mm-2559 === Subject: Re: Ania Elzbieta Slusarczyk - December 23rd 1982 [snip standard Kabatoff irrational rant about his favorite subject - penises] Aha! There's the pitch. Just as I predicted. -- Few things are more satisfying to a zealot than condemning everyone else to hell. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Another Integration Problem How do I integrate e^ ((1/3200)x^2) between 0 and inf? Havn't done integration by parts in many a year so might need to be written in Help for Dummys format :-) === Subject: Re: Another Integration Problem Originator: mtx014@linux.services.coventry.ac.uk (Robert Low) >How do I integrate >e^ ((1/3200)x^2) between 0 and inf? You don't... that function gets bigger and bigger as x does, so the integral from 0 to infinity doesn't converge. If it should have been e^ (-(1/3200)x^2) then consider substituting some multiple of x and then looking up a table for the integral of e^(-x^2), which is a standard result. -- Rob. http://www.mis.coventry.ac.uk/~mtx014/ === Subject: Re: Another Integration Problem Sorry should have been e^ ((-1/3200)x^2) between 0 and inf? > How do I integrate > e^ ((1/3200)x^2) between 0 and inf? > Havn't done integration by parts in many a year so might need to be written > in Help for Dummys format :-) === Subject: Re: Another Integration Problem <33e026c6ab406b67a91411950a8f80b6@TeraNews> > Sorry should have been > e^ ((-1/3200)x^2) between 0 and inf? Look up the Gamma function Gamma(n) = integral(0,oo) x^(n-1) e^-x dx Prove Gamma(n) Gamma(1-n) = pi/sin pi.n Show Gamma(1/2) = 2 integral(0,oo) e^(-t^2) dt Conclude integral(0,oo) e^(-t^2) dt = (sqr pi)/2 Transform variable t = x(sqr 2)/80 > How do I integrate > e^ ((1/3200)x^2) between 0 and inf? === Subject: Re: Another Integration Problem >>Sorry should have been >>e^ ((-1/3200)x^2) between 0 and inf? > Look up the Gamma function > Gamma(n) = integral(0,oo) x^(n-1) e^-x dx There's an even easier function to look up - without wanting to give it away, it's relevant that this was posted to a stats NG. Bob -- Bob O'Hara Rolf Nevanlinna Institute P.O. Box 4 (Yliopistonkatu 5) FIN-00014 University of Helsinki Finland Telephone: +358-9-191 23743 Mobile: +358 50 599 0540 Fax: +358-9-191 22 779 WWW: http://www.RNI.Helsinki.FI/~boh/ === Subject: Re: Another Integration Problem >>Sorry should have been >>e^ ((-1/3200)x^2) between 0 and inf? >> > Look up the Gamma function > Gamma(n) = integral(0,oo) x^(n-1) e^-x dx > There's an even easier function to look up - without wanting to give it > away, it's relevant that this was posted to a stats NG. > Bob > -- > Bob O'Hara > Rolf Nevanlinna Institute > P.O. Box 4 (Yliopistonkatu 5) > FIN-00014 University of Helsinki > Finland > Telephone: +358-9-191 23743 > Mobile: +358 50 599 0540 > Fax: +358-9-191 22 779 > WWW: http://www.RNI.Helsinki.FI/~boh/ Now I am really confused.... === Subject: Re: Another Integration Problem : Sorry should have been : e^ ((-1/3200)x^2) between 0 and inf? There's a short way and a long way. The short way is to think about what this integral is - it's very nearly the same as somethig which is easy to find tabulated. The long way is to find the integral of exp(-x^2) over the same range. Which is a wee bit of a pain to do in the general case, but in this case is traditionally done by noting that if I = int(exp(-x^2), x = 0 .. inf) dx then I^2 = int(exp(-x^2), x = 0 .. inf) dx int(exp(-y^2), y = 0 .. inf) dy And by not being too fussy about orders and limits I^2 = int(exp(-(x^2 + y^2) dx dy which is a surface integral. Put it into polar coordinates, note that dx dy = r dr dtheta, fiddle around a bit, get another I to appear and orff yer goes. Ian -- === Subject: Re: Another Integration Problem <33e026c6ab406b67a91411950a8f80b6@TeraNews> > : Sorry should have been > : e^ ((-1/3200)x^2) between 0 and inf? > I = int(exp(-x^2), x = 0 .. inf) dx > then > I^2 = int(exp(-x^2), x = 0 .. inf) dx int(exp(-y^2), y = 0 .. inf) dy > And by not being too fussy about orders and limits > I^2 = int(exp(-(x^2 + y^2) dx dy > which is a surface integral. Put it into polar coordinates, note that > dx dy = r dr dtheta, fiddle around a bit, get another I to appear and Indeed, the easy way. I^2 = integral(0,oo) integral(0,oo) e^-(x^2 + y^2) dx dy = integral(0,pi/2) integral(0,oo) e^(-r^2) r dr dt = 1/2 * integral(0,pi/2) integral(0,oo) e^-u du dt = 1/2 * integral(0,pi/2) (-e^-u)|0,oo dt = 1/2 * integral(0,pi/2) dt = pi/4 I = (sqr pi)/2 integral(-oo,oo) e^(-t^2) dt = sqr pi = Gamma(1/2) = (-1/2)! === Subject: Re: Another Integration Problem > How do I integrate > e^ ((1/3200)x^2) between 0 and inf? OK, we've gone off track here. If you want an answer, it does help to ask the right question. It may no longer matter, but let me see if I can clear up the confusion. To recap, we want the expected value of the density f(x) = (x/1600) exp(-(1/3200) x^2). See: The expected value is integral(0,inf) x f(x) dx, i.e., integral(0,inf) (x^2)/1600 exp(-(1/3200) x^2) dx For convenience, let a = 1/3200, so the integrand is 2 a x^2 exp(-a x^2). Let t = x sqrt(a). Then dt = dx sqrt(a), so the integral (call it E) is E = integral(0 sqrt(a), inf sqrt(a)) 2 t^2 exp(-t^2) dt/sqrt(a) = (1/sqrt(a)) integral(0,inf) 2 t^2 exp(-t^2) dt Let u = t, and let dv = 2 t exp(-t^2) dt. Then du = dt, and v = -exp(-t^2). So putting the factor 1/sqrt(a) aside for the moment, the integral is u(inf) v(inf) - u(0) v(0) - integral(0,inf) v du = 0 - 0 + integral(0,inf) exp(-t^2) dt = sqrt(pi)/2 as pointed out elsewhere in this thread. So the overall result is E = (1/sqrt(a)) sqrt(pi)/2 = sqrt(3200) sqrt(pi)/2, which is approximately 50. Well, I'm pretty sure I'm answering the right question, and I hope I didn't make any elementary mistakes in working out the solution. For what it's worth, Robert an integral a day Dodier -- Science may be described as the art of systematic over-simplification. -- Karl Popper === Subject: another proof I need to show that limsup(s(n)+t(n))<=limsup s(n) + lim sup t(n) can I do this by proving that they can be equal or the lhs can be smaller than the rhs or do I need to prove these two ideas together? === Subject: Re: another proof > I need to show that limsup(s(n)+t(n))<=limsup s(n) + lim sup t(n) > can I do this by proving that they can be equal or the lhs can be smaller > than the rhs or do I need to prove these two ideas together? Do both together - there is really not much choice. You might want to give an example where equality holds and one where it does not. -- Paul Sperry Columbia, SC (USA) === Subject: associative property question I have a solution manual that states this problem is associative: Determine whether the binary operation * defned is commutative and whether * is associative... * defined on Z by a*b=a-b I can clearly see how to check to see if this problem is commutative because I have 2 numbers. 1-2 != 2-1 therefore not commutative My question is, how do you check if its associative without 3 numbers? The solution manual said: 2=1-(2-3) != (1-2)-3=-4 ...where did the 3rd number (or c) come from? Marcos === Subject: Re: associative property question > I have a solution manual that states this problem is associative: Never heard of such, would you explain what an associative problem is? > Determine whether the binary operation * defned is commutative and > whether * is associative... > * defined on Z by a*b=a-b > I can clearly see how to check to see if this problem is commutative > because I have 2 numbers. > 1-2 != 2-1 therefore not commutative > My question is, how do you check if its associative without 3 numbers? > The solution manual said: 2=1-(2-3) != (1-2)-3=-4 > ...where did the 3rd number (or c) come from? I suppose it comes with 1 & 2 but until you tell us what set the binary operator operates upon, how are we to know? Who's c? You've yet to introduce her to us. As for a two number counterexample, presuming - is substraction of integers. 1-(1-1) = 1-0 = 1 (1-1)-1 = 0-1 = -1 === Subject: Re: associative property question > I have a solution manual that states this problem is associative: > Determine whether the binary operation * defned is commutative and > whether * is associative... > * defined on Z by a*b=a-b > I can clearly see how to check to see if this problem is commutative > because I have 2 numbers. > 1-2 != 2-1 therefore not commutative > My question is, how do you check if its associative without 3 numbers? > The solution manual said: 2=1-(2-3) != (1-2)-3=-4 > ...where did the 3rd number (or c) come from? > Marcos An operation * on set R is defined to be associative if it satisfies the following property: For all numbers a,b,c in R, (a*b)*c = a*(b*c) -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: associative property question >> I have a solution manual that states this problem is associative: >> Determine whether the binary operation * defned is commutative and >> whether * is associative... >> * defined on Z by a*b=a-b >> I can clearly see how to check to see if this problem is commutative >> because I have 2 numbers. >> 1-2 != 2-1 therefore not commutative >> My question is, how do you check if its associative without 3 numbers? >> The solution manual said: 2=1-(2-3) != (1-2)-3=-4 >> ...where did the 3rd number (or c) come from? >> Marcos > An operation * on set R is defined to be associative if it satisfies the > following property: > For all numbers a,b,c in R, (a*b)*c = a*(b*c) Sorry I wasn't specific enough. I didn't know how to ask the question. I understand what the associative property says. I was looking for an operational procedure to figure out if the problem was associative. Lucky for me, I've figured out how to work the problem. Binary operation * defined on set Z by a*b=a-b where a, b are members of Z. Commutative if and only if a*b=b*a: a-b != b-a // therefore not commutative Associative if (a*b)*c=a*(b*c): (a-b)*c ?= a*(b-c) // where a-b and b-c represents a*b (a-b)-c ?= a-(b-c) // where ()-c and a-() represents a*b a-b-c != a-b+c // therefore not associative Marcos === Subject: basic algebra if xif x if x x+a < y+a twice. With that hint can you prove x < y, r < s ==> x+r < y+s ? === Subject: Re: basic algebra > if x if x y - x is positive. Similarly z < q <=> q - z is positive. So ( that the sum of positives is positive is part of the definition of positive), (y - x) + (q - z) = (y + q) - (x + z) is positive so x + z < y + q. -- Paul Sperry Columbia, SC (USA) === Subject: Re: basic algebra >Yes, you can even prove it. >By definition, x < y <=> y - x is positive. >Similarly z < q <=> q - z is positive. >So ( that the sum of positives is positive is part of the definition of >positive), (y - x) + (q - z) = (y + q) - (x + z) is positive >so x + z < y + q. >-- >Paul Sperry >Columbia, SC (USA) Brilliant and simple. Amazing how some of the proving had been missing from some of the elementary algebra courses; only a few of them asked in each elementary course. Other aspects were fine, but proving was usually expected only in Geometry and some in Pre-Calculus, and briefly in Trigonometry (identities). Questions like that are probably what separates the Mathematicians from the studied-math-as-prerequisite-for-certain-courses type. Applying Math well is one thing; proving stuff within the subject is sometimes another thing. G C === Subject: Re: basic algebra > if xif xR^3, T(x_1,x_2)=(x_1, x_2, 2x_1 + x_2) __________________________________________________ Ker T={(x_1,x_2) | T(x_1,x_2)=(0,0,0)} ={(x_1,x_2) | (x_1, x_2, 2x_1 + x_2)=(0,0,0)} ={(0,0)} Clearly, Ker T = Null Space of T and, I think it is impossible to knowing dimension of Ker T by counting # of basis. And I think maybe there exist a definition about a basis & dimension about Null Space. But, I don't think the definition of dimension and basis about Null Space. If anyone know the definiton or have suggestion, please post reply. === Subject: Re: basis of Null Space > ___________________________ > Find a basis of Ker T, > where T:R^2 ->R^3, T(x_1,x_2)=(x_1, x_2, 2x_1 + x_2) > __________________________________________________ > Ker T={(x_1,x_2) | T(x_1,x_2)=(0,0,0)} > ={(x_1,x_2) | (x_1, x_2, 2x_1 + x_2)=(0,0,0)} > ={(0,0)} Yes, so far so good. > Clearly, Ker T = Null Space of T and, > I think it is impossible to knowing dimension of Ker T by counting # of > basis. What is bothering you I think is dim( {0} ). But, think of it this way: what is the size of the largest linearly independent set of vectors? Answer: 0. So, dim( {0} ) = 0. > And I think maybe there exist a definition about a basis & dimension about > Null Space. > But, I don't think the definition of dimension and basis about Null Space. > If anyone know the definiton or have suggestion, please post reply. There is a Theorem which says that the dimiension of the null space plus the dimension of the range space (_not_ the codomain) equals the dimension of the domain but I don't think that is very useful here. -- Paul Sperry Columbia, SC (USA) === Subject: Re: basis of Null Space > I think it is impossible to knowing dimension of Ker T by counting # > of basis. And I think maybe there exist a definition about a basis > & dimension about Null Space. But, I don't think the definition of > dimension and basis about Null Space. If anyone know the definiton > or have suggestion, please post reply. Remember, the definition of a vector space is the number of elements of a maximal linearly independent set. (It is a standard result that all finite such have the same number of elements.) Can you find a maximal linearly independent set inside the null space? /olov -- I'm no believer in a random massacre of theatreland's sacred cows. I just believe in chasing them down King's Parade on a bright orange spacehopper. -- Ed Richardson, The Cambridge Student, 14/2/2002 === Subject: Re: basis of Null Space >>___________________________ >>Find a basis of Ker T, >>where T:R^2 ->R^3, T(x_1,x_2)=(x_1, x_2, 2x_1 + x_2) >>__________________________________________________ >>Ker T={(x_1,x_2) | T(x_1,x_2)=(0,0,0)} >> ={(x_1,x_2) | (x_1, x_2, 2x_1 + x_2)=(0,0,0)} >> ={(0,0)} >Yes, so far so good. >>Clearly, Ker T = Null Space of T and, >>I think it is impossible to knowing dimension of Ker T by counting # of >>basis. >What is bothering you I think is dim( {0} ). But, think of it this way: >what is the size of the largest linearly independent set of vectors? >Answer: 0. So, dim( {0} ) = 0. >>And I think maybe there exist a definition about a basis & dimension about >>Null Space. >>But, I don't think the definition of dimension and basis about Null Space. >>If anyone know the definiton or have suggestion, please post reply. > There is a Theorem which says that the dimiension of the null space >plus the dimension of the range space (_not_ the codomain) equals the >dimension of the domain but I don't think that is very useful here. Not to the extent that you've already got all the information you need. However, it's easy to find two linearly independent vectors in the range space, so that theorem says (also) that the dimension of the kernel is 0. Such things often make a difference to those of us who are unsure of ourselves -- independent confirmation makes us more secure. Jon Miller === Subject: Bhopinder Singh Bolaria - January 10th 1936 B O L A R I A 2 15 12 1 18 9 1 = 58 Singh Bolaria was out walking his dog and provided stats, he is a professor of Sociology at the University of Saskatchewan. There are four obelisks (Egyptian representations of penises) in the front gates of the U of S. Starting in 1988 I was repeatedly detained and tortured at the U of S, and they would bring me past these front phallic gates and then tell me that I think too much about penises and too much about math. 206 Singh 10 1 36 10/356 +7709 Bhopinder 91 Singh 57 Bolaria 58 Singh was born on the 10th day of the month and on the 10th day of the year. The 10 consonants in his given names add to 110. His first 10 letters add to 110. His last 10 letters add to 87 (3 times the 10th prime). His given names differ in value by 34, it's the first 10 primes (129) minus the first 10 non-primes (95). The primes, squares and cubes in his given names add together for 87, it is an average of 29 (the 10th prime) per category. His 13 different letters add to 130 (10x13). The Lucas valued letters in his given names add to 29 (10th prime). Singh provides the stats on the 16th (10th non-prime). Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 <-10th-> 16 --- --- 129 95 His middle name adds to 57, his given names add together for 148 (the 57+57th non-prime), chapter 148 brings Numbers up to 1114 (557+557) verses. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 --- 108 Singh's first name adds to 91 (1 through 13). His middle name adds to 57, it's the 41st non-prime (while 41 in turn is the 13th prime). His given names add to the 67th and 41st non-primes, together for 108 (the primes in prime positions up to the 13th prime). The 11 different letters in his given names add to 117 (3x3x13). His 13 different letters add to 130 (Numbers 13), the 13 letters he is missing add to 221 (13x17), it's a difference of 91 (1 through 13). The even valued letters in his given names add to 84 (the first 13 primes minus the first 13 non-primes). He has 13 consonants, they exceed his vowels by 78 (6x13). The repeating letters in his given names add to 62, or the 13th prime (41) plus the 13th non-prime (21). In all his unrepeated letters add to 63 (Exodus 13) while his repeating letters add to 143 (11x13). He has 21 letters in all and is 21 years older than me (13th non-prime). Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 --- --- 238 154 <-Lamentations Bolaria (58) was born on the 13158th day of the century. In the mid 1980's I attended an annual meeting of the Western Canadian Anthropology and Sociology Society and this meeting was attended by a few Sociologists from the States bordering Saskatchewan. The Canadian Sociologists were all Marxists while the American Sociologists were not. The Marxist Canadian Sociologists mocked the American non-Marxist Sociologists when these American non-Marxist Sociologists presented papers. Sociologists consider Sociology to be a science, but it isn't a good one if they strongly hold preconceived notions. The Marxists are overly concerned about who owns the means of production, but I think it more important what is produced, and I also believe that other things still are yet more important than that. I was originally arrested and tortured at the U of S when Saskatchewan had a right wing government, then when the left wing party came to power they immediately passed legislation giving additional rights to the psychiatrists and less rights to me. Under the right wing government, I could appeal my sanity at the beginning of the six weeks of detainment, then again 3 weeks into my six week detainment, and then on the final day of the six weeks I would have drugs injected into me and this would leave me in a state of nauseous horror for months after my release. Then when the NDP Socialists came to power provincially, they passed laws allowing the psychiatrists to force medications upon me even after I am released from their torture facilities. It matters little to me if capitalists or Marxists are producing and/or forcing drugs upon me. Anyway, all these Sociologists think themselves wise, but all have zero compassion for me. Singh Bolaria and other Sociologists hear that I was repeatedly detained and tortured at the U of S, and perhaps they sometimes thinks about me when they go on their vacations. Left wing or right wing, you people are the of the earth and I am on my knees begging God to honor Exodus 20:5 and Hosea 4:6 as promises, and terminate your lives. I lost summer after summer after summer after summer after summer after summer after summer to psychiatric torture, and then subsequent summers were lost as well for I remained frantic to flee the country, and not a single person has the compassion to pay me even minimum wage for my work so that I could try to leave this country on my own dime. Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === Subject: Bipartite graph? I was just wondering if someone could this graph to verify that it is bipartite (answers/solutions aren't in the back of the textbook for all questions and I got the first one wrong that it has a solution for). I've uploaded it to www.geocities.com/bjj_freak/bipartite.jpg (might need to be copied & pasted into your browser). I chopped off e1 and the start of v of v5 in the scan. Bon === Subject: Re: Bipartite graph? >I was just wondering if someone could this graph to verify that it is >bipartite (answers/solutions aren't in the back of the textbook for all >questions and I got the first one wrong that it has a solution for). >I've uploaded it to www.geocities.com/bjj_freak/bipartite.jpg (might need to >be copied & pasted into your browser). I chopped off e1 and the start of v >of v5 in the scan. I don't see how I can give hints to this. What's a bipartite graph? It's got two sets of vertices, and all the edges go between vertices in opposite sets. So, if your graph is bipartite, you ought to be able to list the two sets of vertices. Alternatively, you could draw the graph so that there's a dotted line down the middle, and all the edges cross the dotted line. (This will force the two disjoint vertex sets to be on opposite sides of the line.) Jon Miller === Subject: Re: Bipartite graph? >I was just wondering if someone could this graph to verify that it is >bipartite (answers/solutions aren't in the back of the textbook for all >questions and I got the first one wrong that it has a solution for). >I've uploaded it to www.geocities.com/bjj_freak/bipartite.jpg (might need to >be copied & pasted into your browser). I chopped off e1 and the start of v >of v5 in the scan. It's fine. Brian === Subject: Re: Bipartite graph? [...] >So, if your graph is bipartite, you ought to be able to list the two >sets of vertices. (S)he did. [...] Brian === Subject: Re: Bipartite graph? >[...] >>So, if your graph is bipartite, you ought to be able to list the two >>sets of vertices. >(S)he did. >[...] I guess my slider bar doesn't work. Requires thinking enough to click on it and move it. I apologize to group and OP for being snippy. Jon Miller === Subject: Re: Bipartite graph? Bon > It's fine. > Brian === Subject: Books recommendations? Hi all, I have a 4-year B. Sc. degree in math. I obtained it in a non-English speaking country, but now I work in a predominantly English-speaking environment. Can anybody recommend some standard undergrad textbooks (in US) for 1) linear algebra, 2) mathematical analysis (1-d and N-d), 3) discrete mathematics and 4) probability and statistics? I am a predominantly geometrical thinker (I tend to visualize everything I read in terms of visual machinery), thus the books should be structured correspondingly. (Emphasis on geometrical presentation/rationale [think Leibnitz instead of Newton], clear layout, multicolor fonts, rich diagrams.) Excessive rigor (proof-wise) is absolutely not required.Thx, === Subject: Re: Books recommendations? Help! Standard US undergrad textbook recommendations wanted: 1) linear algebra, 2) mathematical analysis (1-d and N-d), 3) discrete mathematics and 4) probability and statistics. What do you guys use? === Subject: Re: Books recommendations? What level of Linear algebra are you lookin for? You could try Howard Anton's book Linear algebra Analysis: Principle of mathematical analysis by Walter Rudin Discrete: Kenneth Rosen's Discrete Math and its applications Prob. & Stats: Jay L. DeVore Probability and Statistics for Scientists and Engineers > Help! Standard US undergrad textbook recommendations > wanted: > 1) linear algebra, > 2) mathematical analysis (1-d and N-d), > 3) discrete mathematics and > 4) probability and statistics. > What do you guys use? === Subject: cardinality exercise This problem is rather long, so please bear with me as I try to explain it. I would really appreciate any help on this. It's not for a class, I'm just trying to work all of the problems in a book. For two sets X and Y, assume that there exist two injective functions f and g such that f:X->Y and g:Y->X. Note that the inverse functions f^{-1}:f(X)->X and g^{-1}:g(Y)->Y are both bijective. For arbitrary x in X, define the chain C_x as below: C_x = ...,f^{-1}(g^{-1}(x)),g^{-1}(x),x,f(x),g(f(x)),f(g(f(x))),... Note that the number of elements to the left of x in the chain may be zero, finite, or infinite, depending on x. Also, note that any two chains are either completely disjoint or identical. Define the sets A and B as: A = {C_x : (C_x intersection Y) is a subset of f(X)} B = {C_x : C_x is not a member of A} Exercise (2 parts): ------------------ i) Prove that for all C_x in B, there exists y in Y, such that y not in f(X) --> C_x = y,g(y),f(g(y)),g(f(g(y))),... ii) Let X_1 = A intersection X, X_2 = B intersection X, Y_1 = A intersection Y, and Y_2 = B intersection Y. Show that f:X_1->Y_1 is surjective and g:Y_2->X_2 is surjective Use this to prove that there exists a bijective function h:X->Y, and hence X~Y. === Subject: Re: cardinality exercise === Subject: cardinality exercise >This problem is rather long, so please bear with me as I try to >explain it. I would really appreciate any help on this. It's not >for a class, I'm just trying to work all of the problems in a book. Ha, as a self learner I tried the same. However you've the web while I didn't. >For two sets X and Y, assume that there exist two injective >functions f and g such that f:X->Y and g:Y->X. Note that the inverse >functions f^{-1}:f(X)->X and g^{-1}:g(Y)->Y are both bijective. This is called the Cantor-Bernstein's theorem which is quite a bit to bite off for as you see, it's famous enuf to be named by those who first proved it. You could find where this was discussed at sci.math by doing an googling the web. >For arbitrary x in X, define the chain C_x as below: > C_x = >...,f^{-1}(g^{-1}(x)),g^{-1}(x),x,f(x),g(f(x)),f(g(f(x))),... Huh? Don't understand, not liking this approach. >Note that the number of elements to the left of x in the chain may >be zero, finite, or infinite, depending on x. Also, note that any >two chains are either completely disjoint or identical. Define the >sets A and B as: A = {C_x : (C_x intersection Y) is a subset of f(X)} B = {C_x : C_x is not a member of A} Here's from my notes upon the topic. Have you see the notation: g(X) = { g(x) | x in X } ? injections f:A -> B, g:B -> A ==> A,B equinumerous let p:P(A) -> P(A), X -> X - g(B - f(X)) p ascending function complete lattice; some Z with p(Z) = Z Indeed p(X) = X - g(B - f(X)) has a fixed point or set. first prove p is ascending, ie X subset Y ==> p(X) subset p(Y) Now use theorem, an ascending function over a complete lattice has a fixed point. To prove that for the complete lattice of sets, show Z = /{ X | X subset p(X) } is fixed point of p, ie p(Z) = Z where / is great union of all the so described X's. Notice the collection of sets isn't empty as A subset p(A) and the exercise isn't dependent upon p other than being ascending over P(A). As the rest is just notes, you may want some clarification or details. Z = Z - g(B - f(Z)); A-Z = g(B - f(Z)) let h(x) = f(x) if x in Z, = g^-1(x) if x in A-Z h surjection. h(Z) = f(Z); h(A-Z) = g^-1(A-Z) = B - f(Z) h injection. If h(x) = h(y): x in Z, y in A-Z not possible if x,y in Z; f(x) = f(y); x = y if x,y in A-Z: g^-1(x) = g^-1(y); x = y Corollary surjections f:A->B, g:B->A ==> A,B equinumerous g':B -> A, x -> choose f^-1(x) is injection Choose is axiom of choice function that takes one element out of f^-1(x) = { y | y = f(x) }. g' is just a hint, to use CB thm. You'll want a f' also. Cantor's diagonal theorems no surjection f:S -> P(S). Otherwise let A = { x in S | x not in f(x) } some a in S with f(a) = A; a in A iff a not in f(a) iff a not in A surjection f:S -> A^S ==> |A| <= 1. Otherwise: let g:S -> A, s -> choose { a in A | a /= f(s)(s) } some s in S with f(s) = g; f(s)(s) = g(s) /= f(s)(s) >Exercise (2 parts): >i) Prove that for all C_x in B, there exists y in Y, such that > y not in f(X) --> C_x = y,g(y),f(g(y)),g(f(g(y))),... >ii) Let X_1 = A intersection X, X_2 = B intersection X, >Y_1 = A intersection Y, and Y_2 = B intersection Y. Show that > f:X_1->Y_1 is surjective > g:Y_2->X_2 is surjective >Use this to prove that there exists a bijective function h:X->Y, and >hence X~Y. Perhaps you'll find the fixed point method more pleasant. ---- === Subject: Re: cardinality exercise >This problem is rather long, so please bear with me as I try to >explain it. I would really appreciate any help on this. It's not for >a class, I'm just trying to work all of the problems in a book. >For two sets X and Y, assume that there exist two injective functions >f and g such that f:X->Y and g:Y->X. Note that the inverse functions >f^{-1}:f(X)->X and g^{-1}:g(Y)->Y are both bijective. For arbitrary x >in X, define the chain C_x as below: >[*] C_x = ...,f^{-1}(g^{-1}(x)),g^{-1}(x),x,f(x),g(f(x)),f(g(f(x))),... >Note that the number of elements to the left of x in the chain may be >zero, finite, or infinite, depending on x. Also, note that any two >chains are either completely disjoint or identical. Define the sets A >and B as: > A = {C_x : (C_x intersection Y) is a subset of f(X)} > B = {C_x : C_x is not a member of A} >Exercise (2 parts): >------------------ >i) Prove that for all C_x in B, there exists y in Y, such that > y not in f(X) --> C_x = y,g(y),f(g(y)),g(f(g(y))),... Start with a C_x in B. For convenience let Y_x be the intersection of C_x with Y; by definition of B, Y_x is not a subset of f(X), so there must be some y in Y_x f(X). This means that y is not f(x) for any x in X, i.e., that f^(-1)(y) is empty. On the other hand, y is in C_x. If you look at the definition of C_x, you'll see that this is possible only if y is the 'first' element of C_x in the order listed at [*] above, i.e., that C_x is precisely {y,g(y),f(g(y)),g(f(g(y))),...}. >ii) Let X_1 = A intersection X, X_2 = B intersection X, Technically this is nonsense: elements of A are sets of points, while elements of X are points, so that A intersect X is empty. What's meant is that X_1 should be {z in X : z in C_x for some C_x in A}, i.e., the intersection of X with the union of the collection A. The definitions of X_2, Y_1, and Y_2 are marred by the same sort of sloppiness. >Y_1 = A intersection Y, and Y_2 = B intersection Y. Show that > f:X_1->Y_1 is surjective This is pretty straightforward. Start with an x in X_1 and show that f(x) must be in Y_1; this is trivial, since if x is in some C_z, then f(x) is in the same C_z. Then start with a y in Y_1 and show that it's f(x) for some x in X_1; this is pretty immediate from the definition of A. > and > g:Y_2->X_2 is surjective Start with y in Y_2 and show that g(y) is in X_2; then start with x in X_2 and show that it's g(y) for some y in Y_2, for which you'll probably want part (i). >Use this to prove that there exists a bijective function h:X->Y, and >hence X~Y. Paste together f | X_1 and g^(-1) | X_2. Brian === Subject: Re: cardinality exercise > This problem is rather long, so please bear with me as I try to > explain it. I would really appreciate any help on this. It's not for > a class, I'm just trying to work all of the problems in a book. > > For two sets X and Y, assume that there exist two injective functions > f and g such that f:X->Y and g:Y->X. Note that the inverse functions > f^{-1}:f(X)->X and g^{-1}:g(Y)->Y are both bijective. For arbitrary x > in X, define the chain C_x as below: > C_x = ...,f^{-1}(g^{-1}(x)),g^{-1}(x),x,f(x),g(f(x)),f(g(f(x))),... > Note that the number of elements to the left of x in the chain may be > zero, finite, or infinite, depending on x. Also, note that any two > chains are either completely disjoint or identical. Define the sets A > and B as: > A = {C_x : (C_x intersection Y) is a subset of f(X)} > B = {C_x : C_x is not a member of A} > Exercise (2 parts): > ------------------ > i) Prove that for all C_x in B, there exists y in Y, such that > y not in f(X) --> C_x = y,g(y),f(g(y)),g(f(g(y))),... Is that question entirely correct? Let f(X) not be empty, and let C_x be any chain in B; then there exists y in Y such that y in f(X); and so the statement y not in f(X) --> C_x = y, g(y),... is trivially true for that y, since F-->T is always a true statement. Maybe instead of --> they meant and? === Subject: Re: cardinality exercise Corrections ensue. >This problem is rather long, so please bear with me as I try to >explain it. I would really appreciate any help on this. It's not >for a class, I'm just trying to work all of the problems in a book. > Ha, as a self learner I tried the same. > However you've the web while I didn't. >For two sets X and Y, assume that there exist two injective >functions f and g such that f:X->Y and g:Y->X. Note that the inverse >functions f^{-1}:f(X)->X and g^{-1}:g(Y)->Y are both bijective. > This is called the Cantor-Bernstein's theorem which is quite a bit to bite > off for as you see, it's famous enuf to be named by those who first proved > it. You could find where this was discussed at sci.math by doing an > googling the web. >For arbitrary x in X, define the chain C_x as below: > C_x = >...,f^{-1}(g^{-1}(x)),g^{-1}(x),x,f(x),g(f(x)),f(g(f(x))),... > Huh? Don't understand, not liking this approach. >Note that the number of elements to the left of x in the chain may >be zero, finite, or infinite, depending on x. Also, note that any >two chains are either completely disjoint or identical. Define the >sets A and B as: > A = {C_x : (C_x intersection Y) is a subset of f(X)} > B = {C_x : C_x is not a member of A} > Here's from my notes upon the topic. > Have you see the notation: g(X) = { g(x) | x in X } ? > injections f:A -> B, g:B -> A ==> A,B equinumerous > let p:P(A) -> P(A), X -> X - g(B - f(X)) > p ascending function complete lattice; some Z with p(Z) = Z P is the power set of A and p is a function from P(A) to P(A). > Indeed p(X) = X - g(B - f(X)) has a fixed point or set. > first prove p is ascending, ie > X subset Y ==> p(X) subset p(Y) > Now use theorem, > an ascending function over a complete lattice has a fixed point. > To prove that for the complete lattice of sets, show > Z = /{ X | X subset p(X) } > is fixed point of p, ie p(Z) = Z where / is great union of all the so > described X's. Notice the collection of sets isn't empty as A subset p(A) > and the exercise isn't dependent upon p other than being ascending over > P(A). > As the rest is just notes, you may want some clarification or details. > Z = Z - g(B - f(Z)); A-Z = g(B - f(Z)) > let h(x) = f(x) if x in Z, = g^-1(x) if x in A-Z > h surjection. h(Z) = f(Z); h(A-Z) = g^-1(A-Z) = B - f(Z) > h injection. If h(x) = h(y): x in Z, y in A-Z not possible > if x,y in Z; f(x) = f(y); x = y > if x,y in A-Z: g^-1(x) = g^-1(y); x = y > Corollary > surjections f:A->B, g:B->A ==> A,B equinumerous > g':B -> A, x -> choose f^-1(x) is injection > Choose is axiom of choice function that takes one element out of > f^-1(x) = { y | y = f(x) }. g' is just a hint, to use CB thm. > You'll want a f' also. f^-1(x) = { y | x = f(y) } perhaps better stated f^-1(y) = { x | y = f(x) } > Cantor's diagonal theorems > no surjection f:S -> P(S). Otherwise let A = { x in S | x not in f(x) } > some a in S with f(a) = A; a in A iff a not in f(a) iff a not in A > surjection f:S -> A^S ==> |A| <= 1. Otherwise: > let g:S -> A, s -> choose { a in A | a /= f(s)(s) } > some s in S with f(s) = g; f(s)(s) = g(s) /= f(s)(s) >Exercise (2 parts): >i) Prove that for all C_x in B, there exists y in Y, such that > y not in f(X) --> C_x = y,g(y),f(g(y)),g(f(g(y))),... >ii) Let X_1 = A intersection X, X_2 = B intersection X, >Y_1 = A intersection Y, and Y_2 = B intersection Y. Show that > f:X_1->Y_1 is surjective > g:Y_2->X_2 is surjective >Use this to prove that there exists a bijective function h:X->Y, and >hence X~Y. > Perhaps you'll find the fixed point method more pleasant. > ---- === Subject: Cherie Marie Peterson - February 20th 1975 P E T E R S O N 16 5 20 5 18 19 15 14 = 112 In the afternoon I went to the Broadway Roastery on 8th Street, Cherie came and sat at the table I was facing and sat pretty much as near to me as she could. She is completing her doctorate in Psychology at the U of S and sports a huge diamond engagement ring. I was more attracted to Cherie's cute friend but received a sign telling me to request stats from Cherie instead. Cherie joined me at my table and I had pleasant conversation her, and when Cherie's friend approached she did not feel comfortable enough to sit down with us, I'll have to work a little harder on my interviewing skills. 112+ Dad 22 3 49 81/284 +2889 112+ Mom 30 6 52 152/214 +1723 112+ Bro 14 4 73 104/261 5900 206 Cherie 20 2 75 51/314 6577 Cherie 48 Marie 46 Peterson 112 112+ Sis 8 2 77 39/326 7296 112+ Bro 10 2 79 41/324 8028 Mom was born on the 152nd day of 52. The kids were born on days of the month adding to 52. The first of the kids arrived on the 104th (52+52nd) day of the year. The sisters were born in years adding to 152 and also the brothers were born in years adding to 152. The first and last kids were born in years adding to 152 and the middle kids were born in years adding to 152. The family was born on days and in months and years adding to 528. Dad was born in 49 (Ephesians with 155 verses). The first two kids were born on days of the year adding to the 155 verses of Ephesians, Book 7x7. The first of the kids was born on day 104 (77th non-prime). I am 6577 days older than the second of the kids, and these first two kids are separated by 677 days. The third of the kids was born in 77. The sisters are separated by 719 days, it's the 128th (2 to the 7th) prime and is the number of verses in Bible Book 12 (7th non-prime). The family was born on days of the month adding to 104 (77th non-prime). The kids were together born 1006 days after dad's birthdays (the 77th chapter of The New Testament). Dad was born on day 81 (59th non-prime while 59 in turn is the 17th non-prime) and with 284 days remaining in the year (Second Samuel 17). The females were born on days of the month adding to 58 (the 7 primes up to 17) and in years adding to 204 (Joshua 17). The brothers were born on days of the month adding to 145, it's a combination of the 17th, 17+17th and the 17+17+17th non-primes (26, 49 and 70). The brothers were together born 440 days closer to the beginning of their years than to the end of their years, it's a combination of the first 17 primes. The kids were born on days 14, 20, 8 and 10, together these Bible Books contain 2517 verses. Cherie's first 17 letters add to 177 (3 times the 17th prime). Her common name adds to 160 (Deuteronomy 7). She was born on the 51st (17+17+17th) day of the year. Her vowels add to 59 (the 17th prime). Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 <-17 is the 7th prime 5 11 <- 11 while the primes up 6 13 to 7 add to 17 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 <-the 7th prime in --- prime position 167 Esther Book 17 <-the 7th prime The parents are separated by 166.57 weeks. The first of the kids was born on day 104 (Leviticus 14 with 57 verses), and he was born 157 days closer to the beginning of the year than to the end of the year. Cherie was born in 75 and I am 6577 days older than her. The 57's are at Genesis 41, Leviticus 14, Judges 9 and John 11, together for Cherie's 75th year of birth. Cherie was born with 314 (157+157) days remaining in the year. The little sister was born in 77, pretty as 77 plus the 77th prime (389) plus the 77th non-prime (104) adds to 570. And that little sister was born 576 days after her parent's birthdays. The family was born on days of the month adding to 104 (Leviticus 14 with 57 verses). The family was together born with 1723 days remaining in their years (the 269th prime while 269 in turn is the 57th prime), it is the 57th prime in prime position. Daniel (in part about 666) contains 357 verses while Book 66 Revelation (also in part about 666) contains 404 verses, or 57 plus the 57th prime (269) plus the 57th non-prime (78). The brothers were on average born 220 days closer to the beginning of their years than to the end of their years (Judges 9 with 57 verses), and the 57's are at chapters 41, 104, 220 and 1008, together for 1373 (the 220th prime). 389 <-77th prime 104 <-77th non-prime 77 <-77 --- 570 The Four 57's Genesis 41 -> 41 Leviticus 14 -> 104 Judges 9 -> 220 <-I dreamt of 220 roofs blown John 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th prime Chapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime <-together for 57-> Major Books of End-Times Prophecy (Daniel and Revelation are in part about 666 while Isaiah contains 66 chapters): Daniel - 357 verses Revelation - 404 verses <-57 plus the 57th prime plus the 57th non-prime Isaiah - 1292 verses <-an average of 19.575757... verses per chapter The kids were born on days of the year adding to 235, the 184th prime (1097) and the 184th non-prime (235) averages 666. The sisters were born on days of the year adding to 90 (66th non-prime), Exodus contains 1213 verses (the 66+66+66th prime) and brings the Bible up to 90 chapters (66th non-prime). Cherie is a 6 lettered name adding to a multiple of 6, her middle name adds to 46, her full name adds to 206, she is coming out of a family of 6. The females were born on days of the year adding to 242 (First Samuel 6). The brothers were on average born 220 days closer to the beginning of their years than to the end of their years, it's 66.666...% of the 6th and 66th primes (13 and 317). The last two kids are separated by 732 (666+66) days. The middle kids are together 13 days closer in age than the last two kids (the 6th prime). Dad and Cherie were born on days of the year averaging 66. Cherie was born a total of 600 days after her parent's birthdays. The first of the kids was born 23 days after dad's birthday (6th prime plus the 6th non-prime), corresponding to Isaiah with 66 chapters, see that Isaiah 4, 12 and 20 (adds to 6x6) each contain 6 verses. The kids were together born 1006 days after dad's birthdays. The males were born on days of the year adding to 226. The parents are separated by 1166 days (166.57 weeks). 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - Romans 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 66 1062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 6 1070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6) 1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime) 1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) The parents were together born 265 days closer to the beginning of their years than to the end of their years (First Samuel 29) while Cherie was born 265 days after mom's birthday (First Samuel 29). The parents were born on days 22 and 30, these Bible Books differ in length by 29 verses. Dad was born exactly 29 weeks closer to the beginning of the year than to the end of the year. The first two kids are together 42 days closer in age than the middle two kids (29th non-prime), Bible Book 29 is Joel (42). The first of the kids was born in 73 (the Lucas numbers up to 29 add to the 73 verses of Bible Book 29). The brothers were born on days of the year adding to 145 (5x29). The females were born on days of the month adding to 58 (29+29, and is the 29th non-prime in non-prime position). The last of the kids was born an average of 290 days after his parent's birthdays. Bible Book 13 (6th prime) contains 29 chapters and Bible chapter 666 contains 29 verses, pretty as 29 is 6 plus the 6th prime (13) plus the 6th non-prime (10). Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime Copper and Zinc are elements 29 and 30 (together for 59), and together they make Brass (59): B R A S S 2 18 1 19 19 = 59 Cherie seemed to like the patterns and so I told her that I she would find them at alt.dear.whitehouse and on 14 other usenet discussion groups, and that she would be used as an example for others. She thought I needed her permission to be goink and doink that, but I have absolutely zero respect for her, her family and friends, and so needed no permission whatsoever. I told Cherie of some of the problems I faced with doctors, Hindus and Christians in the community, I explained that they tortured me year after year and that I lost summer after summer to psychiatric torture, and begged desperately but to no avail for assistance from anyone to get out of the country, and so Cherie responds by asking what she could do for me. Well why don't you keep on giving the U of S thousands of dollars year after year, and why don't you and your friends go buy another diamond?!!! You filthy God-damned assholes annually and collectively spend billions of dollars on turning trees into decorated idols, you collectively spent millions detaining and torturing me for saying so, and then I meet with your kids in restaurants and show them evidence that their very names are a gift from God and they are so cheap and ignorant that they can't even offer to buy me a cookie for my labor, they are so cheap and ignorant that they don't even have the decency to send me a cheap letter in the mail expressing thanx for showing them evidence that their very names are a gift from God. There is not a person on this earth that respects me or the work I do enough to even pay me minimum wage for it, so I could at least attempt to leave this country on my own dime. Indeed, when I try to buy and sell items such as a woman's fur coat, rather than offer a fair price for any goods I have for sale, people instead mock me and put up web pages claiming that I must have stolen the goods, and since I have a mink coat I am rich and in need of no assistance from anybody. Presently you people are only wise enough to pay me reparations for torture with the lives of your children, you forget His Commandments, now may God forget your children (Exodus 20:5 and Hosea 4:6). I told Cherie that if I hear of the deaths of any of her family members then I would cheer, it would be in accordance to Scripture (Psalm 137:9) and I would post her stats again, and she seemed pretty happy with that. 206 Cherie 20 2 75 51/314 6577 Cherie 48 Marie 46 Peterson 112 186 Sarah 20 2 87 51/314 10960 Sarah 47 Auralia 63 Thomas 76 Earlier today I met Sarah Thomas, who shares the same birthday as Cherie Peterson, they were both born on the 20th and their names differ in value by 20. Their given names add together for 204. They are separated by 12 years (20 is the 12th non-prime). Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === Subject: Conditional Probability Hi: Could someone help me understand how to model the following problem with conditional probability? Suppose a box has 10 balls, 6 black and 4 white. Find the probability that all three balls removed are black if at least one of the removed balls is black. I attempted to model this problem letting B_i = a black ball is removed on the i'th removal for i = 1,2,3 The conditional probability I came up with is P( (B_1)(B_2)(B_3) | B_1 U B_2 U B_3 ) = P( (B_1)(B_2)(B_3) ) / P( B_1 U B_2 U B_3 ) Is this model right? -- Jayme === Subject: Re: Conditional Probability > Hi: > Could someone help me understand how to model the following problem with > conditional probability? > Suppose a box has 10 balls, 6 black and 4 white. Find the probability that > all three balls removed are black if at least one of the removed balls is > black. I assume that three balls are removed without replacement. Let the sample space, S, consist of all possible sets of three balls removed (where each ball is taken to be distinct from the rest). Let E be the event that all three balls removed are black, and let F be the event that at least one of them is black. Then the number of outcomes in E is n(E) = C(6,3) = 20. We also have n(S) = C(10,3) = 120. Let F' be the complement of F, i.e., F' is the event that all three balls are White. Then n(F') = C(4,3) = 4. So, n(F) = n(S) - n(F') = 116. So, the desired conditional probability is n(E)/n(F) = 20/116. > I attempted to model this problem letting > B_i = a black ball is removed on the i'th removal for i = 1,2,3 > The conditional probability I came up with is > P( (B_1)(B_2)(B_3) | B_1 U B_2 U B_3 ) = P( (B_1)(B_2)(B_3) ) / P( B_1 U > B_2 U B_3 ) > Is this model right? I don't understand your notation, but I believe that keeping track of which ball was removed first, which second, and which third is unnecessarily complicated. === Subject: Re: Conditional Probability > Could someone help me understand how to model the following problem with >conditional probability? >Suppose a box has 10 balls, 6 black and 4 white. Find the probability that >all three balls removed are black if at least one of the removed balls is >black. >I attempted to model this problem letting >B_i = a black ball is removed on the i'th removal for i = 1,2,3 >The conditional probability I came up with is >P( (B_1)(B_2)(B_3) | B_1 U B_2 U B_3 ) = P( (B_1)(B_2)(B_3) ) / P( B_1 U >B_2 U B_3 ) >Is this model right? I don't know, but it's unnecessarily complicated. If you remove three balls, of which at least one is known to be black, from a container of 6 black + 4 white, that is identical to removing two balls (about which you know nothing) from a container of 5 black + 4 white. The probability of 2 black is 5/9 * 4/8 = 5/18, and therefore that is the probability of three black in the original problem. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. I try not to fly in the face of public opinion. === Subject: Re: Conditional Probability >If you remove three balls, of which at least one is known to be >black, from a container of 6 black + 4 white, that is identical to >removing two balls (about which you know nothing) from a container >of 5 black + 4 white. No it's not identical. That would be true only if you knew _which_ ball was black. Apologies for any confusion I created. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ You find yourself amusing, Blackadder. I try not to fly in the face of public opinion. === Subject: Re: Conditional Probability >Hi: > Could someone help me understand how to model the following problem with >conditional probability? >Suppose a box has 10 balls, 6 black and 4 white. Find the probability that >all three balls removed are black if at least one of the removed balls is >black. >I attempted to model this problem letting >B_i = a black ball is removed on the i'th removal for i = 1,2,3 >The conditional probability I came up with is >P( (B_1)(B_2)(B_3) | B_1 U B_2 U B_3 ) = P( (B_1)(B_2)(B_3) ) / P( B_1 U >B_2 U B_3 ) >Is this model right? Yes. The numerator is then C(6, 3)/C(10, 3), where C(n, m) is the binomial coefficient n choose m, and the denominator is 1 - C(4, 3)/C(10, 3) = [C(10, 3) - C(4, 3)]/C(10, 3), making the desired probability C(6, 3)/[C(10, 3) - C(4, 3)] = 20/(120 - 4) = 20/116 = 5/29. Brian === Subject: constructing some infinite series, and think about how you are goint to use it? converge uniformely to some functions. All of the examples of the infinite series that I saw had very simple forms. For example, 1) s(x)=1+x^2+x^3+...+x^n+.... 2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+... When I was reading the definition of what power series are, the book said: if all of the coefficients in the infinite series are independent of x, the series is called a power series. If this is the case, how about the infinite series like this one? s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x^5)/(1^3)}+{(x ^6)/(2^4)}+... The series above has the following properties. 1) whenever x^(even numbers, like 0,2,4..),then the denominators of each term is in the form: 2^(some number). 2) whenever x^(odd numbers), then the denominators of each term is in the form: 1^(some numbers) Since all of the coefficients of the terms in the infinite series is independent of x, could I call this series a power series? If so is there any way for me to know what this power series converges to? The reason why I am asking this question is that it is relatively easy to make many kinds of infinite series, but then I don't know which infinite series I come up with is more useful than the others. I also would like to know if there is any general method to check that the infinite series that I am interested in can be converged to some function of a simple form? When I was looking at the way book explained how some of the infinite series can be changed to a simpler forms, it seems that they strongly depended on the form of the infinite series. However, if there is some general way to tackle problems of simplifying infinite series, that would help me very much. in my infinite series. I could not use some math text editor to write more neatly than that. === Subject: Re: constructing some infinite series, and think about how you are goint to use it? Infinite series can be thought of as polynomials that go on forever (no infinite zero coefficients) 3x^5+2x^7-8x^16+.... is an infinite series for example. > converge uniformely to some functions. All of the examples of the > infinite series that I saw had very simple forms. For example, > 1) s(x)=1+x^2+x^3+...+x^n+.... > 2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+... > When I was reading the definition of what power series are, the book said: > if all of the coefficients in the infinite series are independent of > x, the series is called a power series. > If this is the case, how about the infinite series like this one? s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x^5)/(1^3)}+{(x ^6)/(2^4)}+... > The series above has the following properties. > 1) whenever x^(even numbers, like 0,2,4..),then the denominators of each > term is in the form: 2^(some number). > 2) whenever x^(odd numbers), then the denominators of each term is in > the form: 1^(some numbers) > Since all of the coefficients of the terms in the infinite series is > independent of x, could I call this series a power series? If so is > there any way for me to know what this power series converges to? > The reason why I am asking this question is that it is relatively easy > to make many kinds of infinite series, but then I don't know which > infinite series I come up with is more useful than the others. I also > would like to know if there is any general method to check that the > infinite series that I am interested in can be converged to some > function of a simple form? When I was looking at the way book explained > how some of the infinite series can be changed to a simpler forms, it > seems that they strongly depended on the form of the infinite series. > However, if there is some general way to tackle problems of simplifying > infinite series, that would help me very much. > in my infinite series. I could not use some math text editor to write > more neatly than that. === Subject: Re: constructing some infinite series, and think about how you are goint to use it? Yes, s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x^5)/(1^3)}+{(x ^6)/(2^4)}+... is an infinite series s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x^5)/(1^3)}+{(x ^6)/(2^4)}+...= (1/2)+(1/1)x+(1/2^2)x^2+(1/1^2)x^3+(1/2^3)x^4+... The point is, an infinite series can be expressed as the sum and difference of an infinte number of (combined) terms of the form-- a constant times x raised to a non-negative integer. If the meaning of integer, constant and/or non-negative integer are not clear then you MUST look them up. Good luck > converge uniformely to some functions. All of the examples of the > infinite series that I saw had very simple forms. For example, > 1) s(x)=1+x^2+x^3+...+x^n+.... > 2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+... > When I was reading the definition of what power series are, the book said: > if all of the coefficients in the infinite series are independent of > x, the series is called a power series. > If this is the case, how about the infinite series like this one? s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x^5)/(1^3)}+{(x ^6)/(2^4)}+... > The series above has the following properties. > 1) whenever x^(even numbers, like 0,2,4..),then the denominators of each > term is in the form: 2^(some number). > 2) whenever x^(odd numbers), then the denominators of each term is in > the form: 1^(some numbers) > Since all of the coefficients of the terms in the infinite series is > independent of x, could I call this series a power series? If so is > there any way for me to know what this power series converges to? > The reason why I am asking this question is that it is relatively easy > to make many kinds of infinite series, but then I don't know which > infinite series I come up with is more useful than the others. I also > would like to know if there is any general method to check that the > infinite series that I am interested in can be converged to some > function of a simple form? When I was looking at the way book explained > how some of the infinite series can be changed to a simpler forms, it > seems that they strongly depended on the form of the infinite series. > However, if there is some general way to tackle problems of simplifying > infinite series, that would help me very much. > in my infinite series. I could not use some math text editor to write > more neatly than that. === Subject: Re: constructing some infinite series, and think about how you are goint to use it? > converge uniformely to some functions. All of the examples of the > infinite series that I saw had very simple forms. For example, > 1) s(x)=1+x^2+x^3+...+x^n+.... > 2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+... > When I was reading the definition of what power series are, the book said: > if all of the coefficients in the infinite series are independent of > x, the series is called a power series. > If this is the case, how about the infinite series like this one? > s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x^5)/(1^3)}+{(x ^ 6) > /(2^4)}+... > The series above has the following properties. > 1) whenever x^(even numbers, like 0,2,4..),then the denominators of each > term is in the form: 2^(some number). > 2) whenever x^(odd numbers), then the denominators of each term is in > the form: 1^(some numbers) > Since all of the coefficients of the terms in the infinite series is > independent of x, could I call this series a power series? Yes > If so is > there any way for me to know what this power series converges to? Yes. Note: (1) 1/2 + x^2/2^2 + x^4/2^3 + x^6/2^4 +... = 1/2(1 + (x^2/2)^1 + (x^2/2)^2 + (x^2/2)^3 +...) The series is geometric in x^2/2. Also (2) x + x^3 + x^5 +....= x(1 + (x^2) + (x^2)^2 +...). The series is also geometric in x^2. The two series both converge for -1 < x < 1, you know what they converge to and your series is the sum of (1) and (2). > The reason why I am asking this question is that it is relatively easy > to make many kinds of infinite series, but then I don't know which > infinite series I come up with is more useful than the others. That's backwards. You don't make up a series and then figure out what it's good for; basically, you come up with a series when all else fails. (I may get some argument about that.) > I also > would like to know if there is any general method to check that the > infinite series that I am interested in can be converged to some > function of a simple form? Simple form? Well you know series for several common functions so you could try to turn your series into one of those. As a rule, if you are lucky, ingenuity might do it. In general, it is a non-trivial problem and even though the power series may converge with non-zero radius of convergence, it doesn't converge to an elementary function. > When I was looking at the way book explained > how some of the infinite series can be changed to a simpler forms, it > seems that they strongly depended on the form of the infinite series. Sure. > However, if there is some general way to tackle problems of simplifying > infinite series, that would help me very much. I'm afraid there is none. Also, to make matters worse, not all series are power series. [...] -- Paul Sperry Columbia, SC (USA) === Subject: Re: constructing some infinite series, and think about how you are goint to use it? > When I was reading the definition of what power series are, the book said: > if all of the coefficients in the infinite series are independent of > x, the series is called a power series. > If this is the case, how about the infinite series like this one? > s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{(x^5)/(1^3)}+{(x ^ 6)/(2^4)}+... > The series above has the following properties. > 1) whenever x^(even numbers, like 0,2,4..),then the denominators of each > term is in the form: 2^(some number). > 2) whenever x^(odd numbers), then the denominators of each term is in > the form: 1^(some numbers) > Since all of the coefficients of the terms in the infinite series is > independent of x, could I call this series a power series? If so is > there any way for me to know what this power series converges to? Yes & yes. Try seperating the two parts into two series and see if each converges. > The reason why I am asking this question is that it is relatively easy > to make many kinds of infinite series, but then I don't know which > infinite series I come up with is more useful than the others. I also > would like to know if there is any general method to check that the > infinite series that I am interested in can be converged to some > function of a simple form? Wishful thinking. > When I was looking at the way book explained > how some of the infinite series can be changed to a simpler forms, it > seems that they strongly depended on the form of the infinite series. > However, if there is some general way to tackle problems of simplifying > infinite series, that would help me very much. Yup, practice, experience, insight, repeat as necessary. > in my infinite series. I could not use some math text editor to write > more neatly than that. Don't bother, often they do not show up as you intend. x + (x^2)/(2^2) + ... is sufficient x + x^2 / 2^2 + ... is loose tho possible Please use some spaces for easier reading. Compare: x^2=3y+7=27+6z x^2 = 3y+7 = 27+6z === Subject: Re: constructing some infinite series, and think about how you are goint to use it? > The reason why I am asking this question is that it is relatively easy >to make many kinds of infinite series, but then I don't know which >infinite series I come up with is more useful than the others. I also >would like to know if there is any general method to check that the >infinite series that I am interested in can be converged to some >function of a simple form? When I was looking at the way book explained >how some of the infinite series can be changed to a simpler forms, it >seems that they strongly depended on the form of the infinite series. >However, if there is some general way to tackle problems of simplifying >infinite series, that would help me very much. To rephrase the question: given a convergent series, how can you find the sum in closed form (if indeed a closed form exists)? There is no completely general answer AFAIK. There are certain tricks, e.g.: 1) Check for telescoping series, i.e. series that can be written in the form sum_n (f(n+1)-f(n)) for some function f. 2) Write the series as a power series, or as a special case of a power series, i.e. if your series is sum_n f(n) look at sum_n f(n) z^n. Then a) recognize the series for some of the standard elementary functions, or even some of the non-elementary ones (e.g. hypergeometric series cover a lot of territory) b) see if differentiation or integration will simplify the form of the series c) see if the series satisfies some differential equation that can be solved 3) If the coefficients are rational functions of n times an n'th power, use a partial fraction decomposition. 4) If the sum is numeric, evaluate it numerically to high precision and see if it is recognized by the Inverse Symbolic Calculator at 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: constructing some infinite series, and think about how you are goint to use it? >> The reason why I am asking this question is that it is relatively easy >>to make many kinds of infinite series, but then I don't know which >>infinite series I come up with is more useful than the others. I also >>would like to know if there is any general method to check that the >>infinite series that I am interested in can be converged to some >>function of a simple form? When I was looking at the way book explained >>how some of the infinite series can be changed to a simpler forms, it >>seems that they strongly depended on the form of the infinite series. >>However, if there is some general way to tackle problems of simplifying >>infinite series, that would help me very much. > To rephrase the question: given a convergent series, how can you find the > sum in closed form (if indeed a closed form exists)? > There is no completely general answer AFAIK. There are certain tricks, > e.g.: > 1) Check for telescoping series, i.e. series that can be written in the > form sum_n (f(n+1)-f(n)) for some function f. > 2) Write the series as a power series, or as a special case of a power > series, i.e. if your series is sum_n f(n) look at sum_n f(n) z^n. Then > a) recognize the series for some of the standard elementary functions, > or even some of the non-elementary ones (e.g. hypergeometric series > cover a lot of territory) > b) see if differentiation or integration will simplify the form of the series > c) see if the series satisfies some differential equation that can be > solved > 3) If the coefficients are rational functions of n times an n'th power, > use a partial fraction decomposition. > 4) If the sum is numeric, evaluate it numerically to high precision and > see if it is recognized by the Inverse Symbolic Calculator at > > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 Also, take a look at A=B at http://www.cis.upenn.edu/~wilf/AeqB.html Absolutely amazing stuff. Martin Cohen === Subject: Re: constructing some infinite series, and think about how you are goint to use it? [...] > Also, take a look at A=B at http://www.cis.upenn.edu/~wilf/AeqB.html > Absolutely amazing stuff. > Martin Cohen You can even download the whole book. === Subject: Correction: Error in NOTICE My message NOTICE: Error in Currently Taught Mathematics posted June Here's the relevant portion from the post: That definition depends on a polynomial P(x) of degree n with integer coefficients being monic so that you have P(x) = (x + a_1)...(x + a_n) but that is unbalanced as a complete ring must handle all non monic polynomials of degree n with integer coefficients to get P(x) = (a_1 x + b_1)...(a_n x + b_n) but in fact the a's and b's cannot here always be algebraic integers as I've shown in my paper Advanced Polynomial Factorization... However the paper does not show that all the a's and b's cannot be algebraic integers for that factorization as it shows a coprimeness result for a particular family of polynomials, which then can be used to show a problem in the ring as explained in multiple postings. My apologies for the error and the lateness in issuing a correction. === Subject: Re: Correction: Error in NOTICE > My message NOTICE: Error in Currently Taught Mathematics posted June > Here's the relevant portion from the post: > > That definition depends on a polynomial P(x) of degree n with integer > coefficients being monic so that you have > P(x) = (x + a_1)...(x + a_n) > but that is unbalanced as a complete ring must handle all non monic > polynomials of degree n with integer coefficients to get For the sake of those of us who have not been following in detail . . . can you define complete ring. please. Also, what does it mean for a ring to handle a polynomial? === Subject: Re: Correction: Error in NOTICE > My apologies for the error and the lateness in issuing a correction. > Please expand the apology to cover posting your error-ridden attempt at a proof in the first place. Then take it down. It is toast. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Correction: Error in NOTICE Bait 'n' switch. > My message NOTICE: Error in Currently Taught Mathematics posted June > Here's the relevant portion from the post: > > That definition depends on a polynomial P(x) of degree n with integer > coefficients being monic so that you have > P(x) = (x + a_1)...(x + a_n) > but that is unbalanced as a complete ring must handle all non monic > polynomials of degree n with integer coefficients to get > P(x) = (a_1 x + b_1)...(a_n x + b_n) > but in fact the a's and b's cannot here always be algebraic integers > as I've shown in my paper Advanced Polynomial Factorization... > > However the paper does not show that all the a's and b's cannot be > algebraic integers for that factorization as it shows a coprimeness > result for a particular family of polynomials, which then can be used > to show a problem in the ring as explained in multiple postings. I may be misreading this, but it seems as though the JSHter may be attempting to distort my correction to his mistaken claim regarding divisibility properties of the coefficients of a certain factorization the same impression as I, of the above paragraph) is that JSH takes a certain form of argument, to a manifestly incorrect conclusion (the divisibility result). I point out that his conclusion is incorrect, therefore his argument is incorrect. In the above paragraph, JSH appears (to me) to be stating that the multiple postings highlight a problem in the ring. > My apologies for the error and the lateness in issuing a correction. > Apologize all you like, but your claim to have proven that any a's are coprime to 5 in the factorization 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) is quite obviously mistaken, as I have shown on multiple occasions. Any claim that this represents a problem in the ring of algebraic integers, any attempt to return to the original family of polynomials, anyh other discussion is simply your attempt at diverting attention from your own inability. This so-called apology is, as are virtually all of your missives, thoroughly misguided and irrelevant. If you can't even answer my example showing your so-called Primary argument is incorrect, then you are admitting that your whole argument is a fraud. It's really too bad you didn't have what it took to address a mathematical argument. Perhaps that Java coding will earn you the fame you're so desperately seeking. Look what it did for Al Hirt. Dale === Subject: Re: Correction: Error in NOTICE >My message NOTICE: Error in Currently Taught Mathematics posted June No, really? ************************ David C. Ullrich === Subject: Re: Correction: Error in NOTICE > My message NOTICE: Error in Currently Taught Mathematics posted June > > Here's the relevant portion from the post: > > > That definition depends on a polynomial P(x) of degree n with integer > coefficients being monic so that you have > > P(x) = (x + a_1)...(x + a_n) > > but that is unbalanced as a complete ring must handle all non monic > polynomials of degree n with integer coefficients to get > For the sake of those of us who have not been following in detail . . . > can you define complete ring. please. > Also, what does it mean for a ring to handle a polynomial? I'll make one reply in this thread, which is this one. The definition of algebraic integers allows one to find algebraic integers a_1, a_2, and a_3, such that they are roots of a monic cubic irreducible over Q with integer coefficients, where one is provably coprime to a prime factor of the last coefficient. That coprimeness results leads inevitably to the conclusion that all must be coprime to that prime factor in the ring of algebraic integers, which is the contradictory result which proves the incompleteness of the ring. By incomplete I mean that elements that should be in the ring are not such that the contradiction arises. By handle all non-monic polynomials above I meant that the factorization for all non-monic polynomials should be in the ring. After that I asserted that my paper Advanced Polynomial Factorization proved they did not, which is the error corrected by this post, as actually I prove a coprimeness result. === Subject: Re: Correction: Error in NOTICE >My message NOTICE: Error in Currently Taught Mathematics posted June >Here's the relevant portion from the post: > >That definition depends on a polynomial P(x) of degree n with integer >coefficients being monic so that you have > P(x) = (x + a_1)...(x + a_n) >but that is unbalanced as a complete ring must handle all non monic >polynomials of degree n with integer coefficients to get >>For the sake of those of us who have not been following in detail . . . >>can you define complete ring. please. >>Also, what does it mean for a ring to handle a polynomial? > I'll make one reply in this thread, which is this one. > The definition of algebraic integers allows one to find algebraic > integers a_1, a_2, and a_3, such that they are roots of a monic cubic > irreducible over Q with integer coefficients, where one is provably > coprime to a prime factor of the last coefficient. That coprimeness > results leads inevitably to the conclusion that all must be coprime to > that prime factor in the ring of algebraic integers, which is the > contradictory result which proves the incompleteness of the ring. Not so. Your claim that one of the a's is provably coprime to a prime factor of the last coefficient is misleading. You should state that you have developed an argument that purports to prove as much. The fact of the matter is that this argument that you continue to promote is flawed. If you had a proof, using standard techniques, you might have something worth talking about. As it is, you exercise a flawed method of argument, fail to see its inadequacy, and arrive at an incorrect conclusion (your so-called contradictory result). What is amazing is your inability to recognize that the first obligation after finding results that appear to have been contradictory is *not* to attack the foundations of the field [in this case, algebraic number theory], but to validate the method used to locate the contradictions. You refuse to do this, relying instead on a well-refuted sense of intuition. > By incomplete I mean that elements that should be in the ring are > not such that the contradiction arises. Elements that should be in the ring are not such that the contradiction arises? Let's be specific. In the case 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) where the a's are defined below (following Let ...), please state *what elements* should be in the ring of algebraic integers, but are not. Let u = (63 + i*sqrt(12415))/2, ubar = (63 - i*sqrt(12415))/2, zeta = (-1 + i*sqrt(3))/2, zetabar = (-1 - i*sqrt(3))/2. Then the values a1,a2,a3 can be given as follows: x1 = -(u^(1/3) + ubar^(1/3) + 4) x2 = -(zeta*u^(1/3) + zetabar*ubar^(1/3) + 4) x3 = -(zetabar*u^(1/3) + zeta*ubar^(1/3) + 4) where the (...)^(1/3) above are the values with argument = 1/3 times the argument of (...). Note that this convention forces the three roots to be real (as they must be, for this polynomial). The values are approximately: a1 ~ -11.50930 a2 ~ 2.14375 a3 ~ -2.63445 > By handle all non-monic polynomials above I meant that the > factorization for all non-monic polynomials should be in the ring. > After that I asserted that my paper Advanced Polynomial Factorization > proved they did not, which is the error corrected by this post, as > actually I prove a coprimeness result. You seem to be back to your periodic denial that Magidin & McKinnon were incorrect in their rediscovery of the result of Cohn, McAdam, and Rush. Why not pick one side of that issue and remain on it? For your sake you might try to choose the side that has an actual proof going for it. As far as the claim of a coprimeness result, you should say that you have an argument that you claim proves such a result. I have refuted your claim of coprimeness in the case I cite above, but you are just ignoring me. That is your prerogative, of course, but ignoring the truth is not the same as proving your case. What *is* true is that your argument is flawed. Why is it flawed? Who knows, maybe your mom didn't treat you right. What is the evidence that it is flawed? It produces incorrect conclusions. Don't those conclusions really mean that mathematics itself has a problem? I mean, mathematicians are human, aren't they? Don't flatter yourself. It is overwhelmingly more likely that you have made a mistake than hundreds of thousands, if not hundreds of millions, of independent observations have made *exactly* the same error in not seeing the contradictions that you claim. Why can't I see the errors in my argument? Many people have asked the same question. The consensus of opinion is that you refuse to believe that you can be wrong. Your notion that mathematics is like some battle of wills, where perseverence makes right, and ones critics are like enemies that need to be denied endlessly, is a perversion. Why can't anyone else see the errors in my argument? Several reasons: (1) it is impenetrable, and idiosyncratic in the extreme, (2) statements in the argument are not fixed in meaning, having one interpretation at one stage and another interpretation later on (3) hand-waving (4) argument by example rather than construction (5) people *have* pointed out errors, which you deny. For the readers: I don't expect JSH to address these points, or any actual points. I'm not as dumb as I look. However, the present thread is an affront to the honesty and competence of a number of people who have participated in the periodic debunking of JSH and his style of argument [both mathematical and personal]. As such, it shouldn't stand unanswered. > Dale. === Subject: Re: Correction: Error in NOTICE > My apologies for the error and the lateness in issuing a correction. > We don't deserve your apologies. After all we're only humans. Jamis Harres === Subject: Re: Correction: Error in NOTICE Visiting Assistant Professor at the University of Montana. [.snip.] >You seem to be back to your periodic denial that Magidin & McKinnon were >incorrect in their rediscovery of the result of Cohn, McAdam, and Rush. Too many negations! You mean, You seem to be going back to your periodic CLAIM that [...] were incorrect in [...] or do you mean You seem to be going back to your periodic denial that [...] were CORRECT in [...] ? Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Correction: Error in NOTICE > [.snip.] >>You seem to be back to your periodic denial that Magidin & McKinnon were >>incorrect in their rediscovery of the result of Cohn, McAdam, and Rush. > Too many negations! > You mean, > You seem to be going back to your periodic CLAIM that [...] were > incorrect in [...] > or do you mean > You seem to be going back to your periodic denial that [...] were > CORRECT in [...] I had meant his periodic denial of your re-discovery, et cetera. Still working on that proofreading business, Dale > Why do you take so much trouble to expose such a reasoner as > Mr. Smith? I answer as a deceased friend of mine used to answer > on like occasions - A man's capacity is no measure of his power > to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more; > and a long purse, which does most of all. He has made at least > ten publications, full of figures few readers can critize. A great > many people are staggered to this extend, that they imagine there > must be the indefinite something in the mysterious all this. > They are brought to the point of suspicion that the mathematicians > ought not to treat all this with such undisguised contempt, > at least. > -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan > Arturo Magidin > magidin@math.berkeley.edu === Subject: Re: Correction: Error in NOTICE > My message NOTICE: Error in Currently Taught Mathematics posted June > Here's the relevant portion from the post: > > That definition depends on a polynomial P(x) of degree n with integer > coefficients being monic so that you have > P(x) = (x + a_1)...(x + a_n) > but that is unbalanced as a complete ring must handle all non monic > polynomials of degree n with integer coefficients to get > P(x) = (a_1 x + b_1)...(a_n x + b_n) > but in fact the a's and b's cannot here always be algebraic integers > as I've shown in my paper Advanced Polynomial Factorization... > > However the paper does not show that all the a's and b's cannot be > algebraic integers for that factorization as it shows a coprimeness > result for a particular family of polynomials, which then can be used > to show a problem in the ring as explained in multiple postings. > My apologies for the error and the lateness in issuing a correction. > This is not good enough, for three reasons: 1. As an apology it is lame. You have called me and others here liars for saying the conclusions of Advanced Polynomial Factorization are wrong. You owe us a direct personal apology. I am not a liar. 2. Your APF web page has not changed. You are thus flouting your own apology and admission of error. People who continue to access that site are going to think either that it is correct, and thus learn some erroneous math (admittedly the chances of this are slim), or that you are too proud or stupid to retract it even after you have admitted it is wrong. 3. You persist in claiming the following: The definition of algebraic integers allows one to find algebraic integers a_1, a_2, and a_3, such that they are roots of a monic cubic irreducible over Q with integer coefficients, where one is provably coprime to a prime by you today in this same thread] Let G(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are integers and c = p*v, where p is a prime and v is another integer. G(x) is clearly monic. Assume G(x) is irreducible over the rationals. Let u1, u2, and u3 be roots of G(x). Note that by definition, u1, u2, and u3 are algebraic integers. You are claiming that at least one of u1, u2, or u3 is coprime to p. Assume u1 is coprime to p. By standard theory***, there exists an automorphism F12 of the field of algebraic numbers such that: 1. F12 leaves the subfield of rational numbers fixed, i.e., if q is rational, F12(q) = q. 2. F12(u1) = u2. 3. If t is an algebraic integer, F12(t) is also an algebraic integer. Now since u1 is relatively prime to p, there exist algebraic integers r and s such that [1] r*u1 + s*p = 1. Now apply the automorphism F12 to both sides of [1]: F12(r)*F12(u1) + F12(s)*F12(p) = F12(1). By the properties above, F12(p) = p and F12(1) = 1. Moreover, r' = F12(r) is an algebraic integer, and s' = F12(s) is an algebraic integer. Finally, F12(u1) = u2. Thus one obtains: r'*u2 + s'*p = 1, which says: u2 and p are coprime in the algebraic integers. Similarly one shows that u3 and p are coprime. Therefore if one of u1, u2, or u3 is coprime to p, then they all are. But u1 * u2 * u3 = - p * v. That is, p divides the product of u1, u2, and u3. Therefore p cannot be coprime to each of u1, u2, and u3. Putting all this together, one concludes that NONE of u1, u2, or u3 can be coprime to p. This directly contradicts your statement which was quoted above. Please feel free to point out any errors in the proof I just gave. You need to do more than just retract your June 10 NOTICE. You need to retract all of Advanced Polynomial Factorization. There is a lot more wrong with it than just a few misprints and algebraic slips. The whole underlying idea is a crock. I have pointed out previously where your logic goes astray but you have given no hint of having understood it. Worse yet: Consider the following snippet from your web page purporting to prove Fermat's Last Theorem: ... exactly two of the a's must have a factor of f^j. That is, you need exactly the same kind of conclusion that you had for APF in your FLT argument. And you absolutely do not have it. You no more have a proof for your ill-defined objects than you do for algebraic integers. But look on the bright side. You get a lot more free space on your website. Nora B. ***: For discussion of such automorphisms, see: http://www.math.niu.edu/~beachy/aaol/galois.html, especially Prop. 8.6.2. Or see the excellent textbook Abstract Algebra by John Beachy and William D. Blair. === Subject: Re: Correction: Error in NOTICE > I'll make one reply in this thread, which is this one. > The definition of algebraic integers allows one to find algebraic > integers a_1, a_2, and a_3, such that they are roots of a monic cubic > irreducible over Q with integer coefficients, where one is provably > coprime to a prime factor of the last coefficient. That coprimeness > results leads inevitably to the conclusion that all must be coprime to > that prime factor in the ring of algebraic integers, which is the > contradictory result which proves the incompleteness of the ring. > By incomplete I mean that elements that should be in the ring are > not such that the contradiction arises. > By handle all non-monic polynomials above I meant that the > factorization for all non-monic polynomials should be in the ring. > After that I asserted that my paper Advanced Polynomial Factorization > proved they did not, which is the error corrected by this post, as > actually I prove a coprimeness result. > Give up, . You are wrong! You have been proven wrong repeatedly, and spawning new threads won't cover your tracks. Go back to your sandbox. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Correction: Error in NOTICE David C. Ullrich >My message NOTICE: Error in Currently Taught Mathematics posted June > No, really? As soon as I saw the subject line, I knew this was a JSH thread :) Larry === Subject: Re: Correction: Error in NOTICE >You seem to be back to your periodic denial that Magidin & McKinnon were >incorrect in their rediscovery of the result of Cohn, McAdam, and Rush. >> Too many negations! >> You mean, >> You seem to be going back to your periodic CLAIM that [...] were >> incorrect in [...] >> or do you mean >> You seem to be going back to your periodic denial that [...] were >> CORRECT in [...] >> ? >Darn it all, anyhow. I outworded myself. I couldn't fail to disagree with you less! === Subject: Re: Correction: Error in NOTICE >You seem to be back to your periodic denial that Magidin & McKinnon were >incorrect in their rediscovery of the result of Cohn, McAdam, and Rush. >> >> Too many negations! >> You mean, >> You seem to be going back to your periodic CLAIM that [...] were >> incorrect in [...] >> or do you mean >> You seem to be going back to your periodic denial that [...] were >> CORRECT in [...] >> ? >Darn it all, anyhow. I outworded myself. > I couldn't fail to disagree with you less! Is the following statement about you true or false: You are not the kind of person who wouldn't oppose the idea of not taking a negative stance against those who do not fail to protest the opposition to legislation that would legalize child pornography? from Scott Campisi, Wake Village, Tex. Contest: A bogus question you would like to sneak into the interview portion of the Miss Universe Pageant) === Subject: cross product I don't understand the proof in my textbook that A, B and AxB form a right-handed triad. Little help? -- Peace, EJ === Subject: Re: cross product , > I don't understand the proof in my textbook that A, B and AxB form a > right-handed triad. Little help? > -- > Peace, > EJ Assuming that you have a list of three linearly independent vectors in a real 3-space, do you know how to tell whether they form a right handed triad or a left handed triad? === Subject: Re: cross product >I don't understand the proof in my textbook that A, B and AxB form a >right-handed triad. Little help? You realize, I hope, that we don't know what textbook you have, so we don't know exactly what proof you're referring to. Can you give us more details of what the proof is, and what you don't understand about it? 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: cross product > I don't understand the proof in my textbook that A, B and AxB form a > right-handed triad. Little help? > -- > Peace, > EJ My understanding of a right-handed triad is that if you use your right thumb, index and middle fingers to represent the directions of A, B and AxB respectively, it tells you in which direction AxB is. So A = thumb, B = index finger (make an L with these fingers), and AxB is you middle finger (bent so it sticks out from your palm). So AxB is perpendicular to the plane that the vectors A, B are in. (ie: AxB is perpendicular to both A and B) Rick === Subject: Re: cross product >I don't understand the proof in my textbook that A, B and AxB form a >right-handed triad. Little help? There is another version, IIRC, which says point thumb in the direction of A, second finger in the direction of B, and then the middle finger, positioned perpendicular to both these, will show you the orientation of A x B. If you are not sure of what sense that makes, try it with both hands and visualise the difference. === Subject: Re: cross product > I don't understand the proof in my textbook that A, B and AxB form a > right-handed triad. Little help? > -- > Peace, > EJ Is it a proof ? I mean, isn't the right-hand rule just the orientation-definition of the cross product. Sounds like one of those stupid prove axiom by using the axiom tasks, e.g Prove that 1+2=2+1. /TV === Subject: Re: cross product I don't know the proof. But I do know that two vectors are perpendicular if their dot product is zero. You presumably know the formula for the crossproduct a = b x c and the formula for the dot product x=a dot b you could just substitute and confirm (b x c) dot b = a dot b = 0 by substitution. > I don't understand the proof in my textbook that A, B and AxB form a > right-handed triad. Little help? > -- > Peace, > EJ === Subject: Re: cross product > My understanding of a right-handed triad is that if you use your right > thumb, index and middle fingers to represent the directions of A, B and AxB > respectively, it tells you in which direction AxB is. > So A = thumb, B = index finger (make an L with these fingers), and AxB is > you middle finger (bent so it sticks out from your palm). > So AxB is perpendicular to the plane that the vectors A, B are in. (ie: AxB > is perpendicular to both A and B) It amounts to the same thing, but my text has it like this: Place your right hand so that your fingers point in the direction of A, and when you bend them they rotate toward the direction of B. Then your thumb will point in the direction of AxB. That's the right-hand rule. === Subject: Re: cross product I know what it means (geometrically), but I don't know how to tell (algebraically) whether it's true or not. | , | | > I don't understand the proof in my textbook that A, B and AxB form a | > right-handed triad. Little help? | > -- | > Peace, | > EJ | | Assuming that you have a list of three linearly independent | vectors in a real 3-space, do you know how to tell whether | they form a right handed triad or a left handed triad? === Subject: Re: cross product The textbook I'm talking about is Calculus of Several Variables by Robert A. Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) cross product UxV is the vector determined by the following (geometric) conditions: 1) perpendicular to U and to V 2) length is |U|*|V|*|sin(x)| 3) U,V,(UxV) form a right-handed triad The first two parts are amenable to simple algebraic verification. Those parts are clear. But when he gets to part (3) he lapses into hand-waving and loses me completely. | >I don't understand the proof in my textbook that A, B and AxB form a | >right-handed triad. Little help? | | You realize, I hope, that we don't know what textbook you have, so | we don't know exactly what proof you're referring to. Can you give | us more details of what the proof is, and what you don't understand | about it? | | 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: cross product > I know what it means (geometrically), but I don't know how to tell > (algebraically) whether it's true or not. Take the three vectors in order, make them the rows of a determinant. If the determinant is positive, it is a right-handed system. If negative, left-handed. If zero, they three vectors are co-planar, so right or left is not defined. === Subject: Re: cross product Elaine says... >The textbook I'm talking about is Calculus of Several Variables by Robert A. >Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) cross >product UxV is the vector determined by the following (geometric) conditions: >1) perpendicular to U and to V >2) length is |U|*|V|*|sin(x)| >3) U,V,(UxV) form a right-handed triad >The first two parts are amenable to simple algebraic verification. Those parts >are clear. But when he gets to part (3) he lapses into hand-waving and loses me >completely. Here's one definition of a right-handed triad: vectors A, B, and C form a right-handed triad if the triple product A . (B x C) is positive, where . is the scalar product. So compute U . (V x (U x V)). It's pretty easy to see that it is positive if you use the permutation rule A . (B x C) = C . (A x B) -- Daryl McCullough === Subject: Re: cross product > The textbook I'm talking about is Calculus of Several Variables by Robert A. > Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) cross > product UxV is the vector determined by the following (geometric) conditions: > 1) perpendicular to U and to V > 2) length is |U|*|V|*|sin(x)| > 3) U,V,(UxV) form a right-handed triad > The first two parts are amenable to simple algebraic verification. Those parts > are clear. But when he gets to part (3) he lapses into hand-waving and loses me > completely. I have two comments: (1) For the geometric picture, you know (1) and (2), so you've identified (from 1) a line in 3-space, together with (from 2) a magnitude. To get a unique vector, since there are at this stage TWO possibilities, all you need is to establish which of the two values is the correct one; that is, all you care about at this point is a direction along that line. The right-handed triad jazz is all about taking a [standard, right-handed] screw, aligning it along that line, and rotating it in the direction from U towards V The direction that the screw would move if you were actually screwing [pardon the unintentional salacious allusion], gives the direction along that line, to the correct value of UxV. You could also imagine a (again, right-hand) threaded rod, aligned with that line, and a nut that you place on the rod at some point. As you rotate the nut from U towards V, it will move in the direction you want to choose. (2) The real proof is most likely a hand-wave because it relies on the right-handedness of your coordinate system; that is (using the i,j,k convention for unit vectors in 3-space) one has this: i x j = k. If the above screw analogy applies to this triple, then your coordinate system is right-handed, and your cross-product yields a right-handed triad. What the cross-product *does* do is to yield a triple with the same handedness as the basic triple i,j,k. > | >I don't understand the proof in my textbook that A, B and AxB form a > | >right-handed triad. Little help? > | You realize, I hope, that we don't know what textbook you have, so > | we don't know exactly what proof you're referring to. Can you give > | us more details of what the proof is, and what you don't understand > | about it? > | Robert Israel israel@math.ubc.ca > | Department of Mathematics http://www.math.ubc.ca/~israel > | University of British Columbia > | Vancouver, BC, Canada V6T 1Z2 Dale. === Subject: Re: cross product >The textbook I'm talking about is Calculus of Several Variables by Robert A. >Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) cross >product UxV is the vector determined by the following (geometric) conditions: >1) perpendicular to U and to V >2) length is |U|*|V|*|sin(x)| >3) U,V,(UxV) form a right-handed triad >The first two parts are amenable to simple algebraic verification. Those parts >are clear. But when he gets to part (3) he lapses into hand-waving In this case that should be OK as long as it's the right hand he's waving... Sorry about that. > and loses me >completely. For the case U=i, V=j, UxV=k you can see it's right-handed. Now imagine starting there and moving U and V into some other positions by a continuous motion, being careful to avoid U and V ever being parallel. As you do so, UxV also moves continuously (because its components are continuous functions of the components of U and V). If your right hand's thumb points in the direction of U and index finger in the direction of V, UxV will either be in the direction of your palm or the back of your hand. But just as UxV moves continuously, so does your hand (the front and back won't suddenly switch). So since it starts out with the palm in the direction of UxV, it will end that way too. Hope this helps. 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: cross product >> I know what it means (geometrically), but I don't know how to tell >> (algebraically) whether it's true or not. > Take the three vectors in order, make them the rows of a determinant. > If the determinant is positive, it is a right-handed system. If > negative, left-handed. If zero, they three vectors are co-planar, so > right or left is not defined. The nice thing about this approach is that it generalizes immediately to n dimensions (unless I am forgetting something obvious). fourierr at fastermail dot com -- === Subject: Re: cross product >> I know what it means (geometrically), but I don't know how to tell >> (algebraically) whether it's true or not. > Take the three vectors in order, make them the rows of a determinant. > If the determinant is positive, it is a right-handed system. If > negative, left-handed. If zero, they three vectors are co-planar, so > right or left is not defined. > The nice thing about this approach is that it generalizes immediately > to n dimensions (unless I am forgetting something obvious). > fourierr at fastermail dot com > -- Iirc, only in dimensions 1, 3 and 7 do cross products exist. GC === Subject: Re: cross product > , Elaine >> I know what it means (geometrically), but I don't know how to tell >> (algebraically) whether it's true or not. > Take the three vectors in order, make them the rows of a > determinant. > If the determinant is positive, it is a right-handed system. If > negative, left-handed. If zero, they three vectors are > co-planar, so > right or left is not defined. >> The nice thing about this approach is that it generalizes >> immediately >> to n dimensions (unless I am forgetting something obvious). >> fourierr at fastermail dot com >Iirc, only in dimensions 1, 3 and 7 do cross products exist. Niel doesn't explicitly mention cross-product. However Spivak does and says that the definition of cross-product is sometimes given only on R^3. I suspect that we are talking about two different things. The following is from Spivak's _Calculus on Manifolds_ : Given v_1, v_2,..., v_(n-1) vectors in R^n we may define a function from R^n to R by f(w) = det(v_1,...,v_(n-1),w). This is multi-linear, alternating function on R^n and therefore is given by for some z in R^n. In this case define the n-dimensional cross-product of n-1 vectors in R^n by v_1 x v_2 x ... x v_(n-1) = z. Notice that the cross-product of n-1 vectors is defined rather than the cross-product of two. Spivak uses Niel's method rather than the cross-product(even after the trouble to define it). The method is used to define an orientation for any basis of R^n. In R^3 bases which share the orientation of {(1,0,0),(0,1,0),(0,0,1)} are said to satisfy the right-hand rule. fourierr at fastermail dot com -- === Subject: Re: cross product >The textbook I'm talking about is Calculus of Several Variables by Robert A. >Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) cross >product UxV is the vector determined by the following (geometric) conditions: >1) perpendicular to U and to V >2) length is |U|*|V|*|sin(x)| >3) U,V,(UxV) form a right-handed triad Just out of interest, which edition of Calculus of Several Variables is that? I have both the first and fourth editions, and neither has a Section 10.3; the fourth edition of Calculus: A Complete Course by Adams does talk about cross products in Section 10.3, but it has conditions 1), 2) and 3) as the definition of cross product and Theorem 2 as giving the algebraic formula. 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: cross product >> , Elaine >> > I know what it means (geometrically), but I don't know > how to tell > (algebraically) whether it's true or not. >> >> Take the three vectors in order, make them the rows of a >> determinant. >> If the determinant is positive, it is a right-handed system. If >> negative, left-handed. If zero, they three vectors are >> co-planar, so >> right or left is not defined. > The nice thing about this approach is that it generalizes > immediately > to n dimensions (unless I am forgetting something obvious). > fourierr at fastermail dot com >>Iirc, only in dimensions 1, 3 and 7 do cross products exist. >>GC > Niel doesn't explicitly mention cross-product. However > Spivak does and says that the definition of cross-product is > sometimes given only on R^3. I suspect that we are > talking about two different things. The following is from > Spivak's _Calculus on Manifolds_ : Given v_1, v_2,..., v_(n-1) > vectors in R^n we may define a function from R^n to R by > f(w) = det(v_1,...,v_(n-1),w). This is multi-linear, alternating Typo: That should read linear rather than multi-linear. > function on R^n and therefore is given by for some z in > R^n. In this case define the n-dimensional cross-product of n-1 > vectors in R^n by v_1 x v_2 x ... x v_(n-1) = z. Notice that > the cross-product of n-1 vectors is defined rather than the > cross-product of two. > Spivak uses Niel's method rather than the cross-product(even > after the trouble to define it). The method is used to define > an orientation for any basis of R^n. In R^3 bases which Another typo: That should be ordered basis rather than just basis. > share the orientation of {(1,0,0),(0,1,0),(0,0,1)} are said > to satisfy the right-hand rule. To clarify and match up with what Niel said: Given 3 vectors put them in order into a matrix and take the determinant. The ones with positive determinant all share one orientation and the ones with negative determinant share the opposite orientation. My only point was that this approach immediately generalizes to n dimensions-simply replace 3 by n. The ordered basis consisting of {(1,0,...,0),...,(0,...,0,1)} is said to have the usual orientation. The other orientation is the negative or opposite of the usual orientation. > fourierr at fastermail dot com fourierr at fastermail dot com -- === Subject: Re: cross product It helps a little. In fact I thought about that proof, but it's too informal for my taste. I also thought about showing that a matrix M that rotates things around in 3-space preserves cross-products (ie: M(VxW)=MVxMW...I'm surprised nobody suggested this). That's much more formal, but it seems to have more moving parts than what's warranted. What I'd really like is some way of translating the condition right-handed triad into an algebraic condition on the coordinates of the vectors in question. | >The textbook I'm talking about is Calculus of Several Variables by Robert A. | >Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) cross | >product UxV is the vector determined by the following (geometric) conditions: | | >1) perpendicular to U and to V | >2) length is |U|*|V|*|sin(x)| | >3) U,V,(UxV) form a right-handed triad | | >The first two parts are amenable to simple algebraic verification. Those parts | >are clear. But when he gets to part (3) he lapses into hand-waving | | In this case that should be OK as long as it's the right hand he's | waving... Sorry about that. | | > and loses me | >completely. | | For the case U=i, V=j, UxV=k you can see it's right-handed. Now imagine | starting there and moving U and V into some other positions by a | continuous motion, being careful to avoid U and V ever being parallel. | As you do so, UxV also moves continuously (because its components | are continuous functions of the components of U and V). If your | right hand's thumb points in the direction of U and index finger in | the direction of V, UxV will either be in the direction of your | palm or the back of your hand. But just as UxV moves continuously, | so does your hand (the front and back won't suddenly switch). So since | it starts out with the palm in the direction of UxV, it will end that | way too. | | Hope this helps. | | 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: cross product What I'd really like is some way of > translating the condition right-handed triad into an algebraic condition on > the coordinates of the vectors in question. Maybe what you're looking for is this: vectors {a,b,c}, {d,e,f}, and {g,h,i} form a right-handed triad if the determinant of [a b c] [d e f] [g h i] is positive. Well, this still has 9 parts which could unexpectedly move. I hope that's not too many. I once had an interesting discussion with the science fiction writer Vernor Vinge, who is (I opine) the most brilliant of all science fiction writers. I asserted that if we could exchange streams of zeros and ones with an intelligent culture on a distant star, then we could eventually communicate everything about ourselves except handedness. Vernor disagreed, pointing out that the electromatic waves that carry the zeros and ones have a built-in handedness. It is with some trepidation that I mention anything philosophical in this den of crackpots. === Subject: Re: cross product What I'm looking for is some kind of proof that the cross product as you define it has, in every case, the same geometric relationship to the vectors whose cross product it is. | Elaine says... | >The textbook I'm talking about is Calculus of Several Variables by Robert A. | >Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) cross | >product UxV is the vector determined by the following (geometric) conditions: | >1) perpendicular to U and to V | >2) length is |U|*|V|*|sin(x)| | >3) U,V,(UxV) form a right-handed triad | >The first two parts are amenable to simple algebraic verification. Those parts | >are clear. But when he gets to part (3) he lapses into hand-waving and loses me | >completely. | | Here's one definition of a right-handed triad: vectors A, B, and C | form a right-handed triad if the triple product A . (B x C) is positive, | where . is the scalar product. | | So compute U . (V x (U x V)). It's pretty easy to see that it is positive | if you use the permutation rule | | A . (B x C) = C . (A x B) | | -- | Daryl McCullough | === Subject: Re: cross product > What I'm looking for is some kind of proof that the cross product as you define > it has, in every case, the same geometric relationship to the vectors whose > cross product it is. I don't quite follow same geometric relationship. Maybe we could do it this way: you state a theorem, and if it's true, somebody in the ng will prove it for you. And for that matter, if it's false, someone will undoubtedly provide a counterexample. The only way I can make sense of same geometric relationship is the theorem that the cross product is invariant under orthogonal transformations of R^3, but I seem to recall that you weren't happy with that. === Subject: Re: cross product Will says... >I once had an interesting discussion with the science fiction writer >Vernor Vinge, who is (I opine) the most brilliant of all science >fiction writers. I asserted that if we could exchange streams of >zeros and ones with an intelligent culture on a distant star, then we >could eventually communicate everything about ourselves except >handedness. Vernor disagreed, pointing out that the electromatic >waves that carry the zeros and ones have a built-in handedness. No, they really don't. He might have been thinking about the right-handed rule for electromagnetic waves, which says that for an electromagnetic wave travelling in direction K, the following form a right-handed triple (or maybe it's left-handed, I don't remember which) E, B, K where E is the electric field, and B is the magnetic field. However, the direction of the B field is not actually measurable; the most you can measure about the B field is: (1) the plane that is perpendicular to B and (2) a sense of circulation about that plane. That sense of circulation is the direction (clockwise or deflected if it is travelling in the plane perpendicular to B. by the magnetic field at all.) Using the right-hand rule, you can compute a direction for B from the plane and the sense of circulation, but without such a rule, the direction of B is not specified. So you can't use E,B,K to compute handedness unless you already have a built-in handedness to compute the right-hand rule. -- Daryl McCullough Ithaca, NY === Subject: Re: cross product > Will says... >I once had an interesting discussion with the science fiction writer >Vernor Vinge, who is (I opine) the most brilliant of all science >fiction writers. I asserted that if we could exchange streams of >zeros and ones with an intelligent culture on a distant star, then we >could eventually communicate everything about ourselves except >handedness. Vernor disagreed, pointing out that the electromatic >waves that carry the zeros and ones have a built-in handedness. > No, they really don't. He might have been thinking about the right-handed > rule for electromagnetic waves, which says that for an electromagnetic > wave travelling in direction K, the following form a right-handed triple > (or maybe it's left-handed, I don't remember which) > E, B, K > where E is the electric field, and B is the magnetic field. > However, the direction of the B field is not actually measurable; But you _can_ communicate handedness with electromagnetic waves by manipulating their polarization: slowly rotate the direction of the E field. 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: cross product >Is it a proof ? I mean, isn't the right-hand rule just the >orientation-definition of the cross product. Sounds like one of >those stupid prove axiom by using the axiom tasks, e.g Prove that >1+2=2+1. But that has a proof (assuming associativity of addition): 1+2=1+(1+1) ... definition of 2 =(1+1)+1 ... associativity =2+1 ... definition of 2 === Subject: Re: cross product >Is it a proof ? I mean, isn't the right-hand rule just the >orientation-definition of the cross product. Sounds like one of >those stupid prove axiom by using the axiom tasks, e.g Prove that >1+2=2+1. > But that has a proof (assuming associativity of addition): > 1+2=1+(1+1) ... definition of 2 > =(1+1)+1 ... associativity > =2+1 ... definition of 2 Are these types of proofs really considered as _proofs_ ? I mean, e.g. Task : Prove that (a+b)+c=a+(b+c). (which is the associativity axiom for addition) By axiom (a+b)+c=a+(b+c) , QED ? I would like to state the task as Show that (a+b)+c=a+(b+c). === Subject: Re: cross product >>Is it a proof ? I mean, isn't the right-hand rule just the >>orientation-definition of the cross product. Sounds like one of >>those stupid prove axiom by using the axiom tasks, e.g Prove that >>1+2=2+1. >> But that has a proof (assuming associativity of addition): >> 1+2=1+(1+1) ... definition of 2 >> =(1+1)+1 ... associativity >> =2+1 ... definition of 2 >Are these types of proofs really considered as _proofs_ ? >I mean, e.g. >Task : Prove that (a+b)+c=a+(b+c). (which is the associativity axiom for >addition) >By axiom (a+b)+c=a+(b+c) , QED ? >I would like to state the task as Show that (a+b)+c=a+(b+c). Where in the above proof do you see commutativity used to prove commutativity? Where do you see the result asserted as an axiom? Two axioms are invoked. One is that 2 = 1+1, the other is that addition is associative. Neither one is the thing being proven. - Randy === Subject: Re: cross product <5o9gfv8hn95ma2h4v7dc9n19elfmnon13k@4ax.com> >>Is it a proof ? I mean, isn't the right-hand rule just the >>orientation-definition of the cross product. Sounds like one of >>those stupid prove axiom by using the axiom tasks, e.g Prove that >>1+2=2+1. >> >> But that has a proof (assuming associativity of addition): >> 1+2=1+(1+1) ... definition of 2 >> =(1+1)+1 ... associativity >> =2+1 ... definition of 2 >> >> >Are these types of proofs really considered as _proofs_ ? >I mean, e.g. >Task : Prove that (a+b)+c=a+(b+c). (which is the associativity axiom for >addition) >By axiom (a+b)+c=a+(b+c) , QED ? >I would like to state the task as Show that (a+b)+c=a+(b+c). > Where in the above proof do you see commutativity used to prove > commutativity? Where do you see the result asserted as an axiom? > Two axioms are invoked. One is that 2 = 1+1, the other is that > addition is associative. Neither one is the thing being proven. My point was (originally) that to prove 1+2=2+1 you simply need to state by the commutative axiom of addition x+y=y+x => 1+2=2+1. In some cases if seems so stupid to prove a property/relation for a formula/theorem that is build up, constructed and based on that very property/relation that should be proven. start to wonder why commutativity of addition is an axiom ? (Eventhough it's a very obvious relation). I thought the point of having axioms is that in order to prove them they prove themself, not that you setup a new set of axioms based on other, older (more fundamental) axioms. What I remember, from class, was that the fundamental axioms could differ quite alot from book to book. TV === Subject: Re: cross product > , > I don't understand the proof in my textbook that A, B and AxB form a > right-handed triad. Little help? > -- > > Peace, > EJ > > > Assuming that you have a list of three linearly independent > vectors in a real 3-space, do you know how to tell whether > they form a right handed triad or a left handed triad? My opinion is the following: it's better to teach, in this order, following notions : dot (scalar) product , cross product and then vector-product. Consider three (free) vectors A=(a_1,a_2,a_3), B=(b_1,b_2,b_3) , C=(c_1,c2,c_3)=c_1*i+c_2*j+c_3*k where i=(1,0,0) ,j=(0,1,0) , k=(0,0,1). 1) The dot (scalar) product A*B=(A,B):=a_1*b_1+a_2*b_2+a_3*b_3:= ||A||.||B||cos(phi) where phi is the angle ,from [0,pi], between vectors A and B. 2) The cross-product (A,B,C) is by definition |a_1 a_2 a_3| (A,B,C) := |b_1 b_2 b_3| |c_1 c_2 c_3| DEFINITION : the vectors A,B,C form a right handed triad, if and only if (A,B,C) > 0 . 3) The (vector) -product A x B , A=/=0,B=/=0, is defined to be the unique vector D with -the norm ||D||=||A||.||B||sin(phi) -the direction of D is such that A*D=0 , B*D=0, that is D is perpendicular on the plane determined by suports of A and B . -the sense of D is such that (A,B,D) >= 0 . Moreover, when A=0 or B=0 the one define A x B = 0 . |a_2 a_3| |a_3 a_1| |a_1 a_2| A x B= | |*i + | |*j + | |*k = formaly= |b_2 b_3| |b_3 b_1| |b_1 b_2| |i j k | = |a_1 a_2 a_3| |b_1 b_2 b_3| THEOREM 1 . (A,B,C)=A*(B x C) . Proof. According to above formulas it's very easy. THEOREM 2 . (A,B,C)=(B,C,A)=(C,A,B)=-(A,C,B)=-(B,A,C)=-(C,B,A) . Proof. Nothing to prove taking into account properties of determinants. THEOREM 3. Suppose that the non-zero vectors A,B does not have the same direction ( B=/=k*A , k in R). Then A, B, AxB (in this order) form a right handed triad. Proof. Use above formulas,or theorem 2, and you find that (A,B,A xB)=(AxB,A,B)=(AxB)*(AxB)=||AxB||^2 >0 . THEOREM 4. A,B,C with (A,B,C)=/=0 is a vectorial basis, namely any vector X may be written as X=a*A+b*B+c*C , a,b,c in R .More precisely a=(X,B,C)/(A,B,C) , b=(A,X,C)/(A,B,C) , c=(A,B,X)/(A,B,C) . It's possible to use also the terminology: Definition. A triplet A,B,C is a positive basis iff (A,B,C) > 0. A,B,C form a negative basis when (A,B,C) < 0 . Alex === Subject: Re: cross product >Is it a proof ? I mean, isn't the right-hand rule just the >orientation-definition of the cross product. Sounds like one of >those stupid prove axiom by using the axiom tasks, e.g Prove that >1+2=2+1. > But that has a proof (assuming associativity of addition): > 1+2=1+(1+1) ... definition of 2 > =(1+1)+1 ... associativity > =2+1 ... definition of 2 > Are these types of proofs really considered as _proofs_ ? > I mean, e.g. > Task : Prove that (a+b)+c=a+(b+c). (which is the associativity axiom for > addition) > By axiom (a+b)+c=a+(b+c) , QED ? > I would like to state the task as Show that (a+b)+c=a+(b+c). If you go back to foundations, it's possible to formally construct a system which the 'axioms' of a complete ordered field. _Foundations of Mathematics_ by Ian Stewart and David Tall (OUP) goes into this in a fairly accessible manner. -- MHW === Subject: Re: cross product > If you go back to foundations, it's possible to formally construct a system > which the 'axioms' of a complete ordered field. That should read which *satisfies* the axioms of a complete ordered field. > _Foundations of Mathematics_ by Ian Stewart and David Tall (OUP) goes into > this in a fairly accessible manner. > -- > MHW === Subject: Re: cross product If you take the cross product of two linearly independent vectors, you get a third vector which is perpendicular to the other two vectors. The right-hand rule, if you take the thumb (pointing up) and pointerfinger (pointing forward) as the first two vectors, your middle finger (pointing to the left) will be the direction of the vector formed from the cross product. - Stephen http://pages.prodigy.net/stephenmorais > I don't understand the proof in my textbook that A, B and AxB form a > right-handed triad. Little help? > -- > Peace, > EJ === Subject: Re: cross product > If you take the cross product of two linearly independent vectors, you > get a third vector which is perpendicular to the other two vectors. > The right-hand rule, if you take the thumb (pointing up) and > pointerfinger (pointing forward) as the first two vectors, your middle > finger (pointing to the left) will be the direction of the vector formed > from the cross product. Sounds obscene. === Subject: Re: cross product > If you take the cross product of two linearly independent vectors, you > get a third vector which is perpendicular to the other two vectors. > The right-hand rule, if you take the thumb (pointing up) and > pointerfinger (pointing forward) as the first two vectors, your middle > finger (pointing to the left) will be the direction of the vector formed > from the cross product. > Sounds obscene. No worse than self-injective modules. === Subject: Differential equation solution? How do I solve the differential equation: a (2 + 3y + y^2) = x y ( y' + y^2 + 3y + 2) where y of course means y(x), and a is a constant, if it has any solutions? Jeremy === Subject: Re: Differential equation solution? Jeremy scribe: > How do I solve the differential equation: > a (2 + 3y + y^2) = x y ( y' + y^2 + 3y + 2) > where y of course means y(x), and a is a constant, if it has any solutions? (...) Some experiments show, that there exist solutions. A closed solution I have not got, but it is also an interesting task to look for the asymptotic behaviour of solutions. First some special solutions should be mentioned: y =(ident.)=-1 and -2. Theses are the zero points of 2+3y+y^2=0. Solve Your equation explicit to yÇ=(a-x*y)*(2+3y+y^2)/(x*y) and show, that lim((a-x*y)/(x*y)=-1 for x-->00. With this substitution you will get an equation with separated variables, which easy should be solved. Hope, there is no mistake, because I am in a hurry. Another chance seems to be the development of [(1/2)*y^2(x^2/2)]Ç and to compare it with x*y*yÇ. Alfred === Subject: Easy proof of mathematician lies I think I've figured out a way to show basically all of you, including people who think they don't know any math that mathematicians have been lying about my work. It's so trivial you *should* wonder why they thought they could get away with it. Here goes. My paper Advanced Polynomial Factorization depends on considering a factor of a polynomials that I call g. (Paper linked to at http://groups.msn.com/AmateurMath as usual.) And in my paper I start by showing that I can write that as g = r + c where either r=0, or r changes as the polynomial's value changes, while c does not. Now you can consider all factors of a given polynomial using g's, with something like g_1...g_k = P(x) where you have k factors. For instance, for P(x)=x^2 + 2x + 1, g_1 = x+1, g_2 = x+1, gives you 2 factors. Those are polynomial factors, but I'm generalizing in a simple way to say that for the factors g, in general, you have an element I call r, which changes as the independent variable changes, and you have another element I call c, which does not. For my example up above it's easy, as with g_1 = x + 1, x varies as x varies, while 1 does not. Now that's enough that the proof in the paper is straightforward, but posters have argued with me anyway, with some trying to argue over the definition of polynomial, amazingly enough. However, consider that the g's have an important feature, which is that when x=0, I have g_1...g_k = P(0). For instance, with my simple example, with P(x) = x^2 + 2x + 1, with x=0, P(0) = 1 = g_1(g_2), where g_1 = g_2 = x+1 = 1. You see, P(0) gives the constant term, so at x equal 0, the g's must multiply to give the constant term. So then, maybe you still want to believe the mathematicians and question that I can write g = r + c. Well consider that substituting gives me g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x), which gives r_1...r_k +...+c_1...c_k = P(x), which is r_1...r_k +...+P(0) = P(x), which means that if you believe the mathematicians then they've convinced you to doubt algebra itself, as then you must believe that everything to the left of P(0) above can *maybe* be constant, but also *maybe* vary as x varies. So why would mathematicians argue against such a simple result? Two reasons I suggest. First because they wish to disagree with me. Second because they probably believe that they can get away with it. That is, MOST of you will doubt algebra itself rather than consider that mathematicians, whom you probably don't even personally know, would lie. So where does this lead? Well the polynomial I show in the paper is P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf which seems to be just complicated enough to give mathematicians room to lie. For instance, you may be saying, HEY, what's with the 'm' when you had 'x' before??!!! Well, there's no rule that says that you have to use the letter x as the variable label for a polynomial. Also, there are historical reasons for my usage as it goes back to my work with FLT where x, y and z are used with x^p + y^p = z^p. Finally, the weirder thing is that one poster in particular got a lot of mileage out of questioning my finding the constant term with an expression like the above by using m=0, as that gives me P(0) = 3xy^2 + y^3 and he got a lot of mileage for YEARS (before I had discovered proof I want to add and after) by emphasizing that two of the ROOTS of such an expression considered as a polynomial with respect to x are not defined at that point. Well that's easy enough to see as the original expression is (v^3+1)x^3 - 3vxy^2 + y^3 which if you *wish* to see it as a polynomial with respect to x, is of degree 3, but when v=-1, it's of degree 1, so if you solve for the roots, you'll get funky stuff. Now when I was finding the proof of FLT...remember the process took some years...at times I'd talk of polynomials with respect to other than m, but I refined my discourse as my understanding improved. However, people arguing with me did not. You may *choose* to believe that they did not because they don't know enough mathematics to follow, but we're talking about actual mathematicians here. What's more rational? I say it's more rational to suppose that they *did* figure out that it worked as described, but also noticed that as long as they disagreed, no one seemed to call them on making false statements, except me, and they knew my credibility wasn't so great. For most of you, there's probably the belief that there's some funky higher math involved that your pitiful brain can't follow or you don't know about, as you may suppose that mathematicians either wouldn't lie, or they wouldn't lie in such a dumb way where I could catch them so easily. But consider what's in the balance: 1. I discovered a proof of Fermat's Last Theorem that's more available to people in general than most of what mathematicians have been producing lately. 2. Worse I did so having said I'd find it years ago, and having spent years looking for it posting a lot of my ideas, and getting in insult battles with posters, quite a few who happened to be mathematicians. 3. Then to add insult to injury, I keep questioning mathematicians in terms of their ethics, and maybe *extremely importantly* I doubt that Wiles found a proof of Fermat's Last Theorem. And those are just highlights as there's more but I think that kind of gives my point. For mathematicians the situation could be considered one of the worst possible disasters they can imagine. EXCEPT, it looks like all they have to do is either stay quiet, or *claim* I'm wrong. Many of you simply believe them, and question algebra itself, which is rather sad. I'd think at least some of you valued your educations. Others of you may figure it doesn't matter, maybe because Western civilization seems to be based on lying anyway, and maybe you figure I should just grow up, accept that everybody lies and move on. And I don't have to talk about Enron or pedophile Catholic priests or things in that vein. I mean, look at George W. Bush and Iraq. If people can be *killed* over lies, without consequences to the liars, then what's with some freaking stupid math? Good point. Mathematicians *are* a part of society after all. Why should they tell the truth now? It'd be like Bush owning up. They can just sit tight, and be quiet, like so many American citizens or they can out and out lie, like so many other patriots. After all, that's so easy, now isn't it? Which is why you need to understand why I talk about mathematicians potentially being prosecuted. Liars don't just stop because the gig is up, as then, they wouldn't necessarily be liars, then eh? It'd be against their *true* natures. === Subject: Re: Easy proof of mathematician lies > I think I've figured out a way to show basically all of you, including > people who think they don't know any math that mathematicians have > been lying about my work. It's so trivial you *should* wonder why > they thought they could get away with it. Mr. Harris, -X === Subject: Re: Easy proof of mathematician lies I've looked at your paper... now looking at this I'll bite. I may be getting in over my head but I'll point out the things that don't make sense to me. I'm open to being enlightened. > I think I've figured out a way to show basically all of you, including > people who think they don't know any math that mathematicians have > been lying about my work. It's so trivial you *should* wonder why > they thought they could get away with it. Either you aren't being clear, or they missed something. It is a strong charge to suggest that ALL mathematicians are in this great conspiracy to teach incorrect math. Sort of goes against what they stand for. > Here goes. > My paper Advanced Polynomial Factorization depends on considering a > factor of a polynomials that I call g. > (Paper linked to at http://groups.msn.com/AmateurMath as usual.) > And in my paper I start by showing that I can write that as > g = r + c > where either r=0, or r changes as the polynomial's value changes, > while c does not. Questions: is g the polynomial to be factored or a factor? Is r a polynomial, variable, or constant? Is c a polynomial or constant? What do you mean about a polynomial's value changing? A polynomial is in the form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0. It simply _is_, it doesn't change. Note: at this point I'm fairly confused, but I'll read on. > Now you can consider all factors of a given polynomial using g's, with > something like > g_1...g_k = P(x) > where you have k factors. For instance, for P(x)=x^2 + 2x + 1, > g_1 = x+1, g_2 = x+1, gives you 2 factors. This part made sense, except you'd usually note them as g_1(x), g_2(x). > Those are polynomial factors, but I'm generalizing in a simple way to > say that for the factors g, in general, you have an element I call r, > which changes as the independent variable changes, and you have > another element I call c, which does not. So, r is a function, and c is a constant? Why are you talking about independent variables changing? > For my example up above it's easy, as with g_1 = x + 1, x varies as x > varies, while 1 does not. so in this case, r=x, c=1? > Now that's enough that the proof in the paper is straightforward, but > posters have argued with me anyway, with some trying to argue over the > definition of polynomial, amazingly enough. Question: does each factor g_i(x) get its own r_i(x)? I'm not looking at the proof right now, but simply trying to understand what you are claiming to have prooved. The terminology you are using is not clear to me, and I can't easily comment on the value or validity of your work until that is made clear. If you read math papers, you will note that people specify what each object is, whether it's an integer, polynomial over the rationals, etc. I don't see these details. > However, consider that the g's have an important feature, which is > that when x=0, I have > g_1...g_k = P(0). Don't you mean (g_1...g_k)(0)=P(0)? > For instance, with my simple example, with P(x) = x^2 + 2x + 1, with > x=0, > P(0) = 1 = g_1(g_2), where g_1 = g_2 = x+1 = 1. x+1 is not 1. x+1 at x=0 is 1. > You see, P(0) gives the constant term, so at x equal 0, the g's must > multiply to give the constant term. > So then, maybe you still want to believe the mathematicians and > question that I can write g = r + c. Just questioning what it means. > Well consider that substituting gives me > g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x), which gives This seems to suggest that each g_i gets it's own r_i, c_i. > r_1...r_k +...+c_1...c_k = P(x), which is This step is NOT clear. What all goes between the product of r's and product of c's? > r_1...r_k +...+P(0) = P(x), > which means that if you believe the mathematicians then they've > convinced you to doubt algebra itself, as then you must believe that > everything to the left of P(0) above can *maybe* be constant, but also > *maybe* vary as x varies. Well, based on how you're defining things it appears that all of your r's should be written as r_i(x), not simply r_i. What am I missing? If r depends on x, please clearly indicate it. If r does not depend on x, then your initial definition of r doesn't make sense. There is ambiguity in here that appears to be the source of your *maybe*s, but I think it came from you, not mathematicians. > So why would mathematicians argue against such a simple result? Because it's not clearly stated. > Two reasons I suggest. First because they wish to disagree with me. > Second because they probably believe that they can get away with it. Right now, I don't understand what you're trying to say because there is too much that you have not defined. I cannot disagree with you because I do not understand you. I cannot accept your claims until I understand them, however. I've worked through enough math to have headaches from the strain of maintaining precision in language. Your paper lacks that precision. > That is, MOST of you will doubt algebra itself rather than consider > that mathematicians, whom you probably don't even personally know, > would lie. No, I just want to know what you're trying to say. > So where does this lead? > Well the polynomial I show in the paper is > P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf > which seems to be just complicated enough to give mathematicians room > to lie. I'd like to know why you introduced v, then defined it in terms of m and f, rather than do the substitution yourself. Also, why are you apparently leaving x,y,f as unaccounted for variables? What type of polynomial is P supposed to be? > For instance, you may be saying, HEY, what's with the 'm' when you > had 'x' before??!!! Couldn't care less what your independent variable is. I would like to know how x,y,f depend on m. > Well, there's no rule that says that you have to use the letter x as > the variable label for a polynomial. Also, there are historical > reasons for my usage as it goes back to my work with FLT where x, y > and z are used with x^p + y^p = z^p. FLT specifically defines x,y,z, and p. > Finally, the weirder thing is that one poster in particular got a lot > of mileage out of questioning my finding the constant term with an > expression like the above by using m=0, as that gives me > P(0) = 3xy^2 + y^3 > and he got a lot of mileage for YEARS (before I had discovered proof I > want to add and after) by emphasizing that two of the ROOTS of such an > expression considered as a polynomial with respect to x are not > defined at that point. > Well that's easy enough to see as the original expression is > (v^3+1)x^3 - 3vxy^2 + y^3 > which if you *wish* to see it as a polynomial with respect to x, is of > degree 3, but when v=-1, it's of degree 1, so if you solve for the > roots, you'll get funky stuff. Based on this statement, what do you mean by a polynomial? I suspect you are using a non-traditional meaning. [anti-mathematician discussion deleted] > But consider what's in the balance: > 1. I discovered a proof of Fermat's Last Theorem that's more > available to people in general than most of what mathematicians have > been producing lately. I'd like to read that proof. Where is it located? > 2. Worse I did so having said I'd find it years ago, and having spent > years looking for it posting a lot of my ideas, and getting in insult > battles with posters, quite a few who happened to be mathematicians. > 3. Then to add insult to injury, I keep questioning mathematicians in > terms of their ethics, and maybe *extremely importantly* I doubt that > Wiles found a proof of Fermat's Last Theorem. Then read his proof and find a flaw in it. I believe it's publicly available. [anti-mathematician discussion deleted] Notes: I am a mathematician. I have only recently started reading this board, so I am not familiar with your previous difficulties with mathematicians. What I can see here is that your mathematics is unclear. It looks like what you're doing should be easy, except I'm not sure what it is. Also, what is your background? Having studied logic, abstract algebra, etc... I find some of your accusations difficult to believe because I am aware of how far mathematicians have gone to make sure what is being taught actually works. Most of that involves precisely defined terms and precise use of notation. This level of precision takes a long time to learn and is part of why peer reviews are common. It's easy to overlook a detail. If we ALL overlooked one, all we ask is that you point it out so that we can fix the mistake. -- Will Twentyman email: wtwentyman at copper dot net