)*n then G has a matching with at least m edges. approach for a) My approach was to use induction on n. for n = 1 we would get. One vertex in X, one in Y, the minimum degree would have to be 0 . so |A| -|J(A)| <= 1-0 = 1. Using the job analogy, there is only one person competing for at least one job, so only one person can get dissapoin. Now we assume true for n: |A| -|J(A)| <= n-k and show it is true for n +1: |A| -|J(A)| <= n+1-k. And that would complete the proof ? :). Yes i know i am probably out in the blue here, but at least i tried. Help me out please approach for b) NO approach, have absolutely no idea how to prove that one. Comments on my ramblings and solution for these two problems are greatly apprecia :) === Subject: Re: categories and set theory >.. It is no harder (and to my mind >a bit easier) to define categories without having sets around than to >define sets without having sets around. There ought to be references somewhere where this is worked out in formal metamathematics and explicitly in terms of say a first order theory with equality. I am working on a theorem prover, and it might be useful as an input, as it is hard to get a computer to understand handwaving. Hans Aberg === Subject: Re: When was zero inven? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0UIL0m30909; > One is a number for all practical purposes. > Because with one, comes minus one. > Zero is not even a number. Zero is not even a Ônumber'?!?!?! ... so when we include zero in the set of integers we're being idiots? Oh I see.... > It's a set of NOTHING, idiots. I enjoy you're sense of wonder and the unimaginable idiocy you just pos to this wonderful forum. ----------------------------------- I think we're looking at the subject in a wrong way. I don't doubt the original poster that a merchant would have no concept of zero (Ônothing'), since he probably faced bankruptcy at one point or another. In fact, at some points he may have been in Ôdebt' as well (Ônegative')! I think the fact is, is that most people were illiterate, and so hadn't the same formalization of algebra and mathematics as we do (this was reßec in the way mathematicians did math also). The first Ôrecord' of the use of zero was a triumph because someone decided to break out of the norm, and physically attribute a symbol for the concept of zero. I think we're chasing our own tails: the concept of zero (not written) must have been around for a LONG time. However, the first account of it being used algebraicaly is relatively modern, maybe something we take for gran. MM. === Subject: Can the TI-89 do 3D vector field plots? Is there software for this? Can the TI-89 do 3D vector field plots? Is there software for this? What is the difference between Derive, Maple, Mathematica, MathCad, and other mathematical software? Casey === Subject: Re: A question about a prime divisor of a number. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0UJ8fe02822; This is an interesting question. It is NOT true for q = 3. (67^3 - 1)/(67-1) = 3 . 7^2 . 31 (79^3 - 1)/(79-1) = 3 . 7^2 . 43 Let r be a divisor. We have N = (p^q-1)/(q-1). (q | p - 1) Further, r must be a quadratic residue, otherwise r | (p^q + 1)/(p+1). In order for all of the divisors of N to be less than p we will need a sequence of q.r. primes r_i such that r_i = 1 mod q. The GRH places a limit on the number of sequential quadratic residues a number can have. I suspect, but do not know that a similar result would apply for primes in an AP. It is likely that we can not get enough small primes all equal to 1 mod q and less than p such that their product exceeds p^(q-1) except perhaps for a very thin set of p's. Add in the requirement that r | N and it becomes extremely improbable to get enough primes of the right form unless p is large and q is *very* small relative to p. Theorems about the least prime in an AP will also probably play a role here. They would place a bound on the smallest such r. For q >= 5, we have N ~p^4. It is probably impossible to get enough primes equal to 1 mod 5 smaller than p and a q.r. such that their product is big enough. Only 1 prime in 4 is 1 mod 5 and they all must divide N. I suspect that a full proof would require some deep results in analytic no. theory, and probably requires GRH as well. >********************************************** >p, q : prime numbers >p > q > 3 >Prove that 1 + p + p^2 + ... + p^(q-1) has a prime divisor which is >greater than p. >********************************************** >Is it possible to prove this question? Or are there papers or theorems >or properties whatever that is rela or helpful in this question? === Subject: Re: A question about a prime divisor of a number. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0UKSmp10554; I already pos on this subject. Some further computation has shown that 903125^5 - 1 has all of its prime factors less than 903125. True, p is not prime, but it gives some hope that a prime p might be found. >********************************************** >p, q : prime numbers >p > q > 3 >Prove that 1 + p + p^2 + ... + p^(q-1) has a prime divisor which is >greater than p. >********************************************** >Is it possible to prove this question? Or are there papers or theorems >or properties whatever that is rela or helpful in this question? === Subject: simple equation solving, with a twist? This is a problem which origina in chemistry, but being that math lover that I am I quickly tried to find a way to turn it into a pure math problem. I succeeded, but now I don't know how to solve it (other than trial and error or similar... you'll understand what I mean), so any help would be apprecia. I am particularly interes in whether this can be programmed into a calculator--I have a TI-83+--so that would also be of interest. Say we have some equations with a number of variables. For example: 5x - 3y = 212 3y - z = 7 Here's the twist: we don't want to solve the set of equations (infact, in this case they are not solvable). Instead, we want to find what, say, 5x - z is equal to. Now, in this case it is easy, we can simply add the bottom to the top, and we get 5x - z = 219. We can do similar things for other numbers etc. In general, you are allowed to multiply any equation by a constant, add any two equations, and subtract any two equations. But, with, say, 5 equations and 7 variables, it can get tricky. Each equation doesn't have every variable in it; most of them have 3 or 4, which may make it easier. But, any hints on how to solve this in a general case (even though this isn't necessarily solvable for any expression)? One idea I had for programming my calculator for doing it was using a matrix, e.g. we could represent the above example by 5 -3 0 212 0 3 -1 7 We can add another row if we want, for the expression we want to get... 5 0 -1 ?? But I don't know what to do with it after that. Any hints etc. would be greatly apprecia. Jonathan Christensen --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Extremely easy algebra question, but... ...I can't figure it out! My question is if we have,say, the following general expression: a = 2n^3 + p(n) where p is some function, and we know that p(n) < 2n^4, does this imply that a < 2n^4 + 2n^3 or in fact that a < 2n^4 - 2n^3?? What if the plus sign were a multiplication sign? I'm basically having one of those doing an advanced problem but getting stuck on the trivial basics kind of day. It's been quite a few (too many) years since I did inequalities! Thanks in anticipation! === Subject: Re: Extremely easy algebra question, but... Adjunct Assistant Professor at the University of Montana. I can't figure it out! >My question is if we have,say, the following general expression: >a = 2n^3 + p(n) >where p is some function, and we know that p(n) < 2n^4, does this imply that >a < 2n^4 + 2n^3 or in fact that a < 2n^4 - 2n^3?? Why would you have the latter? and 2n^3+p(n) is equal to a, so a = 2n^3 + p(n) < 2n^3 + 2n^4. >What if the plus sign were a multiplication sign? i.e., if a= 2n^3*p(n)? Then it is going to depend on whether n is positive or negative. When you take an inequality and multiply by a positive number, the inequality sign doesn't change. But if you multiply by a negative number, the inequality sign has to be ßipped . If n>0, then 2n^3>0, so from p(n) < 2n^4, mutliplying both sides by 2n^3, you get 2n^3*p(n) < (2n^3)*(2n^4) = 8n^7. And so a = 2n^3*p(n) < 8n^7. But if n<0, then 2n^3 < 0, so when you multiply p(n) < 2n^4 by 2n^3, you must exchange the < for a > . You get (2n^3)*p(n) > (2n^3)*(2n^4) = 8n^7. So you have a = 2n^3*p(n) > 8n^7. -- ============ === Subject: Re: minesweeper Another consideration is how many mines remain undetec. In the extreme case, if one mine remains to be found, that alone may become important in choosing the next move. In fact such information may render safe an ordinarily risky move. As you know, Minesweeper gives this count. > It seems to me that the fact that minesweeper is NP-Complete > focuses on the wrong question. Instead, let me propose some > alternative questions that are (perhaps) more relevant to winning > the game. It's not so much that we need to know if a configuation > is Ôlegal', but rather how to compute the probability that there is a > mine in a given cell. > (1) Suppose one has a configuration where, with respect to an > established safe cell, one of two neighbors (X and Y) must contain > a mine. Assuming that the distribution of mines is Poisson, then the > probability is 1/2 for each of the neighbors containing a mine. > However, this analysis is done in isolation from other nearby cells. > Suppose in addition there are nearby cells for which X and Y are > two of say 6 cells of and that these 6 cells only have two mines. > X or Y contains a mine. > This shows that in order to establish the probability that a cell > contains a mine one must compute the JOINT pdf from known > cells. I have tried without success to do this. Here's an example. > M = known mine, x,y,z = unknown cells. > w 1 1 > y 2 1 > y 3 M > z z M > x 4 M > x x 2 > the point of view of the cell marked Ô3', the probability should be 1/2 > whether the mine is in one of the cells marked y and 1/2 whether it is in one > of the cells marked z. w, y, > and y, thus the probability that there is a mine in either y is 2/3 and hence > there should only be a 1/3 probability of there being a mine in either z. > (since there must be one in y or z) > Now consider the cell marked Ô4'. There must be two mines among the 5 adjacent > any given space is 2/5. Thus, the probability that there is no mine in either > z space is 9/25, and thus the probability that there is a mine in one of the > z spaces is 16/25. This is substantially higher than the 1/2 we get from > cell Ô3'. > So how do we compute an exact probability that there is a mine in one of > the two z spaces. Clearly we must consider the joint probabilities, but > I don't know how to compute the joint pdf. > IDEAS? > I have tried to do this without success. > (2) Suppose one reaches a configuation where a guess MUST be made. Can an > optimal strategy be determined? One could of course choose the guess with > highest probability of success, but this may not be optimal. It could be the > case that a correct guess of a lower probability configuration reveals more > information about how to then proceed. > Also, there might be several guess of equal highest probability. All of > them might not be equal in the sense of information revealed from a successful > guess. === Subject: Re: minesweeper Speaking of minesweeper, I heard a variant for two-players: The objective is to find the mines by clicking on them. The players take turns and the mechanics work as usual (except instead of ending the game you get a point for every mine found). The player who finds more mines wins. Do the rules favor aggressive searching around places where mines are known to exist? -- I'm not interes in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: What is the difference between Derive, Maple, Mathematica, MathCad,and other mathematical software? What is the difference between Derive, Maple, Mathematica, MathCad, and other mathematical software? Casey === Subject: Norm of an operator basics If A is a bounded linear operator, then if we define ||A|| = sup ||Ax|| (over ||x|| <= 1), why does sup ||Ax|| (over ||x|| <= 1) = sup ||Ax|| (over ||x|| = 1) That is, I am assuming this equivalently means that sup ||Ax|| (over ||x|| < 1) < sup ||Ay|| (over ||y|| = 1) Why is this true? Moshe === Subject: Re: Norm of an operator basics >If A is a bounded linear operator, then if we define >||A|| = sup ||Ax|| (over ||x|| <= 1), why does >sup ||Ax|| (over ||x|| <= 1) >= sup ||Ax|| (over ||x|| = 1) Hint: sup {r: 0 <= r < 1} = 1. BTW, your statement would be false in the trivial case where the domain is {0}. >That is, I am assuming this equivalently means that >sup ||Ax|| (over ||x|| < 1) >< sup ||Ay|| (over ||y|| = 1) No, it doesn't. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Norm of an operator basics >If A is a bounded linear operator, then if we define >||A|| = sup ||Ax|| (over ||x|| <= 1), why does >sup ||Ax|| (over ||x|| <= 1) >= sup ||Ax|| (over ||x|| = 1) > Hint: sup {r: 0 <= r < 1} = 1. Yes, I know this. So, sup ||x|| (over ||x||<=1) is equal to 1. I'm trying to play around with this but I must be going the wrong way (although I think I have a proof) ||Ax|| = || A ||x|| x/||x|| || = || Au||x|| || (where u is of norm 1) = ||x|| ||Au|| So, sup ||Ax|| (over ||x|| <= 1) = sup ||Au|| (where u is of norm 1) since sup ||x|| where ||x||<=1 is equal to 1. Did I get this right? === Subject: Please advise! what kind of training I am lackfor math? Dear all, I have long been headache about this problem... please help me! let me try to explain my problem to you: For years I have been headache about reading math notations. A concept, if it is put in straightforward way, I can understand. But if it is put in a math notation, or even advanced math notation, I will feel dizzy when I read it. I cannot avoid feeling dizzy if I meet math formulars having more than 3 lines. But I am in graduate school and must accquaint myself with maths. For some reason, in my research field, information theory and signal processing, if there is no math, the research may be regarded as low quality; hence high quality journals are full of maths. I can never fully understand a paper full of maths. I guess my problem is my lack of exposure to math and lack of systematical education in math. I thought I will be a programmer and all my past time was devo to programming. But later I found programming is kind boring so I end up need to use math. In fact I am quite good in terms of grades in math classes; but I have only studied very few math courses: calculas, linear algebra, complexity analysis, probability, all in undergraduate and introductory graduate level. So I am at a very akward stage now: if give me undergraduate math to read, I feel too easy and boring, if give me difficult math to read as those in Information Theory Transactions, I feel dizzy... Can anybody recommend some procedures/reference books to give me a treatment for my current sickness? Can anybody give a booklist that any big guy in the field would recommend as a must-read as a systematic treatment to math for researchers in math-rela engineering/science area? Thank you very much, -Walala === Subject: Re: Please advise! what kind of training I am lackfor math? > Dear all, I have long been headache about this problem... please help me! let me try > to explain my problem to you: For years I have been headache about reading math notations. A concept, if > it is put in straightforward way, I can understand. But if it is put in a > math notation, or even advanced math notation, I will feel dizzy when I read > it. I cannot avoid feeling dizzy if I meet math formulars having more than 3 > lines. The way to read math is slowly, one line at a time. When presen with three lines, just read the first one. Understand it. If need be, take pencil and paper and work out to your own satisfaction what they are saying. Then go on to the next line. The trick is to learn to read math very slowly. But I am in graduate school and must accquaint myself with maths. For > some reason, in my research field, information theory and signal processing, > if there is no math, the research may be regarded as low quality; hence high > quality journals are full of maths. I can never fully understand a paper > full of maths. > Signal processing? Yup. All math. === Subject: Re: Please advise! what kind of training I am lackfor math? >For years I have been headache about reading math notations. A concept, if >it is put in straightforward way, I can understand. But if it is put in a >math notation, or even advanced math notation, I will feel dizzy when I read >it. I cannot avoid feeling dizzy if I meet math formulars having more than 3 >lines. But I am in graduate school and must accquaint myself with maths. For >some reason, in my research field, information theory and signal processing, >if there is no math, the research may be regarded as low quality; hence high >quality journals are full of maths. I can never fully understand a paper >full of maths. >I guess my problem is my lack of exposure to math and lack of systematical >education in math. I thought I will be a programmer and all my past time was >devo to programming. But later I found programming is kind boring so I >end up need to use math. >In fact I am quite good in terms of grades in math classes; but I have only >studied very few math courses: calculas, linear algebra, complexity >analysis, probability, all in undergraduate and introductory graduate level. >So I am at a very akward stage now: if give me undergraduate math to read, I >feel too easy and boring, if give me difficult math to read as those in >Information Theory Transactions, I feel dizzy... >Can anybody recommend some procedures/reference books to give me a treatment >for my current sickness? Can anybody give a booklist that any big guy in the >field would recommend as a must-read as a systematic treatment to math for >researchers in math-rela engineering/science area? The notation of mathematics is merely its *language*. As with a natural language, one learns it best through active use. That is, you will get used to notation by writing it down yourself. Start by using more notation in your exercises, or in the writing up of other proofs that you know (e.g., from textbooks). In fact, when you read through a paper the second time (the first time should be just an attempt to get the basic idea), do so with pen and paper at hand. Work through proofs to learn to understand the paper. And try to use notation in your own writing more and more. Eventually, you will become more comfortable with notation. However, I suspect that you (as most) will need to use this approach throughout your career. By the way, a well-written paper should also explain the gist of the argument without the obscurity of mathematical notation; it seems to me that authors are often too lazy to take this step, or they are incapable of doing so. And academic culture tends tolerate this trend. In particular, when I make a presentation (which is usually in the field of operations research), I often consciously try to *avoid* the use of notation in my slides: the important part is concepts and results I am trying to convey. Proofs can be outlined in the presentation; parties interes in details can read the paper. I have found that graduate students in particular appreciate this approach, but I suspect the appeal is more general than that. Always, the important contribution of a work is its content, not its notation. -- === Subject: Re: Please advise! what kind of training I am lackfor math? > Can anybody recommend some procedures/reference books to give me a treatment > for my current sickness? Can anybody give a booklist that any big guy in the > field would recommend as a must-read as a systematic treatment to math for > researchers in math-rela engineering/science area? One fact is that everyone learns and understands in a different way. Some people understand things by reading. Some people understand things when they listen, or when they do them. I personally understand things as soon as I write them down. It may be as simple as that: Find out what is your method of learning things and use that method. === Subject: Re: Please advise! what kind of training I am lackfor math? > Dear all, I have long been headache about this problem... please help me! let me try > to explain my problem to you: For years I have been headache about reading math notations. A concept, if > it is put in straightforward way, I can understand. But if it is put in a > math notation, or even advanced math notation, I will feel dizzy when I read > it. I cannot avoid feeling dizzy if I meet math formulars having more than 3 > lines. ... > Can anybody recommend some procedures/reference books to give me a treatment > for my current sickness? Can anybody give a booklist that any big guy in the > field would recommend as a must-read as a systematic treatment to math for > researchers in math-rela engineering/science area? Thank you very much, -Walala OK, Buddy: we're much alike. My problem with math is not doing it -- I find that straightforward -- but reading it. I tend to skim, and that's fatal. When I need details and not just gist (which is most of the time, unfortunately), I decipher it instead of reading it. I use the symbols as a guide to the pathway that the writer tries to lead me down, and I walk it, every step. Most of the time, that seems to work. Jerry -- Engineering is the art of making what you want from things you can get. ¿¿¿¿¿ ¿¿¿ [OS lash]¿¿¿¿[DownQuest ion]¿¿¿ ¿¿¿¿¿ ¿¿¿ [OSl ash]¿¿¿¿[DownQuesti on]¿¿¿ ¿¿¿¿¿ ¿¿¿ [OSl ash]¿¿¿¿[DownQuesti on]¿¿¿ ¿¿¿¿¿ ¿¿¿ [OSl ash]¿¿¿¿[DownQuesti on]¿¿¿ ¿¿¿ === Subject: Smolin's & Rovelli's Three Roads to Quantum Gravity Lecture 5: RovelliÍs History of Quantum Gravity to 1999 pp. 287 [CapitalEth] 301 Preamble to SmolinÍs and RovelliÍs ñThree Royal Roads to Quantum Gravity.î What does it mean to quantize a theory? This is a top -> down idea. You start with a classical physics theory. What are the key classical theories? infinite speed of light, EinsteinÍs 1905 special relativity with a finite speed of light barrier but without gravity and EinsteinÍs geometrodynamic 1915 general relativity with gravity. (ii) MaxwellÍs electromagnetic field theory that cannot be consistently formula in Galilean relativity because electromagnetic waves propagate at the finite speed of light that is the same number for all observers independent of their speed or motion or their acceleration. MaxwellÍs theory is automatically special relativistic and can be extended to general relativity with gravity as seen in the cosmological red shifts of our expanding universe accelera by repulsive exotic vacua ñdark energyî, and in the gravitational lensing showing clumps of attractive exotic vacuum ñdark matter.î (iii) Yang-Mills field theories of the weak and strong short-range forces. (iv) EinsteinÍs geometrodynamic field theory of gravity as the curving of a dynamical space-time that is not merely a rigid stage, like in special relativity where the full action-reaction principle is viola and on which all the other fields play, but is, rather, itself a player in ñtwo-way relationî (Bohm and Hiley) obeying the full action-reaction principle. field ñobservablesî by square arrays of complex numbers called ñmatricesî that are ñrepresentationsî or faithful images of certain mathematical groups of transformations of different kinds of ñframes of referenceî that represent configurations of macroscopic detectors making ñmeasurementsî of these observables. The measured numbers are real ñeigenvaluesî of the square matrix arrays corresponding to columns (or rows) of complex numbers called ñeigenstatesî. These matrices are ñquantum computersî or ñgatesî. The set of eigenstates are strings of a new kind of non-classical information called ñqubitsî and they form a ñbasisî in which the square matrices are ñdiagonalî i.e. the real eigenvalues are on the diagonal positions and all the off-diagonal positions in the matrix array are zero. A key property is that these eigenstates can be coherently superposed posed to form a set of eigenstates that are a basis for a different set of matrices that represent a different incompatible configuration of detectors. The ñbasisî spans a ñHilbert spaceî or ñhouseî or ñcontainerî (the ñBaytî of Qabala), as it were, where all the quantum strings ñlive.î This is where the Heisenberg uncertainty principle comes from -- that not all observable properties of a system can be measured simultaneously to arbitrarily high precision, i.e vanishingly small errors. There is also another important counter-intuitive property called ñentanglementî (technically a ñtensor product of Hilbert spacesî [CapitalEth] one space for each part of the entangled whole) in which several quantum systems share a common pool of quantum information and do not have qubit strings of their own. Quantum wholes are greater than the simple sum of their parts. The quantum information comes in two forms called ñactiveî and ñinactive.î This is explained in detail in David BohmÍs and Basil HileyÍs ñThe Undivided Universe.î Entanglement is important in quantum computing applications like ñteleportation of qubitsî and ñuntappable cryptography.î It also is the key idea in ñenvironmental decoherenceî, which is an attempt, only partly successful, that tries to explain the irreversible thermodynamic ßow or ñarrow of timeî and the ñcollapse of the quantum stateî or ñVon Neumann projectionî, i.e. why the large-scale world of ordinary experience seems ñdefiniteî without us being alive and dead at the same time in the same place as a na .95ve extrapolation of the quantum properties of tiny simple objects suggests. Note that while our inner perceptions of the outer world seem definite without the same object being in two places at the same time, our pure inner conscious thought has quantum superposition in which we hold two, or more, incompatible ideas in our mindÍs eye simultaneously as in HamletÍs speech ñTo be, or not to be. That is The Question .83î On the other hand John Archibald Wheeler says ñThe Question is: What is The Question.î ñQuantum logicî is the study of ñquantum binary questionsî arranged in a non-Boolean partially-ordered lattice. This is very different from our computers whose logic is that of a Boolean lattice. The non-Boolean lattice, where each question is a node on a graph, has ñIsles of Boolean latticesî corresponding to compatible questions that can be answered definitely simultaneously with the same configuration of detectors. There is an approach to quantum gravity called ñconsistent historiesî which consists of a ñstory lineî of questions (Lee SmolinÍs ñThe Three Roads to Quantum Gravityî) the problem is that knowing which questions to ask to tell the story is an insolvable problem, or perhaps I should say, is an undecidable question in the sense of GodelÍs ñincompleteness theoremî of 1931 and itÍs corollaries like the ñHalting Problemî of computer theory asking the question ñWhen exactly will the program stop?î So this is what Wheeler is alluding to in his cryptographic remark: ñThe Question is: What is The Question?î Now first thing we need to understand in looking at RovelliÍs version of Lee SmolinÍs ñThree Roads to Quantum Gravityî is that all three roads are top -> down. But there is a Gurdjieffian ñFourth (bottom -> up) Wayî of ñemergenceî, ignored completely by Smolin and Rovelli and All The KingÍs Men trying to put Humpty Dumpty together again, due to the great Soviet physicist Andrei Sakharov and also the Princeton physicist P.W. Anderson. Sakharov called it ñmetric elasticityî. Anderson called it ñMore is differentî with an information-rich (low thermodynamic entropy) giant quantum coherence field that is local without entanglement with ñgeneralized phase rigidityî making it immune to ñenvironmental decoherenceî that in a special case is SakharovÍs metric elasticity and also the ñtensionî of ñstring theoryî. The fact that the giant quantum coherence field is local automatically explains why the outer world is definite without the same object being in two places at the same time and why we are not alive and dead at the same time like ñPhysics is simple when it is local.î P.W. AndersonÍs ñMore is differentî explains why large [CapitalEth]scale physics is local without needing OxfordÍs David DeutschÍs ñexcess metaphysical baggageî (WheelerÍs term) of the quantum ñMultiverseî of ñshadow objectsî. This is not to exclude ñparallel universesî, but one must be vary careful on how ñuniverse is defined. It all depends on what you mean by is . It's the ontology stupid! J === Subject: JSH: Short argument, modified Decker example In an attempt at questioning an important conclusion of mine a Rick Decker, a professor at Hamilton College, made a post with his own example of a non-polynomial factorization: (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). Making the substitution a_2(x) = b_2(x) - 1 to get (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) allows both a_1(0) = 0 and b_2(0) = 0, so let that be a requirement. Therefore, both a_1(x) and b_2(x) have some factor of x itself, which implies a_1(x) has a factor of 7, so using a_1(x) = 7b_1(x) you have (5(7) b_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) and dividing both sides by 7 gives (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2. At this point it's all straightforward except that you can actually check at x=1, you find that doesn't work in the ring of algebraic integers. === Subject: Re: JSH: Short argument, modified Decker example > In an attempt at questioning an important conclusion of mine a Rick > Decker, a professor at Hamilton College, made a post with his own > example of a non-polynomial factorization: (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). Making the substitution a_2(x) = b_2(x) - 1 to get (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) allows both a_1(0) = 0 and b_2(0) = 0, so let that be a requirement. Therefore, both a_1(x) and b_2(x) have some factor of x itself, which > implies a_1(x) has a factor of 7, so using a_1(x) = 7b_1(x) you have The part up to so makes no sense. Certainly there is no reason given or implied why a_1(x) should be divisible by 7. > (5(7) b_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) and dividing both sides by 7 gives (5b_1(x) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2. At this point it's all straightforward except that you can actually > check at x=1, you find that doesn't work in the ring of algebraic > integers. > Correct. In fact, there are only two values of x for which a_1(x) is divisible by 7 in the algebraic integers. In all other cases you find that 7 divides the LHS but that 7 doesn't divide either one of the two factors on the left. Rick === Subject: Re: JSH: Short argument, modified Decker example Here comes a proper example of a short argument: Me: Harris, you are an idiot. Harris: No, I'm not. Everyone else: Yes, you are. Argument over. === Subject: Derived group I can't sleep because of one problem: We have nonabelian group G of order 24, and we know that it has a normal subgroup H of order 8. Problem: Find derived subgroup G' (aka [G:G] - commutator subgroup) Any ideas? Is it possible to definitely answer this question at all? (Of course without checking by hand all 12 nonabelian groups of order 24 :) ) Surely it must be subgroup of H, (|G/H|=3, so G/H is abelian) but I think it's really smaller, probably cyclic group of order 2. Are we able to say something about structure of such group G ? Thanks for any help and comments. -- Best Rafal Kucharski. === Subject: GCD-Product = LCM-Product First, let r be shorthand for ßoor(min(m/(k_1), m/(k_2), m/(k_3),..., m/(k_n))), the integer-part of the lowest of the m/(k_j)'s, for 1<= j <= n. ( k_j is k-sub-j, of course.) Then, for m and n = positive integers: m m m --- --- --- | | | | | | | | | | ... | | GCD(k1,k2,k3,...,kn) k1=1 k2=1 kn=1 = m m m --- --- --- | | | | | | | | | | ... | | LCM(1, 2, 3,..., r) , k1=1 k2=1 kn=1 which is in linear-mode: product{k_1 =1 to m} product{k_2 =1 to m}.... product{k_n =1 to m} GCD(k_1,k_2,k_3,...,k_n) = product{k_1 =1 to m} product{k_2 =1 to m}.... product{k_n =1 to m} LCM(1, 2, 3,..., r) (An intriguing result, if only for its aesthetic appeal.) Additionally, at: http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off& threadm=b4be2fdf .0401271302.6a3b77ed%40posting.google.com&rnum=7&prev= I give the prime-factorization of the GCD-multiple-product, and also the limit of the geometric-average of this product's terms (as m approaches infinity, and for fixed n). So, these results obviously also apply to the LCM-multiple-product. ;) (obviously) (By the way...the n = 1 case is the topic of my sci.math post at: http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off& threadm=b4be2fdf .0205251719.1dc3b82e%40posting.google.com&rnum=3&prev=) === Subject: JSH: Factor of x I'll admit that I'm still hoping on that math journal or some contacts that I have out there who are supposed to get back to me soon, but I'm bored, so I'll see what happens here. Recently Rick Decker, a professor at Hamilton College, apparently trying to refute my research came up with a quadratic example, which I like because it's a quadratic, and easier to manipulate than the cubics I've used before. If you wish to see his original post here are some headers which also show that he posts from Hamilton College: Subject: Re: Mathematical consistency, courage Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of non-polynomial factors. Notice that despite not being polynomials they are algebraic integers if x is an algebraic integer because a_1(x) and a_2(x) are the two roots of a^2 - (x - 1)a + 7(x^2 + x). However, there's something odd here as if you let a=0, you have one of the roots equals 0, but the other equals -1, so it makes sense to use b_2(x), where a_2(x) = b_2(x) - 1 which gives (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) where a_1(0) = b_2(0) = 0. But that implies that a_1(x) and b_2(x) *both* have some factor in common with x with algebraic integer x. Now given that a_1(x) a_2(x) = 7(x^2 + x) = 7x(x+1) that's especially odd, as what about the one that does versus the one that doesn't? Or doesn't it? Do *both* somehow have factors in common with x? It might seem too esoteric with x there, so let's put in a value for x, and let x=13. Then a^2 - 12a + 7(13)(14) and we already know that *one* of the a's, is coprime to 13, or wait, do we? Can any of you explain how it all works? Note that the a's are (12 + sqrt(-4952))/2, and I didn't put indices on them as how do you know? === Subject: Re: Factor of x > I'll admit that I'm still hoping on that math journal or some contacts > that I have out there who are supposed to get back to me soon, but I'm > bored, so I'll see what happens here. > Recently Rick Decker, a professor at Hamilton College, apparently > trying to refute my research came up with a quadratic example, which I > like because it's a quadratic, and easier to manipulate than the > cubics I've used before. > If you wish to see his original post here are some headers which also > show that he posts from Hamilton College: > Subject: Re: Mathematical consistency, courage > Decker put forward the quadratic > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). > The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of > non-polynomial factors. > Notice that despite not being polynomials they are algebraic integers > if x is an algebraic integer because a_1(x) and a_2(x) are the two > roots of > a^2 - (x - 1)a + 7(x^2 + x). > However, there's something odd here as if you let a=0, you have one of > the roots equals 0, but the other equals -1, so it makes sense to use > b_2(x), where > a_2(x) = b_2(x) - 1 > which gives > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > where a_1(0) = b_2(0) = 0. > But that implies that a_1(x) and b_2(x) *both* have some factor in > common with x with algebraic integer x. > Now given that > a_1(x) a_2(x) = 7(x^2 + x) = 7x(x+1) > that's especially odd, as what about the one that does versus the one > that doesn't? Or doesn't it? Do *both* somehow have factors in > common with x? > It might seem too esoteric with x there, so let's put in a value for > x, and let x=13. Then > a^2 - 12a + 7(13)(14) > and we already know that *one* of the a's, is coprime to 13, or wait, > do we? > Can any of you explain how it all works? > Note that the a's are (12 + sqrt(-4952))/2, and I didn't put indices > on them as how do you know? > James, get a life and clean up your filthy mouth. This is the umpteenth time you pos it and you only need to post this one time. === Subject: Re: Filling A Grid By Stepping 1,2,3,4,... I will give the first 5 positions that *I* got below so as to get you star. Actually, if there is any way to solve this at all, there are probably an infinite number of ways. Warning: My 1st 5 positions may just be enough to make the answer obvious... > Start with an infinite (in every direction) rectangular grid/lattice. Put 1 in a square of the grid (or at a vertex of the lattice). Place 2, 3, 4, 5, ....infinity, by some algorithm, into the > grid/lattice so that each integer (1+m) is exactly m squares (in only > directions of either up,down,left,or right) from the square/vertex > with the integer m, for all m's. And only one, no more/ no fewer, integer per vertex/square. > I think I found a simple pattern for placing the integers so as to > completely fill the infinite grid with unique positive integers. It might actually be more difficult to fill the half-plane (grid has > one edge) or the infinite quarter-plane (ie. one quadrant of an > infinite cartesian graph), than to fill an infinite grid. (Do not > place any m's off the grid!) Is it possible to fill any n-by-n grids for FINITE n, aside from n = > 1?? > | | V | | V | | V | | V | | V *3 ** 12 *4***5 (Well, perhaps revealing the sixth position would have made the solution even more obvious...) === Subject: Re: Filling A Grid By Stepping 1,2,3,4,... > I have rewritten the puzzle as to make it more charming . > Assume we have an infinitely large ßat planet on which is an > infinite meadow, each blade of grass in the meadow at a unique > vertex of a uniformly spaced rectangular lattice, no lattice-points > without a blade of grass....to start. > So, there is a grasshopper who ends up somehow at one of the > blades of grass. Every time the grasshopper eats a blade of grass, > it can jump one unit (distance between each blade) farther than > before it ate the blade of grass. > As a matter of fact, it can ONLY jump this far. > And, for some reason, it can only jump east, west, north, > or south (which is also how the meadow's lattice is orien, > coincidently). > But, if the grasshopper lands at a lattice-point where it has already > eaten the blade there, it dies of starvation immediately. > But if it keeps eating, and keeps jumping farther and farther > (by exactly one more unit each time), it can live forever. > So, show that it is possible (as I guess is true...) for it to live forever. Just let it keep jumping straight north each time? I think you forgot to say that each blade of grass should eventually be eaten? This is a non-intersecting space filling walk on the graph with vertices = the lattice points in the plane with vertex x adjacent to vertex y if x is one unit north, south, east, or west of y. An obvious solution is to just spiral out from the center. In your other problem you had jumps of distance m instead of just one didn't you? So before you needed to have an algorithm to accomodate any m? This should take more thought. In fact, starting at (0,0) it seems you could only reach the vertices (mk,mj), so you couldn't eat all the grass if m>1? So the corresponding graph isn't connec! Am I misunderstanding something? One could ask this for any graph and the answer may depend on the starting vertex. As I said before it is a type of hamiltonian path problem. (which makes sense whether the graph is finite or not.) Some people are obsessed with such problems. See, for example, Frans Fasse's site http://home.wxs.nl/~faase009/counting.html where he describes how his interest star with writing computer programs to draw snakes in the plane. > .. > And, even though the quarter-plane version of this puzzle is harder, > as I believe could be, I think it might be possible for the GH to live > forever in this case also. > (Here, the planet has two edges, meeting perpendicularly as the x-axis > and the y-axis, where the GH would fall off its planet into an abyss > if it jumps past any of these edges.) > By the way, as the puzzle is rewritten, it now seems familiar to me. > Did I steal this puzzle from someplace?? > Start with an infinite (in every direction) rectangular grid/lattice. Put 1 in a square of the grid (or at a vertex of the lattice). Place 2, 3, 4, 5, ....infinity, by some algorithm, into the > grid/lattice so that each integer (1+m) is exactly m squares (in only > directions of either up,down,left,or right) from the square/vertex > with the integer m, for all m's. And only one, no more/ no fewer, integer per vertex/square. > I think I found a simple pattern for placing the integers so as to > completely fill the infinite grid with unique positive integers. It might actually be more difficult to fill the half-plane (grid has > one edge) or the infinite quarter-plane (ie. one quadrant of an > infinite cartesian graph), than to fill an infinite grid. (Do not > place any m's off the grid!) Is it possible to fill any n-by-n grids for FINITE n, aside from n = > 1?? > Ignore the finite n question at the > bottom of the original post! Since a grid has n^2 cells, but there can only be n, at most, filled > squares, there is no solution for finite n >= 2. My original question may be a Duh! , as well. > But I wonder still how many alternative solutions repliers can come up > with. > thanks again, > Leroy === Subject: Re: Filling A Grid By Stepping 1,2,3,4,... > I have rewritten the puzzle as to make it more charming . Assume we have an infinitely large ßat planet on which is an > infinite meadow, each blade of grass in the meadow at a unique > vertex of a uniformly spaced rectangular lattice, no lattice-points > without a blade of grass....to start. So, there is a grasshopper who ends up somehow at one of the > blades of grass. Every time the grasshopper eats a blade of grass, > it can jump one unit (distance between each blade) farther than > before it ate the blade of grass. > As a matter of fact, it can ONLY jump this far. > And, for some reason, it can only jump east, west, north, > or south (which is also how the meadow's lattice is orien, > coincidently). But, if the grasshopper lands at a lattice-point where it has already > eaten the blade there, it dies of starvation immediately. > But if it keeps eating, and keeps jumping farther and farther > (by exactly one more unit each time), it can live forever. So, show that it is possible (as I guess is true...) for it to live > forever. > Just let it keep jumping straight north each time? I think you forgot to say that each blade of grass should eventually be > eaten? Yes, I did... I realized I had ealier today. DUH!... :/ (When in doubt, the rules of the grasshopper game and more abstract version of my original post are the same.) This is a non-intersecting space filling walk on the graph with vertices > = the lattice points in the plane with vertex x adjacent to vertex y if x > is one unit north, south, east, or west of y. An obvious solution is to just spiral out from the center. In your other > problem you had jumps of distance m instead of just one didn't you? So > before you needed to have an algorithm to accomodate any m? This should take > more thought. In fact, starting at (0,0) it seems you could only reach the > vertices (mk,mj), so you couldn't eat all the grass if m>1? So the > corresponding graph isn't connec! > Am I misunderstanding something? > The size of the jumps increases by one unit each step. So, the grasshopper moves to an adjacent blade (after he eats the first grass-blade) on the 1st move. Then, since he is energized, he jumps 2 positions (over 1 intervening blade) on the 2nd move, then 3 positions on the 3rd move, etc... > One could ask this for any graph and the answer may depend on the starting > vertex. As I said before it is a type of hamiltonian path problem. (which > makes sense whether the graph is finite or not.) Some people are obsessed with such problems. See, for example, Frans Fasse's > site http://home.wxs.nl/~faase009/counting.html where he describes how his interest star with writing computer programs > to draw snakes in the plane. > .. And, even though the quarter-plane version of this puzzle is harder, > as I believe could be, I think it might be possible for the GH to live > forever in this case also. > (Here, the planet has two edges, meeting perpendicularly as the x-axis > and the y-axis, where the GH would fall off its planet into an abyss > if it jumps past any of these edges.) > By the way, as the puzzle is rewritten, it now seems familiar to me. > Did I steal this puzzle from someplace?? > Start with an infinite (in every direction) rectangular grid/lattice. Put 1 in a square of the grid (or at a vertex of the lattice). Place 2, 3, 4, 5, ....infinity, by some algorithm, into the > grid/lattice so that each integer (1+m) is exactly m squares (in only > directions of either up,down,left,or right) from the square/vertex > with the integer m, for all m's. And only one, no more/ no fewer, integer per vertex/square. > I think I found a simple pattern for placing the integers so as to > completely fill the infinite grid with unique positive integers. It might actually be more difficult to fill the half-plane (grid has > one edge) or the infinite quarter-plane (ie. one quadrant of an > infinite cartesian graph), than to fill an infinite grid. (Do not > place any m's off the grid!) Is it possible to fill any n-by-n grids for FINITE n, aside from n = > 1?? > Ignore the finite n question at the > bottom of the original post! Since a grid has n^2 cells, but there can only be n, at most, filled > squares, there is no solution for finite n >= 2. My original question may be a Duh! , as well. > But I wonder still how many alternative solutions repliers can come up > with. > thanks again, > Leroy > === Subject: Basic set theory question (indexed sets) This came up in a discussion with a friend, he found it in a topology textbook: cap_k { A_k: k in I } should have the meaning { x | forall k in I, x in A_k } which seems ok, but for I = O {the empty set), reads { x | forall k in O, x in A_k } obviously there is something absurd about it. We think (and I think the book says so), that every x satisfies it. I think the reason is, the opposite of it is: { x | exists k in O, x not in A_k } which is manifestly false. My friend easily concludes then, the desired set is U, the universal set, but having read (the first third of) Foundations of Set Theory , I am confused what U would mean in this context. But the *basic* question which I know no answer of is: is what we did indeed legitimate? Sta alternatively, does this kind of treatment invites any of the known antimonies (paradoxes)? Pretty ßattered you read this far. Now please reply. :D === > 1+1=9-9=56-2= 545 starbuckscardss i don't think so little eeeee Subject: perfect square === Hey, can anyone help me with a proof of why 2,3,7,8 can't be the last digits of a perfect square? I'm pretty sure that I should start of with the fact that any number can be written in the form 10a+ b where 0<=b<=9. But that means that b can't = 2,3,7, or 8. Here, I get stuck, thanks. -Greg === Subject: Re: perfect square > Hey, can anyone help me with a proof of why 2,3,7,8 can't be the last digits > of a perfect square? I'm pretty sure that I should start of with the fact > that any number can be written in the form 10a+ b where 0<=b<=9. But that > means that b can't = 2,3,7, or 8. Here, I get stuck, thanks. Calculate 0*0, 1*1, 2*2, ... 9*9 mod 10. === Subject: Lebesgue measure... Let E be a Lebesgue measurable subset of R with m(E) < INFTY; and let f(x) = m((E+x) INTERSECTTION E), where E+x = {x+y : y IN E} Show that L.M.T f(x) = 0 x -> INFTY Could you please give me any hint...? Thanx! === Subject: Re: SVD? > Dear Sylvain Thanks so much for posting this video series. Paul See the video lectures > http://ocw.mit.edu/OcwWeb/Mathematics/18- 06Linear-AlgebraFall2002/VideoLectur es/index.htm > http://ocw.mit.edu/OcwWeb/Mathematics/18- 085Mathematical-Methods-for-Engineer s-IFall2002/VideoLectures/index.htm Ditto! These are awesome. === Subject: Is this series converge ? 1+1/8+1/4+1/32+1/16+1/128+1/64+... I try to check this by detecting n-th term and applying root test. But, I don't know what is n-th term. If someone know how can I know convergence(or divergence) of this series, please post reply. thanks for reading. === Subject: Re: Regular Transversality Content-Length: 3922 Originator: rusin@vesuvius > but my question relates the following situation: >>g:A->B , f:[0,1]->B such that f is transverse to the set of critical >>values of g. >>my question is if g^{-1}(im(f)) is a smooth sub-manifold of A. > Not necessarily. Let z=x+iy and h(z,t)=(z^3,t), then h:R^3->R^3 has a > critical set z=0. Let further p be the projection p(x,y,t)=(x,t) and > consider the composition g=ph. Then g:R^3->R^2 has a critical values > set x=0 (a line). Now, let L be another line in R^2 transverse, but > non orthogonal to x=0. (It may be realized as the image of a map > f:R->R^2 transverse to line x=0.) Then it is easy to see that > g^{-1}(L) consists of 3 planes intersecting in a single point, so it > is not a manifold. I don't quite believe this. Take the diagram h p > R^3 ---> R^3 ---> R^2 and consider p^(-1) L. It's simply a plane that sits obliquely > to the t-axis (in your terminology). Off that axis, h is a 3-fold > covering map, so h: h^(-1) (p^(-1) L {t-axis}) --> L {t-axis} is a 3-fold covering map. However, it can't be a set of three punctured > planes, since there is no arrangement of three planes that meet at a > single point in R^3. Instead, if you take a right circular Z cylinder > about the t-axis, the intersection of L with Z is an ellipse; the > preimage h^(-1) Z is again a right circular cylinder surrounding the > t-axis, In fact, the same one - h^(-1) Z=Z. and the lift of that ellipse is a curve that oscillates up > and down (i.e., parallel to the axis) in 3 periods as it encircles > the cylinder once. The amplitude of these oscillations (for the map > you've described, (z,t) |--> (z^3,t), appears to be proportional to > r^(1/3), where r is the radius of the cylinder h^(-1)Z. If I now > draw radii in the plane p^(-1)L, they pull back (via h^(-1)) to curves > that look like r^(1/3) in the appropriate plane containing the t-axis. > In particular, it appears that (with the exception for the two points > in the ellipse that sit on the plane orthogonal to the t-axis, or > rather their preimages via h^(-1)), all those radii pull back to > curves that are tangent to the t-axis. should be a single surface, with a point on the t-axis where there > is no tangent plane, due to the oscillation of nearby points I've > just described. Right. I was mistaken by the fact that h^(-1)(p^(-1)L) has to be invariant with respect to the rotation about the t-axis at angle 2pi/3. Of course, it doesn't imply that h^(-1)(p^(-1)L) is an union of 3 planes, as the plane p^(-1)L is not invariant under h. In any case, this is a counterexample to the question, isn't it? As you poin out, the surface h^(-1)(p^(-1)L) cannot have a tangent plane at the point P, where it intersects the t-axis. Let me give another argument: Suppose that h^(-1)(p^(-1)L) is a smooth surface, as noticed above, it has to be invariant with respect to the rotation about the t-axis at angle 2pi/3. Therefore, the tangent plane at point P is orthogonal to t-axis. Let now l be a line in p^(-1)L non orthogonal to axe t, then h^(-1)(l) is lying in the cone K with vertex P and generatrix l. On the other hand, it is lying on the surface , which is impossible - the cone K has to intersect our surface in single point P in a small neighbourhood of it, though this intersection is infinite with any neighbourhood of P. Simeon Maybe this is true for stable singularities, for example in case of > fold type singularity set, it seems obvious. Simeon >>thanks again >>nir Dale. === Subject: Re: stable singularities? Content-Length: 726 Originator: rusin@vesuvius > what are those stable singularities?, my map is a quadratic g:R^n->r^m ; (n>m) Is there a simple way of knowing which kind of singularity I have? A stable singularity is not changing locally its look under small (smooth) perturbations of the map. For example, z^2 has an unstable singularity at z=0; it is enough to take a perturbation of the type z^2+eps*z', where eps is a small real number and z' denotes the conjugate to z. In small dimensions there are lists of stable singularities (see any book on the topic). In case of quadratic singularity, I suppose it is possible to recognize easily its type. Simeon === Subject: Re: Relations between Orthogonal Vectors Content-Length: 2984 Originator: rusin@vesuvius Consider two Sets A and B of n-dimensional vectors. A contains ALL the >vectors with exactly Ôi' 1s (and Ôn-i' 0s), and B contains all the >vectors with Ôj' 1s (and Ôn-j' 0s). WLOG, assume i < j <= n. More to the point, assume i + j <= n so it is possible for vectors > in A and B to be orthogonal. Now consider the folowing game: at each round, you remove one random >vector from A (if it is not exhaus yet), and all of its orthogonal >vectors (vecotrs whose position wise boolean AND yeild the vecotr >000...000) from B (if there is any left). For example. for i=2, j=3, >and n=6, if you remove 110000 from A, you also remove 001110, 000111, >001011, 001101 from B. Note that some of these vectors from B might >have already removed by some earlier removal from A. Is it possible to find a closed form of the function F(r), # of >vectors remaining in the set B as a function of the number of removals >r from A? Of course not, since this depends on which vectors are removed from A, > not just the number of them. In case it is not possible, a reasonably tight upper bound >would suffice. ... > It seems likely that (as in this example) a larger number > of survivors will occur when the chosen members of A share as many > members as possible. Take the largest m<=i such that > binom(n-m,i-m) >= r. Then it is possible to choose r members of A > which all contain {1,...,m}. Any member of B that intersects > {1,...,m} will survive. There are binom(n,j) - binom(n-m,j) members > of B that intersect {1,...,m}. Thus U(r) >= binom(n,j) - binom(n-m,j) > whenever binom(n-m,i-m) >= r. Of course, if m = 0 this bound is trivial. An improvement: after taking m_0 = m as above, if m_0 < i, we take the greatest m_1 <= i+1 so that binom(n-m_1,i-m_1) + (m_1-m_0) binom(n-m_1,i+1-m_1) >= r. Then it is possible to choose r members of A which all contain {1,...,m_0} and lack at most one of {m_0+1,...,m_1}. Any member of B that intersects {1,...,m_0} or contains at least two members of {m_0+1,...,m_1} will survive. So U(r) >= binom(n,j) - binom(n-m_1,j) - (m_1 - m_0) binom(n-m_1,j-1) Further improvements along these lines may be possible, if we can choose r members of A which lack none of {1,...,m_0}, at most one of {m_0+1,...,m_1}, at most two of {m_1+1,...,m_2}, ..., at most s of {m_{s-1}+1,..., m_s}. Any member of B that contains at least t+1 members of {m_{t-1}+1,...,m_t} for some t in {0,1,...,s} will survive, and we can count the number that do so (although I think it will get rather complica). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Axiomatization of metacategories Content-Length: 311 Originator: rusin@vesuvius Can somebody provide a reference to (a metamathematical) axiomatization of metacategories in terms of say a first order theory with equality (in a similar way axiomatic number theory and set theory are developed)? -- For example, MacLane's book only speaks about metacategories with a handwaving. Hans Aberg === Subject: Mathcad 2001 and PDEs? I am considering purchasing Mathcad 2001 (not the latest release, Version 11, but the one before that), and I have pos a few questions regarding that purchase. I have one more question... I wan to know if Mathcad 2001 can provide closed form solutions for at least some partial differential equations. I pos this question several places including Mathsoft's own collaboratory (their user-helping-user Web site). One or two people responded in the affirmative. However, Mathsoft's own people claim it does not. I know that sometimes tech support people are not always 100% accurate even about their own products (especially one that they no longer support or sell). And I'm just cynical enough to think that they might even deliberately mislead me, in an effort to get me to buy the latest version of the product. (I will not buy MC 11, because it requires Product Activation, and only allows installation on two computers. I find that entire technology unacceptable in principle, and in practise I have three computers -- I'm certainly not buying a third copy of the software for the third computer...) Anyway, the question is, can someone confirm that MC 2001 (not 2001i, but just 2001) does or does not provide closed-form solutions to PDEs? (I understand it won't solve all such equations, but it would help if it at least has the ability to solve some such equations, and rela boundary value problems.) Thanks in advance for all replies. Steve O. Standard Antißame Disclaimer: Please don't ßame me. I may actually *be* an idiot, but even idiots have feelings. === Subject: Journals and my papers by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i13MBLW19213; === Dear newsgroup: computing. Sincerely Dr.M.Basti Dear Professor Basti, I am writing you regarding four papers 043909-1 Polynomials and Riccati differential equations 043910-1 New Methods Of Solving Riccati Differential Equations 043911-1 New Methods Of Solving Riccati Differential Equations II 043939-1 A Class Polynomial II Computing. After consulting with members of the editorial board, I am afraid that I must reject these papers. We believe that the paper is not in the traditional scope of our journal which focuses on nonnumeric methods. We recommend that you submit your papers to a journal focused more on Scientific Computing seems the most appropriate. I am sorry that the outcome of the editorial process has not been more favorable. Computing. Sincerely, Eva Tardos === Subject: Journals and my papers by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i13MYvd20980; === Dear newsgroup: The following is an e-mail by Professor Tardos. Sincerely Dr.Mehran Basti Dear Dr.Mehran Basti, I am sorry that you do not find the email sufficient. Let me try to explain. Your papers propose methods for solving differential equations. I admit that I did not read your paper in detail, and well enough to understand the details. However, I did look at your paper enough to see what the methods and topic is about. This, and rela topics are trea by many journals. Some are more journals Scientific algorithms for solving differential equations, finding roots, etc. The difference of Numertical Analysis considers more theoretical advances, while the focus of you wrong advice about which of the two seems more appropriate for your research. I suggest that you should check papers in both of these journals, and feel free to decide yourself which is a better match. computer science, typical papers consider discrete and not continouos problems. The distinctions between the journals is more a matter of tradition, than a well defined boundary. Maybe the best way to topics rela to issues in courses like Algorithms, and Complexity. While the other two journals I mentioned are focused on research issues rela to topics traditionally taught in courses on Numerical Analysis or Scientific Computing. I suggest you should take a look at a few issues for all of these journals, and I hope you will agree with me about this difference, and that the the topic of your apper is not a good fit for our journal. I hope this clarification helps. I did not mean to express any negative judgement about your for this work. The researchers who would find the topic of your research of greatest interest, tend to not read send your paper to refereed, but I don't think it would do any favor to you to send your of Computing referees. I strongly believe such referees, would tell me the same: this topic of Computing. By discussing the paper directly with members of the editorial board, I simply wan to help you, so your paper can be considered by a more appropriate journal in a speedy fashion. Sincerely you Eva Tardos === Subject: Journals and my papers by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i14M6c407683; === Dear Newsgroup: Today I have received e-mail from the Journal of Algebra. Sincerely M.Basti Dear Mr Basti, A class polynomial III , that you have submit to Professor Michel Brou Ôe for a publication in the Journal of Algebra. I will let you know as soon as we receive the referee's report. Best Carine ippe assistant of Michel Brou Ôe, Editor-in-Chief Journal of Algebra Institut Henri Poincar Ôe 11 rue Pierre et Marie Curie F-75231 Paris Cedex 05 France === Subject: Re: Journals and my papers Dear newsgroup: I have sent my papers solving polynomials with differential equations and A class polynomial to the Journal of Algebra. The paper on A class polynomial II was sent to the Journal of pure and applied algebra . I hope that the papers will be suitable for Journals since it is very unconventional. These are solving polynomials with differential equations and I am introducing class polynomials to enhance the Algebra and Analysis of the 21st century. I have some more of these classes and will be sent upon publishing these papers. Since the methods are new I hope they can address any questions they may have to me and I will kindly answer. I believe in the future, applications to group theory will be found. Sincerely Dr.Mehran Basti === Subject: General Factorization Tool Consider n functions f_1(x), f_2(x),..., f_n(x), where f_1(x) f_2(x)...f_n = F(x), where F(x) is a polynomial where F(0) = 0, and numbers g_1, ..., g_n, k_1, ..., k_n, G and K where k_1 g_1...k_n g_n = KG, where also (f_1(x) + k_1 g_1)...(f_n(x) + k_n g_n) = K[F(x) + H(x) + G] and H(x) is a polynomial where H(0) = 0. Here you have the factorization k_1 g_1...k_n g_n = KG, balanced against the factorization (f_1(x) + k_1 g_1)...(f_n(x) + k_n g_n) = K[F(x) + H(x) + G], where the point is that it's *unknown* how factors of K split between f_1(x) through f_n(x). So you can think of my method as a solution for an unknown factorization. Here the solutions vary dependent on the ring. For instance, in the ring of algebraic integers dividing K from both sides *must* give g_1...g_n = G. That may seem trivial, but let n=2, k_1 = 3, k_2 = 1, and consider what can happen in some other ring, like the field of algebraic numbers: (sqrt(3) g_1)(g_2/sqrt(3)) = G. Notice there are an *infinity* of solutions in the field of algebraic numbebut just one in the ring of algebraic integers. Now then given that you have (f_1(x) + k_1 g_1)...(f_n(x) + k_n g_n) = K[F(x) + H(x) + G], it's forced that f_a(x) *should* have g_a, where Ôa' is a counting number such that 1<=a<=n, as a factor in any ring where you have the single factorization g_1...g_n = G. I'll call such a ring a complete ring. Now then, intriguingly, it can be shown that the ring of algebraic integers is not a complete ring! Notice that you can get a simple result where everything works fine even in the ring of algebraic integers. For instance, with n=2, f_1(x) = 3x, f_2(x) = x, H(x) = 2x, g_1=1, g_2=1, k_1=3, k_2 = 1, you have (3x + 3)(x + 1) = 3(x^2 + 2x + 1). That's trivial but demonstrates the basic idea. The extraordinary power of the tool comes with functions where you can't just *see* what the factors are where it is excellent for peering into mathematical regions *inside* of non-polynomial functions. === Subject: Re: C routines for Special Functions >Error Handling ... >>GSL does have Ônatural form' calling conventions available (more >>code on their side to maintain) but they do not permit error >>checking. I prefer returning NaN's that have embedded error >>information. If you don't check the return codes, at least your >>program is more likely to return gibberish than to crash. >Which is a bad thing, of course. (I mean returning gibberish.) >The proper thing to do in a real time show must go on scenario >(like radar tracking of a missile ms before impact) is to >catch the error and substitute a reasonable value, and hope for >the best. Not to send your defense missile astray. Any error protocol requires coordination between the library writer and the library user. People often will simply ignore error codes no matter how many lectures you deliver on the proper thing to do. My point was that by defalt GSL calls abort on error and provides no information in the case of natural form calling. I'm open to better solutions than the one I sugges. >In the context of mathematical functions I just want to know >that the programmer goofed. If you as a project leader dele >the programmers assert's you would have a very bad time with me. >If your head of IT agrees with you, I would take my business >elsewhere. You sound like the kind of guy I wouldn't hire in the first place. FWIW, here is an implementation of my proposed error protocol. It is not portable, but it gives the idea. /************************************************************ *************** *** FILE ensure.h DATE 2002-10-1 AUTHOR Keith A. Lewis [kal@kalx.net] SYNOPSIS #include ensure.h ensure (expr); bool ensure_is_error(double); const char* ensure_message(double); DESCRIPTION Simple error protocol using NaNs. The macro ensure(expr) works much like assert(expr) but it does not call abort and it can only be used in functions that return a double. If expr is false, a NaN is returned. Use ensure_is_error() to detect ensure() failures and ensure_message() to extract the error string. The function ensure_make_nan() embeds a char* into a NaN. It is your responsibility to make sure that the argument to ensure_make_nan() exists until ensure_message() is done using it. Define USE_ENSURE or _DEBUG before including ensure.h in order to make ensure does its job. Unlike assert, the default behavior is to define ensure with an empty body. BUGS: Only works on 32-bit machines. ************************************************************* *************** **/ #ifndef ENSURE_H #define ENSURE_H #include #include <ßoat.h> #if defined(_WIN32) && !defined(isnan) #define isnan _isnan #endif #define D_(x) #x #define S_(x) D_(x) #if defined(USE_ENSURE) || defined(_DEBUG) // TODO: offer more ßexibility. (no return???) #define ensure(e) { if (!(e)) return ensure_make_nan( file: __FILE__ line: S_(__LINE__) error: #e); } #else #define ensure(x) #endif // Typedefs and constants typedef double ensure_error; const ensure_error ensure_ok = 0; typedef union { long l[2]; double d; } ensure_union; /* Endian tests */ inline int ensure_lo() { static int i = 1; return *(char*)&i != 1; } inline int ensure_hi() { static int i = 1; return *(char*)&i == 1; } /* Equivalent to isnan() function. */ inline bool ensure_is_error(double x) { ensure_union u; u.d = x; return u.l[ensure_lo()] == 0x7FFFFFFF || 0 != isnan(x); } /* Convert NaN to error message string. */ inline const char* ensure_message(double x) { ensure_union u; u.d = x; #pragma warning(push) #pragma warning(disable: 4312) return u.l[ensure_lo()] == 0x7FFFFFFF ? (const char*)u.l[ensure_hi()] : isnan(x) ? NaN : ensure_message: argument is not a NaN ; #pragma warning(pop) } /* Insert char* into NaN. You are respsible for the lifetime of s. */ inline double ensure_make_nan(const char* const s) { ensure_union u; if (0 == s) return ensure_ok; u.l[ensure_lo()] = 0x7FFFFFFF; #pragma warning(push) #pragma warning(disable: 4311) u.l[ensure_hi()] = (long)s; #pragma warning(pop) return u.d; } #ifdef isnan #undef isnan #endif #endif /* ENSURE_H */ === Subject: Re: numerical boundary conditions u_t+a*u_x=0, a>0, so you cannot specify the outßow boundary (at x=L) condition. Extrapolation should be OK as long as it is at least first-order so the global 2nd-order accuracy is maintained. You can do extrapolation along the outgoing characteristic, i.e., essentially use a forward time backward space scheme on the last node. Since LW is disipative the stability issue is no so critical but it pays to be well-educa so I suggest you consult J.W. Thomas' two-volume set on finite differences (stability of numerical BCs is analyzed in volume 2) or Strikwerda's excellent book on the subject. Peter i am trying to solve a hyperbolic PDE, using the Lax-Wendroff method. > I have an initial condition, and a boundary condition at one edge. > However, with the stencil used, i need to know the solution at the > other edge as well. How would i best calculate this? My teacher suggests a linear extrapolation, like this: u(L) = 2u(L-1) - u(L-2) But isn't this only first degree accuracy? As i see it, this error > will creep in and eventually destroy the whole solution? (lax-wendroff > is 2nd order) I would prefer something like second order > extrapolation: u(L) = 3u(L-1) - 3u(L-2) + u(L-3) Please can someone tell me, which is more correct. If i wasn't clear > enough, i can explain in more detail. The PDE is modeling a heat > exchanger. Paramo === Subject: Point me in the right direction I have a fun programming problem, but I am too weak in the math to get star. If you know what kind of problem I am solving, please tell me so I can research it. Electricity generation companies provide secret bids to sell electricity. Their bids are pairs of price and quantities in a step function. For example, one generator have an offer as such (MW is mega Watt): $30 0-10 MW $40 11-50 MW $50 51-90 MW $100 91MW - 100MW You never know what the Generator bids look like, but the market price and generator outputs are published for every hour. My goal is to collect and analyse enough data (market prices and generator outputs) and decompose it into each generator's bid stack. Has this type of problem ever been solved before. It doesn't appear to be a simple best-fit algorithm because of the step function. Jarson === Subject: Re: Intel Fortran for Windows: student edition $29 While browsing Programmer's Paradise, I saw that a student edition of >Intel Visual Fortran for Windows 8.0 is available for $29. The link is http://www.programmersparadise.com/Product.pasp?txtCatalog= Paradise&txtCateg ory=&txtProductID=I23+0D03 > This product should give Matlab a run for its money, and > run your problems in seconds instead of days! You mean people actually *pay* for FORTRAN compilers??? Only the *smart* ones, the dumb ones pay for a stripped (Fortran) subset stuffed under the Matlab cover. === Subject: MathML runtime C++ to a more open standard - MathML to be more specific. Is anyone aware of a piece of software that can read in MathML and create code (C, C++, C# etc.. ) from it? This would mean we could remove any gaps between the algorithm we say we're using, and the actual implementation. 4Space === Subject: Re: Five Parameter Logistics It is not clear whether you are asking for the five parameter model, or a way to fit the coefficients given the five parameter model. Since you pos in the numerical analysis newsgroups, the previous replies presume that you know the model and just need to know how to fit it. Since you only give the four paramter model, I presume that you are farther back and don't know the five parameter model. The only five parameter logistic that I have seen raises the entire denominator to a power to give an asymmetric sigmoidal curve. When that power is one, then it reduces to the four parameter model. As far as I know, it is a pragmatic way to remove symmetry from the curve that has no physical interpretation. Jerry I'm looking for a formula to fit a set of data to a so called five > parameter logistic curve. I've already done this using a four > parameter logistic using > the following formula: A - D > y = ( ------------- ) + D > 1 + (x/C)^B ...And now I'm urgently looking for the 5 param. version. > Joerg === Subject: Re: fraccional combinatorial numbers Content-Length: 851 Originator: rusin@vesuvius > Consider the following number, where m and n are positive integers: B(m, n) = (5m)!(5n)!/(m!n!(3m + n)!(3n + m)!) I am trying to show it is an integer (something that must be true > since it is true for n and m less than 1000, pari says it). My problem is: how to show this (and for similar combinatorial > numbers involving quotients)?. Quite a few cases. Fortunately, a picture is worth a thousand words. > Another example of the picture method and some references are given in Advanced Problem 6514 (solution by Gregg Patruno), Amer. Math. Monthly (94) 1987, pp.1012--1014. === Subject: Re: Complexity of solving N non-linear equations Content-Length: 2086 Originator: rusin@vesuvius > Can anyone tell me what is the complexity of solving a set of N > non-linear equations (simultaneous equations - having all variables in > each equation). The equations themselves are very complex - contains > integrals of a sample maximum functions. I know the question is somehow vague, however an answer of type > generally it is exponential is also very helpful for me. It depends more on how nonlinear or nonmonotone your functions are (if you plot the values along random lines) than on the complexity of the defining expression. For well-behaved systems F(x)=0 (e.g., those which are C^1 and globally 1-1), the work is O(N*NF+N^3), where NF is the number of operations needed for an evaluation of all equations. (One can use Broyden's method, or with automatic differentiation tools Newton's method, generally applied in a damped fashion that reduces the residual ||F(x)||^2 in each step.) For sparse systems - not your case - the work is somewhat less is one uses special methods. For highly nonlinear systems, the work grows but has typically the same complexity (with a larger factor hidden in the O) if the method succeeds to reduce the residual to zero. (There are theoretical complexity estimates for polynomial systems which say that the cost depends on the closeness to a singular point of the Jacobian F'(x), in a suitable sense.) If not, the problem has a global nature, and the work is in the worst case exponential. Moreover, it might be that there are many solutions - in certain applications exponentially many. In this case you probably want to single out which solution you want to see by imosing additional criteria like Ôthe solution with largest f(x)' - in this case, your problem becomes a nonlinear program, more precisely a global constrained optimization problem. Independent of the complexity issue, many large nonlinear sytems are solved successfully and routinely, and there is lots of excellent software available. See, e.g., http://www.mat.univie.ac.at/~neum/glopt/software_l.html [~ is a tilde] Arnold Neumaier === Subject: Re: Complexity of solving N non-linear equations Content-Length: 1247 Originator: rusin@vesuvius >I must admit I know a little about it, and I really nead some help. >Can anyone tell me what is the complexity of solving a set of N >non-linear equations (simultaneous equations - having all variables in >each equation). The equations themselves are very complex - contains >integrals of a sample maximum functions. >I know the question is somehow vague, however an answer of type > generally it is exponential is also very helpful for me. In any great degree of generality, it is not computable at all. The solution to Hilbert's 10th problem says that there is no algorithm to solve diophantine equations in several variables: given a polynomial P(x_1,...,x_n) with integer coefficients, there is no way to tell whether there are positive integers x_1,...,x_n such that P(x_1,...,x_n) = 0. This is equivalent to saying there is no way to tell whether there are real numbers t_1,...,t_n such that P(t_1^2, ..., t_n^2)^2 + sum_{j=1}^n sin^2(pi t_j^2) = 0 Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Complexity of solving N non-linear equations Content-Length: 1284 Originator: rusin@vesuvius In general, solving systems of non-linear equations are difficult not because of the complexity (i.e., running-time) of algorithms that solve them. They are difficult because the algorithms which solve them usually are iterative and do not always converge to a soltuion - It is very possible that they will diverge and never reach a solution - for instance, look up Newton's method for solving just one equation of one variable and see. However, if you can get an initial guess close enough to the solution, then the running-time of the algorithm is usually real-time (i.e., polynomial of small degree). The book Numberical Recipes in C has, in fact, an algorithm for solving such equations. Craig > Dear forum membe I must admit I know a little about it, and I really nead some help. > Can anyone tell me what is the complexity of solving a set of N > non-linear equations (simultaneous equations - having all variables in > each equation). The equations themselves are very complex - contains > integrals of a sample maximum functions. I know the question is somehow vague, however an answer of type > generally it is exponential is also very helpful for me. David === Subject: Re: Partitioning by hyperplanes > Given m hyperplanes in R^n (m>n), what is the maximal number of > non-empty disjoint convex segments that may be crea? I can offer you a keyword. The partition of R^n by hyperplanes is called an arrangement. Mathworld doesn't offer much more information than that http://mathworld.wolfram.com/Arrangement.html though they do reference a book chapter that may be worth checking out. I expect the result you're looking for probably shows up in algebraic topology somewhere. hth, Rick === Subject: Re: Partitioning by hyperplanes >> Given m hyperplanes in R^n (m>n), what is the maximal number of >> non-empty disjoint convex segments that may be crea? >I can offer you a keyword. The partition of R^n by hyperplanes is called >an arrangement. Mathworld doesn't offer much more information than that >http://mathworld.wolfram.com/Arrangement.html >though they do reference a book chapter that may be worth checking out. >I expect the result you're looking for probably shows up in algebraic >topology somewhere. Yes, there are topological questions to answer, both in R^n and in C^n. I know for example of a number of papers by Orlik, Solomon, and others. But I think the OP's question is answered by this: %N A004070 Table of Whitney numbers W(n,k) read by antidiagonals, where W(n,k) is maximal number of pieces into which n-space is sliced by k hyperplanes, n >= 0, k >= 0. Example: Table W(n,k) begins: 1 1 1 1 1 1 1 ... 1 2 3 4 5 6 7 ... 1 2 4 7 11 16 22 ... 1 2 4 8 15 26 42 ... Here's the long URL: http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?An um=A004070 dave === Subject: Tricky integration - a silly error? I'm trying to integrate sin3x/(1+cosx) This is how I've done it this is the same as integrating (2sinx(cos x)^2)/(1+cosx) + (cos2xsinx)/(1+cosx) Now the numerator is a derivative of the denominator so the answer is -2(cosx)^2ln(1+cosx) + -cos2xln(1+cosx) Is this correct because numerical integration say's it wrong? Sarah === Subject: Re: Tricky integration - a silly error? I'm trying to integrate sin3x/(1+cosx) This is how I've done it this is the same as integrating (2sinx(cos x)^2)/(1+cosx) + (cos2xsinx)/(1+cosx) Now the numerator is a derivative of the denominator so the answer is -2(cosx)^2ln(1+cosx) + -cos2xln(1+cosx) Is this correct because numerical integration say's it wrong? Sarah Yeah, it's wrong. Try the identity sin3x = 3*sinx - 4 * (sinx)^3 and then let u = 1 + cosx be your new variable of integration. -- Julian V. Noble Professor Emeritus of Physics jvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything and thereby take away our free will and that share of glory that rightfully belongs to us. -- N. Machiavelli, The Prince . === Subject: Re: Tricky integration - a silly error? I'm trying to integrate sin3x/(1+cosx) > This is how I've done it > this is the same as integrating (2sinx(cos x)^2)/(1+cosx) + (cos2xsinx)/(1+cosx) > Now the numerator is a derivative of the denominator so the answer is > -2(cosx)^2ln(1+cosx) + -cos2xln(1+cosx) > Is this correct because numerical integration say's it wrong? Sarah Yes, silly error(s): you trea the factors -(cos(x))^2 and cos(2*x) as constants, which they are not. You had a promising start, just pull it a little further: sin(3*x) = (4*(cos(x))^2 - 1) * sin(x) after some algebra, then sin(3*x)/(1+cos(x)) dx = (1 - 4*(cos(x))^2) / (1+cos(x)) * (-sin(x)) dx substitute cos(x) = u, perform long division, integrate, and get back to original variable x. ZVK(Slavek) === Subject: cyclotomic polynomials Hi there!!! SOS!!! Help me !!!! I Know nothing about cyclotomic polynomials but I have to make a small research for a course in number theory. Can anyone give me some help? Any sites I could study for this? Thanks!!! === Subject: Re: Intel Fortran for Windows: student edition $29 While browsing Programmer's Paradise, I saw that a student edition of >Intel Visual Fortran for Windows 8.0 is available for $29. The link is http://www.programmersparadise.com/Product.pasp?txtCatalog= Paradise&txtCateg ory=&txtProductID=I23+0D03 > This product should give Matlab a run for its money, and > run your problems in seconds instead of days! You mean people actually *pay* for FORTRAN compilers??? Only the *smart* ones, the dumb ones pay for a stripped (Fortran) subset > stuffed under the Matlab cover. Oooh meow. Let me see, saying not X in an X newsgroup makes you a... PLONK. === Subject: Re: Intel Fortran for Windows: student edition $29 >While browsing Programmer's Paradise, I saw that a student edition of >>Intel Visual Fortran for Windows 8.0 is available for $29. The link is >>http://www.programmersparadise.com/Product.pasp?txtCatalog= Paradise&txtC ategory=&txtProductID=I23+0D03 >>This product should give Matlab a run for its money, and >>run your problems in seconds instead of days! >>You mean people actually *pay* for FORTRAN compilers??? > Only the *smart* ones, the dumb ones pay for a stripped (Fortran) subset > stuffed under the Matlab cover. > Take a look at http://www.openwatcom.org Unfortunately, Windows, DOS, or OS/2 only. === Subject: Automatic Problem Solver 3.0 New Release Dear All, The third edition of APS has been released and a free demo of APS 3.0 is now available for download: http://www.gepsoft.com/gepsoft/ General features of version 3.0: o Translates the evolved models into 8 different languages (C, C#, C++, Java, JavaScript, Visual Basic, VB.NET, and Fortran). o Draws the parse trees of the evolved models. o Translates the evolved models virtually into any programming language through User Defined Grammars. o A total of 70 different built-in mathematical functions and comparison rules plus Dynamic UDFs and Static UDFs for modeling. o A total of 11 built-in fitness functions for Function Finding. o A total of 10 built-in fitness functions for Classification. o A total of 11 built-in fitness functions for Times Series Prediction. o User Defined Fitness Functions for all problem categories. o Implements a new algorithm for handling random numerical constants. o Data screening engine for preprocessing. o Time series transformation engine. o Evolution from seed models. o Change seed utilities. o Saves all the best-of-generation models of a run. o Plots the evolutionary dynamics of the run. o Complexity increase engine. o Supports Databases and Text Files both for loading input data and scoring. o Recursive testing and prediction for Time Series. o Implements an extensive package of statistical indexes for model evaluation. o and much more. The demo allows you to use your own datasets and all the features are operational except scoring/prediction and visualization of the models. However, for the included datasets and sample runs all APS features are fully operational in the demo. Enjoy! All the best, Candida Ferreira ----------------------------------------------------------- Candida Ferreira, Ph.D. Chief Scientist, Gepsoft 73 Elmtree Drive Bristol BS13 8NA, UK ph: +44 (0) 117 330 9272 http://www.gepsoft.com/gepsoft http://www.gene-expression-programming.com/author.asp ----------------------------------------------------------- === Subject: Re: Journals and my papers Dear newsgroup: I am inviting the journals of mathematics to contact me for my possible submission of my papers. My research has been about closed form solutions of differential equations and problems of solving polynomials (symbolic solutions). This is a new science (I have been working on them for more than 2 decades). I have a lot of materials and eventually I would like to publish them.. If your Journal is interes please contact me with the above e-mail. I would like to initially submit the manuscripts in PDF format, once it is accep in the form reques such as Latex. This is the mathematics of the 21st century. So please let me know, I will keep your Journal in mind, particularly electronic Journals and the recent ones from around the world. Sincerely Dr.Mehran Basti === Subject: Re: Decker Quadratic, algebraic integedone One more correction... > why do I have a quartic and what are the other roots? Well if I used (1 - sqrt(-167)) = (b + sqrt(b^2 + 28))z I'd get the same result, as eliminating the square roots at each > points creates *two* possible solutions that will work. I handled two > square roots, so I have four solutions, which are z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28)) z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28)) Now then, imagine that there exists some algebraic integer b for which > it is reducible over Q, then the root will be a fraction with a 7 in > the denominator. There wouldn't necessarily have to be a rational root for a solution reducible over Q. I was thinking about reducing to linear polynomials. === Subject: Re: Decker Quadratic, algebraic integedone > Turns out there's a rather direct approach to showing a problem with > the old concepts about the ring of algebraic integers. > I have some corrections to make, typos. > and working to eliminate square root terms gives 28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0 and working still further I get 196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0 Should be 196 z^4 + 28b z^3 + (168b^2 + 2324) z^2 - 168bz + 7056 = 0. > and I can divide both sides by 28 to finally get 7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0. Should be 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0. > Importantly, for any integer b, such that it is irreducible over Q, > *none* of the solutions for z can be an algebraic integer! > The conclusion, of course, is not changed by typos. Also I forgot to include the reference information for the Decker quadratic post, though I said I would, so this time it is at bottom. Decker Quadratic Source Information --------------------- Recently Rick Decker, a professor at Hamilton College, apparently trying to refute my research came up with a quadratic example, which I like because it's a quadratic, and easier to manipulate than the cubics I've used before. If you wish to see his original post here are some headers which also show that he posts from Hamilton College: Subject: Re: Mathematical consistency, courage Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). === Subject: Re: symbol > Does anyone know what this symbol is That's probably a lowercase Greek xi . --