m-1045 ==== Subject: How to find this joint distribution of function of random ariables?i all, am wondering how can I find the joint pdf of random variables X, Y.here X=cos(THETA), Y=sin(THETA), where THETA is uniformly distributed over 0, 2*pi].t is difficult to me because all the tricks I know for this kind of problem s the Jacobian.ut this is one R.V (THETA) mapping to two R.V(X, Y)... so I don't know how o write out the Jacobian...anybody knows how?aybe I should define dummy variable?lease help me! ==== Subject: Re: How to find this joint distribution of function of random ariables? I am wondering how can I find the joint pdf of random variables X, Y. where X=cos(THETA), Y=sin(THETA), where THETA is uniformly distributed ver [0, 2*pi].ow about I[x^2 + y^2 = 1]/(2 pi), where I[.] is the indicator functionhat returns 1 if its argument is true, and 0 otherwise. ==== Subject: Re: How to find this joint distribution of function of random ariables? I am wondering how can I find the joint pdf of random variables X, Y. where X=cos(THETA), Y=sin(THETA), where THETA is uniformlyistributed over [0, 2*pi]. It is difficult to me because all the tricks I know for this kind ofroblem is the Jacobian. and Y are not jointly continuous, hence they do not have a joint pdf. ==== SSubject: How to find the autocorrelation function of this random process?i all,he random process is Y(t)=(-1)^(X(t)),here X(t) is a Poisson process with rate lamda. The R.P. Y(t) starts at (0)=1. This is called semirandom telegraph signal because Y(0)=1 is not andom initial value...he problem asks for the autocorrelation of this R.P. did the following:(Y(t)Y(s))=E((-1)^(X(t)+X(s)))=1*P(X(t)+X(s)=even number)-1*P(X(t)+X(s)=odd umber)ince X(t) is Poission distribution with rate lamda*t, X(s) is Poission istribution with rate lamda*s.(t)+X(s) is Poission distribution with rate lamda*(t+s).hen I found the(X(t)+X(s)=even number)-P(X(t)+X(s)=odd number)=exp(-2*lamda*(t+s))ence this E(Y(t)Y(s)) depends on both s and t.ut the solution gives E(Y(t)Y(s))=exp(-2*lamda*(|t-s|))... which is ierd... please help me! Anybody sees what's wrong with my derivation? ==== Subject: Re: How to find the autocorrelation function of this random rocess? Hi all, The random process is Y(t)=(-1)^(X(t)), where X(t) is a Poisson process with rate lamda. The R.P. Y(t)tarts at Y(0)=1. This is called semirandom telegraph signal because Y(0)=1 is not random initial value... The problem asks for the autocorrelation of this R.P. I did the following: E(Y(t)Y(s))=E((-1)^(X(t)+X(s)))=1*P(X(t)+X(s)=even number)-1*P(X(t)+X(s)=odd number) since X(t) is Poission distribution with rate lamda*t, X(s) is Poission distribution with rate lamda*s. X(t)+X(s) is Poission distribution with rate lamda*(t+s).o. This is only true if you are working with a case where X(t), X(s)efer to independent realisations, or independent sections of the samerocess. For example, consider t=s. Then X(t)=X(s) always, so(t)+X(t) is twice a Poisson r.v. with rate lamda*t, NOT a Poissonith rate lamda*(2t).ou can probably make progress by considering, if s>t,X(t)=X(t)-X(0)X(s)={X(s)-X(t)} + {X(t)-X(0)}here the two terms in curly brackets are independent.avid Jones ==== Subject: re:How to find the autocorrelation function of this random proc haven't done the calculation, but your answer just looks wrong. If=t, the autocorrelation has to be 1.-----------------------* www.GroupSrv.com-----------------------* ==== Subject: Re: The Size of Graham's Number*(y+1) = x + (x*y) multiplication^(y+1) = x * (x^y) exponentiation^^1 = x^^(y+1) = x ^ (x^^y) tetration^^^1 = x^^^(y+1) = x ^^ (x^^^y) hyper-tetration^^3 = 3^3^3 = 3^27 = 7,625,597,484,987^^4 = 3^(3^^3) = 3^(3^27) = 3^(7,625,597,484,987)Has about 8.4 x 10^12 digits.]raham's number is between 2^^^66 and 3^^^131.erc ==== Subject: Re: The Size of Graham's Number x*(y+1) = x + (x*y) multiplication x^(y+1) = x * (x^y) exponentiation x^^1 = x x^^(y+1) = x ^ (x^^y) tetration x^^^1 = x x^^^(y+1) = x ^^ (x^^^y) hyper-tetration 3^^3 = 3^3^3 = 3^27 = 7,625,597,484,987 3^^4 = 3^(3^^3) = 3^(3^27) = 3^(7,625,597,484,987) [Has about 8.4 x 10^12 digits.] Graham's number is between 2^^^66 and 3^^^131.o, Graham's number is the 64th number in a rapidly growing equence whose first element already far exceeds 3^^^131.n the hierarchy you've indicated above, let x{k} denote he operator with k ^'s , e.g. x^^^y = x{3}y.hen Graham's number is g_64 in the sequence defined by _1 = 3{4}3_n = 3{g_(n-1)}3 (n>1).hus, _1 = 3{4}3 _2 = 3{3{4}3}3_3 = 3{3{3{4}3}3}3.._64 = 3{3{...3{4}3...}3}3 (64 pairs of {}) = Graham's numberote that the g-sequence *starts* with the number _1 = 3{4}3 = 3^^^^3 = 3^^^(3^^^3) >> 3^^^131.-r.e.s. ==== Subject: Re: The Size of Graham's Number ompletely different definition in Dave Renfro's reference.ttp://mathforum.org/discuss/sci.math/m/394644/394645ere is how Smorynski defines Graham's number (p. 149) --->et N be the natural numbers. First, we define K: N^2 --> N.(0,n) = n^n(m+1, n) = K(m, K(m,n))ext, we define G: N --> N using K.(0) = K(3,3)(n+1) = K(G(n), 3)morynski then defines Graham's number to be G(64).ill the real Graham's number please stand up!erc ==== Subject: Re: The Size of Graham's Number Completely different definition in Dave Renfro's reference. http://mathforum.org/discuss/sci.math/m/394644/394645 Here is how Smorynski defines Graham's number (p. 149) --- Let N be the natural numbers. First, we define K: N^2 --> N. K(0,n) = n^n K(m+1, n) = K(m, K(m,n)) Next, we define G: N --> N using K. G(0) = K(3,3) G(n+1) = K(G(n), 3) Smorynski then defines Graham's number to be G(64). Will the real Graham's number please stand up!enfro's page discusses both definitions, and apparently none f his other references use Smorynski's version. (Regrettably, didn't notice, in his section How Big is Graham's Number?, ssentially the content of my posting.) It would be nice to now for sure which definition is given in the original paperto which, unfortunately, I have no ready access): Graham, R. L. and Rothschild, B. L. Ramsey's Theorem for n-Parameter Sets. Trans. Amer. Math. Soc. 159, 257-292, 1971.-r.e.s. ==== SSubject: Re: The Size of Graham's Number reply-type=response> Will the real Graham's number please stand up! Renfro's page discusses both definitions, and apparently none of his other references use Smorynski's version. (Regrettably, I didn't notice, in his section How Big is Graham's Number?, essentially the content of my posting.) It would be nice to know for sure which definition is given in the original paper (to which, unfortunately, I have no ready access): Graham, R. L. and Rothschild, B. L. Ramsey's Theorem for n-Parameter Sets. Trans. Amer. Math. Soc. 159, 257-292, 1971.pdate: I checked on this myself ... Apparently, practically ll online sources have got the definition wrong. In theriginal paper by Graham & Rothchild, the number now known s Graham's number is defined as = F(F(F(F(F(F(F(12,3),3),3),3),3),3),3)here F(m,n) is defined recursively as F(1,n) = 2^n, F(m,2) = 4, m >= 1, n >= 2 F(m,n) = F(m-1,F(m,n-1)), m >= 2, n >= 3.lease post any replies to the separate thread I've started:raham's Number -- Practically Everyone Has It Wrong!?-r.e.s. ==== Subject: Re: The Size of Graham's NumberThe number of digits it would take to write down Graham's number is just as unfathomable as the number itself.hil had a good point though - couldn't Graham's number concievably beertain precision in a certain volume V? Or is Grahams number so hugehe precision of the Planck length is less than Graham's number? ==== Subject: Re: The Size of Graham's Number>The number of digits it would take to write down Graham's number is just >as unfathomable as the number itself. phil had a good point though - couldn't Graham's number concievably be certain precision in a certain volume V? Or is Grahams number so huge the precision of the Planck length is less than Graham's number? light year ~ 10^16m ~ 10^51 Planck lengthspproximate the universe as a cube of side 10^10 ly = 10^61 Planck lengthsigger than a googolplex but nowhere near Graham's number. ==== Subject: Re: The Size of Graham's Number1 light year ~ 10^16m ~ 10^51 Planck lengthsApproximate the universe as a cube of side 10^10 ly = 10^61 Planck lengthsbigger than a googolplex but nowhere near Graham's number. thought Graham's Number was some sort of (significant) upper boundo a problem in combinatorics. If the number of combinatoric solutionsositions in the universe (i.e. the ultimate combinatorics problem)hen what possible relevance can it have? (Excuse my literalness, Inow maths as such is uninterested in the real world, it justnterests me how you can pose a problem that theoretically could haveo many solutions.) ==== Subject: Re: The Size of Graham's Number <9e6de$41b0d29b$5397ce19$23738@nf2.news-service.com> <271ff$41b1bcd7$5397ce19$29009@nf2.news-service.com> I thought Graham's Number was some sort of (significant) upper boundto a problem in combinatorics. If the number of combinatoric solutionspositions in the universe (i.e. the ultimate combinatorics problem)then what possible relevance can it have? (Excuse my literalness, Iknow maths as such is uninterested in the real world, it justinterests me how you can pose a problem that theoretically could haveso many solutions.)ight! in actuality all these big numbers are from the same class ofunctions, exponential growth. (in between exponential growth and nonomputable functions whose growth cannot be considered at all).ut 10^64 is hardly the largerst. What is useful are some constantso measure different exponential growths, how sharp they shoot upwards.ts not so much the height of the curve but the shape that determinesifferent behaviours. 10^64 does not have the property that takingtself to its own power 10^64 times has relatively little change to theesult, so in that respect its small.hat's bigger : G(64) or BusyBeaver(64) ?erc ==== Subject: Re: The Size of Graham's Number <9e6de$41b0d29b$5397ce19$23738@nf2.news-service.com> <271ff$41b1bcd7$5397ce19$29009@nf2.news-service.com> What's bigger : G(64) or BusyBeaver(64) ?t would be close, G is a small algorithm to code but would it fit into 64 state TM?t could easily be coded in a few hundred states, so BB(500) > G(500).usy Beaver is the BIGGEST POSSIBLE size of output from a fixed sizedomputer program. All you have to do is program your G function in say00 states, then (pseudo) input the number 64, this will require amall TM to run first and put 64 1's on the tape, only about 50 moretates to do this, and you have Graham's number with 450 states.bviosly a TM with 451 states could output a larger number than a TMith 450 states.he fact tesselation is simple to code is one reason why theoncomputable BB function grows as fast as it does.erc ==== Subject: Re: The Size of Graham's Number <9e6de$41b0d29b$5397ce19$23738@nf2.news-service.com> <271ff$41b1bcd7$5397ce19$29009@nf2.news-service.com> What's bigger : G(64) or BusyBeaver(64) ?t would be close, G is a small algorithm to code but would it fit into 64 state TM?t could easily be coded in a few hundred states, so BB(500) > G(500).usy Beaver is the BIGGEST POSSIBLE size of output from a fixed sizedomputer program. All you have to do is program your G function in say00 states, then (pseudo) input the number 64, this will require amall TM to run first and put 64 1's on the tape, only about 50 moretates to do this, and you have Graham's number with 450 states.bviosly a TM with 451 states could output a larger number than a TMith 450 states.he fact tesselation is simple to code is one reason why theoncomputable BB function grows as fast as it does.erc ==== Subject: Re: The Size of Graham's Number <9e6de$41b0d29b$5397ce19$23738@nf2.news-service.com> <271ff$41b1bcd7$5397ce19$29009@nf2.news-service.com> etration not tesselation.erc ==== Subject: Re: Abstract Algebra I'm missing something on this seemingly trivial problem. Let G be a group with 3000 elements and H be a subgroup with 1000 elements. If x is in G, show that either x^2 or x^3 is in H. Hint: If [G:H] = 3, then either H is normal in G, or H has a subgroup of index 2 that's normal in G. I've never seen this before - I'm not sure I believe it - and I don't think anything like it is necessary for the problem at hand.ell, of course it's not *necessary*. It just makes the problemasy to solve.nd it's certainly true. The action of G on the set of cosets of gives a surjective homomorphism from G onto a transitiveegree-3 subgroup of S_3, whose kernel lies in H (since theatter is the point stabilizer of the coset H in this action)._3 has only two transitive subgroups in its natural action, soither H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2ith K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3with the latter 3 cosets squaring to K).- im Heckman ==== Subject: Re: Abstract AlgebraI'm missing something on this seemingly trivial problem. Let G be a group with 3000 elements and H be a subgroup with 1000 elements. If x is in G, show that either x^2 or x^3 is in H.Hint: If [G:H] = 3, then either H is normal in G, or H has a subgroup of index 2 that's normal in G. I've never seen this before - I'm not sure I believe it - and I don't think anything like it is necessary for the problem at hand. Well, of course it's not *necessary*. It just makes the problem easy to solve. And it's certainly true. The action of G on the set of cosets of H gives a surjective homomorphism from G onto a transitive degree-3 subgroup of S_3, whose kernel lies in H (since the latter is the point stabilizer of the coset H in this action). S_3 has only two transitive subgroups in its natural action, so either H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2 with K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3 (with the latter 3 cosets squaring to K).ut. suspect that anyone having trouble with the original problem s not going to understand any of what you've written here. nowing that a problem on the second floor is easy to solve rom the perspective of the 12th floor doesn't help the guy ho has been struggling to reach the mezzanine.- erry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) ==== Subject: Re: Abstract Algebra I'm missing something on this seemingly trivial problem. Let G be a group with 3000 elements and H be a subgroup with 1000 elements. If x is in G, show that either x^2 or x^3 is in H....] And it's certainly true. The action of G on the set of cosets of H gives a surjective homomorphism from G onto a transitive degree-3 subgroup of S_3, whose kernel lies in H (since the latter is the point stabilizer of the coset H in this action). S_3 has only two transitive subgroups in its natural action, so either H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2 with K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3 (with the latter 3 cosets squaring to K). But. I suspect that anyone having trouble with the original problem is not going to understand any of what you've written here. Knowing that a problem on the second floor is easy to solve from the perspective of the 12th floor doesn't help the guy who has been struggling to reach the mezzanine.ctually, I'm having trouble thinking of a way to solve theriginal problem that *doesn't* essentially use the action of Gn the cosets of H to show that G has a surjective homomorphismo C_3 or S_3 whose kernel lies in H. I notice the OP said heigured it out, so I wonder just what floor he's on. Or maybe'm missing some more elementary approach...- im Heckman ==== Subject: Re: Abstract Algebra...]I'm missing something on this seemingly trivial problem. Let G be a group with 3000 elements and H be a subgroup with 1000 elements. If x is in G, show that either x^2 or x^3 is in H....] Actually, I'm having trouble thinking of a way to solve the original problem that *doesn't* essentially use the action of G on the cosets of H to show that G has a surjective homomorphism to C_3 or S_3 whose kernel lies in H. I notice the OP said he figured it out, so I wonder just what floor he's on. Or maybe I'm missing some more elementary approach...bvious elementary approach. D'oh!- im Heckman ==== Subject: Re: Abstract Algebra Actually, I'm having trouble thinking of a way to solve the original problem that *doesn't* essentially use the action of G on the cosets of H to show that G has a surjective homomorphism to C_3 or S_3 whose kernel lies in H. I notice the OP said he figured it out, so I wonder just what floor he's on. Or maybe I'm missing some more elementary approach...ell, I don't know about a method that avoids cosets entirely, butere's an outline of what I came up with:ny two cosets aH and bH are either disjoint or equal. There is also aijection between them (making them the same size if H is finite). Wean see that if aH = bH, a^-1b must be in H.hen consider the 4 cosets H = x^0H, xH = x^1H, x^2H, x^3H. Since G is times the size of H, the pigeonhole principle forces at least one pairf equal cosets; call them x^mH and x^nH where m < n. Since x^mH =^nH, we have that x^(n-m) is in H. If n-m is 2 or 3, we are done. If-m is 1, we use the fact that H is a group one last time to get theesired result. - aniel W. Johnsonanoptes@iquest.netttp://members.iquest.net/~panoptes/39 53 36 N / 086 11 55 W ==== Subject: Re: Abstract Algebrariginator: grubb@lola I'm missing something on this seemingly trivial problem. Let G be a group with 3000 elements and H be a subgroup with 1000 elements. If x is in G, show that either x^2 or x^3 is in H.Actually, I'm having trouble thinking of a way to solve theoriginal problem that *doesn't* essentially use the action of Gon the cosets of H to show that G has a surjective homomorphismto C_3 or S_3 whose kernel lies in H. I notice the OP said hefigured it out, so I wonder just what floor he's on. Or maybeI'm missing some more elementary approach...t's easy to do this on a case by case basis: If x is in H,hen x^2 is in H, so we're done.f x is not in H, then H and xH are disjoint. Now consider x^2.f x^2 is in H, we're done. If x^2 is in xH, then x is in H ande're done. If x^2 is in neither, then x^2H is disjoint fromoth H and xH. Furthermore, since [G:H]=3, these three cosetsover G. Now look at x^3. If x^3 is in H, we're done. If its in xH, then x^2 is in H and we're done. If x^3 is in x^2H,hen x is in H and we're done.-Dan Grubb ==== Subject: Re: How to measure randomness of a deck of cards? ake the difference between consecutive cards,tep 2ake the sum of squares.ompare it to a shuffled deck.lso do it for suits, 0 1 2 3 in mod 4.ould also do it for every second card, every 3rd card.r a neural net could be trained to give you a yes or no if itshuffled.here's no absolute answer unless you're using a naive interpretationf random,ay the deck is 314159, looks stacked to me!tep 2: subtract that from the expected difference (I think)erc ==== Subject: Re: How to measure randomness of a deck of cards? take the difference between consecutive cards, step 2 take the sum of squares. compare it to a shuffled deck. also do it for suits, 0 1 2 3 in mod 4. could also do it for every second card, every 3rd card. or a neural net could be trained to give you a yes or no if its shuffled. there's no absolute answer unless you're using a naive interpretation of random, say the deck is 314159, looks stacked to me! step 2: subtract that from the expected difference (I think) Herchat is this method called? And what is its main purpose? ==== Subject: Re: How to measure randomness of a deck of cards? he sum of squares (of the differences) is standard for measuring totalrror of a sample.'m not sure how it will go using the heuristic of an 'expectedifference' betweenonsecutive cards, but you should be able to detect a fresh deck from ahuffled one.ay you calculate the average difference between consecutive cards is.hen a fresh deck will have differences ofards =1,2,3,4,5,6,7,8,9,10,11,12,13,1,2,3,4,5,6,7,8,9,10,11,12,13,1,2..>elta =1,1,1,1,1,1,1,1,1,1,1,1,1,1,13,1,1,1,1,1,1,1,1,1,1,1,1,1,13,1,1....>rrors = <6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,...> random deck will haveelta = <2,4,1,10,12,4,2,6...>rrors = <5,1,3,5,1,5,1....>he standard deviation will pick it up (on Delta), that's a function ofhe sum of squares, so you don't need an expected difference or Errors.ere's another technique :se GPS (General Problem Solver AKA Search algorithm)o sort the cards *into* orderount the number of card swaps and you have a measure of the *distance*rom an ordered listou could weight the cards swaps, *distance, *sqrt(distance), or justse adjacent swaps.eck one took 5 card swaps to be ordered, deck two took 20, so deck twoas more shuffled.ome problems... a reverse ordered deck will appear shuffled!erc ==== Subject: Re: How to measure randomness of a deck of cards? say the deck is 314159, looks stacked to me!ctually, the digits of pi look not too bad when you do PRNG stats tests... ot as good as SHA-1, but not too bad... ==== Subject: random 1 to 3 to random 1 to 5ne of my friend asked me this question which he was asked during aob interview. Couldn't figure it out and very curious about theolution (or this just mpossible? )The question is, given a function that returns evenly distributedandom number from 1 to 5, design a new function based on thisunction, and return evenly distributed random number from 1 to 3. ==== Subject: Re: random 1 to 3 to random 1 to 5> One of my friend asked me this question which he was asked during a> job interview. Couldn't figure it out and very curious about the> solution (or this just mpossible? )> The question is, given a function that returns evenly distributed> random number from 1 to 5, design a new function based on this> function, and return evenly distributed random number from 1 to 3. would invoke the thing three times in a row and associatehe first selection with the numbers 1-5, the second selectionith 6-10 an the third with 11-15 and divide the result byhree.K, I figured this out in a minute but would NEVER have beenble to do so during a job interview with mr/mrs personnelfficer watching me, waiting for my defeat.here is no end to the abuse employers will go these days: Iave had to sign a PISS form prior to even being interviewed.ell maybe I am an alcoholic! do you really care if I solveour problem? view all of this as abuse: and it doesn't matter for theutcome anyway since the personnel person will hire peoplehom he(she) likes anyway but has to rationalize his(her)ecisions with some pseudo-scientific rationalization.n the end he/she will hire: people ;ike him/her self! Theseould not be programmers of course but they would make greatersonnel officers.ow to the core of the matter: Is the person who can solveuzzles like the above in a minute the better candidate? don't think so. Maybe this person has trained in goat wolfnd cabbage puzzles all his life! Or maybe he is a member ofensa which means that he can deduce the general term of anyeries with complete confidence from a mere three terms! will never forget the scathing sarcasm of my algebra professorhen he ridiculed certain puzzles in the newspaper asking theeader to complete a certain given series! have heard that MICROSOFT uses puzzles like the above tovaluate potential job candidates so I assume they can crossny river without problems, maybe that's why their softwares so popular.ow, there is such a thing as deep thinking and it requiresime and an environment free of disturbances, much differentrom the environment of the job interview.here are also KNOWLEDGE and EXPERIENCE which one can actuallyeasure in a scientific manner - provided one has some oneself.inally, how a person will behave or fit in within a group cannote deduced from superficial interpersonal behaviour in any way.entje Goslinga ==== Subject: Re: random 1 to 3 to random 1 to 5 Now to the core of the matter: Is the person who can solve puzzles like the above in a minute the better candidate? believe this trend got started because Microsoft interviews rogrammers with logic puzzle, why are manhole covers round, and the ike. But this is bad logic. Microsoft makes terrible software! What eople should do is find out how Microsoft interviews their MARKETING eople, and emulate that! ==== Subject: Re: random 1 to 3 to random 1 to 5 ETAsAhRige9JUWtymOvTm2G2qmtDPVFBJgIUUUFNAcz2vfIa4AVGxplK5kUosEE= h, f(x) =(x+1)/2?-OL ==== Subject: Re: random 1 to 3 to random 1 to 5 uh, f(x) =(x+1)/2?hat will upset the ditribution. In the simplest example, if the 1-5utputs all occur at equal frequency over a sufficiently long time nterval,hen the frequencys of the 1-3 outputs will be: 40%: 40%: 20% ==== Subject: Re: random 1 to 3 to random 1 to 5 ETAtAhUAxBJIWxM51G2i0uLGKaCpuJshQ2UCFHBkr2Qp/3fEVg4RMWkdEskPHtQo phenry, I could not reach you via email. See what I subsequentlyosted for the discrete case.-OL ==== Subject: Re: random 1 to 3 to random 1 to 5aving read Robert Vienneau's and Peter Webb's solution, I am gettingnterested in another similar question.he question is:iven a function F() that returns evenly distributed random number from1,2,3,4,5} in constant time, design a new function G() that returns evenlyistributed random number from {1,2,3}in constant time. coudn't design such G(). ==== Subject: Re: random 1 to 3 to random 1 to 5ot long ago there was a problem of designing a fair 7-sided die. Oneof the solutions found for this was to roll a 6-sded die repeatedly,dd up the numbers and take the residue mod 7. The result is theairest possible division for the given number of rolls.o design your G() function: Start with a seed value, then defineecursively:() = mod(G()+F(),3)+1s time progresses the probability distribution for G() approaches theniform one.he function defined in this way is autocorrelated. When G() is 3 theext iterate is more likely to be 2 or 3 than 1. You can reduce theutocorrelation by storing the last several values of F. Calculate theum modulo 3, calling it S(), then iterate according to:() = mod(G()+S(), 3)+1-OL ==== Subject: Re: random 1 to 3 to random 1 to 5Having read Robert Vienneau's and Peter Webb's solution, I am gettinginterested in another similar question.The question is:Given a function F() that returns evenly distributed random number from{1,2,3,4,5} in constant time, design a new function G() that returns venlydistributed random number from {1,2,3}in constant time.omething similar was discussed here a while ago, except it was aboutontructing a uniform distribution in {1,2,3} from a uniformistribution in {1,2}:- 'm not interested in mathematics that might have anythingo do with reality. -- Russell Easterly, in sci.math ==== Subject: Re: random 1 to 3 to random 1 to 5 One of my friend asked me this question which he was asked during a job interview. Couldn't figure it out and very curious about the solution (or this just mpossible? ) The question is, given a function that returns evenly distributed random number from 1 to 5, design a new function based on this function, and return evenly distributed random number from 1 to 3.and1to3 = (rand1to5 - 1)/2 + 1 ==== Subject: Re: random 1 to 3 to random 1 to 5 One of my friend asked me this question which he was asked during a job interview. Couldn't figure it out and very curious about the solution (or this just mpossible? ) The question is, given a function that returns evenly distributed random number from 1 to 5, design a new function based on this function, and return evenly distributed random number from 1 to 3. rand1to3 = (rand1to5 - 1)/2 + 1'm guessing the OP meant to say integer rather than numberotherwise it's a kind of silly question). ==== Subject: Re: random 1 to 3 to random 1 to 5 One of my friend asked me this question which he was asked during a job interview. Couldn't figure it out and very curious about the solution (or this just mpossible? ) The question is, given a function that returns evenly distributed random number from 1 to 5, design a new function based on this function, and return evenly distributed random number from 1 to 3.et f() be the first function. The new function is defined by theollowing pseudo-code:x := f();while ( x > 3 ) do x := f();end;return x;- ostly economics: cv s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or ealth e a e of a man is a question fit perhaps to be discussed y n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau ==== Subject: Some Propositions Of Mathematical Economics. Because of price Wicksell effects, the interest rate is not equal,n equilibrium, to the marginal product of capital.. The wage is not determined by the value of the marginal product ofabor.. Consider a so-called perverse switch point. Around such a point, higher wage is associated with a cost minimizing technique inhich firms hire more labor to produce a given level of net output.hese propositions use technical terms, and I have explained thoseerms in different ways on many occasions. If you do not like myxplanations, you can always look up some of the literature I haveited. For some reason, mainstream economists posting to Usenetave argued against these true statements. Of course, they areot consistent with certain erroneous teaching found in certainextbooks.- ostly economics: cv s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or ealth e a e of a man is a question fit perhaps to be discussed y n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau ==== Subject: Re: talk science and mathematics, you faggots.> i m supreme. i alone command you. still boring...unacy is never boring. BTW CHAOS stands for Could (do to) Have Another lanzapine Soon. ==== Subject: about semialgebraic sety definition of semialgebraic subset is: A subset V of R^n is called semi-algebraic f it admits some representation of the form V = cup_{i=1}^s cap_{j=1}^{r_i} {x in R^n | P_{i,j}(x) s_{ij} 0 },here ''cup'' stands for ''union'' ''cap'' stands for ''intersection'' s_{ij} in {>,=,<} P_{ij}(X) in R[X], X = (X_1,...,X_n).ay you help me to show that the complementary in R^n f a semialgebraic set is a semialgebraic set itself, lease?.T. ==== Subject: Re: about semialgebraic set My definition of semialgebraic subset is: A subset V of R^n is called semi-algebraic if it admits some representation of the form V = cup_{i=1}^s cap_{j=1}^{r_i} {x in R^n | P_{i,j}(x) s_{ij} 0 }, where ''cup'' stands for ''union'' ''cap'' stands for ''intersection'' s_{ij} in {>,=,<} P_{ij}(X) in R[X], X = (X_1,...,X_n). May you help me to show that the complementary in R^n of a semialgebraic set is a semialgebraic set itself, please?tart by showing the intersection of two semialgebraic sets is emialgebraic.- . A. Edgar ttp://www.math.ohio-state.edu/~edgar/ ==== Subject: Proving the cross product is orthogonal hope I don't seem like one of those louts that only turn up at ssignment time!! However I am struggling with a proof. My answer is not oming out as expected, and I think I am making a silly mistake omewhere. If anyone could have a glance over it, that would be ppreciated.) a =(1, -2, 1) b =(3, 1, 0)ind a x b and prove that your cross-product vector is perpendicular to ach of the vectors a and b.) [I have worked out the cross-product, and I have proven that a is erpendicular to the cross-product, however I am struggling to prove hat b is perpendicular to the cross-product] x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k = -3i + 2j + 7ket w = -3i + 2j + 7krove vector a is perpendicular to w.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6)w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)os&= (a.w)/(|a||w|) = 0rove vector a is perpendicular to w.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10)w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)os&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) know that is wrong, can anyone see what my mistkae is??assandra ==== Subject: Re: Proving the cross product is orthogonal I hope I don't seem like one of those louts that only turn up at assignment time!! However I am struggling with a proof. My answer is not coming out as expected, and I think I am making a silly mistake somewhere. If anyone could have a glance over it, that would be ppreciated. Q) a =(1, -2, 1) b =(3, 1, 0) Find a x b and prove that your cross-product vector is perpendicular to each of the vectors a and b. A) [I have worked out the cross-product, and I have proven that a is perpendicular to the cross-product, however I am struggling to prove that b is perpendicular to the cross-product]o you know about the dot product? ==== Subject: Re: Proving the cross product is orthogonal>I hope I don't seem like one of those louts that only turn up at >assignment time!! However I am struggling with a proof. My answer is not >coming out as expected, and I think I am making a silly mistake >somewhere. If anyone could have a glance over it, that would be ppreciated.>Q) a =(1, -2, 1) b =(3, 1, 0)>Find a x b and prove that your cross-product vector is perpendicular to >each of the vectors a and b.>A) [I have worked out the cross-product, and I have proven that a is >perpendicular to the cross-product, however I am struggling to prove >that b is perpendicular to the cross-product] Do you know about the dot product?es, I like the dot product...much easier then the cross product I think!t turned out my main mistake was silly mistakes in my workings (ie 2*0 2 etc) I did this two places, I think it was only fluke that my orkings for a being perpendicular to a x b, after the silly errors here fixed both proofs worked.assandra ==== Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGEou simply overlooked the figures 0 in -2*0-1*1 and in 1*3-1*0. I guess hat you were writing too small print too tightly in the left upper orner of your paper sheet. Check: (1, -2, 1) X (3, 1, 0) = (-1, 3, 7); tc., etc.HTH - Johan E. Mebius I hope I don't seem like one of those louts that only turn up at assignment time!! However I am struggling with a proof. My answer is not coming out as expected, and I think I am making a silly mistake somewhere. If anyone could have a glance over it, that would be appreciated. Q) a =(1, -2, 1) b =(3, 1, 0) Find a x b and prove that your cross-product vector is perpendicular to each of the vectors a and b. A) [I have worked out the cross-product, and I have proven that a is perpendicular to the cross-product, however I am struggling to prove that b is perpendicular to the cross-product] a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k = -3i + 2j + 7k Let w = -3i + 2j + 7k Prove vector a is perpendicular to w a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) cos&= (a.w)/(|a||w|) = 0 Prove vector a is perpendicular to w b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) I know that is wrong, can anyone see what my mistkae is?? Cassandra ==== Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE You simply overlooked the figures 0 in -2*0-1*1 and in 1*3-1*0. I guess that you were writing too small print too tightly in the left upper corner of your paper sheet. Check: (1, -2, 1) X (3, 1, 0) = (-1, 3, 7); etc., etc. IHTH - Johan E. Mebius> I hope I don't seem like one of those louts that only turn up at > assignment time!! However I am struggling with a proof. My answer is > not coming out as expected, and I think I am making a silly mistake > somewhere. If anyone could have a glance over it, that would be > appreciated.> Q) a =(1, -2, 1) b =(3, 1, 0)> Find a x b and prove that your cross-product vector is perpendicular > to each of the vectors a and b.> A) [I have worked out the cross-product, and I have proven that a is > perpendicular to the cross-product, however I am struggling to prove > that b is perpendicular to the cross-product]> a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k> = -3i + 2j + 7k> Let w = -3i + 2j + 7k> Prove vector a is perpendicular to w> a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0> |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6)> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)> cos&= (a.w)/(|a||w|) = 0> Prove vector a is perpendicular to w> b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7> |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10)> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62)> cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620)> I know that is wrong, can anyone see what my mistkae is??> Cassandrauch appreciated. I have been writing my equations in MS Word using the quation editor. Perhaps I should write them on paper 'big and large' irst!!assie ==== Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE Much appreciated. I have been writing my equations in MS Word using the equation editor. Perhaps I should write them on paper 'big and large' first!!ith a pencil. he trolls in this newsgroup post all kinds of nonsense, sometimes on urposend sometimes because they don't check their work. ==== Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE>Much appreciated. I have been writing my equations in MS Word using the >equation editor. Perhaps I should write them on paper 'big and large' >first!! With a pencil. The trolls in this newsgroup post all kinds of nonsense, sometimes on urpose and sometimes because they don't check their work. es, one of my biggest downfalls in life is that I make lots of istakes. I once had to do a speed & accuracy test as part of a pysch est. I scored full marks (not sure how they calculated it), but I emember the pysch saying something about me getting alot further hrough the test then most, but making more mistakes then most. used to have a turtle on my desk to remind myself to slow down and ouble check things. I find it so difficult, but it is something I hould really work on. I have made two very elementary mistakes in the ourse of this thread...I do need to improve. ==== Subject: Re: Proving the cross product is orthogonalQ) a =(1, -2, 1) b =(3, 1, 0)Find a x b and prove that your cross-product vector is perpendicular to each of the vectors a and b.A) [I have worked out the cross-product, and I have proven that a is perpendicular to the cross-product, however I am struggling to prove that b is perpendicular to the cross-product]a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k = -3i + 2j + 7kuh? Check your algebra.- 'm not interested in mathematics that might have anythingo do with reality. -- Russell Easterly, in sci.math ==== Subject: Re: question about Banach spaces Is there an accessible example of a linear space endowed with two non-equivalent complete norms? jennynfinite-dimensional only. And AC required. How about this:anach spaces l^2 and l^1 are isomorphic as linear spacesboth having Hamel dimension c).- . A. Edgar ttp://www.math.ohio-state.edu/~edgar/ ==== Subject: Re: question about Banach spaces Is there an accessible example of a linear space endowed with two non-equivalent complete norms? jenny Infinite-dimensional only. And AC required. How about this: Banach spaces l^2 and l^1 are isomorphic as linear spaces (both having Hamel dimension c).ne has to necessarily appeal to AC? In other words, is ACquivalent to that statement? am asking that, since I have a somehow realated problem:t is well known that the dual of L^infty strictly contains L^1this can be directly argued from separability arguments; clearly, onlyn infinite dimension)).n spite of that, I am not aware af any example of a continuous linearunctional on L^infty which is not L^1 without appealing tohe Hahn-Banach theorem, which is to say, without appealing to AC.enny ==== Subject: Re: question about Banach spaces Is there an accessible example of a linear space endowed with two non-equivalent complete norms? jennyInfinite-dimensional only. And AC required. How about this: Banach spaces l^2 and l^1 are isomorphic as linear spaces (both having Hamel dimension c). one has to necessarily appeal to AC? In other words, is AC equivalent to that statement?quivalent, probably not, but in fact it cannot be proved in ZF alone.n Solovay's model where every set of reals (and every set in a Polishpace) has the property of Baire, it follows (proved by Christensen, Iuess) that any linear map of separable Banach spaces is automaticallyontinuous. So, in particular, if a given vector space admits twoomplete separable norms, then the identity is a homeomorphism.And since, in metric spaces, a map is continuous if and only ifts restriction to every separable subspace is, we can get rid ofhe separable assumptions I made above.)hat's in Solovay's model. But there is some principle that goes withhis saying any spaces and maps THAT YOU CAN ACTUALLY WRITE DOWNXPLICITLY also work like this, merely using ZF. So, depending onhat accessible means in the original question, the answer maye no. I am asking that, since I have a somehow realated problem: It is well known that the dual of L^infty strictly contains L^1 (this can be directly argued from separability arguments; clearly, only in infinite dimension)). In spite of that, I am not aware af any example of a continuous linear functional on L^infty which is not L^1 without appealing to the Hahn-Banach theorem, which is to say, without appealing to AC.gain, this existence cannot be proved in ZF. But it can be provedsing HB. It is known that HB is strictly weaker than AC, so thisnswers the equivalent question above.- . A. Edgar ttp://www.math.ohio-state.edu/~edgar/ ==== Subject: Re: question about Banach spacesIs there an accessible example of a linear space endowed with two non-equivalent complete norms? jenny Infinite-dimensional only. And AC required. How about this:Banach spaces l^2 and l^1 are isomorphic as linear spaces (both having Hamel dimension c). one has to necessarily appeal to AC? In other words, is AC equivalent to that statement? Equivalent, probably not, but in fact it cannot be proved in ZF alone.also for David C. Ullrich)hen you say that it cannot be proved what do you exactly mean? mean, if it is provable with AC but not provable without AChy it is not automatically equivalent to AC.y point is: one cannot prove this within ZF because he/she isot able to, but in principle he/she could, or there is a way tohow this impossibility?enny ==== Subject: Re: question about Banach spaces> Is there an accessible example of a linear> space endowed with two non-equivalent> complete norms?> jenny> Infinite-dimensional only. And AC required. How about this:> Banach spaces l^2 and l^1 are isomorphic as linear spaces> (both having Hamel dimension c).> one has to necessarily appeal to AC? In other words, is AC> equivalent to that statement?> Equivalent, probably not, but in fact it cannot be proved in ZF alone.(also for David C. Ullrich)When you say that it cannot be proved what do you exactly mean?I mean, if it is provable with AC but not provable without ACwhy it is not automatically equivalent to AC.My point is: one cannot prove this within ZF because he/she isnot able to, but in principle he/she could, or there is a way toshow this impossibility?here is a way to show this impossibility. There was a big hintow that works in the paragraph you snipped, where Edgar saidomething about a certain model of ZF. Any statement that cane proved from the axioms of ZF will be true in every model ofF, so if there is a model of ZF in which P is false it followshat P _cannot_ be proved in ZF.ou need to study a small bit of mathematical logic to reallynow exactly what that means. For now an analogy: Say GT ishe axioms for a group: (ab)c = a(bc), etc. So for exampleab(cd) = (a(bc))d is a theorem of GT - it can be provedrom the axioms. Now a model of GT is precisely the samehing as a _group_. There exists a group in which it isot true that ab = ba for all a, b, and the existence ofuch a group shows that ab = ba for all a, b cannot beroved from the axioms of GT.jenny***********************avid C. Ullrich ==== Subject: Re: question about Banach spaces (also for David C. Ullrich) When you say that it cannot be proved what do you exactly mean? I mean, if it is provable with AC but not provable without AC why it is not automatically equivalent to AC. My point is: one cannot prove this within ZF because he/she is not able to, but in principle he/she could, or there is a way to show this impossibility?n example is the Hahn-Banach theorem, HB.ssuming consistency when needed, it has been shown that:1) ZF does not prove HB,2) ZF+AC proves HB,3) ZF+HB does not prove AC.n that sense, HB is beyond ZF, but is not as strong as AC.n the other hand, Zorn's Lemma, ZL, is equivalent to AC, meaning:4) ZF+AC proves ZL5) ZF+ZL proves ACn most cases, a cannot be proved result is shown by exhibiting model. For example, a model of ZF+HB where AC fails will show3), above.- . A. Edgar ttp://www.math.ohio-state.edu/~edgar/ ==== Subject: Re: question about Banach spaces> Is there an accessible example of a linear> space endowed with two non-equivalent> complete norms?> jenny> Infinite-dimensional only. And AC required. How about this:> Banach spaces l^2 and l^1 are isomorphic as linear spaces> (both having Hamel dimension c).one has to necessarily appeal to AC? In other words, is ACequivalent to that statement?ot that I know the answer to either question, but I'lloint out that the in other words isn't right - sayingomething requires AC is not the same as saying it'squivalent to AC. If ZF does not prove P but ZFC doeshen P requires AC; saying P is equivalent to AC is much stronger statement.For example: If ZFC proves P but, say, ZF + AD (Axiom ofeterminacy) proves not P then AC is required for P,lthough this doesn't say that P is equivalent to AC.his actually happens if P is every set of reals isebesgue measurable.)I am asking that, since I have a somehow realated problem:It is well known that the dual of L^infty strictly contains L^1(this can be directly argued from separability arguments; clearly, onlyin infinite dimension)).In spite of that, I am not aware af any example of a continuous linearfunctional on L^infty which is not L^1 without appealing tothe Hahn-Banach theorem, which is to say, without appealing to AC.jenny***********************avid C. Ullrich ==== Subject: Group presentations'm struggling to understand group presentations.or example, Mathworld gives a group presentation for the Dihedral Group _2n (the symmetry group of the regular n-gon) as:_2n = < x, y | x^2 = 1, y^2 = 1, (xy)^n = 1 >hat are the x's and y's? I assume they are elements in the group of sometries of the plane since D_2n is a subgroup thereof? Can we thus ick *any* x's and y's which satisfy the given relations, for example, hy can't we pick x = 1 and y = 1? They satisfy the relations but learly don't generate D_2n.as Mathworld missed something out? Am I missing the point?ichard Hayden. ==== Subject: Re: Group presentations days. My association with the Department is that of an alumnus.I'm struggling to understand group presentations.For example, Mathworld gives a group presentation for the Dihedral Group D_2n (the symmetry group of the regular n-gon) as:D_2n = < x, y | x^2 = 1, y^2 = 1, (xy)^n = 1 What are the x's and y's? I assume they are elements in the group of isometries of the plane since D_2n is a subgroup thereof?o. They are telling you that x and y are elemnts of D_2n, that everylement of D_2n can be written as a product of x's, y's, and theirnverses, and that the only thing you know about x and y are that x^2s trivial, that y^2 is trivial, and that (xy)^n is trivial. o, for example, you know that y^{-1} = y, because y^2 = 1. You knowhatxy)y = y(xy)^{-1}, because (xy)y = xyy = x1 = x, and y(xy)^{-1} =(y^{-1}x^{-1}) = x^{-1} = x (the last because x^2 = 1, so x=x^{-1}. Can we thus pick *any* x's and y's which satisfy the given relations, for example, why can't we pick x = 1 and y = 1? They satisfy the relations but clearly don't generate D_2n._2n is the LARGEST group which contains elements x and y that satisfyhese conditions. If you pick ANY group with elements a and b whichatisfy these conditions, there will be a group homomorphism from D_2no that group, mapping x to a and y to b. The ONLY thing you knowbout x and y in D_2n is the x^2 =1 , y^2=1, (xy)^n=1, and the thingsou can deduce from this. No equality which cannot be deduced fromhese holds in D_2n.- t's not denial. I'm just very selective aboutwhat I accept as reality. --- Calvin (Calvin and Hobbes)rturo Magidinagidin@math.berkeley.edu ==== Subject: Re: Group presentations For example, Mathworld gives a group presentation for the Dihedral Group D_2n (the symmetry group of the regular n-gon) as: D_2n = < x, y | x^2 = 1, y^2 = 1, (xy)^n = 1 What are the x's and y's? I assume they are elements in the group of isometries of the plane since D_2n is a subgroup thereof?he standard choice consists in taking x and y to be reflections whose axesre neighbours, i.e. form an angle of pi/n. The relations are then uitebvious (recall that the product of two reflections with intersecting axess a rotation whose angle is twice the angle between these axes).his presentation is inspired by what you see in a kaleidoscope - and hat'slso why dihedral groups are called dihedral. Can we thus pick *any* x's and y's which satisfy the given relations, for example, why can't we pick x = 1 and y = 1? They satisfy the relations but clearly don't generate D_2n.he full answer to that entails the notion of a free group and quotientshereof. Roughly speaking, the idea is that x and y should be picked inuch a way thata) relations x^2=1, y^2=1 and (xy)^n=1 hold, *and*b) any other relation between x and y should be an *algebraic* consequencef these three relations.D ==== Subject: Re: Group presentations I'm struggling to understand group presentations. For example, Mathworld gives a group presentation for the Dihedral Group D_2n (the symmetry group of the regular n-gon) as: D_2n = < x, y | x^2 = 1, y^2 = 1, (xy)^n = 1 What are the x's and y's? I assume they are elements in the group of isometries of the plane since D_2n is a subgroup thereof? Can we thus pick *any* x's and y's which satisfy the given relations, for example, why can't we pick x = 1 and y = 1? They satisfy the relations but clearly don't generate D_2n. Has Mathworld missed something out? Am I missing the point?here's two ways (equivalent, of course) to look at this. The first is that _2n is the _maximal_ group satisfying those relations; there are no xtra elations beyond those. Of course, this doesn't always preclude the group eing rivial even so; it might happen that the relations given imply that x = y = . But in this case you can exhibit a nontrivial group that works.he other way is that D_2n is the quotient of the free group on two enerators y the normal subgroup generated by all the words which appear in the elations that is, x^2, y^2, (xy)^n = xyxy...xy). That is, the intersection of all ormal subgroups containing those words. You can see that this gives no xtra elationships; it uses just enough to make all those words 1 and the subset hat emains a group.n either case, x and y are not any particular things; they are symbols. You an draw an isomorphism between D_2n as written here and a certain group of lane isometries, of course, but you can also just use the symbols (and here re other interpretations I'm forgetting).- yan Reichyanr@uchicago.edu ==== Subject: BibTexhy can LaTeX not process this entry, if i cite with cite{ENG82}?ARTICLE{ENG82, AUTHOR = Robert Engle, TITLE = Autoregressive Conditional Heteroskedasticity with stimates of the Variance of UK Inflation, JOURNAL = Econometrica, YEAR = 1982, volume = 50, pages = 987-1008 ==== Subject: Re: BibTex 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: why can LaTeX not process this entry, if i cite with cite{ENG82}? @ARTICLE{ENG82, AUTHOR = Robert Engle, TITLE = Autoregressive Conditional Heteroskedasticity with Estimates of the Variance of UK Inflation, JOURNAL = Econometrica, YEAR = 1982, volume = 50, pages = 987-1008ou need quotes around the pages, I think. Anyway you should be using n en-dash (--) not a hyphen (-) to separate the page numbers.- avid Eppsteinomputer Science Dept., Univ. of California, Irvinettp://www.ics.uci.edu/~eppstein/ ==== Subject: Re: BibTex why can LaTeX not process this entry, if i cite with cite{ENG82}? @ARTICLE{ENG82, AUTHOR = Robert Engle, TITLE = Autoregressive Conditional Heteroskedasticity with Estimates of the Variance of UK Inflation, JOURNAL = Econometrica, YEAR = 1982, volume = 50, pages = 987-1008his question should be posted at the comp.text.tex newsgroup. Besides,ou should provide more details. For instance, what do you mean when youay that LaTeX cannot process this entry? (Don't reply! This isomething you should say while posting at comp.text.tex.)ose Carlos Santos ==== Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere ETAtAhUApQL6jYNcW0hWib4EVPgc7y0QsTUCFDITqMFsndm0APzGwrAYok6z6Keo t's actually a closed curve, which seems very differentfrom what we getn the plane.magine that AB is fixed on the Equator of a spherical Earth, say inouth America with A in northern Ecuador and B in far nothern Brazil.ut C initally in the Carribean Sea and start moving it west, adjustingts latitude so that the triangle keeps the same area. Eventually,ngle A reaches 180 degrees and C accordingly reaches the Equator -- buthe triangle does not collapse into a straight line. Conserving theonzero area has forced the formation of a lune with B and C as theoles (C is now in eastern Indonesia, opposite B). Sides AB and ACogether form a 180-degree arc on the Equator, while side BC is anlternate great circular arc that byapsses the Equator to the north.ow you can flip this lune through line AB, so that BC is now in theouthern Hemisphere, and continue to rotate C around the base AB. Coes through Bolivia and Brazil, all the way around to westernndonesia, opposite A, where another lune is formed. Flip the lunegain to complete tracing out the locus. The net result is a closedurve, symmetric about AB, with cusps a the points opposite A and B.f you specify a smnall lenght ofr AB and a small are for the triangle,he locus in the vicinity of AB looks like the pair of parallel linesou get on the plane.-OL ==== Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere It's actually a closed curve, which seems very differentfrom what we get on the plane. Imagine that AB is fixed on the Equator of a spherical Earth, say in South America with A in northern Ecuador and B in far nothern Brazil. Put C initally in the Carribean Sea and start moving it west, adjusting its latitude so that the triangle keeps the same area. Eventually, angle A reaches 180 degrees and C accordingly reaches the Equator -- but the triangle does not collapse into a straight line. Conserving the nonzero area has forced the formation of a lune with B and C as the poles (C is now in eastern Indonesia, opposite B). Sides AB and AC together form a 180-degree arc on the Equator, while side BC is an alternate great circular arc that byapsses the Equator to the north. Now you can flip this lune through line AB, so that BC is now in the Southern Hemisphere, and continue to rotate C around the base AB. C goes through Bolivia and Brazil, all the way around to western Indonesia, opposite A, where another lune is formed. Flip the lune again to complete tracing out the locus. The net result is a closed curve, symmetric about AB, with cusps a the points opposite A and B.K, ready on our tracks for a symbolic quantificating trigonometry... ==== Subject: Re: Equal Area Spherical Triangles Apex Locus on Sphere ETAuAhUAl7G/DoN3ibTTTtFAwwKX0m7AnQcCFQCUOVX9ZFzKEYVRs9Nd4y62DEDKNg ==== K, I'll start that.et LtA be the latitude of A, north positive, LoA be the longitude of A,ast positive, with similar nomenclature for points B and C. Set N touantities we have:os(AB) = sin(LtA)sin(LtB)+cos(LtA)cos(LtB)cos(LoA-LoB)ikewise draw Triangles NAC and NBC and get analogous expressions foros(AC) and cos(BC). With the sides of Triangle ABC thus characterized,btain the angles in that triangle by using the Law of Cosines solvedor a vertex angle; for example:os C = (cos(AB)-cos(AC)cos(BC))/(sin(AC)sin(BC))here angle C is taken to be within Triangle ABC (no longer involving). To get the sines in the denominator you must take square roots (sin = (+/-) sqrt(1-cos^2 x)); give the square roots positive signs becausehe arcs measure 180 degrees or less. Take the inverse cosines of yourngles, add them up and set the sum to the desired value.he locus may be a simple closed curve, but its algebra is not veryimple!-OL ==== Subject: Re: Delta-eps proofet q be a real number between 0 and 1, that is, let 0oo. the slope goes to oo as x->0, this suggest that weave to separate the domain [0,oo) into [0,r] and (r,oo), where r is a malleal. The slope is bounded on (r,oo) and so f is uniformly continuous onr,oo). For [0,r], the slope is too big, but luckily, the values of f aremall on [0,r], so we can use this fact to overcome the difficulty ofnfinite slope. A bit of computation shows that we have to take r to be theelta. So we are separating the region [0,oo) into [0,delta] and [delta,oo)n the following proof. a Formal Proof @heorem: f(x)=x^q is uniformly continuous on the domain [0,oo)roof:For convenience of the writing, we are going to use d and e, instead ofelta and epsilon.)et e>0 be arbitrarily given.hen set d to be the real number such that (2d)^q=e, d>0.uppose x and y are arbitrary reals such that |x-y|=0 and y>=0.ow, We are going to show that | f(y)-f(x) | < e.here are two cases.ase I: one or both of x and y is in [0,d].or this case, since |x-y|d.herefore z^(q-1) < d^(q-1). (Note that (q-1) is *negative*)o | f(y)-f(x) | = q z^(q-1) |y-x| < d^(q-1) |y-x| < d^(q-1) d < d^q < e.o in both cases, | f(y)-f(x) | < eED. ==== Subject: Re: Delta-eps proof I need to show that x^(1/m), m in pos integers, is uniformly continuous n its domain. That is and for x =c in domain ,I must show that given eps >0 there exist delta > 0 s/t | x^(1/m) - a^(1/m) | < eps whenever 0< |x - a| < delta. <--> |x - a| |x^(1/m - 1) +x^(1/m - 2)a +..+x a^(1/m -2) + a ^(1/m -1)| < eps whenever 0< |x - a| < delta I do not see how to bound (x^(1/m - 1) +x^(1/m - 2)a +..+x a^(1/m -2) + a ^(1/m -1)) Any hints....please.qr x = x^(1/2) is not uniformly continuous on it's domain [0,oo). ==== Subject: Re: Delta-eps proof> I need to show that x^(1/m), m in pos integers, is uniformly continuous n> its domain. That is and for x =c in domain ,I must show that given eps 0> there exist delta > 0 s/t> | x^(1/m) - a^(1/m) | < eps whenever 0< |x - a| < delta.> <--> |x - a| |x^(1/m - 1) +x^(1/m - 2)a +..+x a^(1/m -2) + a ^(1/m -1)| > eps whenever 0< |x - a| < delta> I do not see how to bound (x^(1/m - 1) +x^(1/m - 2)a +..+x a^(1/m -2) + > ^(1/m -1))> Any hints....please.sqr x = x^(1/2) is not uniformly continuous on it's domain [0,oo).roof?int: It's uniformly continuous on [0,1] by compactness. Andts derivative is bounded on [1,infinity), making it uniformlyontinuous there as well.***********************avid C. Ullrich ==== Subject: Re: Delta-eps proof 0 and take x and y such that |x - y| < delta^2.hen |x^(1/2) - y^(1/2)|*(x^(1/2) + y^(1/2)) < delta^2, and therefore ateast one of the numbers |x^(1/2) - y^(1/2)| and x^(1/2) + y^(1/2) ismaller than delta. In the first case, we're done. Otherwise, if^(1/2) + y^(1/2) < delta, then you also have |x^(1/2) - y^(1/2)| < delta.ose Carlos Santos ==== Subject: Re: Delta-eps proof<31dkrtF3ad2qvU1@individual.net sqr x = x^(1/2) is not uniformly continuous on it's domain [0,oo). Yes, it is! Take delta > 0 and take x and y such that |x - y| < delta^2. Then |x^(1/2) - y^(1/2)|*(x^(1/2) + y^(1/2)) < delta^2, and therefore at least one of the numbers |x^(1/2) - y^(1/2)| and x^(1/2) + y^(1/2) is smaller than delta. In the first case, we're done. Otherwise, if x^(1/2) + y^(1/2) < delta, then you also have |x^(1/2) - y^(1/2)| < < delta.m, an example where a divergent derivative doesn't ruin uniformity.he great tragedy of science - the slaying of beautiful hypothesisy an ulgy fact. -- Thomas H. Huxley. ==== Subject: Re: toplogically equevalent... hello.....doctor~ Let (X,d) be a metric space. Define a function d' :XxX -> (R+) U {0} by d'(x,y) = min {1, d(x,y)}, the minumum of 1 and d(x,y), for all x,y in X. show that (1) d' is a bounded metric for X. (2) The metric space (X,d) is topologically equivalent to the bounded metric space (X,d') ---------------------------------------------------- i can do (1) by metric conditions. but...in the (2) let x in X and e>0 then y in B_d_(x,e) => d(x,y) < e => d'(x,y) <= d(x,y) < e => y in B_d'_(x,e) y in B_d'_(x,e) => d'(x,y) < e =>min {1, d(x,y)} < e => 1 < e or d(x,y) < e if d(x,y) < e => y in B_d_(x,e) else 1 < e => (***) i can't deduce y in B_d_(x,e) in the (***) step. so, i need your advice. thank you very much for your advice.et N_e(x) = {y in X | d(x,y)0, B_L = { N_1/n(x) | x in X, n in N, 1/N (R+) U {0} by d'(x,y) = min {1, d(x,y)}, the minumum of 1 and d(x,y), for all x,y in X. show that (1) d' is a bounded metric for X. (2) The metric space (X,d) is topologically equivalent to the bounded metric space (X,d')xercise: Start with a metric d and f:R -> R, f(x) = 0 iff x = 0, f(x+y) <= f(x) + f(y)how fd = f o d is a metric.how when f is ascending and continuous at 0 that f is uniformly continuous and d,fd are equivalent metricsor your problem f(x) = min{ 1,x } and d' = fd. but...in the (2) let x in X and e>0 then y in B_d_(x,e) => d(x,y) < e => d'(x,y) <= d(x,y) < e => y in B_d'_(x,e) y in B_d'_(x,e) => d'(x,y) < e =>min {1, d(x,y)} < e => 1 < e or d(x,y) < e if d(x,y) < e => y in B_d_(x,e) else 1 < e => (***) i can't deduce y in B_d_(x,e) in the (***) step. think you want to show in B_d'(x, min(1,r)) ==== > y in B_d(x,r)ere's from my notes for the exercise:et F(a,r) = { x | fd(a,x) < r }(a,f(r)) subset B(a,r). If x in F(a,f(r)): fd(a,x) < f(r) if r <= d(a,x): f(r) <= fd(a,x) < f(r) which cannot be d(a,x) < r; x in B(a,r)ome s > 0 with B(a,s) subset F(a,r). Some s > 0 with f(s) < r if x in B(a,s): d(a,x) < s; fd(a,x) <= f(s) < r; x in F(a,r)--- ==== Subject: Re: need proof!!!k, so why is the oppside function exist??? A partial answer to some of your questions: The set R-Q is a G_delta, i.e. a countable intersection of open subsets of R. This is because R-Q= cap_{qin Q} (R-{q}). Here the complement of a singleton is obviously open, and as Q is countable, so is this intersection. Obviously this generalizes to the statement that the complement of a countable set is a G_delta. Q is not a G_delta as explained by William Elliott (the complement of a G_delta is an F_sigma, i.e. a countable union of closed sets). I don't know, whether being a G_delta is a sufficient condition for a set to be equal to the points of contuinity of some function. It is relatively easy to see that it is a necessary condition, though. This is an exercise (together with some hints) in Royden's book Real Analysis. I'm sure many other first year graduate texts in real analysis have related material as well. Jyrki Lahtonen, Turku, Finlandut i have one more problemou are based on the fact that The set of points at which f isiscontinuous must be an F_sigma set and thats o.kut i only know that: function f(x) in a single variable x is said to be continuous at point y fim x-->y f(x) = f(y)an you proof basing on the lim of a function????? ==== Subject: Re: need proof!!! but i have one more problem you are based on the fact that The set of points at which f is discontinuous must be an F_sigma set and thats o.k but i only know that: a function f(x) in a single variable x is said to be continuous at point y f lim x-->y f(x) = f(y) can you proof basing on the lim of a function????? did this as an exercise as a first year graduate student.t was good for me to work it out myself. I'm sure it's good for yous well. If you get stuck, look at Dennis May's reply:)yrki ==== Subject: Re: need proof!!! I don't know, whether being a G_delta is a sufficient condition for a set to be equal to the points of contuinity of some function. It is relatively easy to see that it is a necessary condition, though. This is an exercise (together with some hints) in Royden's book Real Analysis. I'm sure many other first year graduate texts in real analysis have related material as well.t is also sufficient.he book Counterexamples in Analysis shows a construction of a unction which is discontinuous at precisely the points of an arbitrary _sigma set. ==== Subject: Re: need proof!!! i resently asked the following question I'm looking for a continuous function f:R->R with discontinuity on irrational domain and continuous on Q. the good people who responded, told me that the answer is NO. now, does anyone knows a formal proof for this... but if you dont know just tell me whyuppose f:R->R is an arbitrary real function.or n in N, n>=1 and y in R let_n(y) = f^-1[(y-1/n, y+1/n)]_n(y) = int A_n(y) where int = topological interior_n = UNION {U_n(y) | y in R} = INTERSECTION {U_n | n in N, n>=1}f f is continuous at x: in int(f^-1[O]) whenever O is open and f(x) in Oo x in U_n(f(x)) for all no x in U_n for all no x in Cf x in C:or all n, x in U_no for each n we can find y_n in R with x in U_n(y_n)o for all n, x in int f^-1[(y_n-1/n, y_n+1/n)]hus for all n, |f(x)-y_n| < 1/nnd also for all n there is d_n>0 such thatx-z| < d_n => |f(z)-y_n| < 1/nuppose e>0 and pick N such that 1/N < e/2x-z| < d_N => |f(z)-f(x)| <= |f(z)-y_N| + |y_N-f(x)| < 1/N + 1/N < eence f is continuous at x.o C is precisely the set of points at which f is continuous.ach U_n is open since it is a union of open sets and so C is a G_delta et as required.f Q were a G_delta set, R-Q would be an F_sigma set, so there would be sequence of closed sets F_0, F_1, ... with R-Q = UNION {F_n | n in N}ach F_i contains no rational and so is nowhere dense.et (q_n) be an enumeration of Q. Each {q_n} is also a nowhere dense et. Then R = F_0 U F_1 U ... U q_0 U q_1 U ... is a countable union of owhere dense sets, which contradicts Baire's category theorem.ence Q is not a G_delta set. ==== Subject: Re: need proof!!! i resently asked the following question I'm looking for a continuous function f:R->R with discontinuity on irrational domain and continuous on Q. the good people who responded, told me that the answer is NO. now, does anyone knows a formal proof for this... but if you dont know just tell me whyhy do you ask? When you've asked your previous question, Dennis Mayiscontinuous must be an F_sigma set. The irrationals are not an F_sigmaet. If that's not a formal proof, then I have no clue about what aormal proof is. Of course, if you had written that you do notnderstand it, that would have been different. But it's a proof.nd, please, stop putting those exclamation marks at the sSubjects ofour posts.ose Carlos Santos ==== Subject: Re: Subbases, compactness ==== Subject: Re: Subbases, compactness Let S be a subbase for a topological space X. Suppose every cover of X by sets in S has a finite subcollection which covers X. Is X necessarily compact?> Alexander's Theorem.> Requires the Axiom of Choice for the proof. Rather than a laborious set theory proof, is there a simpler filterroof? The proof in Kelley's General Topology does not seem so laborious to me. It is a rather nifty application of Tuckey's Lemma, which is a version of Zorn's lemma.hat's Tuckey's Lemma? Zorn's Lemma for P(S) with subset order? The theorem we use here is more easily found as Alexander's Subbase Theorem; sometimes you will see Sub-base hyphenated.ell as included below, I did find an EasyProofOfTychonoffttp://www.theoryandpractice.org/kyle/Wiki/ EasyProofOfTychonoffovering both Alex's Lemma and Tychonov's Theorem. However for all it'slarity, it is lacking a crucial step where it is claimed G = H / Sovers X (marked **). I'm not even seeing why a maximal cover H of X witho finite subcover, would contain any subbase sets S, much less why allhe subbase sets of H would cover X. What steps are needed to show this?n the Tychonov section I corrected a couple of apparent typos ... x_i in X - / G_i for every i. ... and x in X_i - / G_i.nd seeing no need, removed the one time occurrence of bar{U} =- Alexander subbasis lemmaf X is a topological space and S is a subbasis of X, then is compact if and only if every S-cover of X has a finite subcover.roof: Let B be the collection of all open covers of X that do not haveinite subcovers. Assume, by way of contradiction, that B is nonempty.artial order B by set inclusion. Let C be a chain in B.et D be the union of all elements of C.f D has a finite subcover, then those finitely many open setsould be contained in finitely many elements of the chain C.very finite subset of the chain has a maximum element, and so each ofhose finitely many open sets that together cover X are contained inhat maximum element. This is impossible since every element of C isn element of B, and thus has no finite subcover. Thus D has no finiteubcover, so is in B, and is an upperbound for C inside B.ince C was an arbitrary chain in B, the conditions of Zorn'semma are satisfied, and thus B has a maximal element, call it H.* Consider G = H / S. We show that G covers X, so that by hypothesis itas a finite subcover, but that finite subcover is a subcover of H asell. This is a contradiction, and so the assumption that B is nonempty isntenable. Thus every open cover of X has a finite subcover, and X isompact.- Theorem (Tychonoff): Suppose that I is a set and for every i in I suppose X_i is a compact topological space.ndow X = prod_{i in I} X_i with the product topology. X is compact.roof of Tychonoff: Let S be the standard subbasis of the productopology, S = { pi_i^{-1}(U) : U is open in X_i }.y Alexander's lemma it suffices to consider S-covers.et G be an S-cover, that is let G consist of sets pi_i^{-1}(U)or various i in I and U open in X_i.efine G_i = { U : U is open in X_i, pi^{-1}_i(U) in G }.ssume BWOC, that for every i in I that / G_i /= X_i.sing the axiom of choice, choose x_i in X_i - / G_i for every i.ince x = (x_i)_{i in I} is in X, in pi_i^{-1}(U) in G for some i in I and U open in X_i.his is a contradiction since x in U in G_i and x_i in X_i - / G_i.herefore there must exist an i in I such that / G_i = X_i.ince X_i is compact, choose a finite subcover, U_1,... U_n.et C = { pi_i^{-1}(U_1),... pi_i^{-1}(U_n) }. is a finite subcover of X. Since G was an arbitrary S-cover, theonditions of Alexander's subbase lemma are satisfied, and X is compact.--- ==== Subject: Re: JSH: Look at it backwardssnipThere is only one variable Nora Baron . . .ere I agree with Mr. Harris. I once posted that there are evidently number of real people living in the United States that really areamed Nora Baron, but those individuals are constant while our Noraaron may indeed be the only one that is a variable -- although thatroperty has not yet been established for sure. ==== Subject: Re: JSH: Look at it backwardssnipThere is only one variable Nora Baron . . . Here I agree with Mr. Harris. I once posted that there are evidently a number of real people living in the United States that really are named Nora Baron, but those individuals are constant while our Nora Baron may indeed be the only one that is a variable -- although that property has not yet been established for sure.OL. I've been having a good time today. Lots of fun stuff, and someeat accomplishments.nd sci.math posters are actually getting a LOT more entertaining withome of their replies. It's a hoot.he real work though is in working through the details of *explaining*y result. Which is an interesting problem, which is also why I needo many posts as I just throw a bunch of ideas at the problem until Iind something that sticks.t's basically how I found my math results, brainstorming, but withxpository style. think I'm almost done. I've got that mathematicians ignoring myork are passive-aggressive, and using a passive-aggressive style. Iave that the work itself resolves down to acceptance of some VERYASIC mathematics known for thousands of years. And I have thexplanation for why some posters obsessively reply to my posts ashey've learned that simply disagreeing is their most potent tactic.he information can be played out in many different combinations, ands each one plays out, I use various tools to look for impact. ventually I'll get the right combination.he key is what I'm looking for, and I'm hot on the hunt now. It'sust so, terribly exciting!ames Harris ==== Subject: Re: JSH: Look at it backwards The key is what I'm looking for, and I'm hot on the hunt now. It's just so, terribly exciting!e sure and keep your zipper up.-here are two things you must never attempt to prove: the unprovable -- and he obvious.-emocracy: The triumph of popularity over principle.-ttp://www.crbond.com ==== Subject: JSH: Without a traceome of you may have noticed that I'm talking about abstractions nowt higher and higher levels, which I think should help.or instance, given a multiple of a polynomial it has been wellccepted that you can divide that multiple off without leaving arace.onsider (x) = 5(x+1)(x+2) = 5x^2 + 15x + 10nd it is true that you can divide 5 from that factorizaton givingx+1)(x+2) = x^2 + 3x + 2ithout leaving a trace. That is, there is no indication left thathe polynomial was ever multiplied by 5, as how could there be?ationally there are an *infinity* of potential multiples that youould use, so why mathematically should a factorization of x^2 + 3x +, have traces of one particular multiple, like traces of 5.ow that's obvious with polynomial factors but some sci.math'erslearly have many of you convinced that things change if you have(x) = (a_1(x) + b_1)(a_2(x) + b_2) = 5x^2 + 15x + 10nd NOW divide the 5 off, as many of you seem convinced that NOW ifhe a's and b's are somehow complex, or weird, or otherwise differentrom what you get with polynomial then maybe, hmmm, possibly, younow? Maybe there IS a trace, right?ut how?f I divide the 5 off, then I have a factorization of^2 + 3x + 2ust as before, and there is no rational reason to suppose that arace of 5 is left, as why 5? Why not 7? Or 293874983?ogically, it doesn't matter how complicated that a's and b's are inhat example, when the 5 is divided off, it goes--without a trace.he situation now where posters argue with me, with arguments thatave to boil down to some trace being left by a multiple, so that theyan argue that the multiple divides through dependent on someariable, is not unlike an argument between a scientist and people whoon't believe in evolution, but worse.n my case I have precedent from thousands of years of mathematicshat a multiple can be divided off, absolute logic, and just the plainddity of the notion that a multiple has to leave a trace when it'sivided off, but STILL people have argued with me quite successfully,nd I figure many of you, despite what I say here, remain unconvincedhat I'm right.nd you are no different than Creationists arguing with scientistsgainst evolution. Or people who don't believe man landed on theoon.ou are no different, and in fact worse, as here it's mathematics,ith absolute proof.ou people who can't accept mathematics are no different from thosether people who can't accept science. You may think you like or evenove mathematics, but you cannot when you refuse to accept even theost basic concepts in mathematics, with a result that clearly youon't like. understand the *desire* to have certain things be true, but inathematics that doesn't matter. At the end of the day, you put awayour emotions, and go with what's true--if you truly valueathematics.ames Harristtp://mathforprofit.blogspot.com/ ==== Subject: Re: JSH: Without a trace Some of you may have noticed that I'm talking about abstractions now at higher and higher levels, which I think should help. For instance, given a multiple of a polynomial it has been well accepted that you can divide that multiple off without leaving a trace. Consider P(x) = 5(x+1)(x+2) = 5x^2 + 15x + 10 and it is true that you can divide 5 from that factorizaton giving (x+1)(x+2) = x^2 + 3x + 2 without leaving a trace. That is, there is no indication left that the polynomial was ever multiplied by 5, as how could there be? Rationally there are an *infinity* of potential multiples that you could use, so why mathematically should a factorization of x^2 + 3x + 2, have traces of one particular multiple, like traces of 5. Now that's obvious with polynomial factors but some sci.math'ers clearly have many of you convinced that things change if you have P(x) = (a_1(x) + b_1)(a_2(x) + b_2) = 5x^2 + 15x + 10 and NOW divide the 5 off, as many of you seem convinced that NOW if the a's and b's are somehow complex, or weird, or otherwise different from what you get with polynomial then maybe, hmmm, possibly, you know? Maybe there IS a trace, right? But how? If I divide the 5 off, then I have a factorization of x^2 + 3x + 2 just as before, and there is no rational reason to suppose that a trace of 5 is left, as why 5? Why not 7? Or 293874983? Logically, it doesn't matter how complicated that a's and b's are in that example, when the 5 is divided off, it goes--without a trace. The situation now where posters argue with me, with arguments that have to boil down to some trace being left by a multiple, so that they can argue that the multiple divides through dependent on some variable, is not unlike an argument between a scientist and people who don't believe in evolution, but worse. Here yet again, lacking a proof, you try to convinceeople with an oversimplified toy example. Examples areot proofs. As usual, your example is a reducible polynomial.s you well know, at the heart of this controversy is theifference between reducible and irreducible polynomials. So consider an *irreducible* quadratic: a little simplerhan your cubic example, but more complicated than the uadratic example you just gave. Q(x) = 3 * (4*x^2 + 2*x - 1). = 3 * (4*x^2 + 2*(x - 2) + 3). Consider factoring this in the form Q(x) = (2*a1(x) + 3)*(2*a2(x) + 3). I will call this a Harristotelian factorization:hat is, you are factoring as if 2 is a polynomialariable. The a's satisfy the equation a^2 - (x - 2)*a + 3 x^2 = 0. Note that the roots of this equation arelways algebraic integers. For example, when x = 0, the a's are a = 0 and a = -2. This gives the factorization Q(0) = 3 * (-4 + 3) = -3. The obvious way to factor 3 out of both sides ofhis is thus: Q(0)/3 = (2*0 + 3)*(2*(-2) + 3)/3 = (0/3 + 3/3)*(-4 + 3) = -1. That is, a1(0) = 0 and is divisible by 3,nd a2(0) = -1 and is NOT divisible by 3. Now consider Q(1). In this case, the a'satisfy the equation a^2 + a + 3 = 0.his is irreducible. The two roots are a_1(1) = (-1 + sqrt(-11))/2 and a_2(1) = (-1 - sqrt(-11))/2. You can easily check that a_1(1) * a_2(1) = 3,s it should. Therefore both a_1(1) and a_2(1) are not coprime o 3. Moreover, neither a_1(1) nor a_2(1) are divisible y 3. You can easily check that too. So how should Q(1) be factored? Recall from above that Q(1) = (2 a_1(1) + 3)*(2 a_2(1) + 3). Now I will divide both sides by 3. But I m NOT going to divide the first term by 3 nd the second term by 1. That will not work,ecause 2 a_1(1)/3 is not an algebraic integer. Instead I am going to divide the first termy w_1(1) = a_1(1) and the second term by _2(1) = a_2(1) ! Does this work? Let's see --- The first term becomes (2 a_1(1) + 3)/a_1(1) = (2 + a_2(1)). The second term becomes (2 a_2(1) + 3)/a_2(1) = (2 + a_1(1)). Both of these are algebraic integers! That'sood! But I know from the definition of Q(x) that Q(1) = 15, that is, Q(1)/3 = 5. o, is it true that (2 + a_2(1)) * (2 + a_1(1)) = 5? Check it out! It works! Everything comes outight! So what's the moral of all this? The moral is, contrary to what Harris says,he RIGHT way to divide both sides of such equations,o that you get algebraic integer factors after theivision, has to be dependent on x . When x = 0,ou divide the first and second terms respectivelyy 3 and 1. When x = 1, you divide the first andecond terms respectively by w_1(1) = a_1(1) and _2(1) = a_2(1), that is, by two DIFFERENT factorsf 3. All the terms after the division are algebraicntegers. Finally: no, examples are not proofs. Examples are isproofs. [non-math blather deleted] Nora B. ==== Subject: Re: JSH: Without a trace Some of you may have noticed that I'm talking about abstractions now at higher and higher levels, which I think should help. For instance, given a multiple of a polynomial it has been well accepted that you can divide that multiple off without leaving a trace. Consider P(x) = 5(x+1)(x+2) = 5x^2 + 15x + 10 and it is true that you can divide 5 from that factorizaton giving (x+1)(x+2) = x^2 + 3x + 2 without leaving a trace. That is, there is no indication left that the polynomial was ever multiplied by 5, as how could there be? Rationally there are an *infinity* of potential multiples that you could use, so why mathematically should a factorization of x^2 + 3x + 2, have traces of one particular multiple, like traces of 5.ithout a trace has no precise meaning, (and not much meaningn any sense). You are retreating into not even wrong territory.owever, something that has disappeared without a trace is anyention of the constant term. This makes perfect sense. Eachime you brought up the constant term, everyone would ask:sn't the constant term of (a(x)/w(x) + 7/w(x)) equal to 7?nfortunately for you, you could not answer this question withoutknowleging that your entire argument was fallacious. nd too many people were asking the question for you to comfortablygnore it. So you could either admit you were wrong or retreatnto not even wrong territory. No one is surprised by your choice. - William Hughes Now that's obvious with polynomial factors but some sci.math'ers clearly have many of you convinced that things change if you have P(x) = (a_1(x) + b_1)(a_2(x) + b_2) = 5x^2 + 15x + 10 and NOW divide the 5 off, as many of you seem convinced that NOW if the a's and b's are somehow complex, or weird, or otherwise different from what you get with polynomial then maybe, hmmm, possibly, you know? Maybe there IS a trace, right? But how? If I divide the 5 off, then I have a factorization of x^2 + 3x + 2 just as before, and there is no rational reason to suppose that a trace of 5 is left, as why 5? Why not 7? Or 293874983? Logically, it doesn't matter how complicated that a's and b's are in that example, when the 5 is divided off, it goes--without a trace. The situation now where posters argue with me, with arguments that have to boil down to some trace being left by a multiple, so that they can argue that the multiple divides through dependent on some variable, is not unlike an argument between a scientist and people who don't believe in evolution, but worse. In my case I have precedent from thousands of years of mathematics that a multiple can be divided off, absolute logic, and just the plain oddity of the notion that a multiple has to leave a trace when it's divided off, but STILL people have argued with me quite successfully, and I figure many of you, despite what I say here, remain unconvinced that I'm right. And you are no different than Creationists arguing with scientists against evolution. Or people who don't believe man landed on the moon. You are no different, and in fact worse, as here it's mathematics, with absolute proof. You people who can't accept mathematics are no different from those other people who can't accept science. You may think you like or even love mathematics, but you cannot when you refuse to accept even the most basic concepts in mathematics, with a result that clearly you don't like. I understand the *desire* to have certain things be true, but in mathematics that doesn't matter. At the end of the day, you put away your emotions, and go with what's true--if you truly value mathematics. James Harris http://mathforprofit.blogspot.com/ ==== Subject: Re: JSH: Without a trace Some of you may have noticed that I'm talking about abstractions now at higher and higher levels, which I think should help.'m begging for enlightnment. For instance, given a multiple of a polynomial it has been well accepted that you can divide that multiple off without leaving a trace.ou are obviously referring to thoseicked but rational criminal ringsho go about stripping integer polynomials of heir factors leaving no finger prints,NA or other forensic evidence.t's always a shock when your polynomials areampered with in this way.owever, more brutal irrationals tendo leave polys hideously mutilatednd because of the resistance of the victimhe perpetrator's blood and other bodily luids can be left at the scene. Let's hope theorces of law and order can at least capture theserrational miscreants. ==== Subject: Re: JSH: Without a trace Some of you may have noticed that I'm talking about abstractions now at higher and higher levels, which I think should help. For instance, given a multiple of a polynomial it has been well accepted that you can divide that multiple off without leaving a trace.an't we just agree on this one bit of terminology?f you consider the number 12, say ... is a FACTOR of 12, and4 is a MULTIPLE of 12.urely it's not against your principles to use that bit of correct erminology? ==== Subject: Re: JSH: Without a tracef you really understood mathematics, you'd see where your errors are.ou're just too pompous to see that.ave ==== Subject: Re: JSH: Without a trace If you really understood mathematics, you'd see where your errors are. You're just too pompous to see that. Daveet I've made quite a few mistakes over the years trying out differentdeas, and dropping those that failed.ere posters need only do one thing: prove me wrong.nd it's a sad lie for some of them to just repeat that they have orhat I don't answer objections as I've answered objections many, many,any times over a period of years.hese people simpy will not listen to reason.hat do they appear to accomplish? My guess is that leadingathematicians who are aware of my work, pay attention to them, asvidence that they can rely on the passive-aggessive strategy ofaiting.o then sci.math may actually be having an impact after all.ou may be convincing certain people who do know that my work isorrect they have the luxury of waiting.ames Harris ==== Subject: Re: JSH: Without a trace Here posters need only do one thing: prove me wrong. And it's a sad lie for some of them to just repeat that they have or that I don't answer objections as I've answered objections many, many, many times over a period of years.es, but your answers are consistently off base. First, you misrepresent the bjections; second, you ignore theounter-examples or disproofs entirely; third, you repeat your own reviously flawed argument. In your mind, that disposesf the matter. Everyone else, however, sees that you have *not* answered the bjection, but only substituted a loud noise.f all the p-baked cranks, where 'p' equals 1/2, you take the q, where 'q' quals cake.-here are two things you must never attempt to prove: the unprovable -- and he obvious.-emocracy: The triumph of popularity over principle.-ttp://www.crbond.com ==== Subject: Re: Unstoppable Force vs Immovable Object What's that supposed to mean, I address a physical concept in a non-physical way? I conclude that it's logically impossible for an unstoppable force to meet an immovable object. Do you disagree? Do you think it's logically possible?f you consider what the word force means you would conclude there is no nstopable (i.e. infinite) force in the physical universe. Likewise, an mmovable object would have infinite momentum (physically impossible). n either case you are talking about things which not only don't exist, ut can't exist.ob Kolker ==== Subject: Re: Unstoppable Force vs Immovable Objecthat's that supposed to mean, I address a physical concept in a non-physical way? conclude that it's logically impossible for an unstoppable force to meet an immovable object. Do you disagree? Do you think it's logically possible? If you consider what the word force means you would conclude there is no unstopable (i.e. infinite) force in the physical universe. Likewise, an immovable object would have infinite momentum (physically impossible). In either case you are talking about things which not only don't exist, but can't exist. Bob Kolkero you conclude that it's logically impossible for an unstoppableorce or an immovable object to exist. I'm not sure that's correct,ut in any event it's consistent with my conclusion. So what's torgue about?