mm-2679 === Subject: Re: Lipschitz, derived numbers & all that > If the derived numbers of a function f on an interval [a,b] are all > bounded by a common constant K show that f is Lipschitz, hence of > bounded variation. What's the derived numbers of f? > Is this okay? : > let h(x) be the derived number. So |h(x)| =< K for all x in [a,b] given. > => there exists a seq. g_n -> 0 s.t. > |lim n-> infinity (f(x+g_n) - f(x) )/g_n| =< K for all x in [a,b] > Find N s.t |( f(x+g_n) - f(x) )/g_N| =< K > => |f(x+g_n) - f(x)| =< K*|g_N| > => |f(x+g_n) - f(x)| =< K*|x + g_N -x| > => |f(y) - f(x)| =< K*|y - x| if y = x + g_N > This is for any x in [a,b] but y depends on x. > But some |(f(x+g_n) - f(x))/g_n| =< K for all n >= N > then the above is true for any x an any y within g_N of x > (since h_n -> 0) > => f is Lipschitz on [x-g_N, x+g_N] for all x in [a,b] > => f is Lipschitz on [a,b] === Subject: Re: differential equation >>It could be a constant vector, but my guess is that the equation is >>actually >>r'' = -k r/|r|^3 >Well of course that was my guess as well, which is why I >didn't decide right away that k was a constant vector... >>Well, of course it has many closed-form solutions. It's >>not possible to express the general solution in closed >>form, is it? >Not as far as I know (for r as a function of t). Of course >you can eliminate t to get Kepler's conics etc. >(...) Once you have |r| in terms of the angle w, you also have r^2 dw/dt = c dt = (c/r^2)dw So integration gives t in tems of w, rather than vice versa. Still, this can be considered a solution, no? Also, it might be possible that Maple can integrate and then find an inverse function - does anybody know? === Subject: integration Is integration a one way function? More exactly Is integration a one way functional? So if I ask how to integrate complicated expression am I actually asking for help to break a code? ;-) === Subject: Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]? Ray Steiner > I came up with another way of showing that x/tan x has no elementary antiderivative. > We need only one result from Wiener's 1997 paper: > arcsin(x)/x does not have an elementary antiderivative. > Let I = int(x/tan x dx)= int (x cot x dx) > Use integration by parts to get > I = x ln(sin x) - int( ln(sin x) dx) > Let I2= int( ln(sin x) dx) > Let u= sin x, x = arcsin(u), dx = 1/sqrt(1-u^2) du > Then > I2= int ( ln(u)/sqrt(1-u^2) du) > Finally, use parts again to get > I2= ln(u) arcsin(u) - int(arcsin(u)/u du). > So, by Wiener's result, the original integral is not elementary. > More results: > By exactly the same method one can show that > I3 = int (x tan x dx) is not elementary. > Now, let's substitue u= tan x, x = arctan u, dx= 1/ (u^2 + 1) du in I3. > Then it reduces to > I4 = int( u*arctan(u)/(1+u^2) du). > so the second integral of my previous post is non-elementary. > Finally, consider > I5= int ( (arctan(x))^2 dx). > By parts, one can reduce it to integrating I4, so I5 is also non-elementary. LH === Subject: Re: Ex(~x=x), counterpart theory, and contingent identity > In another post I suggested that John Correy's ideas about non-reflexive > identity might have a rational formulation with respect to something > called counterpart theory. (I think categoreal would be the right term here--in my humble > opinion, the apparent self-contradictions associated with John's > intuitions arise from vagueness and ambiguity associated with standard > presuppositions rather than irrational error on his part. For the > record, Langholm's investigation of determinability and > indeterminability in first-order contexts includes incoherent formulas. > Exclusion negation is not informationally well-behaved.) The link, http://www.sussex.ac.uk//Users/muralir/kct_final.pdf > <>((x=x) & ~(x=x)) is discussed as well as what is actually done in counterpart theory to > exclude such an incoherent result. :-) mitch http://www.sussex.ac.uk//Users/muralir/kct_final.pdf > <>((x=x) & ~(x=x)) is discussed as well as what is actually done in counterpart theory to > exclude such an incoherent result. :-) mitch > The tie-in with contingent identity is as asserted by (1). > (1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y))) > That is, identical(x,y) are necessarily identical if, ond only if, > N(x=x) and N(y=y). Thus, granted that the Morning Star and the > Evening Star are each necessarily self-identical, if the Morning Star > and the Evening Star are identical, they are necessarily identical. > Hence the Morning Star and the Evening Star are necessarily > identical. The same is not true, however, for Benjamin Franklin > and the inventor of bifocals. For although Benjamin Franklin is > necessarily self-identical, the inventor of bifocals is not > necessarily self-identical. This is why Benjamin Franklin > and the inventor of bifocals, although identical are not > necessarily identical. > --John > I am not sure that my remarks are welcome here, I have had enough of the > remarks from (G. Frege)(II) > : stupid asshole, ing bitch, and all of that childish rhetoric, I cannot > reply to him at all. > I assume that you guys are more mature than He. > As to John's claim that: (John Correy = John Correy) is necessarily true, I > disagree. > It is not sufficient to say, x=y <-> AF(Fx <-> Fy). > Rather, it is sufficient to say, x=y <->. E!x & E!y &AF( Fx <-> Fy). > After all, (ix: x=John Correy) is (John Correy). > E!(John Correy) is just as doubtful as E!(The poster who claims that > ~Ax(x=x)). > We cannot assume that E!(John Correy) any more than we can assume > E!(Vulcan)! > (x=y) -> [](x=y) and E!x -> [](E!x), are consequences of Leibnitz's Law. > There are no contingent existences nor contingent identities. > Exisence and Identity (and Membership) are analytic properties. > Can we assume that: []Exists(George W. Bush)? > I don't think so, do you? > Surely, the existence of, George W. Bush, is contingent. > There cannot be an assumption that all names of purported physical entities > refer! > Santa dosen't work, Vulcan doesn't work, Pegasus dosen't work, etc. > For although Benjamin Franklin is necessarily self-identical, the inventor > of bifocals is not > necessarily self-identical. > Benjamin Franklin = Benjamin Franklin, is not necessarily true. > (the inventor of bifocals)=(the inventor of bifocals), is also not > necessarily true, imo. > Necessary identity and existence, applies to logical/mathematical objects, > not to emprical objects, imho. > Witt (1) states necessary and sufficient conditions for the necessity of (material) identity: (1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y))) If identicals x and y are necessarily self-identical, then--and only then--is their identity a necessary one. Beyond this, (1) states nothing more: from (1) it neither follows that John Correy is necessarily self-identical nor that John Correy is contingently self-identical--or indeed that John Correy is self-identical at all. (1) does not say which of the foregoing is the case. We can sit around and argue about whether you or I are necessarily self-identical. However, although logicians do argue about such matters--Who else would bother?--it is not as logicians that they argue but as metaphysicians, or as pataphysicians, or as what have you? So, when I claim that (e.g.) Benjamin Franklin is necessarily self-identical while the inventor of bifocals is not, my main warrant for this claim is that if Benjamin Franklin is necessarily self-identical but the inventor of bifocals is not, then Benjamin Franklin and the inventor of bifocals are not necessarily identical (although they are identical). In other words, I take the necessary self-identity of the former and the contingent self-identity of the latter to constitute, together with (1), an *explanation* for the contingent identity of Benjamin Franklin and the inventor of bifocals. To this you might object that these would be also be contingently identical if both Benjamin Franklin and the inventor of bifocals were contingently self-identical. To which I would respond that identities involving what linguistically oriented analytical philosophers refer to as rigid designators, are identities whose terms are necessarily self-identical; whereas identities involving what such philosophers refer to as non-rigid designators, are identities whose terms are contingently self-identical. Therefore, granted that I take rigid and non-rigid designation as the linguistic marks of necessary and contingent self-identity--putting the cart back behind the horse, rather than approaching the matter bass ackwards as it is fashionable to do these days--and granted that I take Benjamin Franklin and John Correy to be 'rigid' designators and 'the inventor of bifocals' to be 'non-rigid', I conclude that the contingent identity of Benjamin Franklin and the inventor of bifocals has as its basis the contingent self-identity of the inventor of bifocals, while Benjamin Franklin is necessarily self-identical. As to whether physical or mathematical objects are contingently self-identical or necessarily so, some sort of metaphysical argument (rather than a logical one) warranting one or the other of these conclusions would have to be made. I suspect that mathematical properties are both essential in, and necessary to, mathematical objects--but this is an intuition and nothing more. John PS It won't surprise me if Paul Holbach or G. Frege bring in talk about 'scope', which I think is only peripherally relevant to discussions of necessary and contingent identity. === Subject: Re: A 3rd Grade Word Problem---HELP === >Subject: Re: A 3rd Grade Word Problem---HELP >Message-id: Day #of new people >----------------------- >Sunday 1 >Monday 2 >Tuesday 4 >Wednesday 8 >Thursday 16 >Friday 32 Wormhole! >Sunday 64 >------------------------ >Sum to 127 >Thus, the problem should have stated Sunday, not Thursday. Thus, the answer should have stated Saturday, not Sunday. >I also worked the problem based on the assumption that each of the >people you told the story to told it to more people in a similar >progression, but the sum grew past 127 on the Sund-Thursday range. >Mathematically, the only remotely intersting thing is the common >observation that: >$$ sum_{i=0}^{n-1} 2^i = 2^n -1 $$ >-Daniel Simeone >> My 3rd grade son brought this word problem home the other day and he was >> given the answer (127). His job, for extra credit, was to figure out how >> get 127. I'm no genius but not a dope either. I couldn't figure out how to >> get 127. Nobody in the whole neighborhood could figure out how to get 127. >> Is this something of a trick question or is there something in the wording >> that I am missing? Any Help???? >> The Problem: >> You know a very good story. On Sunday you tell the story to a friend. On >> Monday you tell it to two new people. (So far, a total of three people >have >> heard the story). Each day after Monday, you double the number of new >> people you tell the story to. What will be the total number of people that >> will have heard your story after you tell it on Thursday? -- Mensanator Ace of Clubs === Subject: Re: torque T = r x F and basic tensors >I think you might be wrong here. A vector transforms u' = Qu, and thus >so should torque if it were really a (polar) vector.But the cross product >makes it a psuedo-vector, or so all the books say. >So if torque is really the 2nd rank anti-symm >matrix I described above, it must transform as T' = Q^{-1} T Q, right? okay okay okay... it is an _accident_ of working in three dimensions that the cross product of two vectors is also a vector... I.e. There is no reason why the cross product (defined by: C = A X B defined by C_i = eps_{ijk}A_jB_k (eps is the epsilon tensor)) should necessarily transform like a vector. but, it does... although it sure is hard to show that it does...it took me like 45 minutes to remember how to work it out. you need to use the fact that: eps_{ijk}R_{im}R_{jn}R_{kp} = eps_{mnp} this isn't too hard to show since the LHS obviously is zero when m=n or m=p or p=n. A for the case m=1 n=2 p=3 you just have the definition of the determinint which is convieniently equal to one. the other cases follow directly. then use the fact that R is orthogonal to show that: eps_{rjk}R_{jb}R_{kg} = eps_{abg}R_{ra} Finally...finally!... lets consider torque: T = r x F T_i = eps_{ijk} r_j F_k since we know how r and F transform we know T transforms to: T_i --> T'_i = eps_{ijk} R_{jm} R_{kp} r_m F_p = eps_{qmp} R_{iq} r_m F_p = R_{iq} eps_{qmp} r_m F_p = R_{iq} T_i whoo hooo. it transforms like a vector... a pseudo-vector. But yeah, it's a tensor too which means nothing less and nothing more than: T_{i,j} = (r_i F_j) - (r_j F_i) T_{i,j} ---(coordinate transformation R)--> T'_{i,j} where T'_{i,j} = R_{i,k} R_{j,m} T_{k,m} (sum on k,m) = R_{i,k} T_{k,m} R^transpose_{m,j} adam === Subject: Re: Two questions in propositional logic > We're working in propositional logic, with our basic symbols being > implies, falsehood and p_1, ... . The axioms are > p -> (q -> p) > [ p -> (q -> r) ] -> [ (p -> q) -> (p -> r) ] > ååp -> p Ah, you didn't expect us be psychic, you included your logical system. > (Where åp = ( p -> F ) ). ~p shows up better than your 1/4 p with 1/4 squeezed into one character. > 1. Write down an explicit function f(n) such that every tautology of > length n has a proof of at most f(n) lines in length. > I wasn't entirely sure what he meant by length here, Including parenthesis as wf construction with -> is (p->q) len('(p->q)') = 3 + len('p') + len('q') Random thoughts aspiring unto slight of hand beyond mere hand waving: How long does it take to convert a statement into it's disjunctive normal form? How long does it take to convert the disjunctive normal form of a tautology to t, ie f->f ? > 2. Let C be a set of propositions. We say C is a chain if for every p, q > in C either p proves q or q proves p (and not both). If the set of > primitive propositions is allowed to be uncountable, can there be an > uncountable chain? No. For C = { P_r | 0 < r in R } to be a chain you have an uncountable number of P_r |- P_s that is, an uncountable number of |- P_r -> P_s But as the number of theorems are countable... === Subject: Re: JSH: Difficult social problem Question: What does one get when one crosses David Ullrich with the progeny of a moebius strip, a parrot and a pooch? Answer: A one-sided, two-dimensional lapdog that covets crackers. === Subject: Re: torque T = r x F and basic tensors > Yes, I do know what a cross product is and how to compute it. I am > wondering > *why* torque is defined in terms of a cross product. > --Jeff > Because of the geometrical interpretation of cross product as the product of > force and moment arm, or equivalently as the product of displacement and > force perpendicular to displacement. This definition of torque follows > intuition and is mathematically simple. Halliday and Resnick discusses > this. 1. The 2 vectors r and F determine a plane in R^3 2. In this plane, let F_p be the component of F perpendicular to r. Then it makes perfect sense that |T| = |r||F_p| = |r||F| sin(a) where a is the angle between r and F. 3. If we adopt the convention that turning according to the right hand rule is positive, then we have the vector equation T = |r||F| sin(a) n where n is the unit normal to plane and this is exactly r x F. I also see how any cross product of vectors transforms as an axial vector. I have no idea why anyone would want to form a 3 by 3 anti-symmetric matrix from the components of an array of 3 numbers, like the components of the T above... What good does it do, and why does Feynman seem to claim that this is the real quantity of interest? vol 2 p. 31-8 === Subject: Re: NOVA strings and branes | > don't have any easy answer, but I am fairly certain that | > if we were to make beligerent fantasies such as the above | > into policy we would quickly come to regret our excesses. | > And if we don't we will regret them more. It is only a matter of time | > before the martyrs of al Q'uaida blow up tunnels and bridges in New York | > City (that is where the Jews are) or shoot down commercial airlines | > right here in the U.S. Then what do we do? Bite our lip, or slay Wog on | > a grand, grand scale. | > Bob Kolker | | I don't have good answers, but before we get carried away, note | that since Sept.11 2001, the terrors we have envisioned have | simply not materialized. Additionally, it is worth noting | that it appears that not even all the hi-jackers on 9/11 | were aware of the suicidal nature of the plan. It appears | that those who would see blood spilled on U.S. soil are having | difficulty recruiting people who are willing to commit both | murder and suicide and who have enough wits about them to pose a | threat of that magnitude. We should not help with recruiting. | | I suspect that in the immediate future the best thing the U.S. | could do is try to set up some form of reasonably functional | governments in Iraq and Afghanistan. | | There are no easy answers. This is not a repeat of the Second | World War. First of all, what nation are we to fight with? | Second, I don't think the world can stand that sort of devastation | again. | | As this is off topic, I will attempt to refrain from continuing. | I merely wished to point out that the views you have expressed | are not those of all U.S. citizens. Sun. Blah, blah, blah. Man, what does it take to wake people up? We are at war. We did not start it. We will be at war until every last bloody terrorist is dead or captured. It is called The War on Terrorism. It don't have to be with any particular country. You are with us or against us. It is that simple. It is only a matter of time before some of us see that great white flash of a nuke going off over here set off by some asshole terrorist. I would much rather it be them than us. Nukes are physics. This is on topic. Terrorism *has* to be wiped out or some of us will be munching on some pretty nasty radioactive stuff. FrediFizzx === Subject: Re: Two questions in propositional logic >>I've just finished being supervised on an example sheet for our 'Logic, >>Computation and Set Theory' course, and there were two questions which >>neither I, my partner, nor my superviser could actually answer. I was >>wondering if someone could give me some hints so that I could try and >>sort these out. (Preferably not more than hints, as some people haven't >>been supervised on this yet. Don't want to spoil it for them. :) >>We're working in propositional logic, with our basic symbols being >>implies, falsehood and p_1, ... . The axioms are >>p -> (q -> p) >>[ p -> (q -> r) ] -> [ (p -> q) -> (p -> r) ] >>ååp -> p >>(Where åp = ( p -> F ) ). >>Fairly standard, but people use some slightly different axioms so I just >>thought I should say which ones I'm using. > You also need to specify what the inference rules are... Sorry, of course. The only rule of inference is modus ponens. >>2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? > I doubt it. Say p < q if p |- q but not q |- p; then your chain is > totally ordered by <. First, either there exists an increasing > sequence with at least two upper bounds This is presumably under the assumption that C is uncountable? > or a decreasing sequence with at least two lower bounds > (by the traditional proof that any sequence of reals contains > a monotone subsequence: Say you have elements p_a, indexed by the > countable ordinals. Say a is dominant if p_a > p_b for all b > a. > Either there are uncountably many dominant a or not. If there > are uncountably many dominant a then the p_a with a dominant > give a cofinal decreasing set, while if there are only countably > many dominant a then the dominant a are bounded above > and you get a cofinal increasing subset.) So, changing the > notation and replacing all the wffs by their negations if > needed, you have > p_1 < p_2 < ... < q_1 < q_2. > It seems clear to me that this implies q_1 must be a > tautology (it seems like this is by compactness or > something, but I can't quite prove it this second) > and then q_2 is weaker than a tautology, contradiction. > ??? Unless you're using some extra property about the p_i, q_i in saying this that I'm not seeing (which is very possible, on account of it being Early here and I just got up. :), that needn't be true. For example take the following sequence: p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. I thought at one point that you could have a chain isomorphic to any countable ordinal, although in retrospect I'm not entirely convinced by my argument (I'm slightly worried about what happens with some of the bigger limit ordinals), but this doesn't really matter. Of course not every chain is isomorphic to an ordinal, as the reverse of a chain is also a chain. Anyway, as I said in the other post I think I've got this sorted out now (although I'll want to write my own proof of it before I'm fully happy with it). If you're interested, John's proof went roughly as follows: Define an equivalence relation on C by p ~ q if there is a bijection f from the set of primitive propositions to itself, such that q = p with all the p_i in p replaced with f(p_i). He then showed that there can be at most one element of each equivalence class in C, and that there are countably many equivalence classes greater than p (because you can take some countable subset of the whole set of primitive propositions and represent every formula in C as a formula using only this countable subset and the primitives that appear in p). He then went on to show there were only countably many elements less than p in a similar manner. I pointed out that this was rather easier, as you could just reverse the chain by negation and use the previous result. I think I can make that proof slightly shorter by considering the set of propositions in the chain provable in <= n lines and considering how 'new' primitives are introduced. Then, up to equivalence, there should be only countably (finitely?) many of these. Then a countable union of countable sets is countable. Not sure if that's going to work, but it looks plausible. David === Subject: Re: teaching descriptive statistics / proposal > The benefits, you say, are that the chiral index is > (a) simpler to compute by hand, and > (b) equal to zero *only* for symmetry. > And, a 'drawback' of skewness is that in some conceivable > distributions, which no one has ever seen in the wild(?), it is > possible to have a skewness of zero while lacking symmetry. > To me, that does not seem like much to gain. Any parameter-estimation, which is derived by aggregation/summation of some values can be cheaten by appropriate data. With skewness one outlier on the one hand can compensate a cluster of data on the other hand making the additive parameter zero. With for appropriately elaborated cases. So it may be of more interest, *in which way* our parameter can be cheaten, and if the one or the other reflects my intuition better, as I understand you, too: > Now, what I would find valuable is an index that tells me > how badly the outliers are going to disrupt my oneway > ANOVA, and another that shows how the scaling will > effect analyses that are more complicated. - I am pretty > sure that those are different, because the simple ANOVA > is robust to analyzing *ranks* ; whereas ranks are *not* > very compatible with the multi-way ANOVA or regression > where the R-squared is large. Since I didn't find powers of 3 in that computation I assume, that the coefficient is less sensible against outliers. But I didn't understand much and would like to get more information beforehand. Gottfried Helms === Subject: Re: Two questions in propositional logic >>We're working in propositional logic, with our basic symbols being >>implies, falsehood and p_1, ... . The axioms are >>p -> (q -> p) >>[ p -> (q -> r) ] -> [ (p -> q) -> (p -> r) ] >>ååp -> p > Ah, you didn't expect us be psychic, you included your logical system. >>(Where åp = ( p -> F ) ). > ~p shows up better than your 1/4 p with 1/4 squeezed into one character. >>1. Write down an explicit function f(n) such that every tautology of >>length n has a proof of at most f(n) lines in length. >>I wasn't entirely sure what he meant by length here, > Including parenthesis as wf construction with -> is (p->q) > len('(p->q)') = 3 + len('p') + len('q') > Random thoughts aspiring unto slight of hand beyond mere hand waving: > How long does it take to convert a statement into it's > disjunctive normal form? > How long does it take to convert the disjunctive normal form > of a tautology to t, ie f->f ? Yeah, we wondered about that. Unfortunately it doesn't appear to be significantly simpler to do it that way. Or if it is, then we all missed why it was. Another approach we tried was to use the deduction theorem to reduce it to the problem of finding an upper bound on the length of a proof of p_i from X with p_i a primitive proposition and |X| <= n (where n may been relabelled from the previous one here). >>2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? > No. For C = { P_r | 0 < r in R } to be a chain > you have an uncountable number of > P_r |- P_s > that is, an uncountable number of > |- P_r -> P_s > But as the number of theorems are countable... Not so. You missed the hypothesis that the number of primitive propositions was allowed to be uncountable. So, for example, { p_i -> p_i : i in I} is an uncountable set of theorems (although of course not a chain). David === Subject: Extremely bizarre number theoretical property Well, I discovered something rather very odd (at least to me) today. Let n be a positive integer, and let s(n) denote the sum of all the divisors of n. So, for example, s(6) = 1+2+3+6 = 12. It turns out there is a class of numbers with the property that a) n is a perfect square b) s(n) is a perfect square Here's some examples: s(9^2) = 11^2 s(20^2) = 31^2 s(180^2) = 341^2 s(1306^2) = 1729^2 s(1910^2) = 2821^2 s(11754^2) = 19019^2 s(17190^2) = 31031^2 s(32486^2) = 43617^2 s(38423^2) = 43491^2 s(47576^2) = 68961^2 s(48202^2) = 72219^2 s(50920^2) = 82677^2 s(51590^2) = 86583^2 s(83884^2) = 117831^2 (due to Jose Luis Gomez Pardo) What's VERY VERY VERY REALLY SUPER interesting, is that there are numbers x and y such that a) x and y are perfect squares b) s(x) and s(y) are perfect squares c)!!! s(xy)=s(x)s(y). Anyone who has taken some algebra should recognize this property immediately. Here's an example of some numbers with this property (calculation can be verified by using the above table) s(9^2*20^2) = 341^2 = 11^2*31^2 = s(9^2)*s(20^2) It turns out that just using the above table you can easily find more numbers like this. If you write a maple function to calculate the Sqrt(DivisorSum(a^2*b^2)), you can easily find more numbers like this that are not on the table above. This is one of the most amazing things I have ever seen. I am going to investigate this further, but the problem is quite complex, and I doubt I'll discover anything. It's a shame it doesn't always work though, because that would be utterly remarkable. But does anybody have any ideas what might be going on here? === Subject: Re: A Question for James Harris Giggle. >James Harris >http://mathforprofit.blogspot.com/ > ************************ > David C. Ullrich Nothing makes a pig happier than a roll in the wallow, as this giggling porker knows! http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg&width=314&height=4 00# http://www.fetchfido.co.uk/sound_files/giggle.wav === Subject: Re: JSH: Difficult social problem Discussion, linux) > Perhaps if you learned some of the standard terminology and techniques, > you'd have better luck communicating. It's like going to France, > refusing to learn any French, and complaining that the waiters won't > bring you soup. And believe me, if you don't speak any French, the > waiters in France won't bring you soup. Gosh. That means that when I visited Marseilles this summer, the whole place was crawling with imposters posing as waiters and bringing me soup. That's a little freaky. -- Now I'm informing all of you that the people arguing against me are EVIL, yes they are real, live EVIL people as mathematics is that important, so it's important enough for Evil itself to send minions like them. -- James Harris on Evil's interest in Alg. Number Theory === Subject: Re: Extremely bizarre number theoretical property > Well, I discovered something rather very odd (at least to me) today. > Let n be a positive integer, and let s(n) denote the sum of all the > divisors of n. So, for example, s(6) = 1+2+3+6 = 12. > It turns out there is a class of numbers with the property that > a) n is a perfect square > b) s(n) is a perfect square ARe there infinitely many? > What's VERY VERY VERY REALLY SUPER interesting, is that there are > numbers x and y such that > a) x and y are perfect squares > b) s(x) and s(y) are perfect squares > c)!!! s(xy)=s(x)s(y). Don't forget that s is a multiplicative function: s(ab) = s(a)s(b) whenever a and b are coprime. Your VERY VERY VERY REALLY SUPER interesting fact follows from the fact that the set {n in N: n, s(n) are square} contains two coprime numbers. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Fulton/Harris Representation Theory good? > I am trying to begin a master's thesis on representations of lie algebras, > and my background is two semesters of algebra (introduction to group, ring, > module, field, and a little galois theory), as well as analysis, linear > algebra, taking topology now, intro combinatorics, some measure theoretic > advanced probability... > Is this a good enough background to read a good chunk Fulton/Harris's > Representation theory yellow book and to write a Master's thesis on > representations of lie algebras? I noticed that tensor products are > introduced on the second page, and I would have to review alternating > powers, symmetric powers, but I think that my other algebra skills are > fairly good. I learned Algebra from Dummit and Foote... It seems to me that the only background knowledge that you need is some basic knowledge concerning differentiable manifolds. Take a look, for instance, at the first volume of Spivak's Comprehensive Introduction to Differential Geometry. Apart from that, you should be OK. Jose Carlos Santos === Subject: Re: Key Core Error Argument If y is not 0, you can't use the constant term tricks that depend on y > being 0. That's stupid. The constant term once found is distinguished by being > constant. All the trick is doing is finding it. So let me give you the example that I've used elsewhere which is > a_1(x) + 7, which has a constant term that is 7. Do you understand > what it means for it to be constant? Here's a test. If I have x=11, then I necessarily have a_1(11) + 7, right? Notice the constant term is STILL there, do you understand? Now then, if I now divide P(11) by 49, what should the constant term > be? If you answer honestly I'll be shocked. Let me try again. (a_1(y) + 7)/x is not the same as 7/x, unless a_1(y) = 0. > Ok, I can go from there Richard Henry, and note that necessarily > a_1(y) has *some* factor in common with 7, right? > So let's call that factor f, now dividing 49 from P(y) will divide f > off from a_1(y) + 7, understand? f will depend upon y. > Now then, you have a_1(y)/f + 7/f, and if f does not equal 7, what > does that tell you about the *constant* term Richard Henry? You have a_1(y)/f(y) + 7/f(y) and you could have f(0)=7 without every other value of f(y) being equal to 7. > Necessarily, you have 7/f left as the constant term for a factor of > P(11)/49, and if 7/f does not equal 1, you have a contradiction. > Understand? > The trick is to FIND the constant term, as you know that it's not > affected by the value of x, and it sits there like a rock, unaffected > by the value of x, and none of your protestations against mathematical > reality will change that fact. > James Harris === Subject: Re: impossible integrals > I've known for a long time that x^x dx and sin(x^2) dx are > impossible to integrate in elementary terms, but I have > no idea how to prove this. Are there any friendly papers > or books which give the proof? There's a short text (in french) with proofs concerning this subject at http://perso.wanadoo.fr/denis.feldmann/PDF/liou.pdf Jose Carlos Santos === Subject: Re: complex integral.....?? >int [cosz / {1+(z^3)+(z^5)}] dz = ? >contourÇ.8c |z| = 1/2 >-------------------------- > um...... i think that root of 1+(z^3)+(z^5)}] exist between -1 ~ -0.5 Well, of course there is at least one *real* root between -1 and -1/2, but the question is: what about *complex* roots z such that |z| < 1/2? That's what matters here. > thus, f(z) is analysis of |z|<1/2 My guess is that you meant analytic here, not analysis. > thus.......i think that answer is 0 > it is right?? Yes, it is right, but it doesn't seem that you understand why. Jose Carlos Santos === Subject: Re: Does compact+continuum connected+locally connected==>pathwise connected? >>Does compact+continuum connected+locally connected >>==> pathwise connected? >>Continuum connected means that any two points of the space lie >>in a continuum (= compact connected set). Usually continuums are Hausdorff, but as that was omitted... >Out of curiousity, what was the countable cofinite example? Any infinitely countable set with the cofinite topology isn't path connected, hence counterexample. Finite set with cofinite topology, tho not path connected, isn't counterexample. A set with cardinality > c and the cofinite topology is path connected Thus not counterexample. >Let U be the first uncountable ordinal. Give X = U x [0, 1) + >{infinity} the lexographic order on U x [0, 1) and have infinity >greater than all other elements. Then give it the order topology. >Itis complete and densely ordered, with endpoints, so compact >It is a linear continuum, so connected. In particular any interval >in it is connected, and the intervals generate the topology, so it >is locally connected. >> path p from a to b in linear order S, a /= b >> ==> [a,b] order isomorphic [0,1] >all you have to do is you have to show that from such a path you can >construct a path which is strictly increasing. I don't see >immediately how to do that though, if the path is something stupid >with infinitely many local maxima / minima. Oh well,.. Is there a >simple proof? path p from a to b in linear order S, a /= b ==> [a,b] order(iso)morphic [0,1] proof: wlog a < b; retract r:S -> [a,b]; rp:[0,1] -> [a,b] surjection rp([0,1]) connected, convex; a,b in rp([0,1); [a,b] subset rp([0,1) [a,b] = rp([0,1]); [a,b] continuous image separable continuum [a,b] separable multi-point linear continuum; [a,b] ordermorphic [0,1] That last step isn't a step, it's a theorem based upon the theorem a countable dense linear order sans end pts is ordermorphic Q. >> path connected linear order S ==> |S| <= c >Hmm. This one I'm less convinced by. I'm tentatively willing to >believe it to be true - certainly it looks intuitively like it might >well be right - but do you have some sort of reference I could check >for that? Put some end points on it and apply the above. ;-) If you can do that continuously, let me know, it'll save us the hassle of >but I'd like to see a proof of that. Reaching into the jumble jungle of my notes and grabing a choice handful of random coherence: path connected linear order S ==> |S| <= c. proof by assuming c < |S| some s in S with c < |(.,s]| or wlog c < |[s,.)| If { xi | s < s_xi } has last element beta: c < |[s,.)| = |[s,s_beta]| <= c which cannot be Thus let beta = lim { xi | s < s_xi } c < |[s,.)| = |/{ [s,s_xi] | xi < beta }| <= c|beta|; c < |beta| note: from 1st theorem |[s,s_xi]| = c omega_1 <= c < |beta| <= beta; some path p from s to s_(omega_1) [s,s_(omega_1)] homeomorphic [0,1]; omega_1 embeds R, not so! ---- === Subject: Re: Math dependency logic REVISED >> You yourself disputed that fact. Starting from C1-C4, >> by applying correct reasoning, one deduces Ex~(x=x). >> Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point out >> the error. >So C1-C4 |- Ex~(x = x)? I assume that's right -- I won't check. >Ullrich said it didn't? Maybe, though if so, it's likely just a minor >error (perhaps caused by context?). > Caused by context indeed. He quotes that thing I said really > a lot, but he never quotes the post where I actually _said_ it, > it's always a followup to a reply to a response. I'd like to see > the original, where I make that statement with no > at the > start of the lines... Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point out the error. No > at the start of the line! PURE, UNADULTERATED BULLRICH! >Where does this entail a denial of if you start with something true >and apply correct reasoning then the conclusion must be true? >Are you just lying? Or, shall we be charitable and assume that you're >stupid? > It _is_ quite remarkable how slow he's being about this point, > how it is that calling a proof incorrect, (correctly or incorrectly) > does not involve denying that if you start with something true > and apply correct reasoning the conclusion must be true. Of course, you don't deny that if one starts with something true and apply correct reasoning, then the conclusion must be true. But coming from you these are empty words, for you have demonstrated (on more than one occasion!) your inability to reason from C1-C4 to Ex~(x=x). C1 AxAy[x=y -> Az(z in x <-> z in y)] C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) Classification C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] (Weak Extensionality) What sets you apart from a eunuch in a harem? You 'know how it's done'--you've seen it every day--but you're unable to do it yourself! --John === Subject: Re: Math dependency logic REVISED >> Ain't it touchin' the lengths to which Logikoi will go to bail >> one another out? See Camaraderie of the Experts >I did go to that link. Here's how it begins. >Lonely, are you? > If not for the way _he_ tends to speculate on people's personal lives > when he can't refute their arguments I'd say this wasn't very nice, > pointing out what a pathetic character he must be, forced to talk to > himself like this in public where everyone can see. >Anyway, the lengths the Logikoi will go to bail each other out of >*what*? Was there something threatening in this thread? > I know _I've_ been terrified of the possible consequences. I mean > of course everything he's said here has been nonsense, but > regardless, what if someone at OSU found out that there was > someone on the internet saying bad things about me? > (Giggle. Nothing makes a pig happier than a roll in the wallow, as this giggling porker knows! http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg&width=314&height=4 00# http://www.fetchfido.co.uk/sound_files/giggle.wav === Subject: Higher Dimensional Knot Theory Hi! I've recently become interested in knots, and I was just wondering if anyone was able to help me out... I've read that it's possible to generalise knot theory to higher dimensions by embedding n-manifolds in n+2 dimensional space. Is this purely topological, or is there dependence on the metric structure? What I mean is: do you need more dimensions to embed a pseudo-Riemannian manifold in a knotted way? The only reason I ask is that Campbell's theorem from differential geometry says you generally do need more dimensions for embedding pseudo-Riemannian manifolds. I wasn't sure if this was the case for obtaining knottedness. Can anyone recommend a good book on higher dimensional knot theory? from Jonny Evans! === Subject: The Mathematics of the Human Soul boundary=----=_NextPart_000_005B_01C3A5E8.C601AC60 --------------------------------------------------------------------- General strategies of behavior acquisition with mathematically definable filters: What can this hypothesis contribute to a better understanding of man? Do you understand German? Click www2.arnes.si/~mkramb5 and choose the length of the answer (4 levels, from 1 to 270 book pages). You don?t understand German? Click www2.arnes.si/~mkramb5 and go to Level 2 for a brief presentation of the theory in English. === Subject: Re: Math dependency logic REVISED >Ain't it touchin' the lengths to which Logikoi will go to bail >one another out? > Giggle. What's fascinating is the consistency with which you reply > with non-sequiturs when it becomes apparent even to you that > you've been making an idiot of yourself. C1 AxAy[x=y -> Az(z in x <-> z in y)] C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)Classification C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] (Weak Extensionality) Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point out the error. Laud as you will valid reasoning from true premises, this is something you don't know how to do. Imagine! A Ph.D. in mathematics who. instead of pointing out that Ex~(x=x) follows from C3,C4 alone, DENIES that Ex~(x=x) follows from these premises at all!!! --John You concentrated on mathematics because it is predictable, because there is always a right answer you can check in the back of the book, because you like following very precise rules, because it allows you to escape from everyday life into a world that has nothing to do with everyday life, and because mathematics does not require the creativity that you completely lack. Keith Devlin, Commencement address to the mathematics graduating class of UC Berkeley, May 23, 1997 A performance system is designed to work in a defined task domain, accepting particular goals and seeking to reach them by some kind of highly selective search. The system must be told what goal is to be reached and must be given a description of the structure and characteristics of the task domain in which it is to operate: its problem space. [...] In contrast, a learning system, is capable of acquiring a problem space, in whole or part, by interacting with the external environment and without being instructed about it directly. A performance system is designed to work in a defined task domain, accepting particular goals and seeking to reach them by some kind of highly selective search. The system must be told what goal is to be reached and must be given a description of the structure and >characteristics of the task domain in which it is to operate: its problem space. [...] In contrast, a learning system, is capable of acquiring a problem space, in whole or part, by interacting with the external environment and without being instructed about it directly. Herbert Simon === Subject: Re: Math dependency logic REVISED >> [...] >Of course, you don't deny that if one starts with something true and >apply correct reasoning, then the conclusion must be true. Well that's progress - for some reason you've decided not to be entirely stupid today. So of course I didn't deny that. So that example is totally irrelevant regarding your (_repeated_) statement that I was _lying_ when I said nobody was denying that. Which brings us back to the beginning: If I was lying when I said nobody was denying that you should produce an example where someone _has_ denied that. Or admit that you were being either stupid or dishonest when you said I was lying. Of course neither of those is going to happen, because you have more concern for exposing evildoers than for telling the truth. > But >coming from you these are empty words, for you have demonstrated (on more >than one occasion!) your inability to reason from C1-C4 to Ex~(x=x). Oops. Back to the irrelvancies... ************************ David C. Ullrich === Subject: Re: Two questions in propositional logic >[...] >2. Let C be a set of propositions. We say C is a chain if for every p, q >in C either p proves q or q proves p (and not both). If the set of >primitive propositions is allowed to be uncountable, can there be an >uncountable chain? >> I doubt it. Say p < q if p |- q but not q |- p; then your chain is >> totally ordered by <. First, either there exists an increasing >> sequence with at least two upper bounds >This is presumably under the assumption that C is uncountable? Yes. >> or a decreasing sequence with at least two lower bounds >> (by the traditional proof that any sequence of reals contains >> a monotone subsequence: Say you have elements p_a, indexed by the >> countable ordinals. Say a is dominant if p_a > p_b for all b > a. >> Either there are uncountably many dominant a or not. If there >> are uncountably many dominant a then the p_a with a dominant >> give a cofinal decreasing set, while if there are only countably >> many dominant a then the dominant a are bounded above >> and you get a cofinal increasing subset.) So, changing the >> notation and replacing all the wffs by their negations if >> needed, you have >> p_1 < p_2 < ... < q_1 < q_2. >> It seems clear to me that this implies q_1 must be a >> tautology (it seems like this is by compactness or >> something, but I can't quite prove it this second) >> and then q_2 is weaker than a tautology, contradiction. >> ??? >Unless you're using some extra property about the p_i, q_i in saying >this that I'm not seeing No. >(which is very possible, on account of it being >Early here and I just got up. :), that needn't be true. >For example take the following sequence: >p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. ??? Is it true that p_3 v p_4 |- p_1 v p_3 ? I don't think so... Not that I can see how to prove what I said, and it may well be false, but this is not a counterexample, unless _I'm_ missing something - I did think about simple examples like this. In fact assuming that the p_j are propositional _variables_ then if p_3 v ... v p_n |- w for all n then it does follow that w is a tautology: If w is not a tautology then it is falsified by some truth assignment. Since w contains only finitely many variables, it is falsified by some truth assignment that sets some irrelevant variable to T, and that shows that p_3 v ... v p_n |- w is false for some n. I actually figured that out last night - realizing that _that_ chain cannot be followed by anything but a tautology is why I conjecture that no increasing sequence can be followed by anything but a tautology. (Any increasing sequence has the same form as above (identifying wffs that are logically equivalent), except that the p_j are wffs instead of variables. It seems like one should be able to use more or less the same argument, sort of, although I haven't worked out _exactly_ how it goes...) >I thought at one point that you could have a chain isomorphic to any >countable ordinal, although in retrospect I'm not entirely convinced by >my argument (I'm slightly worried about what happens with some of the >bigger limit ordinals), but this doesn't really matter. Of course not >every chain is isomorphic to an ordinal, as the reverse of a chain is >also a chain. >Anyway, as I said in the other post I think I've got this sorted out now >(although I'll want to write my own proof of it before I'm fully happy >with it). >If you're interested, John's proof went roughly as follows: (Sorry if there's something below I should have replied to that I'm appearing not to notice - you've forced me to stop reading at this point..) [snipped with eyes shut] ************************ David C. Ullrich === Subject: Re: Uncle Al is Sadistic . > 3 Africans I met in Computer science department in the last 4 > years were way above average in thier programming skills in > the midst of Chinese and Indian grad students who are the > overwhelming majority in that department. You will always find bright smart people in just about any naturally > occuring group of humans. Race is nothing. Culture is everything. > Bob Kolker That's my point. What point? Culture Shmulture....that is simplistic politically correct > heurist bull of the first kind. >>No, it is not. > If it were that simple >> It is THAT simple where I grew up. >>Children in the poor neighborhood with their parents making ends meet >>and the children having to do the house work like an adult and start >>working while they themslevs are children (rather than play or study, >>do not get a chance to test their brain since there is no books (not >>referring to school text books and notes) lying around the house. >> That's a lot of bull. House work (or farm work) is so boring >> one has nothing else to do but think. It also gives one an incentive >> to go to school and study real hard so that one can get a job that >> doesn't involve either. > Afriend of mine from Wisconsan said justt hat. > He had to get up and helped his father milk the cows in the farm and >it was boring for him. He became a programmer. It was possible for him >because he is the this US of A. A country which was built by people. Nobody gave them anything; they did the work themselves. Since you want us to believe that you are from an underprivileged country that contains no books, no schools, no business opportunities, I suggest that you begin to work to build an infrastructure. However, since you're complaining about having to do housework from dawn to dusk, I conclude that there exist people who are able to hire other people to clean. That implies that they make the money they pay those housecleaners in some other industry than housecleaning. Thus, that country has opportunities available for thousands. >> Housework is a very good lesson in >> eliminating some choices of employment. >Not for those who is left with no time to study when it is not just >helping around the house with chores. Excuse me? Housework has ample opportunity for study. Since it takes no thinking (or not much) to vacuum or wash a floor, one can use their brains to learn at the same time. I would set a book up, read a paragraph, and think about it while doing other work. If I still hadn't got it, I would read it again and continue doing the physical work. When I figured it out, I would read the next part of the text. In the computer biz, we called this flavor of doing many things at once, timesharing. >>It also gives one a fallback >> option for employment. >> The curious will find a way to satisfy their itch. A good resource >> is^Wwas public libraries. The curious also have to learn how to avoid >> spoon-feeding or they will lose that precious commodity. >> public libraries. This thread was talking about US schools and such. All countries may not have 100% good public libraries. Ours sucks (and it's in the USA). If you have no access to any library, go buy a text. If you have no money to buy a text, barter with a professor (or somebody who does have books) to clean his/her house in exchange for a couple of hours access to his books every week. Get a job at your country's university. Now you can even talk to people who are learning; I used to sop up lots and lots of knowledge just by listening to people talking about their work. If the country is in an industrial stage, get a job at a factory. Lots of physics, mechanics, efficiency studies, chemistry, math, accounting, management, etc. can be learned just by noticing how the company does its biz. >> You are a troll. > Not more than you are. I may be a troll, but I'm one that can figure out how to get something done with no resources. It's called work. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Uncle Al is Sadistic . replied to this post of yours. I would not have otherwsie. >Put it in your think head. You cannot demand; you can only request. HUH?! /BAH Subtract a hundred and four for e-mail. === Subject: Re: Uncle Al is Sadistic . > 3 Africans I met in Computer science department in the last 4 > years were way above average in thier programming skills in > the midst of Chinese and Indian grad students who are the > overwhelming majority in that department. You will always find bright smart people in just about any naturally > occuring group of humans. Race is nothing. Culture is everything. > Bob Kolker That's my point. What point? Culture Shmulture....that is simplistic politically correct > heurist bull of the first kind. >>No, it is not. > If it were that simple >> It is THAT simple where I grew up. >>Children in the poor neighborhood with their parents making ends meet >>and the children having to do the house work like an adult and start >>working while they themslevs are children (rather than play or study, >>do not get a chance to test their brain since there is no books (not >>referring to school text books and notes) lying around the house. >> That's a lot of bull. House work (or farm work) is so boring >> one has nothing else to do but think. It also gives one an incentive >> to go to school and study real hard so that one can get a job that >> doesn't involve either. Housework is a very good lesson in >> eliminating some choices of employment. It also gives one a fallback >> option for employment. >I agree. When I was a kid I worked on a local farm. I've decided that a good way to train kids, is to have them work on a farm. A lot can be learned by shoveling bull. A farm is a practical application of the whole range of majors offered at universities. It also teaches kids how to work which is one of the things that is lacking in all areas in the USA. > ..I also helped my >father build several rental properties when I wasn't working. My >weekly home duties included maintaining the lawns at home and at the >rental places, weeding the garden, picking vegetables when they were >ready, making dinner a few times a week, doing dishes, and washing my >own clothes. I managed to find time to read several books a week, >practice the piano, and to play soccer. >My kids (all 5 of them) have nightly and weekly chores. The older kids >(15 & 16) each make dinner at least once a week. Occasionally my wife >or I help the kids with some of the housework but they are expected to >keep it clean. Aha! Mess prevention. >Side note: Last night a parent took one of the kids on my soccer team >home early. The guy said that his son had to finish his homework. >Right. The SOB needed a drink and couldn't manage to wait another 15 >minutes. The kid is getting coping lessons early. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Uncle Al is Sadistic . >> > >> > > > > Children in the poor neighborhood with their parents making ends meet >and the children having to do the house work like an adult and start >working while they themslevs are children (rather than play or study, >do not get a chance to test their brain since there is no books (not >referring to school text books and notes) lying around the house. > That's a lot of bull. House work (or farm work) is so boring >one has nothing else to do but think. It also gives one an incentive >to go to school and study real hard so that one can get a job that >doesn't involve either. Housework is a very good lesson in >eliminating some choices of employment. It also gives one a fallback >option for employment. > >>I agree. When I was a kid I worked on a local farm. I also helped my >>father build several rental properties when I wasn't working. My >>weekly home duties included maintaining the lawns at home and at the >>rental places, weeding the garden, picking vegetables when they were >>ready, making dinner a few times a week, doing dishes, and washing my >>own clothes. I managed to find time to read several books a week, >>practice the piano, and to play soccer. >>My kids (all 5 of them) have nightly and weekly chores. The older kids >>(15 & 16) each make dinner at least once a week. Occasionally my wife >>or I help the kids with some of the housework but they are expected to >>keep it clean. >>Side note: Last night a parent took one of the kids on my soccer team >>home early. The guy said that his son had to finish his homework. >>Right. The SOB needed a drink and couldn't manage to wait another 15 >>minutes. >You and your kids have/had chores. These kids have jobs. >http://www.theatlantic.com/issues/96feb/pakistan/pakistan.htm >not that they get paid. One of the societal problems in the USA is that kids are not allowed to be gainfully employed. /BAH === Subject: Re: complex integral.....?? > int [cosz / {1+(z^3)+(z^5)}] dz = ? contourÇ.8c |z| = 1/2 -------------------------- > um...... i think that root of 1+(z^3)+(z^5)}] exist between -1 ~ -0.5 > thus, f(z) is analysis of |z|<1/2 > thus.......i think that answer is 0 > it is right?? All the roots of a_n*z^n + a_{n-1}z^(n-1) + ... + a_1*z + a0 verify 1/(1 + B/|a_0|) < |z| < 1 +A/|a_n| Where A = max(|a_0|, |a_1|, ..., |a_{n-1}|) and B = max(|a_1|, ..., |a_{n-1}|, |a_n|) Then, you are right ... -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: complex integral.....?? Ignacio Larrosa Ca.96estro escribi.97 > int [cosz / {1+(z^3)+(z^5)}] dz = ? contourÇ.8c |z| = 1/2 -------------------------- > um...... i think that root of 1+(z^3)+(z^5)}] exist between -1 ~ -0.5 > thus, f(z) is analysis of |z|<1/2 > thus.......i think that answer is 0 > it is right?? > All the roots of a_n*z^n + a_{n-1}z^(n-1) + ... + a_1*z + a0 verify > 1/(1 + B/|a_0|) < |z| < 1 +A/|a_n| If a_n * a_0 =/= 0, of course ... > Where A = max(|a_0|, |a_1|, ..., |a_{n-1}|) and B = max(|a_1|, ..., > |a_{n-1}|, |a_n|) > Then, you are right ... > -- > Ignacio Larrosa Ca.96estro > A Coru.96a (Espa.96a) > ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Differentiation Is differentiation a one way function? More exactly Is differentiation a one way functional? So if I ask how to integrate complicated expression am I actually asking for help to break a code? ;-) ---- === Subject: Re: Two questions in propositional logic >>[...] >>2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? > I doubt it. Say p < q if p |- q but not q |- p; then your chain is > totally ordered by <. First, either there exists an increasing > sequence with at least two upper bounds >>This is presumably under the assumption that C is uncountable? >Yes. > or a decreasing sequence with at least two lower bounds > (by the traditional proof that any sequence of reals contains > a monotone subsequence: Say you have elements p_a, indexed by the > countable ordinals. Say a is dominant if p_a > p_b for all b > a. > Either there are uncountably many dominant a or not. If there > are uncountably many dominant a then the p_a with a dominant > give a cofinal decreasing set, while if there are only countably > many dominant a then the dominant a are bounded above > and you get a cofinal increasing subset.) So, changing the > notation and replacing all the wffs by their negations if > needed, you have p_1 < p_2 < ... < q_1 < q_2. It seems clear to me that this implies q_1 must be a > tautology (it seems like this is by compactness or > something, but I can't quite prove it this second) > and then q_2 is weaker than a tautology, contradiction. ??? >Unless you're using some extra property about the p_i, q_i in saying >>this that I'm not seeing >No. >>(which is very possible, on account of it being >>Early here and I just got up. :), that needn't be true. >>For example take the following sequence: >>p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. >??? Is it true that p_3 v p_4 |- p_1 v p_3 ? I don't think so... >Not that I can see how to prove what I said, and it may well >be false, but this is not a counterexample, unless _I'm_ >missing something - I did think about simple examples >like this. In fact assuming that the p_j are propositional >_variables_ then if p_3 v ... v p_n |- w for all n then it >does follow that w is a tautology: If w is not a tautology then >it is falsified by some truth assignment. Since w contains >only finitely many variables, it is falsified by some >truth assignment that sets some irrelevant variable to T, >and that shows that p_3 v ... v p_n |- w is false for some n. >I actually figured that out last night - realizing that _that_ >chain cannot be followed by anything but a tautology is >why I conjecture that no increasing sequence can be >followed by anything but a tautology. (Any increasing >sequence has the same form as above (identifying >wffs that are logically equivalent), except that the p_j >are wffs instead of variables. It seems like one should >be able to use more or less the same argument, >sort of, although I haven't worked out _exactly_ >how it goes...) Never mind - ths above is not a counterexample, but of course what I conjectured is false. Realized that I wanted to show that the intersection of any strictly decreasing sequence of open subsets of the Cantor set had empty interior, realized immediately that that was false, hence the example p_1 & (p_2), p_1 & (p_2 v p_3), ... p_1. (Well, if you had an uncountable chain you could get an increasing chain isomorphic to omega_1 as above, and I bet it's impossible to have a strictly decreasing omega_1-sequence of open subsets of the Cantor set... but never mind, you say you have a proof.) >>I thought at one point that you could have a chain isomorphic to any >>countable ordinal, although in retrospect I'm not entirely convinced by >>my argument (I'm slightly worried about what happens with some of the >>bigger limit ordinals), but this doesn't really matter. Of course not >>every chain is isomorphic to an ordinal, as the reverse of a chain is >>also a chain. >>Anyway, as I said in the other post I think I've got this sorted out now >>(although I'll want to write my own proof of it before I'm fully happy >>with it). >>If you're interested, John's proof went roughly as follows: >(Sorry if there's something below I should have replied >to that I'm appearing not to notice - you've forced me to >stop reading at this point..) >[snipped with eyes shut] >************************ >David C. Ullrich ************************ David C. Ullrich === Subject: Re: torque T = r x F and basic tensors > 2. How is torque a tensor? Following Feynman vol 2 ch 31, a tensor is a > linear map A relating 2 vector quantities, for example p = Ae (where I am > using p for the polatization vector of a dielectric and e for the electric > field vector) and hence if we change the basis of R^3, A must transform as > A' = Q^{-1} A Q where V is the new ordered basis, U is the old and > V = UQ^{-1} > I can see Feynman gets the 3 by 3 anti-symmetric matrix > T_{ij} = x_i F_j - x_j F_i i,j = 1,2,3 > How would this matrix object ever be used for torque? > If it has any relevance, I can derive the formula > Q( a x b ) = 1/(detQ) ( Qa x Qb ) > where Q is the matrix of the change of basis, but I'm not quite seeing > how that matches the A' = Q^{-1} A Q form. As you say, under the change of orthonormal basis, the vectors a and b go to a' = Qa and b' = Qb. In component form, a_i = (sum over m) Q_(im) a_m and b_i = (sum over n) Q_(in) b_n I've used different summation indices in order that the substitutions below makes sense. Because both the new and old bases are orthonormal, Q is an orthogonal matrix, i.e., Q^T = Q^(-1), i.e if M = Q^(-1), M_(ij) = Q_(ji). Thus, det Q = +/- 1. If in addition, right-handed orthonormal bases are taken to right-handed orthonormal bases, then det Q = 1, and Q is called a special orthogonal matrix. The set of all special orthogonal matrices forms the group SO(3), where the notation is fairly self-explanatory - S for special (det +1), O for orthogonal (inverse = transpose), 3 for 3 by 3. Define matrix A by A_(ij) = a_i b_j - a_j b_i. Under a transformation, A'_ij = a'_i b'_j - a'_j b'_i = (sum over m,n) [Q_(im) a_m Q_(jn) b_n - Q_(jn) a_n Q_(im) b_m] = (sum over m,n) [Q_(im) a_m b_n Q_(jn) - Q_(im) a_n b_m Q_(jn)] = (sum over m,n) [Q_(im) (a_m b_n - a_n b_m) Q_jn] = (sum over m,n) [Q_(im) A_(mn) Q^(-1)_nj] A = Q A Q^(-1) This is a little different than A = Q A Q^(-1) because, the way Feynman defines things, a'_i = (sum over n) Q^(-1)_(in) a_n. It's a good exercise to show this. George === Subject: Re: Key Core Error Argument > ... > > Actually it is, as Dik Winter is trying to find a way that 7, 7 and 22 > > become 1, 1 and 22 based on a *varying* x. > > That is, he's trying to make the change in *constants* dependent on a > > variable. Eh? > I want readers to understand that his behavior is crank, while I guess > > many of you may sympathize with his strong desire for me to be wrong, > > remember, it's not about people as the math didn't just decide to > > change. > > What you should be sympathetic to, is the truth. You claim that having P(x) = g1(x).g2(x).g3(x) with g1(0) = g2(0) = 7 > and g3(0) = 22; the *only* way to divide P(x) by 49 is by dividing g1(x) > and g2(x) by 7 and g3(x) by 1. Because now g1(0)/7 = 1, g2(0)/7 = 1 and > g3(0)/1 = 22. I claim there are other ways to do that. Have w1(x), w2(x), w3(x), such > that w1(0) = w2(0) = 7, w3(0) = 1, w1(x).w2(x).w3(x) = 49. Because now > g1(0)/w1(0) = 1, g2(0)/w2(0) = 1 and g3(0)/w3(0) = 22. Where do I change constants? The basic algebra is that if you have 7 and divide it by *something* > and get 1, then you divided by 7. And no tricks will change that fact, and it's simply crank behavior to > try and act like there's some complicated way you can divide 7 to get > 1, without actually just dividing it by 7. > Since nobody is claiming to be doing anything else, your > diatribe seems a bit pointless. > At x=0: You divide by 7, Dik divides by 7 At x<>0: You divide by 7, Dik divides by something other than 7 The two factorizations are thus different. But the constant > terms depend only on what happens at x=0. Thus the constant > terms are the same. -William Hughes That's the kind of odd illogic which shows a crank, and I'll explain > quickly why your approach is specious. So I have the factors (a_1(x) + 7) and people like yourself and Dik > Winter want desperately to believe that some variable factor of 49 > divides through so that the factor itself has a varying factor of 7. I've pointed out that the constant term of its corollary factor is 1. So let's pick x=9 and see what happens, as then you have a_1(9) + 7 This is nonsense. The constant term of (a1(x) +7)/w(x) is > (a1(0) +7)/w(0). The value (a_1(9) + 7)/w(9) has no bearing. Since the factor is a_1(x) + 7, if x=9, it is a_1(9) + 7 as I said. > > At x = 9 Dik divides by w1(9) The constant term doesn't change with x, because it is constant. The constant term does not depend on what Dik divides > by at x=9. Do you understand that in general the factor is a_1(x) + 7? Now then, if you understand that, do you understand that at x=9, it is > a_1(9)+7? Maybe if that's gotten out of the way you can see the rest. The factor is (a_1(x) +7) > The value of the factor is (a_1(x) +7) > The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7 > At x=0 > The value of (a_1(x) +7) is (a_1(0) + 7) > The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7 > At x=1 > The value of (a_1(x) +7) is (a_1(1) + 7) > The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7 > At x=2 > The value of (a_1(x) +7) is (a_1(2) + 7) > The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7 > At x=9 > The value of (a_1(x) +7) is (a_1(9) + 7) > The constant term of (a_1(x) +7) is (a_1(0) + 7) = 7 > Note that the value of (a_1(x) + 7) depends on x > Note that the constant term of (a_1(x) +7) > does not depend on x > Note that the constant term of (a_1(x) + 7) does not > depend on the value (a_1(9) + 7) > Now let's add the factor w(x) > The factor is (a_1(x)/w(x) + 7/w(x)) > The value of the factor is (a_1(x)/w(x) +7/w(x)) > The constant term of (a_1(x)/w(x) +7/w(x)) is > (a_1(0)/w(0) + 7/w(0)) = 7/w(0) You keep writing a ratio because apparently you think a ratio is more powerful or mysterious, capable of doing something that it can't. Now then, if you admit that a_1(x)/w(x) is an algebraic function, it can be replaced by f(x), and if you admit that 7/w(x) is an algebraic function, it can be replaced by g(x), so then you have f(x) + g(x). But the constant term of f(x) + g(x) is 1, so let h(x) + 1 = f(x) + g(x), to isolate constant terms as before. It's that result h(x) + 1 that you're terrified of, which is why you keep writing ratios as if algebra is mysticism! You are a crank William Hughes who has seized upon ratios as a way to promote your illogical position. The important factorization is (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 where you you have a clear separation between constant terms and varying terms, which is why 1(1)(22) equals the constant term of 300125 x^3 - 18375 x^2 - 360 x + 22. What you're trying to do is *hide* the simple reality by using ratios!!! It's *crank* behavior and anti-math, as you're trying to refute a logical, mathematical position by attempting to fool people using a dodge. James Harris === Subject: Re: Usenet Posting Guide? >> With all respect, dear man, you strike me as that type more enamored >> with process rather than function. Most of us out here have far better >> things to do than screw around for weeks configuring newsreaders and >> the like, even if we were so inclined, which most of us clearly are >> not. I suppose we could memorize the phone book, too, but would that >> help us communicate our ideas any better? >> Some of us prefer to view the forest rather than count the trees down >> there. Better view, too. >Well, I find the technical aspects of how USENET functions much more >interesting than most of the discussions that take place on it. By the >same token, I am much more interested in the hardware and operating system >software of the systems I administer than in any of the applications >for which they are used. >Once I hoped that the growing popularity of personal computers meant >that nearly everyone would learn to think like programmers. It never >occurred to me that, sadly, the opposite would happen: That computers >would be designed to be used by people who *can't* program them. But that's always been the case. We made a lot of money doing work that our customers didn't want or have to do. What we didn't do (when we weren't ing idiots) was hide all the warts under pretty pictures. /BAH Subtract a hundred and four for e-mail. === Subject: parallelizability of manifolds In my self-studies of differential topology, I am not sure if I understood the concept of parallelizability correctly. A n-dim. manifold M is said to be parallelizable iff its tangent bundle TM is diffeomorphic to (M x R^n). Is this equivalent to the following: two tangent vectors (at maybe different points on the manifold) are identical in one chart if and only if they are identical in all charts (with the appropriate range). The following argument makes me think that this is false, but I do not quite understand why. I have read that S^1, S^3 and S^7 are parallelizable. Because there are no specific charts mentioned, I assume that this is the case for every atlas equivalent to the atlas consisting of the two stereographic projection charts. But this is not true; even for S^1, it is easy to show that, given any non-constant differentiable curve on S^1, tangent vectors in one chart may be parallel while they are not in the other one. So how to imagine parallelizability? And how can it be proven? Are the tori T^n:=(S^1)^n parallelizable? TIA, Tobias -- reverse my forename for mail! === Subject: Re: Usenet Posting Guide? > Once I hoped that the growing popularity of personal computers meant > that nearly everyone would learn to think like programmers. It never > occurred to me that, sadly, the opposite would happen: That computers > would be designed to be used by people who *can't* program them. >That ought to be positive, it is called job security. I would shudder >when everybody at our institute thought like a programmer. Yep. Even in our neck of the woods, where we were getting paid to be programmers, we had strict rules about what not to do. > ..You would >have umpteen versions of all software around, and hope you could in some >way couple the lot. There would always be two versions that did exactly opposite things. The way it used to be was that enough information was shipped with each computer so that mere users had access to more detailed knowledge about how the system worked. They could get as sophisticated as they wanted to. Thus a system could service those who really weren't interested in how the system did their bidding and provide the hard/software to those who got interested in how a system got things done. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Greek Alphebet .93 Ioannis [NonBreakingSpace].8b.96.87À.8c .97.99.95 .93fi.92.9b.93.87 > .93 Thomas Bushnell, BSG [NonBreakingSpace].8b.96.87À.8c .97.99.95 .93fi.92.9b.93.87 > Nowadays there is only one accent, but in ancient Greek and in its > modern > equivalents, there were two accents: The grave (tilde) and the sharp > (regular accent). What you saw is probably an omega with a grave. > No, ancient Greek had three accents: acute, grave, and circumflex. > The circumflex is sometimes written as a tilde or a macron. > Well, I really don't know their names in English, but upon more careful > inspection, you are probably right. I meant to write circumflex and > acute,which is still missing the grave. > Acute = ' (oxeia) > Circumflex = ~ (perispomeni) > Grave = ? (bareia) > I don't recall what the grave looks like, as in modern Greek it doesn't > exist, at all. It faces the opposite way of the oxeia. To mention that other simbols not in use today are: -psili (pneyma). -dasia (pneyma). -ypogegrameni (under the eta and the omega). -- E' mai possibile, oh porco di un cane, che le avventure in codesto reame debban risolversi tutte con grandi puttane! F.d.A === Subject: Factorization dispute It turns out that I can isolate the current dispute easily enough by focusing on the factorizations: Consider (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 where even by inspection you can see that the constant terms are separated out, so that you have 1(1)(22) = 22, the constant term of the polynomial. I'll add that at x=0, a_1(0) = a_2(0) = b_3(0) = 0. Notice, (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where again you see that the constant terms match as now you have 7(7)(22) = 1078, which is again the constant term of the polynomial. If 22 does not have 7 as a factor, the former factorization is the *only* allowed way for 49 to divide through. (For more detail, like what the a's are, see http://mathforprofit.blogspot.com/ where more is explained.) I have a result that shows a problem with a definition that mathematicians have used for over a hundred years, and rather than face the result which follows from rather basic algebra mathematicians are being pussies and running like scared cowards from the result. Some posters, not professional mathematicians from what I've gathered, are at least trying to stand and fight, but because what I have is a math proof, their claims are necessarily irrational. I need you to stand up for the truth, here and now. Let's chase these mathematician cowards down, and make them face the music. After all physics had its challenges and physicists faced them, while mathematicians seem to believe that they can just ignore problems. Let's take 'em out. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: Two questions in propositional logic >>I've just finished being supervised on an example sheet for our 'Logic, >>Computation and Set Theory' course, and there were two questions which >>neither I, my partner, nor my superviser could actually answer. I was >>wondering if someone could give me some hints so that I could try and >>sort these out. (Preferably not more than hints, as some people haven't >>been supervised on this yet. Don't want to spoil it for them. :) >>We're working in propositional logic, with our basic symbols being >>implies, falsehood and p_1, ... . The axioms are >>p -> (q -> p) >>[ p -> (q -> r) ] -> [ (p -> q) -> (p -> r) ] >>ååp -> p >>(Where åp = ( p -> F ) ). >>Fairly standard, but people use some slightly different axioms so I just >>thought I should say which ones I'm using. > You also need to specify what the inference rules are... Sorry, of course. The only rule of inference is modus ponens. >>2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? > I doubt it. Say p < q if p |- q but not q |- p; then your chain is > totally ordered by <. First, either there exists an increasing > sequence with at least two upper bounds > This is presumably under the assumption that C is uncountable? > or a decreasing sequence with at least two lower bounds > (by the traditional proof that any sequence of reals contains > a monotone subsequence: Say you have elements p_a, indexed by the > countable ordinals. Say a is dominant if p_a > p_b for all b > a. > Either there are uncountably many dominant a or not. If there > are uncountably many dominant a then the p_a with a dominant > give a cofinal decreasing set, while if there are only countably > many dominant a then the dominant a are bounded above > and you get a cofinal increasing subset.) So, changing the > notation and replacing all the wffs by their negations if > needed, you have p_1 < p_2 < ... < q_1 < q_2. It seems clear to me that this implies q_1 must be a > tautology (it seems like this is by compactness or > something, but I can't quite prove it this second) > and then q_2 is weaker than a tautology, contradiction. ??? Unless you're using some extra property about the p_i, q_i in saying > this that I'm not seeing (which is very possible, on account of it being > Early here and I just got up. :), that needn't be true. > For example take the following sequence: > p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. Pointed out a few minutes ago that this is not a counterexample. Of course what I said is wrong - a counterexample is p_1 & (p_2), p_1 & (p_2 v p_3), ... p_1. (Realized I wanted to show that the intersection of a strictly decreasing sequence of open subsets of the Cantor set had empty interior. Realized that that was false. Now, I bet that there's no strictly decreasing omega_1-sequence of subsets of the Cantor set, but never mind, the proof you already have is simpler.) Sorry if I already said this - I just had a PC go up in smoke, not sure where I was when that happened... > I thought at one point that you could have a chain isomorphic to any > countable ordinal, although in retrospect I'm not entirely convinced by > my argument (I'm slightly worried about what happens with some of the > bigger limit ordinals), but this doesn't really matter. Of course not > every chain is isomorphic to an ordinal, as the reverse of a chain is > also a chain. > Anyway, as I said in the other post I think I've got this sorted out now > (although I'll want to write my own proof of it before I'm fully happy > with it). > If you're interested, John's proof went roughly as follows: > Define an equivalence relation on C by p ~ q if there is a bijection f > from the set of primitive propositions to itself, such that q = p with > all the p_i in p replaced with f(p_i). > He then showed that there can be at most one element of each equivalence > class in C, and that there are countably many equivalence classes > greater than p (because you can take some countable subset of the whole > set of primitive propositions and represent every formula in C as a > formula using only this countable subset and the primitives that appear > in p). > He then went on to show there were only countably many elements less > than p in a similar manner. I pointed out that this was rather easier, > as you could just reverse the chain by negation and use the previous > result. > I think I can make that proof slightly shorter by considering the set of > propositions in the chain provable in <= n lines and considering how > 'new' primitives are introduced. Then, up to equivalence, there should > be only countably (finitely?) many of these. Then a countable union of > countable sets is countable. Not sure if that's going to work, but it > looks plausible. > David === Subject: Re: parallelizability of manifolds >In my self-studies of differential topology, I am not sure if I understood >the concept of parallelizability correctly. >A n-dim. manifold M is said to be parallelizable iff its tangent bundle TM >is diffeomorphic to (M x R^n). Is this equivalent to the following: two >tangent vectors (at maybe different points on the manifold) are identical >in one chart if and only if they are identical in all charts (with the >appropriate range). It's hard for me to make sense of this last sentence, but in the best rendering I can give it, the answer is, no, that's not what parallelizable means. >So how to imagine parallelizability? And how can it be proven? Surely the simplest way to understand the concept of parallelizability is this: the n-manifold M is parallelizable if and only if there exist n vectorfields V_i on M with the property that, at each point x of M, the n vectors V_i(x) are a basis of the tangent space of M at x. No imagination required. >Are the >tori T^n:=(S^1)^n parallelizable? Yes, indeed. (And, more generally, the product of parallelizable manifolds is parallelizable.) Lee Rudolph === Subject: Re: Two questions in propositional logic >[...] >>2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? [...] > If you're interested, John's proof went roughly as follows: > Define an equivalence relation on C by p ~ q if there is a bijection f > from the set of primitive propositions to itself, such that q = p with > all the p_i in p replaced with f(p_i). > He then showed that there can be at most one element of each equivalence > class in C, Because if p ~ q and p |- q then a permutation of the variables in the proof shows that q |- p. >and that there are countably many equivalence classes > greater than p (because you can take some countable subset of the whole > set of primitive propositions and represent every formula in C as a > formula using only this countable subset and the primitives that appear > in p). ??? Isn't it clear that there are only countably many equivalence classes altogether, and so we're done? I mean any wff involving only n variables is ~ - equivalent to one involving only p_1,... p_n, and up to logical equivalence there are only finitely many of those. Oh. That's not right with the ~ you defined. But why not instead say p ~ q if there exists a permutation of the variables which makes q into a wff logically equivalent to p, instead of requiring that it literally become p? > He then went on to show there were only countably many elements less > than p in a similar manner. I pointed out that this was rather easier, > as you could just reverse the chain by negation and use the previous > result. > I think I can make that proof slightly shorter by considering the set of > propositions in the chain provable in <= n lines and considering how > 'new' primitives are introduced. Then, up to equivalence, there should > be only countably (finitely?) many of these. Then a countable union of > countable sets is countable. Not sure if that's going to work, but it > looks plausible. > David === Subject: Re: parallelizability of manifolds > Surely the simplest way to understand the concept of parallelizability > is this: the n-manifold M is parallelizable if and only if there exist > n vectorfields V_i on M with the property that, at each point x of M, > the n vectors V_i(x) are a basis of the tangent space of M at x. > No imagination required. Ah, that's interesting! So why are there no such vector fields for S^2? > Yes, indeed. (And, more generally, the product of parallelizable > manifolds is parallelizable.) Ok, this is clear because the tangent bundle of a product is the product of the tangent bundles. -- reverse my forename for mail! === Subject: Re: Two questions in propositional logic > Anyway, as I said in the other post I think I've got this sorted out now > (although I'll want to write my own proof of it before I'm fully happy > with it). > If you're interested, John's proof went roughly as follows: > Define an equivalence relation on C by p ~ q if there is a bijection f > from the set of primitive propositions to itself, such that q = p with > all the p_i in p replaced with f(p_i). > He then showed that there can be at most one element of each equivalence > class in C, and that there are countably many equivalence classes > greater than p (because you can take some countable subset of the whole > set of primitive propositions and represent every formula in C as a > formula using only this countable subset and the primitives that appear > in p). > He then went on to show there were only countably many elements less > than p in a similar manner. I pointed out that this was rather easier, > as you could just reverse the chain by negation and use the previous > result. I found this the most natural approach. However (and IBL will probably kill me for saying anything at all about this sheet in public) you should also be aware there is a one word answer to this question. That is, there is one particular word (familiar to everyone) which instantly makes this question a triviality. I'll leave you to guess what it is. By the way you didn't ask about the last question (if the set of primitive propositions is allowed to be uncountable is it true given a set S of propositions that you can find an independent set of propositions equivalent to S); does that mean you've solved it? I still don't have it, one year after taking the course. No hints please! Michael === Subject: Decimal Expansion If m> Surely the simplest way to understand the concept of parallelizability >> is this: the n-manifold M is parallelizable if and only if there exist >> n vectorfields V_i on M with the property that, at each point x of M, >> the n vectors V_i(x) are a basis of the tangent space of M at x. >> No imagination required. >Ah, that's interesting! So why are there no such vector fields for S^2? There's not even the *beginning* of such a sequence of vectorfields for S^2: no matter what the vectorfield V_1 on S^2, there is always some point x of S^2 at which V_1(x) does not belong to any basis of the tangent space of S^2 at x. That's just a (seemingly more complicated, but ultimately worthwhile) way of saying that every (continuous!) vectorfield on S^2 has at least one zero. And that, in turn, follows (after building the appropriate machinery relating differential topology to algebraic topology) from the fact that the Euler characteristic of S^2 is non-zero. In general, you could (and people do) ask, given an n-manifold M, what is the smallest k such that there is some sequence of k vectorfields V_i on M such that, at every point x of M, the vectors V_i(x) span the tangent space of M at x. Alternatively (somewhat dually) you could (and people do) ask, given a generic sequence of k vectorfields V_i on M, what is the nature of the subset of M at every point of which the vectors V_i(x) are linearly dependent. If you pursue such questions, you eventually find yourself studying characteristic classes (in homology as well as in cohomology); the Euler class is just the tip of that particular iceberg. Lee Rudolph If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2) with the c's algebraic integers, I think few of you would have a problem realizing that only two of the c's have 7 as a factor. But, of course, you're looking at *functions* of x, as you have f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, so I could also write it as (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2). Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2 as long as you're in a ring where 7 is not a factor of 22. I want to emphasize that point as notice there's only *one* way to divide through by 49 if 7 is not a factor of 22. Usually you can *see* the other factors of 7, but I want you to abstract, and generalize. Please pay careful attention to that example. You may see people who reply claiming that the word polynomial has some significance, as if it's a mystical thing which refutes basic logic, so if something isn't polynomial it no longer behaves logically. Now then, in my advanced factorization work, I just use functions of x that are a lot more complicated than f_1(x) = c_1 x, and unfortunately there are people who can use an unfamiliar leap in complexity to confuse others. Some of you have learned various advanced math topics, now imagine if in your classrooms there were some hecklers who continually hollered out at your teacher, or otherwise disrupted the class? What if when there were difficult concepts those hecklers would try to confuse everyone as they sought to discredit the mathematics? If you find that hard to imagine, imagine me in your class with you questioning the professor and calling him names. How much would you have learned? I need those of you interested in mathematics to focus on the basics, so that you can understand the advanced. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: Factorization dispute > It turns out that I can isolate the current dispute easily enough by > focusing on the factorizations: > Consider [snipped] Crank Information http://www.crank.net/harris.html http://www.crank.net/usenet.html === Subject: Re: Difficult social problem > It looks like I'm swinging at tissue paper with a sledgehammer when it > comes to getting acceptance of my work, as while I can get initial > contact with mathematicians they tend to run as soon as I give them > enough information to realize the implications of my work and that I > am correct. Mr. Harris, even if your work were correct (and it is not), it would be as awe inspiring as my morning piss after waking up. It would hardly be worth a mention anywhere, but hey, it has only taken you 7 + years to form a wrong page of nonsense (I dare not refer to it as proof). You are so full of yourself and don't even consider the possibility that you could be wrong. Get a reality check. === Subject: Re: parallelizability of manifolds > There's not even the *beginning* of such a sequence of vectorfields > for S^2: no matter what the vectorfield V_1 on S^2, there is always > some point x of S^2 at which V_1(x) does not belong to any basis of > the tangent space of S^2 at x. That's just a (seemingly more > complicated, but ultimately worthwhile) way of saying that every > (continuous!) vectorfield on S^2 has at least one zero. And that, > in turn, follows (after building the appropriate machinery relating > differential topology to algebraic topology) from the fact that the > Euler characteristic of S^2 is non-zero. Indeed, algebraic topology seems to be an extremely powerful tool: S^2 is simply connected, so every continuous image of it also is; in contrast, R^2{(0,0)} is not. Realizing that this gives the proof for S^2 was a pretty cool heureka-moment! Maybe it is possible to use higher homotopy groups for the n-spheres with n>2, but one has to be careful for n=7 :-) > In general, you could (and people do) ask, given an n-manifold M, > what is the smallest k such that there is some sequence of k > vectorfields V_i on M such that, at every point x of M, the > vectors V_i(x) span the tangent space of M at x. Alternatively > (somewhat dually) you could (and people do) ask, given a generic > sequence of k vectorfields V_i on M, what is the nature of the > subset of M at every point of which the vectors V_i(x) are > linearly dependent. If you pursue such questions, you eventually > find yourself studying characteristic classes (in homology as well > as in cohomology); the Euler class is just the tip of that particular > iceberg. Sounds very interesting, but I'm afraid that it is a lot ahead of me now... -- reverse my forename for mail! === Subject: Re: Usenet Posting Guide? > > Once I hoped that the growing popularity of personal computers meant > > that nearly everyone would learn to think like programmers. It never > > occurred to me that, sadly, the opposite would happen: That computers > > would be designed to be used by people who *can't* program them. > That ought to be positive, it is called job security. I would shudder > when everybody at our institute thought like a programmer. You would > have umpteen versions of all software around, and hope you could in some > way couple the lot. Well, I wasn't talking *just* about programming computers. Primarily I meant that I'd hoped that everyone would get so far into the programming mindset that they'd approach everyday life as a sort of programming exercise. It's much easier to deal with people who approach everything in a series of logical steps and who don't introduce extraneous issues. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: {Group Theory} Confusing group theory conundrum > For instance, an algebraist says: We have a group G. Then it must > have the identity element G_e. But by saying this, she has > implicitly used such a function. > I don't see why she has implicitly used such a function. Well, let's look at an analog. We have a number N. Then it must have the square N_s. Admittedly, this is rather peculiar terminology and notation, but I did that purposefully to show how we can thus throw a veil over the fact that we are, in this case, implicitly using the function f(x)=x^2. When we conjure up the identity of an abstract group, the situation is very similar, with the exception that there is no easy formula for the group's identity like there is for a number's square. I admit I am probably much less experienced than you. Please know I have much respect for your advanced mathematical knowledge and, if you still believe I am mistaken, I look forward to being corrected and set on the path toward increased understanding of these clandestine matters. Sniz Pilbor === Subject: Re: Key Core Error Argument >So let me give you the example that I've used elsewhere which is >a_1(x) + 7, which has a constant term that is 7. Do you understand >what it means for it to be constant? But by your definition, the constant term of b(x) = a_1(x)+7 would be the value of b(0) = a_1(0) + 7. Who says a_1(0) has to be 7? What if a_1(x) = sqrt(x^2+1) + 0.5(x^3-5)? >Here's a test. >If I have x=11, then I necessarily have a_1(11) + 7, right? >Notice the constant term is STILL there, do you understand? Yes, but if a_1(0) is divisible by 7, e.g. a_1(x) = sqrt(x+49), who says that a_1(11) is still divisible by 7? >Now then, if I now divide P(11) by 49, what should the constant term >be? I dunno. What's the constant term of sqrt(x+49)/7 + 7/7? What's the value at x=0? - Randy I'm trying to find an e-version of Church's On the Concept of a Random help? gimma === Subject: Re: Uncle Al is Sadistic . >> >My kids (all 5 of them) have nightly and weekly chores. The older kids >(15 & 16) each make dinner at least once a week. Occasionally my wife >or I help the kids with some of the housework but they are expected to >keep it clean. >Side note: Last night a parent took one of the kids on my soccer team >home early. The guy said that his son had to finish his homework. >Right. The SOB needed a drink and couldn't manage to wait another 15 >minutes. > >>You and your kids have/had chores. These kids have jobs. >>http://www.theatlantic.com/issues/96feb/pakistan/pakistan.htm >>not that they get paid. >> >One of the societal problems in the USA is that kids are not >allowed to be gainfully employed. Well, of course if you want your six year old to support you, you are out luck. OTOH, kids older that 14 can work, and older than 12 in agricultural jobs in the US. The rules, of course, vary from country to country. That you don't see them, is your problem. josh halpern Last I looked there were lots of kids working in lots of jobs. >/BAH === Subject: Re: Ex(~x=x), counterpart theory, and contingent identity > In another post I suggested that John Correy's ideas about non-reflexive > identity might have a rational formulation with respect to something > called counterpart theory. (I think categoreal would be the right term here--in my humble > opinion, the apparent self-contradictions associated with John's > intuitions arise from vagueness and ambiguity associated with standard > presuppositions rather than irrational error on his part. For the > record, Langholm's investigation of determinability and > indeterminability in first-order contexts includes incoherent formulas. > Exclusion negation is not informationally well-behaved.) The link, http://www.sussex.ac.uk//Users/muralir/kct_final.pdf > <>((x=x) & ~(x=x)) is discussed as well as what is actually done in counterpart theory to > exclude such an incoherent result. :-) mitch http://www.sussex.ac.uk//Users/muralir/kct_final.pdf > <>((x=x) & ~(x=x)) is discussed as well as what is actually done in counterpart theory to > exclude such an incoherent result. :-) mitch The tie-in with contingent identity is as asserted by (1). (1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y))) That is, identical(x,y) are necessarily identical if, ond only if, > N(x=x) and N(y=y). Thus, granted that the Morning Star and the > Evening Star are each necessarily self-identical, if the Morning Star > and the Evening Star are identical, they are necessarily identical. > Hence the Morning Star and the Evening Star are necessarily > identical. The same is not true, however, for Benjamin Franklin > and the inventor of bifocals. For although Benjamin Franklin is > necessarily self-identical, the inventor of bifocals is not > necessarily self-identical. This is why Benjamin Franklin > and the inventor of bifocals, although identical are not > necessarily identical. --John > I am not sure that my remarks are welcome here, I have had enough of the > remarks from (G. Frege)(II) > : stupid asshole, ing bitch, and all of that childish rhetoric, I cannot > reply to him at all. > I assume that you guys are more mature than He. > As to John's claim that: (John Correy = John Correy) is necessarily true, I > disagree. > It is not sufficient to say, x=y <-> AF(Fx <-> Fy). > Rather, it is sufficient to say, x=y <->. E!x & E!y &AF( Fx <-> Fy). > After all, (ix: x=John Correy) is (John Correy). > E!(John Correy) is just as doubtful as E!(The poster who claims that > ~Ax(x=x)). > We cannot assume that E!(John Correy) any more than we can assume > E!(Vulcan)! > (x=y) -> [](x=y) and E!x -> [](E!x), are consequences of Leibnitz's Law. > There are no contingent existences nor contingent identities. > Exisence and Identity (and Membership) are analytic properties. > Can we assume that: []Exists(George W. Bush)? > I don't think so, do you? > Surely, the existence of, George W. Bush, is contingent. > There cannot be an assumption that all names of purported physical entities > refer! > Santa dosen't work, Vulcan doesn't work, Pegasus dosen't work, etc. > For although Benjamin Franklin is necessarily self-identical, the inventor > of bifocals is not > necessarily self-identical. > Benjamin Franklin = Benjamin Franklin, is not necessarily true. > (the inventor of bifocals)=(the inventor of bifocals), is also not > necessarily true, imo. > Necessary identity and existence, applies to logical/mathematical objects, > not to emprical objects, imho. > Witt > (1) states necessary and sufficient conditions for the necessity of > (material) identity: > (1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y))) > If identicals x and y are necessarily self-identical, then--and > only then--is their identity a necessary one. Beyond this, > (1) states nothing more: from (1) it neither follows that > John Correy is necessarily self-identical nor that John > Correy is contingently self-identical--or indeed that John > Correy is self-identical at all. (1) does not say which of > the foregoing is the case. > We can sit around and argue about whether you or I are necessarily > self-identical. However, although logicians do argue about such > matters--Who else would bother?--it is not as logicians that they > argue but as metaphysicians, or as pataphysicians, or as what have > you? Good point. That is one of the reasons that I have always known that there is a philosophical element involved with discussions of identity. > So, when I claim that (e.g.) Benjamin Franklin is necessarily > self-identical while the inventor of bifocals is not, my main > warrant for this claim is that if Benjamin Franklin is necessarily > self-identical but the inventor of bifocals is not, then Benjamin > Franklin and the inventor of bifocals are not necessarily identical > (although they are identical). In other words, I take the necessary > self-identity of the former and the contingent self-identity of the > latter to constitute, together with (1), an *explanation* for the > contingent identity of Benjamin Franklin and the inventor of bifocals. > To this you might object that these would be also be contingently > identical if both Benjamin Franklin and the inventor of bifocals were > contingently self-identical. To which I would respond that identities > involving what linguistically oriented analytical philosophers refer > to as rigid designators, are identities whose terms are necessarily > self-identical; whereas identities involving what such philosophers > refer to as non-rigid designators, are identities whose terms are > contingently self-identical. Therefore, granted that I take rigid and > non-rigid designation as the linguistic marks of necessary and > contingent self-identity--putting the cart back behind the horse, > rather than approaching the matter bass ackwards as it is > fashionable to do these days--and granted that I take > Benjamin Franklin and John Correy to be 'rigid' designators > and 'the inventor of bifocals' to be 'non-rigid', I conclude > that the contingent identity of Benjamin Franklin and the inventor > of bifocals has as its basis the contingent self-identity of the > inventor of bifocals, while Benjamin Franklin is necessarily > self-identical. Personally, I still have some doubts about self-identical and not self-identical. But I am now fairly certain that if one accepts Frege's argument for the definition of number as attaching to an object, one should logically accept this distinction. > As to whether physical or mathematical objects are contingently > self-identical or necessarily so, some sort of metaphysical argument > (rather than a logical one) warranting one or the other of these > conclusions would have to be made. I suspect that mathematical > properties are both essential in, and necessary to, mathematical > objects--but this is an intuition and nothing more. > It won't surprise me if Paul Holbach or G. Frege bring in talk > about 'scope', which I think is only peripherally relevant to > discussions of necessary and contingent identity. Well, when you claim that you can define scope so that self-identity is always implicit, you must deal with a Kantian possibility--... The logical determination of a concept by reason is based upon a disjunctive syllogism, in which the major premiss contains a logical division (the division of the sphere of a universal concept), the minor premiss limiting this sphere to a certain part, and the conclusion determining the concept by means of this part. --Immanuel Kant Critique of Pure Reason A577/605 ...--namely, the non-self-identical. Of course, the concept is still a fiction in Kantian epistemology. However, it is also the reason for the apparent complexity of his ideas--he does not trivially assert self-identity as self-evident. :-) mitch === Subject: Re: Usenet Posting Guide? > Some of us prefer to view the forest rather than count the trees down > there The point was this, back in the days when you had to think before you posted, those who, to borrow your analogy, visited the forest actually appreciated it, and contributed to the ecology of the forest, rather than now, where too many people use the forest as a place to hold their drug and alcohol parties, while thrashing the forest. -- Entities are not to be multiplied without necessity. --Bishop William William of Occam === Subject: Re: Factorization dispute [all newsgroups except sci.math snipped] mathematicians have used for over a hundred years, and rather than > face the result which follows from rather basic algebra mathematicians > are being pussies and running like scared cowards from the result. No you don't. And nobody's running. You can't refute the errors others have pointed out and have resorted to ad hominem attacks and cut and pasting of your entire argument assuming that if you repeat it often enough, then people will fail to respond to one of you posts and you can then assume you must finally be right. > Some posters, not professional mathematicians from what I've gathered, > are at least trying to stand and fight, but because what I have is a > math proof, their claims are necessarily irrational. Nonsense. Maybe if your proof was flawless you could claim something like that, but you have a serious error in step 6 of your argument. (think about this - *everybody* who has responded to you has pointed this out). You assume that because you can divide the terms of your polynomial by 49 that you can show that two of the factors must be divisible by 7, but that's just not true. You haven't come close to proving that. And in your new and improved short form argument you start with: > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 You start off with your conclusion and work backward, proving what? Nothing at all. You *assume* that the variable terms of the first two factors are divisible by 7. Nice trick. > I need you to stand up for the truth, here and now. Making appeals to the gallery again... I'm in the gallery, and I'll stand with the mathematicions on this one. You're argument is flawed. > Let's chase these mathematician cowards down, and make them face the > music. Hilarious. Make my music King Crimson's 21st Century Schizoid Man. > After all physics had its challenges and physicists faced them, while > mathematicians seem to believe that they can just ignore problems. HIlarious. > Let's take 'em out. To do that you will need to correct your errors (if you can...). I will echo the advice another poster gave you yesterday. Get a text on Abstract Algebra and work through the theorems. Ask for help here when you get stuck. In a few months you will be a better amateur mathematician. You've already given 6-7 years to this effort. Taking a few months off for education will give you the potential for not wasting the next 7 years. --Stan Gula === Subject: Re: Greek Alphebet by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA876Jr16266; Back in the 1970's when I took a course in astrophysics, the professor told us that this was called a Cambridge pi. He didn't know why it was called that. ---Steve Maye >.91Ä G. A. Edgar .91[CapitalEth].91ñ.93»[EDouble Dot]±.93Ì.91[Micro] .93¡.93.b3.91Ë .91.b9.91¬.91ü.93á[EDo ubleDot].b9.91± >> Here is a web page on Greek letters. >> <http://www .math.ohio-state.edu/~edgar/GreekChartLarge.jpg> Maybe you mean the old-fashioned cursive pi (see the lower right), >> nowadays still used in astronomy for perihelion. >Hmmm, so that explains _why_ some students in my elementary school and high >school were writing this symbol instead of pi on their theses. >Apparently it was used interchangeably with the regular pi. I have no idea >why. >> -- >> G. A. Edgar >http://www.math.ohio-state.ed u/~edgar/http://users.forthnet.gr/ath/jgal /Eventually, _everything_ is understandable === Subject: Re: ELLIPTICAL INTEGRAL PROBLEM by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA876Ju16262; I don't have the handbook,so could u please send me copy of the section 17 if have it thanx Samer >
 Hi All
>> I have a problem in solving the following integrals
>> which falls in to the form of elliptical integrals.
>> K = LImits 0-PI/2 Integral of 1/sqrt(1-k^2sin^2Theta) w.r.t Theta.
>> E = Limits 0-PI/2 Integral of sqrt(1-k^2sin^2Theta) w.r.t Theta.
>> Please help in finding the values of K and E
>> Sairam
>>       http://www.dejanews.com/   
  Search, Read, Post to Usenet
>Sairam,
>Look in Abramowitz and Stegun, _Handbook of Mathematical Functions_ , 
>section 17.
>John
>
=== Subject: Re: (!?) need fast algorithm for calculating large factorials by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA876Nx16348; > Can someone point me to information on the web about fast factorial > calculation? This needs to be for exact value. I can handle the >part >>about > it being too large to represent, but would like to find a faster >method >>than > just multiplying every number. Adam, Try this one http://www.luschny.de/math/ind ex.htm.>It's labview. NI is sporting a contest to code the fastest factorial >>program. I have a few ideas on how to calculate numbers larger than >>representable in normal computer formatting, but wanted to see what >>algorithms might I use that would be quicker than 2x3x4... Just for funs >>tho. >> An interesting property is that for n = 2m, >> n! = 2^m (m!)^2 >I used to think of myself as somewhat of a math wiz but I'm not getting this >equation. Cant seem to make it balance for sample numbers. In fact looking >at {n,m}={4,2}, I dont see how it could work at all. 4! contains a factor >of 3 which I dont see the right side coming up with no matter how I look at >it. Can you clue me in as to what I am missing? As submitted it only works for m = 0 or 1 (2m)! = 2^m(m!)^2 = 2m(2m-1)! = 2^m*m(m-1)!*m(m-1)! = (2m-1)! = 2^(m-1)*m(m-1)!*(m-1)! This implies 2m-1 divides m or m-1 true for 0 and 1 only since m>1 would make 2m-1 > m > m-1 or 2m > m+1 > m. Integers a > b => a does not divide b. === Subject: Re: Logic Question by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA876Lj16323; >How can one use the numbers 1through 9 only in a 3 digit plus three digit problem and come up with a 3 digit answer without using any of the numbers again? >I am stumped...and so are all of my other master's level friends! Makes us wonder how we got through grad school! Please help! Gosh Genie, I know the anwer to that question. You answer mine and I'll answer yours. How about it? Does an MBA count? Or is it just your speechy buds? Grad school sounds good, but sometimes it is just a bunch of crapola-huh? === Subject: Re: The crazy counterfeit coin problem strikes again! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA876Jm16270; What if one doesn't need to find which blocks are Ds? How many less weighings will it take? === Subject: Re: lengths of segments in a pentagram....... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA876IX16250; >i'm trying to find James Choike's modern proof that there are ratios >of lengths of segments in a pentagram that are irrational. i know >the proof was published in the college of mathematics journal in >1980 on pg 312-316 and using the school library is out of the >question because we only carry the journal as far back as 1984, can >anyone point me to a website or maybe another mathematics book that >may lead to this proof. i would like it to describe it to a >class.... >SLH you should have a look at http://www.maa.org/pubs/cmj.html. Hope it helps M. Doerfner === Subject: Re: Fulton/Harris Representation Theory good? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA876K516295; I would use Humphreys as the primary text (Springer GTM 9), with Fulton / Harris for additional examples and some interesting tangents. === Subject: Re: Who contributed most to mathematics? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA876Jb16283; Hi John, have it below. I admit the political statement was offbase and should have not have been included in my origional post. I'm pissed at the French Bashing. I was just trying to see who in fact borned the most mathematicians. My SWAG list was not intended to be complete but just a start. Ok, I blew it after France at least according to MacTutor link below. However I was close with Italy, Germany and former Russia but totally underestimated England and the U.S. and way over estimated India and Greece. This is a quantative analysis. The next step is qualitative analysis which is why I asked you guys the question. My SWAG Italy Germany Russia India Greece England Japan China United States Egypt Hungary Austria Serbia >There sure have been a lot of important French mathematicians; >Fourier, Galois, d'Alembert... Fermat - but his last theorem was published after his death. So during his life time it did not play a major role in mathematical developement. The same can be said of Da Vinci and other Greats whose certain works were suppressed during their lifetime. >but the Greeks started it. Just about every thing the Greeks did was rediscovered by others as it was needed in their time. Much of Archemedies work was for War efforts as they were needed. Sure they have being first name recognition but played no more role than India and other countries to thought today. >However, I really suspect that the Germans have to take top honors. >mathematicians. Germans are third in the list below. >Euler and Gauss. I wonder how much of these guy's work was invented or discovered by >John Savard This was taken from MacTutor on Mathematicians by country. You can visit the site below to get a good biography of some 1468 mathematicians from 350 A.D. to 1997. http://www-history.mcs.st-andrews.ac.uk/history/index.html I delimited it in Excel and copied below. It is interesting that France ranks no. 1. However, if one placed a weighting on each member, a different first place might occur. For example Sir Issac Newton would pull heavily for England. Then again if you add Scotland=44,Ireland=18,Wales=5 Great Britain would take the helm. Similarly, we can combine Russia and the Ukraine to get 118 surpassing Italy and the United States. Then if we look at just Europe, we will have 1163 or 80% of the total. Then we can do it per capita, by country etc. Just joking! Also interesting is that Iraq produced more than China or Japan. This is probably due to the non-availability of written documentation in China and Japan and the pictographic nature of their languages. Perhaps the arabic derived number system we use today played a great part in the evolution of mathematical thought. I suspect there is a lot more that can be learned from this list by simply asking questions as the ones above. China 10? There was no time for math, only for defense.the wall and agricultural subsistence. The math that was done was probably kept secret and destroyed if necessary so the enemy could not use it against them. France,England 415? Sailing ships to conquer the world required mathematical skills so it was an important for the educational system to nurture future navel officers and war planners. War (hot and cold) was probably the single greatest influence on mathematical progress. This is unfortunate. I think the numbers below if done by year would bear that out. Conquering countries demand mathematicians and produce them. Some of you were taken aback by not seeing certain countries on my SWAG list. Please do not be offended by the list below. That is all it is - just a list of Mathematicians by place of birth. It is not by religion or genetics. Isreal shows 2 but Poland and Germany had many many Jews which borned many mathematicians in those countries. Cino The straight list of the number mathematicians by place of birth up to 1997 Algeria 1 Azerbaijan 1 Argentina 1 Gibraltar 1 Haiti 1 Indonesia 1 Jordan 1 Malta 1 Mexico 1 Moldavia 1 Slovenia 1 Tajikistan 1 Morocco 1 Estonia 2 Georgia 2 Israel 2 Lebanon 2 Libya 2 Lithuania 2 Luxembourg 2 New Zealand 2 Pakistan 2 South Africa 2 Croatia 3 Portugal 3 Uzbekistan 3 Latvia 4 Australia 5 Finland 5 Romania 5 Slovakia 5 Syria 5 Wales 5 Canada 7 Belarus 8 Japan 9 China 10 Norway 10 Iran 11 Spain 11 Sweden 11 Czech Republic 13 Egypt 13 Iraq 17 Denmark 18 Ireland 18 Belgium 19 Turkey 22 Greece 25 Netherlands 25 Switzerland 27 Austria 28 Hungary 29 Ukraine 37 India 39 Scotland 44 Poland 68 Russia 81 Italy 101 USA 108 Germany 162 England 204 France 211 Total 1462 === Subject: Re: Uncle Al is Sadistic . >>Sure, the I don't fail you you fail you bull. He didn't >>educate anyone either. He didn't go out and get more >>resources for more students. >When you run a race, you don't automatically award first >place to every runner. > There were six hundred places. What is the difference between the 599th > person in that class and the 601st in terms of ability? The wonderprof > simply abdicated any responsiblity for educating the students. Quite > common. Exactly. That is why a rational society only cares about win, place, and show rather than compassionately about a huge pool of incompetents. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Factorization dispute Nothing. http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.net -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Uncle Al is Sadistic . > Exactly. That is why a rational society only cares about win, place, > and show rather than compassionately about a huge pool of > incompetents. Not necessarily incompetent, but folks of middling ability, as most of us are. Remembers it is the grunts who provide the ballast to keep the ship stable in the water. It is the general labor force that make it possible for capitalists to be rich (in addtion to the genius of the capitalist). On a desert island an enterprising capitalist will survive quite nicely, but he will not be able to build an empire unless he can have it done by Friday. We all have some part to play. Bob Kolker === Subject: Re: Ex(~x=x), counterpart theory, and contingent identity > (1) states necessary and sufficient conditions for the necessity of > (material) identity: > (1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y))) > If identicals x and y are necessarily self-identical, then--and > only then--is their identity a necessary one. Beyond this, > (1) states nothing more: from (1) it neither follows that > John Correy is necessarily self-identical nor that John > Correy is contingently self-identical--or indeed that John > Correy is self-identical at all. (1) does not say which of > the foregoing is the case. > We can sit around and argue about whether you or I are necessarily > self-identical. However, although logicians do argue about such > matters--Who else would bother?--it is not as logicians that they > argue but as metaphysicians, or as pataphysicians, or as what have > you? > So, when I claim that (e.g.) Benjamin Franklin is necessarily > self-identical while the inventor of bifocals is not, my main > warrant for this claim is that if Benjamin Franklin is necessarily > self-identical but the inventor of bifocals is not, then Benjamin > Franklin and the inventor of bifocals are not necessarily identical > (although they are identical). In other words, I take the necessary > self-identity of the former and the contingent self-identity of the > latter to constitute, together with (1), an *explanation* for the > contingent identity of Benjamin Franklin and the inventor of bifocals. > To this you might object that these would be also be contingently > identical if both Benjamin Franklin and the inventor of bifocals were > contingently self-identical. To which I would respond that identities > involving what linguistically oriented analytical philosophers refer > to as rigid designators, are identities whose terms are necessarily > self-identical; whereas identities involving what such philosophers > refer to as non-rigid designators, are identities whose terms are > contingently self-identical. Therefore, granted that I take rigid and > non-rigid designation as the linguistic marks of necessary and > contingent self-identity--putting the cart back behind the horse, > rather than approaching the matter bass ackwards as it is > fashionable to do these days--and granted that I take > Benjamin Franklin and John Correy to be 'rigid' designators > and 'the inventor of bifocals' to be 'non-rigid', I conclude > that the contingent identity of Benjamin Franklin and the inventor > of bifocals has as its basis the contingent self-identity of the > inventor of bifocals, while Benjamin Franklin is necessarily > self-identical. > As to whether physical or mathematical objects are contingently > self-identical or necessarily so, some sort of metaphysical argument > (rather than a logical one) warranting one or the other of these > conclusions would have to be made. I suspect that mathematical > properties are both essential in, and necessary to, mathematical > objects--but this is an intuition and nothing more. > John > PS It won't surprise me if Paul Holbach or G. Frege bring in talk > about 'scope', which I think is only peripherally relevant to > discussions of necessary and contingent identity. Let me bring in a quote: One must distinguish between the claim that identity sentences are contingent and the claim that the identity relation itself is contingent. For the relation to be contingent, there need to be things between which it holds merely contingent. For it to be necessary, it has to be that if the relation obtains between things, it obtains between those very things of necessity. [...] One can consistently say that there are contingent identity sentences, though the relation itself is necessary. Thus one could say that The first Postmaster-General of the US was the inventor of bifocal lenses. is contingent and is an identity sentence, but that if we consider the object, x, which is in fact referred to by the first Postmaster-General of the US and the object, y, which is in fact referred to by the inventor of bifocal lenses, it is necessary that x is identical to y. [Sainsbury, M. (1995). Philosophical Logic. In A. C. Grayling (Ed.), /Philosophy. A Guide Through the Subject/ (pp. 61-122). Oxford: Oxford University Press. (p. 93)] PH === Subject: Re: Uncle Al is Sadistic . Littlemanwearingbigboypants whines: >>Sure, the I don't fail you you fail you bull. He didn't >>educate anyone either. He didn't go out and get more >>resources for more students. >When you run a race, you don't automatically award first >place to every runner. >>There were six hundred places. What is the difference between the 599th >>person in that class and the 601st in terms of ability? The wonderprof >>simply abdicated any responsiblity for educating the students. Quite >>common. > Exactly. That is why a rational society only cares about win, place, > and show rather than compassionately about a huge pool of > incompetents. In your form of rational society you'd have been recycled a long time ago. You're redefining rational society to suit your rather warped fundie worldview. A successful society can afford to care about all its members. A successful society *is* rational. === Subject: Re: Uncle Al is Sadistic . > 3 Africans I met in Computer science department in the last 4 > years were way above average in thier programming skills in > the midst of Chinese and Indian grad students who are the > overwhelming majority in that department. You will always find bright smart people in just about any > naturally > occuring group of humans. Race is nothing. Culture is everything. > Bob Kolker That's my point. What point? Culture Shmulture....that is simplistic politically > correct > heurist bull of the first kind. >>No, it is not. > If it were that simple >> It is THAT simple where I grew up. >>Children in the poor neighborhood with their parents making ends meet >>and the children having to do the house work like an adult and start >>working while they themslevs are children (rather than play or study, >>do not get a chance to test their brain since there is no books (not >>referring to school text books and notes) lying around the house. >> That's a lot of bull. House work (or farm work) is so boring >> one has nothing else to do but think. It also gives one an incentive >> to go to school and study real hard so that one can get a job that >> doesn't involve either. > Afriend of mine from Wisconsan said justt hat. > He had to get up and helped his father milk the cows in the farm and >it was boring for him. He became a programmer. It was possible for him >because he is the this US of A. > A country which was built by people. Nobody gave them anything; > they did the work themselves. Since you want us to believe that > you are from an underprivileged country that contains no books, > no schools, no business opportunities, I suggest that you begin > to work to build an infrastructure. However, since you're complaining > about having to do housework from dawn to dusk, I conclude that > there exist people who are able to hire other people to clean. > That implies that they make the money they pay those housecleaners > in some other industry than housecleaning. Thus, that country > has opportunities available for thousands. Can you read well? When did I ever say I had to work from dawn to dusk? First of all, I wasn't talking about myself. Secondly, here is how my day started. Get up, brush my teeth and wash my face. Have breafast which was already on the the table. Get ready for school. Grab my lunch box and go wait for school bus which arrives around 8:20Am. School starts at 9:15 and releaseed at 3:15. School bus brought me home. Change my clothes. Have some snack and wait for my Math tutor (not everyday; in that case, I did my Math homework, the only homework I was keen to do). Have dinner around 6. Then , it varies how my day ends. That was 6th grade. In 7th garde, my Math tutoring moved to 7Am till 8:30 in the morning and so I had to catch the school bus when it came around the senodn time by walking to a friend's house on the next block. In 8th grade, it was after school. So on. Now, say that I was spoiled or privileged. But notice that there wasn't extracurricular activites sponsored by school like during my older siblings time. They went to Private schools (founded during British time) which did not exist anymore in my time. That's because there was no budget for it except for short term ones, which I joined some times. I did learn swimming one summer at a private place. But before 7th grade, i.e when we were not old enough to be rebellious, in summer, my mother would teach us. With college education during my time, unlike my oldest sister's time, I would not qualify to be considered privileged. I wasn't complaining about that. About your telling me I suggest that you begin to work to build an infrastructure., I tell you this: stay within the context of diuscussion and ... ..do not act so dumb not to know that there are countries where people are not free to do (build) their infrastructure. But mostly, just stay within the context, will you? >> Housework is a very good lesson in >> eliminating some choices of employment. >Not for those who is left with no time to study when it is not just >helping around the house with chores. > Excuse me? Housework has ample opportunity for study. Since it > takes no thinking (or not much) to vacuum or wash a floor, one > can use their brains to learn at the same time. >I would set a book > up, read a paragraph, and think about it while doing other work. > If I still hadn't got it, I would read it again and continue doing > the physical work. When I figured it out, I would read the next > part of the text. In the computer biz, we called this flavor > of doing many things at once, timesharing. Well..I never was so keen to study like a bookworm. In fact, only one of my siblings was. But I happened to pay attention during lectures (and when my mother taught us) and that was mostly enough for me. My hobby was to read all kinds of books - there happened to be many books around the house brought by my much older brothers, cousins, etc. - that are not related to my school work. I was about 10 or 11 when I saw that book, the Biography of Peron (translated version). I advanced from reading story books to such books though I didn't really understand all of it. There were many such books (my oldest brother was keen on politics ) and other kinds of books. Also, my next door neighbor was the president of the office - I don't know how to traslate it - where all books get sensored before they get published. So I had access to those books, cartoons, etc. he brought home since their children and us were very close. Going back to the context, ... the point was that I was NOT talking about me. I was talking about the kids from the poor family. Do you know how poor family lives in 3rd world countries? Obviosuly not. >>It also gives one a fallback >> option for employment. >> The curious will find a way to satisfy their itch. A good resource >> is^Wwas public libraries. The curious also have to learn how to avoid >> spoon-feeding or they will lose that precious commodity. >> Public libraries? You seem to think that all countries have good >public libraries. > This thread was talking about US schools and such. No it was not. It was about Asians. > All countries may not have 100% good public libraries. Ours > sucks (and it's in the USA). If you have no access to any > library, go buy a text. If you have no money to buy a text, > barter with a professor (or somebody who does have books) > to clean his/her house in exchange for a couple of hours > access to his books every week. For cripe's sake, I can see that there is no point in explaining to you. You are a typical avergae American, who doesn't know much about beyond your border. Sad! Very sad! > Get a job at your country's university. Now you can even talk > to people who are learning; I used to sop up lots and lots of > knowledge just by listening to people talking about their work. > If the country is in an industrial stage, get a job at a factory. > Lots of physics, mechanics, efficiency studies, chemistry, math, > accounting, management, etc. can be learned just by noticing > how the company does its biz. Again..GEE!!!!!!!!!! >> You are a troll. > Not more than you are. > I may be a troll, but I'm one that can figure out how to get > something done with no resources. It's called work. I am very sad for you that . In fact, Josh halpern's reply was enough. I shouldn't even have bothered to reply to you. You were so way out of the context of the topic in discussion. > /BAH > Subtract a hundred and four for e-mail. === Subject: Re: Uncle Al is Sadistic . >When you run a race, you don't automatically award first >place to every runner. There were six hundred places. What is the difference between the 599th > person in that class and the 601st in terms of ability? The wonderprof > simply abdicated any responsiblity for educating the students. Quite > common. > Exactly. That is why a rational society only cares about win, place, > and show rather than compassionately about a huge pool of > incompetents. > Uncle Al I wouldn't insist too strongly on your idea, Al. The treatment of the incompetents has to be cared about by society, otherwise they, or many of them, will come to your house and tell you to beat it and kick your ass out without thought nor regrets. I sympathize with you that things should be as simple as you say. But it just isn't in the cards. If it were that straight forward then the solution would have been found long ago and would have been practiced all over the place. Utopia would be at hand. hanson === Subject: Re: Usenet Posting Guide? <1l7rb.46039$BX.34096@bignews5.bellsouth.net> Discussion, linux) > Well, I wasn't talking *just* about programming computers. Primarily I > meant that I'd hoped that everyone would get so far into the programming > mindset that they'd approach everyday life as a sort of programming > exercise. It's much easier to deal with people who approach everything > in a series of logical steps and who don't introduce extraneous issues. Approach everything in a series of logical steps and don't introduce extraneous issues? Doesn't sound like the programmers I know. Certainly doesn't sound like the programmers I've been. -- [T]here's no point in telling any of you what mathematicians I'm in email contact with, just like there's no point in going into detail about my contacts in a major news organization. [...] [P]olite disinterest is what I've found. --JSH on his important contacts. === Subject: Re: Uncle Al is Sadistic . > Littlemanwearingbigboypants whined: >> Exactly. That is why a rational society only cares about win, place, >> and show rather than compassionately about a huge pool of >> incompetents. > Not necessarily incompetent, but folks of middling ability, as most of > us are. Remembers it is the grunts who provide the ballast to keep the > ship stable in the water. It is the general labor force that make it > possible for capitalists to be rich (in addtion to the genius of the > capitalist). > On a desert island an enterprising capitalist will survive quite nicely, > but he will not be able to build an empire unless he can have it done by > Friday. > We all have some part to play. It pays to remember that drones won't work for vinegar. > I'm trying to find an e-version of Church's On the Concept of a Random > help? > gimma === Subject: Re: Why is math so difficult for some people? >> Similarly, Cramer's proof of the Levy-Cramer theorem that >> the sum of two independent random variables is not normal >> unless they both are is probably the only easy proof, but >> it likewise obscures the ideas. Any time characteristic >> functions are used to prove a probability theorem, the >> concepts are not even present. >Do you happen to know of a probabilistic proof of the Cramr-Lvy >theorem? There are entropy proofs. I am not up on all the available proofs, but I doubt there can be a fully probabilistic proof. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: division by general integer using register shifts >You're in trouble. Distributivity works for multiplication but >it doesn't for division. > n/(a+b) /= n/a + n/b You're right. In fact, the characteristic properties are Axiom 1: a/(b/c) = c/(b/a), Axiom 2: a/(a/b) = b, Axiom 3: (a+b)/c = a/c + b/c. >Use > n/(a+b) = 1/(a/n + b/n) = 1/ [(1/n)(a + b)] >Twice use a fast reciprocal algorithm and multiplication shifting. In fact, n/(a+b) = z/(z/(n/(a+b))) by Axiom 2 = z/((a+b)/(n/z)) by Axiom 1 = z/(a/(n/z) + b/(n/z)) by Axiom 3 and different z's other than 1 may be more useful. Other properties that follow are: (a/c)/(b/c) = c/(b/(a/c)) = c/(c/(a/b)) = a/b (a/b)/(a/c) = c/(a/(a/b)) = c/b a/(b/b) = b/(b/a) = a (a/a)/(b/c) = c/(b/(a/a)) = c/b a/(b/(c+d)) = (c+d)/(b/a) = c/(b/a) + d/(b/a) = a/(b/c) + a/(b/d) a/a = b/(b/(a/a)) = b/(a/(a/b)) = b/b respectively by Axioms 1,1,2; 1,2; 1,2; 1,(1,2); 1,3,1-twice & 2,1,2. > If you saw > (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2) > with the c's algebraic integers, I think few of you would have a > problem realizing that only two of the c's have 7 as a factor. > But, of course, you're looking at *functions* of x, as you have > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > so I could also write it as > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2 > as long as you're in a ring where 7 is not a factor of 22. > I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. > Usually you can *see* the other factors of 7, but I want you to > abstract, and generalize. > Please pay careful attention to that example. OK. Let's follow through (paying careful attention). The example you gave has solutions for the constants c_1, c_2 and c_3 as follows: c_1 = 4.45887 - 9.4089I c_2 = 4.45887 + 9.40789I c_3 = 0.452072 Hence, they are *not* functions of 'x'. It doesn't add anything to rewrite 'c_1 x' as 'f_1(x)' since it only consists of a constant times 'x'. Conveniently, if you set 'x' to zero, the equation you gave reduces to: (7)(7)(2) = 49*2 = (7)(7)(2), which is all very tidy. But if you set 'x' to 1, the equation becomes, (c_1 + 7)(c_2 + 7)(c_3 + 2) = 49*11 = (7)(7)(11) and, in view of the values of the 'c's, c_1 + 7 = 11.45887 - 9.4089I c_2 + 7 = 11.45887 + 9.40789I c_3 + 2 = 2.452072 and we have: (11.45887 - 9.4089I)(11.45887 + 9.4089I)(2.452072) = (7)(7)(11) How do the factors distribute here? You have demonstrated a plausible partitioning for the case 'x' = 0, but you suggest that your solution holds for all 'x'. If so, you should be able to explain what goes where for the case 'x' = 1, or any other value. Since the objections posted by your critics specifically challenge the generalization from the case x = 0 to all 'x', this is a crucial point. Please address it. -- 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: Uncle Al is Sadistic . > ..do not act so dumb not to know that there are countries where > people are not free to do (build) their infrastructure. But mostly, > just stay within the context, will you? Awwwwwww. Where's the fun in that? > For cripe's sake, I can see that there is no point in explaining to > you. You are a typical avergae American, who doesn't know much about > beyond your border. Sad! Very sad! Farm life and computer room, in fact. >>You are a troll. >Not more than you are. >>I may be a troll, but I'm one that can figure out how to get >>something done with no resources. It's called work. > I am very sad for you that . > In fact, Josh halpern's reply was enough. I shouldn't even have > bothered to reply to you. You were so way out of the context of the > topic in discussion. Now that you've figured it out......? === Subject: Re: Decimal Expansion > If m contains the digits 2, 5, 1 consecutively. This is an old math contest 32/127 = .2519685039... === Subject: Ideals & Lattices Hi everyone, I need some examples of a _prime ideal_ for an infinite lattice. TIA === Subject: Re: ELLIPTICAL INTEGRAL PROBLEM > I have a problem in solving the following integrals > which falls in to the form of elliptical integrals. > K = LImits 0-PI/2 Integral of 1/sqrt(1-k^2sin^2Theta) w.r.t Theta. > E = Limits 0-PI/2 Integral of sqrt(1-k^2sin^2Theta) w.r.t Theta. > Please help in finding the values of K and E These are definitions of the complete elliptic integrals K and E. They are not elementary functions of k. So what values do you want? table for E(k) from Maple: E(0.0) = 1.570796327 E(0.1) = 1.566861942 E(0.2) = 1.554968546 E(0.3) = 1.534833465 E(0.4) = 1.505941612 E(0.5) = 1.467462209 E(0.6) = 1.418083394 E(0.7) = 1.355661136 E(0.8) = 1.276349943 E(0.9) = 1.171697053 E(1.0) = 1. === Subject: Re: Uncle Al is Sadistic . > replied to this post of yours. I would not have otherwsie. >Put it in your think head. You cannot demand; you can only request. > HUH?! Did you have to use the word 'f***ing'? You could have asked What countries were they from? > /BAH > Subtract a hundred and four for e-mail. === Subject: Re: Key Core Error Argument If y is not 0, you can't use the constant term tricks that depend > on y > being 0. That's stupid. The constant term once found is distinguished by being > constant. All the trick is doing is finding it. So let me give you the example that I've used elsewhere which is > a_1(x) + 7, which has a constant term that is 7. Do you understand > what it means for it to be constant? Here's a test. If I have x=11, then I necessarily have a_1(11) + 7, right? Notice the constant term is STILL there, do you understand? Now then, if I now divide P(11) by 49, what should the constant term > be? If you answer honestly I'll be shocked. Let me try again. (a_1(y) + 7)/x is not the same as 7/x, unless a_1(y) > = 0. > Ok, I can go from there Richard Henry, and note that necessarily > a_1(y) has *some* factor in common with 7, right? > Why? Because a_1(y) + 7 has a factor in common with 7, right? > So let's call that factor f, now dividing 49 from P(y) will divide f > off from a_1(y) + 7, understand? > No. Why? Look at my previous answer Richard Henry. > Now then, you have a_1(y)/f + 7/f, and if f does not equal 7, what > does that tell you about the *constant* term Richard Henry? > I do not understand the question. Please rephrase it. If the constant term of a_1(y) + 7 is 7, since a_1(0) = 0, and now you divide through by some number I've called f, what is the constant term of the new expression? > Necessarily, you have 7/f left as the constant term for a factor of > P(11)/49, and if 7/f does not equal 1, you have a contradiction. > I don't follow your logic to say necessarily. Well I rephrased above so maybe that has changed, has it Richard Henry? > Understand? > No. > The trick is to FIND the constant term, as you know that it's not > affected by the value of x, and it sits there like a rock, unaffected > by the value of x, and none of your protestations against mathematical > reality will change that fact. > The constant term you refer to is found by zeroing out the other terms by > setting x to zero. When x is not zero, and those other terms are not > necessarily zero, you can not always extend the properties found for the > constant term to the entire expression. The property of the constant term is being constant because it's a number. For instance, with f(x) = x+ 2, the constant term is 2, as notice that at x=0, f(0) = 2, but the constant term just sits there like a rock, as you vary x, do you understand that Richard Henry? So you can't assume that if you change x the constant term changes, understand? James Harris === Subject: Modelling the movement of an electro mechanical device. I have a servo system, which moves through a certain angle, depending on the amount of voltage applied to it's input. The response of the movement from when the control signal is input, until the armature actually reaching the desired position varies as the device operates through different angles. This can be measured over a range of angles and frequencies. Does anyone have any suggestions as to what route I should be looking to take if I wanted to produce a mathematical model of the device, so that it is possible to predict the movement of the device? JD === Subject: Re: parallelizability of manifolds | |>> Surely the simplest way to understand the concept of parallelizability |>> is this: the n-manifold M is parallelizable if and only if there exist |>> n vectorfields V_i on M with the property that, at each point x of M, |>> the n vectors V_i(x) are a basis of the tangent space of M at x. |>> No imagination required. |>> |>Ah, that's interesting! So why are there no such vector fields for S^2? | |There's not even the *beginning* of such a sequence of vectorfields |for S^2: no matter what the vectorfield V_1 on S^2, there is always |some point x of S^2 at which V_1(x) does not belong to any basis of |the tangent space of S^2 at x. That's just a (seemingly more |complicated, but ultimately worthwhile) way of saying that every |(continuous!) vectorfield on S^2 has at least one zero. you should at least mention something about hedgehogs or coconuts here. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Two questions in propositional logic >>[...] >>2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? >I doubt it. Say p < q if p |- q but not q |- p; then your chain is >totally ordered by <. First, either there exists an increasing >sequence with at least two upper bounds >>This is presumably under the assumption that C is uncountable? > Yes. >or a decreasing sequence with at least two lower bounds >(by the traditional proof that any sequence of reals contains >a monotone subsequence: Say you have elements p_a, indexed by the >countable ordinals. Say a is dominant if p_a > p_b for all b > a. >Either there are uncountably many dominant a or not. If there >are uncountably many dominant a then the p_a with a dominant >give a cofinal decreasing set, while if there are only countably >many dominant a then the dominant a are bounded above >and you get a cofinal increasing subset.) So, changing the >notation and replacing all the wffs by their negations if >needed, you have > p_1 < p_2 < ... < q_1 < q_2. >It seems clear to me that this implies q_1 must be a >tautology (it seems like this is by compactness or >something, but I can't quite prove it this second) >and then q_2 is weaker than a tautology, contradiction. >??? >>Unless you're using some extra property about the p_i, q_i in saying >>this that I'm not seeing > No. >>(which is very possible, on account of it being >>Early here and I just got up. :), that needn't be true. >>For example take the following sequence: >>p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. > ??? Is it true that p_3 v p_4 |- p_1 v p_3 ? I don't think so... Gah. You're right. I was trying to construct that counterexample from memory way too early in the morning and it didn't work. :) What it is *meant* to say is the following: p_1 ^ p_3, p_1 ^ (p_3 v p_4), ... p_1, p_1 v p_2 If you're interested, John's proof went roughly as follows: > (Sorry if there's something below I should have replied > to that I'm appearing not to notice - you've forced me to > stop reading at this point..) > [snipped with eyes shut] Heh. No problem. :) There wasn't anything below it except a sketch of John's proof and my possible proof approach based loosely on what he did. David === Subject: Re: Two questions in propositional logic >>Anyway, as I said in the other post I think I've got this sorted out now >>(although I'll want to write my own proof of it before I'm fully happy >>with it). >>If you're interested, John's proof went roughly as follows: >>Define an equivalence relation on C by p ~ q if there is a bijection f >>from the set of primitive propositions to itself, such that q = p with >>all the p_i in p replaced with f(p_i). >>He then showed that there can be at most one element of each equivalence >>class in C, and that there are countably many equivalence classes >>greater than p (because you can take some countable subset of the whole >>set of primitive propositions and represent every formula in C as a >>formula using only this countable subset and the primitives that appear >>in p). >>He then went on to show there were only countably many elements less >>than p in a similar manner. I pointed out that this was rather easier, >>as you could just reverse the chain by negation and use the previous >>result. > I found this the most natural approach. However (and IBL will probably > kill me for saying anything at all about this sheet in public) you > should also be aware there is a one word answer to this question. That > is, there is one particular word (familiar to everyone) which > instantly makes this question a triviality. I'll leave you to guess > what it is. Hmm. ... Oh hell. Is it the one-word that you always use to prove obvious statements? It is, isn't it... ::sketches proof in his head:: Excuse me while I go sulk. (To anyone reading this who doesn't understand that line, it's a Leaderism. :) > By the way you didn't ask about the last question (if the set of > primitive propositions is allowed to be uncountable is it true given a > set S of propositions that you can find an independent set of > propositions equivalent to S); does that mean you've solved it? I > still don't have it, one year after taking the course. No hints > please! Well... yes and no. I thought I had a solution. There was a tiny problem in it contained right at the end which I think I can fix (but it may be a way bigger problem than I thought it was. :) That being said, my solution to the last question was only one line. My solution to the previous question did not assume countability. I was lazy and didn't feel like proving it twice... David === Subject: Re: Two questions in propositional logic Unless you're using some extra property about the p_i, q_i in saying >>this that I'm not seeing (which is very possible, on account of it being >>Early here and I just got up. :), that needn't be true. >>For example take the following sequence: >>p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. > Pointed out a few minutes ago that this is not a counterexample. > Of course what I said is wrong - a counterexample is > p_1 & (p_2), p_1 & (p_2 v p_3), ... p_1. Oops. Should have read this post before my last one. That's exactly the counterexample I had in mind when I made the (wrong) counterexample this morning, I was just half asleep and misremembering. :) > (Realized I wanted to show that the intersection of a strictly > decreasing sequence of open subsets of the Cantor set had > empty interior. Realized that that was false. Now, I bet > that there's no strictly decreasing omega_1-sequence of > subsets of the Cantor set, but never mind, the proof you > already have is simpler.) Are you sure that the cantor set approach works? Also, do you mean {0,1}^X for some possibly uncountable X rather than the cantor set? (which is homeomorphic when X is countable). Because I tried that, and I couldn't come to the conclusion that you could get the association to work both ways - you could get a clopen subset of {0, 1}^X for every proposition, but I wasn't sure you could go the other way... it looked like if you chose some sufficiently nasty clopen set it wouldn't correspond to a proposition. (I didn't prove that you couldn't, but it looked like it was going to go badly wrong if I tried to prove that you could, so I gave up on that approach). David === Subject: Re: Uncle Al is Sadistic . > 3 Africans I met in Computer science department in the last 4 > years were way above average in thier programming skills in > the midst of Chinese and Indian grad students who are the > overwhelming majority in that department. You will always find bright smart people in just about any > naturally > occuring group of humans. Race is nothing. Culture is everything. > Bob Kolker That's my point. What point? Culture Shmulture....that is simplistic politically > correct > heurist bull of the first kind. >>No, it is not. > If it were that simple >> It is THAT simple where I grew up. >>Children in the poor neighborhood with their parents making ends meet >>and the children having to do the house work like an adult and start >>working while they themslevs are children (rather than play or study, >>do not get a chance to test their brain since there is no books (not >>referring to school text books and notes) lying around the house. >> That's a lot of bull. House work (or farm work) is so boring >> one has nothing else to do but think. It also gives one an incentive >> to go to school and study real hard so that one can get a job that >> doesn't involve either. Afriend of mine from Wisconsan said justt hat. > He had to get up and helped his father milk the cows in the farm and >it was boring for him. He became a programmer. It was possible for him >because he is the this US of A. A country which was built by people. Nobody gave them anything; > they did the work themselves. Since you want us to believe that > you are from an underprivileged country that contains no books, > no schools, no business opportunities, I suggest that you begin > to work to build an infrastructure. However, since you're complaining > about having to do housework from dawn to dusk, I conclude that > there exist people who are able to hire other people to clean. > That implies that they make the money they pay those housecleaners > in some other industry than housecleaning. Thus, that country > has opportunities available for thousands. > Can you read well? > When did I ever say I had to work from dawn to dusk? First of all, I > wasn't talking about myself. Secondly, here is how my day started. > Get up, brush my teeth and wash my face. Have breafast which was > already on the the table. Get ready for school. Grab my lunch box and > go wait for school bus which arrives around 8:20Am. > School starts at 9:15 and releaseed at 3:15. > School bus brought me home. Change my clothes. Have some snack and > wait for my Math tutor (not everyday; in that case, I did my Math > homework, the only homework I was keen to do). Have dinner around 6. > Then , it varies how my day ends. That was 6th grade. In 7th garde, > my Math tutoring moved to 7Am till 8:30 in the morning and so I had to > catch the school bus when it came around the senodn time by walking to > a friend's house on the next block. In 8th grade, it was after school. > So on. > Now, say that I was spoiled or privileged. But notice that there > wasn't extracurricular activites sponsored by school like during my > older siblings time. They went to Private schools (founded during > British time) which did not exist anymore in my time. That's because > there was no budget for it except for short term ones, which I joined > some times. > I did learn swimming one summer at a private place. But before 7th > grade, i.e when we were not old enough to be rebellious, in summer, my > mother would teach us. > With college education during my time, unlike my oldest sister's time, > I would not qualify to be considered privileged. I wasn't complaining > about that. > About your telling me I suggest that you begin > to work to build an infrastructure., I tell you this: stay within > the context of diuscussion and ... > ..do not act so dumb not to know that there are countries where > people are not free to do (build) their infrastructure. But mostly, > just stay within the context, will you? > Housework is a very good lesson in >> eliminating some choices of employment. Not for those who is left with no time to study when it is not just >helping around the house with chores. Excuse me? Housework has ample opportunity for study. Since it > takes no thinking (or not much) to vacuum or wash a floor, one > can use their brains to learn at the same time. >I would set a book > up, read a paragraph, and think about it while doing other work. > If I still hadn't got it, I would read it again and continue doing > the physical work. When I figured it out, I would read the next > part of the text. In the computer biz, we called this flavor > of doing many things at once, timesharing. Well..I never was so keen to study like a bookworm. In fact, only one > of my siblings was. But I happened to pay attention during lectures > (and when my mother taught us) and that was mostly enough for me. My > hobby was to read all kinds of books - there happened to be many books > around the house brought by my much older brothers, cousins, etc. - > that are not related to my school work. > I was about 10 or 11 when I saw that book, the Biography of Peron > (translated version). I advanced from reading story books to such > books though I didn't really understand all of it. There were many > such books (my oldest brother was keen on politics ) and other kinds > of books. Also, my next door neighbor was the president of the office > - I don't know how to traslate it - where all books get sensored > before they get published. So I had access to those books, cartoons, > etc. he brought home since their children and us were very close. > Going back to the context, ... > the point was that I was NOT talking about me. I was talking about > the kids from the poor family. Do you know how poor family lives in > 3rd world countries? Obviosuly not. >>It also gives one a fallback >> option for employment. >> The curious will find a way to satisfy their itch. A good resource >> is^Wwas public libraries. The curious also have to learn how to avoid >> spoon-feeding or they will lose that precious commodity. >> Public libraries? You seem to think that all countries have good >public libraries. This thread was talking about US schools and such. > No it was not. It was about Asians. I was mistaken with another thread when I said No... but still your response was to my post about the poor Asian kids having to work too much too early in life. Also, you failed to understand poor over there means not enough food, and in extreme cases, not enough wood or charcoal to make a fire to cook, etc. > All countries may not have 100% good public libraries. Ours > sucks (and it's in the USA). If you have no access to any > library, go buy a text. If you have no money to buy a text, > barter with a professor (or somebody who does have books) > to clean his/her house in exchange for a couple of hours > access to his books every week. For cripe's sake, I can see that there is no point in explaining to > you. You are a typical avergae American, who doesn't know much about > beyond your border. Sad! Very sad! I meant to type ... much about things beyond ... > Get a job at your country's university. Now you can even talk > to people who are learning; I used to sop up lots and lots of > knowledge just by listening to people talking about their work. If the country is in an industrial stage, get a job at a factory. > Lots of physics, mechanics, efficiency studies, chemistry, math, > accounting, management, etc. can be learned just by noticing > how the company does its biz. > Again..GEE!!!!!!!!!! > >> You are a troll. > Not more than you are. I may be a troll, but I'm one that can figure out how to get > something done with no resources. It's called work. > I am very sad for you that . > In fact, Josh halpern's reply was enough. I shouldn't even have > bothered to reply to you. You were so way out of the context of the > topic in discussion. /BAH Subtract a hundred and four for e-mail. === Subject: Re: Uncle Al is Sadistic . > It is THAT simple where I grew up. [= Culture is everything per Bob] > Children in the poor neighborhood with their parents making ends meet > and the children having to do the house work like an adult and start > working while they themslevs are children (rather than play or study, > do not get a chance to test their brain since there is no books (not > referring to school text books and notes) lying around the house. > This has nothing to do with culture. It's poverty, which is pancultural. > May be this is a bit too simplistic scenario to apply in every parts > of the world but I would say that it would apply to most parts of the > world. > No, it does not apply because it has nothing to do with culture. > There is no culture which says We are Povertarians, we are > poor and proud of it, because POVERTY is our way of life > No such country or culture exists. Poverty is pancultural. > then the problem [if it would be of cultural origin] would have > been solved a long time ago. Painlessly. > Care to elaborate? > ... because, everybody everywhere would adopt, copy and > assimilate into such a wonderful (nonexisting) utopian culture. > But it does not exist. So any elaboration aint gonna help you. > But one can easily see the problem being enhanced and aggravated > at every step of its emergence and development because, basically... What was thoughtless came by easy. > What was easy became habit. > What were habits became tradition. > What were traditions became culture. > What was culture became religion. > What is religion becomes thoughtless. > Here, think about this hexliner above. The reasons for your > unhappiness are in there. That's how I see the world. > You may see it differently then me. That's cool by me. > End of story. > Diversity will always be with us, whether we like it or not. > again: > ahahahaha...........ahahahahanson I don't know what happened to my response post where I said In some situations, the poverty and culture get intertwined to the extent that it becomes one, the culture. I do not want to spell it out but I hope you know which group I am talking about. Also that group was forcefully formed as a group, despite the different oiginal cultures among them, if you get what I mean. === Subject: Re: Uncle Al is Sadistic . amanda grava .88 la saucisse et au marteau: [SNIP] Could you *please* the lines you are not responding to. Otherwise, this is very hard and unpleasant to read and, besides, it takes more time to download (and it's significative with a low-speed modem). -- Nicolas === Subject: Re: Factorization dispute > I have a result that shows a problem with a definition that > mathematicians have used for over a hundred years, No, you don't. Your claim is false. > and rather than > face the result which follows from rather basic algebra mathematicians > are being pussies and running like scared cowards from the result. They are challenging your (false) claim. No one is running, except in the sense that anyone would run from a maniac flinging manure. > Some posters, not professional mathematicians from what I've gathered, > are at least trying to stand and fight, but because what I have is a > math proof, their claims are necessarily irrational. Your proof is flawed. It has been thoroughly refuted many times. That makes the challenges rational, and your defense irrational. > I need you to stand up for the truth, here and now. Been there, done that. You ignore, or simply repost. > Let's chase these mathematician cowards down, and make them face the > music. ..and dance? > After all physics had its challenges and physicists faced them, while > mathematicians seem to believe that they can just ignore problems. What mathematicians seem to believe that they can just ignore problems? I've seen no evidence of that.You, on the other hand, have a consistent prior record of ignoring counter-examples and refutations of your arguments. > Let's take 'em out. > James Harris > http://mathforprofit.blogspot.com/ Take 'em out? What do you mean by that? Are you back to your previous threat of calling out the military or the CIA and the FBI, or are you advocating the formation of a private militia? -- 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: Factorization dispute > Let's take 'em out. > James Harris > Take 'em out? What do you mean by that? Are you back to your previous > threat of calling out the military or the CIA and the FBI, or are you > advocating the formation of a private militia? I think James wants his vast silent audience to treat mathematicians to dinner and a movie. I suspect that there's an ulterior motive. === Subject: Re: Uncle Al is Sadistic . >>Sure, the I don't fail you you fail you bull. He didn't >>educate anyone either. He didn't go out and get more >>resources for more students. When you run a race, you don't automatically award first >place to every runner. There were six hundred places. What is the difference between the 599th > person in that class and the 601st in terms of ability? The wonderprof > simply abdicated any responsiblity for educating the students. Quite > common. > Exactly. That is why a rational society only cares about win, place, > and show rather than compassionately about a huge pool of > incompetents. With your approach, my sister would have never become a physician if she had not come to US. Back home, there's only once chance to get admitted to medical school. She missed it by a shortage of .3 in her score. Years later in US, she went to Med School in Georgia. === Subject: Re: Probability of a Run >Your formula >u(m + 1) = u(m) + (1 - u(m - n)) (1 - p) p^n >is deducible from mine, for m > n. Just calculate P(n,m+1)-P(n,m) in >my notation, swap P(n,m) to u(m) and q to p and you'll get the same. I >had just about realized this couldn't be the difficult part of the >problem! >I'm slightly puzzled as to why you said I want to calculate the >probability of this if you know the answer, unless you're not a kid >with a TI-83 or a home computer! > Glad to explain! I am working on the subject of betting strategies for > my web site at > http://www.cybcity.com/ranmath/start.htm > A resort city close to me has been inundated by gambling casinos. If > everyone knew as much math as I do or as you do or had studied > psychology under B. F. Skinner as I have, they wouldn't gamble, yet > they do. I am trying to educate them. In a recent paper, Edward O. > expectation for a game with fixed expectation in the slightest, yet I > have found a counterexample. I have found a betting strategy that will > make the player's expectation WORSE, so I feel that the last word on > this subject has not yet been spoken. > To calculate the player's expectation for game + strategy one must > take into account the probability of ruin and for the case of the > player who uses a Martingale betting strategy it is necessary to refer > to the Theory of Runs. It is a deep and difficult subject, or at least > it has been up to now. The above recurrence relation is not mine. I > got it from Burnside and Uspensky. My own attempts to derive such > things generally lead to wrong answers. > I have the same mathematical machinery as Robert Israel does but he > knows how to use it better and even has helped me with it. I don't > feel I can tell my people that all they have to do is to expand this > complicated rational function in powers of s and the coefficients will > be the required probabilities. Robert Israel can do it and maybe I can > do it but they can't do it. I want to give them something they can > use. > I am studying everything you write but am a little behind. I have > gotten as far as your formula >P(n,m) = 0 if m < n, > > = q^n.(1 + (1-q).( (m-n) - sum(i=n..m-n-1, P(n,i) ) ). >The logic is simple. Either the first n trials are losses or the first > trial is a success and then we have n losses or we have no runs of n > or more in i trials followed by a success and then n losses (for i = > 1.. m-n-1). I think this enumerates all the possibile combinations > without duplication. > and have the following observation: > P(n,m) appears to be defined in terms of P(n,n) for the range n > m 2n+2 and there seems to be no definition of P(n,n) except a circular > one. Setting m = n, the limits of the summation are n to -1 and the > first term in the sum is P(n,n). Presumably the other terms would have > the value 0 because m < n. > I construe the character between (m-n) and sum in your formula as > a minus sign. I assume that your formula is not intended to be > executable and will study your programming code. Have you checked the > output against results obtained in other ways? I have no difficulty believing my explanations are difficult to understand. My father was a very good teacher but I never inherited any of those skills. beginning to think I was back at schoool. I'll answer your last question first. The only results I could obtain were hand calculations for very small examples, the simple case where n=1 and m is any value and quite a few results from large simulations. I did say I'd tried to make sure the formula wasn't ludicrous before I put forward any reply. When I said Burnside and Uspensky's formula is deducible from mine, I didn't make it clear that my definition is also deducible from their's (i.e. they are equivalent and I have not discovered anything which was not already known). In my definition the summation is intended to be 0 if the lower limit exceeds to upper limit as is the case for P(n,n). Although I have seen languages in which you could execute the function definitions, this wasn't really the intention here. Besides, it would not be an efficient algorithm if it worked directly from the definition. The first program does work using this definition, you'll be reassured to know. By writing SP(n,m) = 0 if n > m = sum(i=n..m, P(n,i)) otherwise we can get rid of the summations. Basically we calculate P(n,n) and SP(n,n), P(n,n+1) and SP(n,n+1)...P(n,j) {using SP(n,j-n-1) which we have already calculated} and SP(n,j)...P(n,m) We don't really need all the values of SP at any given moment and as old habits die hard I economised on the space. Unfortunately, it makes a simple algorithm look complicated. The second algorithm works by undoing the recursive part of the function definition. I'll rewrite the new definition as there were errors in the ranges for parts of the definition. I'll also replace (1-q) by p just to save space. P(n,m) = 0, if m < n = q^n(1+p((m-n))), if m < 2n+1 = q^n(1+p((m-n)-q^n((m-2n)+p/2((m-2n-1)(m-2n))))), if m < 3n+2 = q^n(1+p((m-n)-q^n((m-2n)+p/2((m-2n-1)(m-2n)/2)-q^n((m-3n-1)(m-3n))+p/3((m-3n - 2)(m-3n-1)(m-3n))))))), if m < 4n+3 ... This has been reorganised for calculation purposes as P(n,m) = 0, if m < n = q^n(1+p(m-n)), if m < 2n+1 = q^n(1+p(m-n)(1-q^n(m-2n)/(m-n)(1+p/2(m-2n-1)))), if m < 3n+2 = q^n(1+p(m-n)(1-q^n(m-2n)/(m-n)(1+p/2(m-2n-1)(1-q^n(m-3n-1)(m-3n)/(m-2n)/(m-2 n -1)(1+p/3(m-3n-2)))))), if m < 4n+3 ... The things to notice are that the new definition is exact. The approx in the function name in the code merely means that the expression for the m < 20n+19 case is similar to that for the case m < 21n+20 and you don't really have to go all the way down to the bottom of the expression to evaluate it accurately. compare this algorithm with that for 1-exp(-x) which can be calculated as x(1-x/2(1-x/3(1-x/3(1-x/4...)))) This formula is ok until x gets too large and then its starts returning nonsense. It might be sensible to limit how big m can be - basically if j (the number of levels inside the brackets it starts the expression from) is too big you probably won't get a sensible answer. You might also have to wait a while due to an inefficient implementation of mnfunc for large values of i - I struggle to pick sensible names! The code I've produced is in VBA because I find spreadsheets convenient for holding disorganised calculations. The code is pretty simple so I don't think it will be hard to translate the code into a language of your choice. Ian Smith === Subject: Re: Factorization dispute It turns out that I can isolate the current dispute easily enough by > focusing on the factorizations: Consider [snipped] > Crank Information > http://www.crank.net/harris.html > http://www.crank.net/usenet.html Readers should please check out *all* the links Sam the Worm listed. As for the math, notice that with (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 no other factorization works as long as 7 is not a factor of 22. That's because I've isolated the factors of 22, which is obvious by inspection. James Harris http://mathforprofit.blogspot.com/ === Subject: 3 x 3 matrix / eigenvalue If A is a nonsingular 3 x 3 matrix with nonnegative entries, then why must A have a positive real eigenvalue? Mike === Subject: Re: Difficult social problem > It looks like I'm swinging at tissue paper with a sledgehammer when it > comes to getting acceptance of my work, as while I can get initial > contact with mathematicians they tend to run as soon as I give them > enough information to realize the implications of my work and that I > am correct. > Mr. Harris, > even if your work were correct (and it is not), it would be as awe inspiring > as my morning piss after waking up. And people that's math society. That's how math people *really* are, so forget the movies. Keep that image of a math person taking his morning piss. James Harris === Subject: Re: Does compact+continuum connected+locally connected==>pathwise connected? >Does compact+continuum connected+locally connected >==> pathwise connected? >Continuum connected means that any two points of the space lie >in a continuum (= compact connected set). > Usually continuums are Hausdorff, but as that was omitted... > >Out of curiousity, what was the countable cofinite example? > Any infinitely countable set with the cofinite topology > isn't path connected, hence counterexample. > Finite set with cofinite topology, tho not path connected, > isn't counterexample. > A set with cardinality > c and > the cofinite topology is path connected > Thus not counterexample. > >Let U be the first uncountable ordinal. Give X = U x [0, 1) + > >{infinity} the lexographic order on U x [0, 1) and have infinity > >greater than all other elements. Then give it the order topology. > >Itis complete and densely ordered, with endpoints, so compact > >It is a linear continuum, so connected. In particular any interval > >in it is connected, and the intervals generate the topology, so it > >is locally connected. > >> path p from a to b in linear order S, a /= b > >> ==> [a,b] order isomorphic [0,1] > >all you have to do is you have to show that from such a path you can > >construct a path which is strictly increasing. I don't see > >immediately how to do that though, if the path is something stupid > >with infinitely many local maxima / minima. Oh well,.. Is there a > >simple proof? > path p from a to b in linear order S, a /= b > ==> [a,b] order(iso)morphic [0,1] > proof: > wlog a < b; retract r:S -> [a,b]; rp:[0,1] -> [a,b] surjection > rp([0,1]) connected, convex; a,b in rp([0,1); [a,b] subset rp([0,1) > [a,b] = rp([0,1]); [a,b] continuous image separable continuum > [a,b] separable multi-point linear continuum; [a,b] ordermorphic [0,1] > That last step isn't a step, it's a theorem based upon the theorem > a countable dense linear order sans end pts is ordermorphic Q. Ah yes, of course. Clever, and way more elegant than my initial attempts at a proof. :) > >> path connected linear order S ==> |S| <= c > >Hmm. This one I'm less convinced by. I'm tentatively willing to > >believe it to be true - certainly it looks intuitively like it might > >well be right - but do you have some sort of reference I could check > >for that? > Put some end points on it and apply the above. ;-) If you can > do that continuously, let me know, it'll save us the hassle of But... that doesn't work. Obviously. X is a counterexample, as the long-line is path-connected. (Of course that's not a counterexample to the original statement, it just shows that proof can't work) > >but I'd like to see a proof of that. > Reaching into the jumble jungle of my notes and > grabing a choice handful of random coherence: > path connected linear order S ==> |S| <= c. > proof by assuming c < |S| > some s in S with c < |(.,s]| or wlog c < |[s,.)| > If { xi | s < s_xi } has last element beta: > c < |[s,.)| = |[s,s_beta]| <= c which cannot be > Thus let beta = lim { xi | s < s_xi } > c < |[s,.)| = |/{ [s,s_xi] | xi < beta }| <= c|beta|; c < |beta| > note: from 1st theorem |[s,s_xi]| = c > omega_1 <= c < |beta| <= beta; some path p from s to s_(omega_1) > [s,s_(omega_1)] homeomorphic [0,1]; omega_1 embeds R, not so! Ok. Makes sense, I think... I'll have to look over it later in a bit more detail. David === Subject: Re: Applications of mathematics >>Suppose you were to tell senior highschool students about applications >>of mathematics that would be interesting and understandable to them. >>What applications would you talk about? >Knot theory and sex, because sex is the only thing interesting to them. That reminds of a quote from Louis-Ferdinand C.8eline's Voyage au bout de la nuit found on the Mathematical Quotations Server (http://math.furman.edu/~mwoodard/mquot.html): ---------------------------------------------------------------------------- ------------------------ Entre le p.8enis et les math.8ematiques... il n'existe rien. Rien! C'est le vide. [Between the penis and mathematics there is nothing. Nothing! The void!] ---------------------------------------------------------------------------- ------------------------ John Mitchell === Subject: Re: parallelizability of manifolds >> There's not even the *beginning* of such a sequence of vectorfields >> for S^2: no matter what the vectorfield V_1 on S^2, there is always >> some point x of S^2 at which V_1(x) does not belong to any basis of >> the tangent space of S^2 at x. That's just a (seemingly more >> complicated, but ultimately worthwhile) way of saying that every >> (continuous!) vectorfield on S^2 has at least one zero. And that, >> in turn, follows (after building the appropriate machinery relating >> differential topology to algebraic topology) from the fact that the >> Euler characteristic of S^2 is non-zero. >Indeed, algebraic topology seems to be an extremely powerful tool: S^2 is >simply connected, so every continuous image of it also is; in contrast, >R^2{(0,0)} is not. Realizing that this gives the proof for S^2 was a >pretty cool heureka-moment! ... I'm not sure your eureka! moment was authentic. For example, S^3 is both simply connected and parallelizable. John Mitchell === Subject: Re: Uncle Al is Sadistic . > >>Genius isn't a monopoly. You consider individuals as individuals, one >>by one. However, you mostly won't mine diamonds out of anything but >>lamproite and kimberlite, plus placer deposits. If resources are >>limited you invest where the odds are demonstrated best. Only a fool >>*continues* to commit a majority of resources in barren ground. You >>can dig down five miles deep and you won't find one natural diamond in >>New Jersey. >>The top 20% of any university graduating class wasn't the top 20% of >>SAT scores on admission. OTOH, the top 20% of SAT scores on admission >>are substantially over-represented in the top 20% of the graduating >>class - especially in subjects containing objective truth. A >>university seeking to fill its sciences, engineering, math, >>computer... departments would be insane to choose diversity (racial >>quotas) over objective performance. Look in sci.physics. What do you >>do to an idiot to fix it? Education isn't relevant. >>If you want metal you mine ore and dispose of the gangue matter of >>course. You sure as Hell don't dig granite if you want copper. It's >>a waste of granite, too. >>UC/Irvine intensely, Liberally, enthusiastically courted nigger and >>spics - guaranteed admission no matter what and full free ride plus >>fat perqs. UC/Irvine intensely, Liberally, enthusiastically excluded >>Asians and Jews. I visit the campus. The real subjects are almost >>exclusively populated by Asians and Jews with astounding credentials >>and demonstrated abilties. Your average White kid looks around and >>drops those courses after the first lecture. Don't entertain >>hallucinations of diversity in linear algebra courses. The >>University of California Regents can't for the life of them figure out >>how Asians and Jews can universally do so well. >> > That' just because of all California is filled with morons, > not just U.C. Since Linear Algrebra was invented in > ancient Eqypt, most us are still trying to figure why *any* > mathematicians anywhere believe that they have the credentials > necessary to grade California Home Economics Homework, > nevermind anything that appears to be politics > that idiots at Harvard invented, rather than a > run down compsci dump like U.C. > It's always useful to check assumptions against reality > http://math.sfsu.edu/hsu/workshops/treisman.html enough to make a comment when I read the following: <<<<<< I have noticed (in grad school) that was exactly how the Chinese students functioned. What I also noticed was that they would not take in those from outside their ethnic group unless it is a Malay Chinese who is more a Chinese than a Malay. It obviously was politics, at least in that particular grad school. Indian students functioned similarly too. Vietnamese too are a strong group but they are at the undergard programs (at the school mentioned above). May be the reason most vietnamese are not in grad school is that they live here while the Chinese come here as foreign studnets and usually join grad programs (better chance to get student visa). > josh halpern === Subject: Re: Modelling the movement of an electro mechanical device. >I have a servo system, which moves through a certain angle, depending on the >amount of voltage applied to it's input. >The response of the movement from when the control signal is input, until >the armature actually reaching the desired position varies as the device >operates through different angles. What do you means by operates through different angles? > This can be measured over a range of angles and frequencies. Frequencies of what? The input signal? How does that affect the movement? >Does anyone have any suggestions as to what route I should be looking to >take if I wanted to produce a mathematical model of the device, so that it >is possible to predict the movement of the device? === Subject: Re: Does compact+continuum connected+locally connected==>pathwise connected?Countable cofinite space counter example > >Continuum connected means that any two points of > >the space lie in a continuum (= compact connected set). > Usually continuum means compact connected Hausdorff. Yes. But I posted quickly, and neglected to mention I was thinking in a METRIC SPACE. --Ron Bruck === Subject: computation of a series is there any formula to compute the following series {sum{a_j,j=0,j=k-1})^k for real a_j's? === Subject: Re: Usenet Posting Guide? > With all respect, dear man, you strike me as that type more enamored > with process rather than function. Most of us out here have far better > things to do than screw around for weeks configuring newsreaders and > the like, even if we were so inclined, which most of us clearly are > not. I suppose we could memorize the phone book, too, but would that > help us communicate our ideas any better? > Some of us prefer to view the forest rather than count the trees down > there. Better view, too. > Well, I find the technical aspects of how USENET functions much more > interesting than most of the discussions that take place on it. By the > same token, I am much more interested in the hardware and operating system > software of the systems I administer than in any of the applications > for which they are used. > Once I hoped that the growing popularity of personal computers meant > that nearly everyone would learn to think like programmers. It never > occurred to me that, sadly, the opposite would happen: That computers > would be designed to be used by people who *can't* program them. thinking here a little more closely. You're seeing the world from a very narrow vantage point. Technology and inventive genius exist to serve those who use and take advantage of them, not merely those who design and invent, otherwise you might just as easily make the argument that the world was a far better place when mostly car builders and those who tinkered all day with Model T Fords were out there driving on the roads. Certainly the roads were safer then, and that those driving them generally knew their cars inside and out, but that ignores the fundamental purpose behind that of automobiles, which is to facilitate transportation. The same exact argument can be made for telephones, television, airplanes and a whole host of other modern conveniences we now take for granted and which represent fundamental shifts in the way we communicate and get around. They were all once the domain of a select few tinkerers and inventors who had absolutely no idea of what they were about to unleash on the world. With all respect, sir, I can guess your politics right across the board. Just for openers, you're a Sierra Club, save-the-whales type. Gotcha, huh? === Subject: Re: Factorization dispute > It turns out that I can isolate the current dispute easily enough by > focusing on the factorizations: > Consider > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > where even by inspection you can see that the constant terms are > separated out, so that you have 1(1)(22) = 22, the constant term of > the polynomial. > I'll add that at x=0, a_1(0) = a_2(0) = b_3(0) = 0. > Notice, > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where again you see that the constant terms match as now you have > 7(7)(22) = 1078, which is again the constant term of the polynomial. > If 22 does not have 7 as a factor, the former factorization is the > *only* allowed way for 49 to divide through. > (For more detail, like what the a's are, see > http://mathforprofit.blogspot.com/ > where more is explained.) > I have a result that shows a problem with a definition that > mathematicians have used for over a hundred years, and rather than > face the result which follows from rather basic algebra mathematicians > are being pussies and running like scared cowards from the result. > Some posters, not professional mathematicians from what I've gathered, > are at least trying to stand and fight, but because what I have is a > math proof, their claims are necessarily irrational. That's not really true. The people with brains stopped fighting mathematicians and their holistic continuum, the day after they invented it. Since it's still well-known throughtout the universe that they're the only people who believe that there are an infinite number of dimensions in a universe that quite obviously only has three. > I need you to stand up for the truth, here and now. It's still impossible to stand up in a mathematics class, since they're only people who actually believe in infinite value logic. And if you did stand up, all they would do is start chanting some some sort of political spirituals about the evil of things that aren't positive. > Let's chase these mathematician cowards down, and make them face the > music. > After all physics had its challenges and physicists faced them, while > mathematicians seem to believe that they can just ignore problems. That's not true. The only statement that has ever universally about science is that mathematicians do G's, physicists do E's, and the people with brains do brainy things. Fission is old, Fusion is new, the NSA is cold, and the U.N. is Blue. > Let's take 'em out. > James Harris > http://mathforprofit.blogspot.com/ === Subject: Re: computation of a series > is there any formula to compute the following series > {sum{a_j,j=0,j=k-1})^k > for real a_j's? Nothing especially nice, no. For a general field (in particular for real numbers) [ sum{j = 1 to m} a_j ] ^n = Sum { x in {1, ..., m}^n } Product {i = 1 to n} a_(x_i) Prove it by induction basically. It's not especially difficult. This doesn't simplify to anything nice in your case, although it might for some specific sequence a_i David === Subject: Re: 3 x 3 matrix / eigenvalue > If A is a nonsingular 3 x 3 matrix with nonnegative entries, then why must > A have a positive real eigenvalue? > Mike This is not true. For example 1 0 0 0 0 1 0 1 0 has eigenvalues 1, 1, and -1. However, it is true if you add the restriction that the matrix be triangular because then the eigenvalues are the diagonal entries. Have a tolerable existence. Eli === Subject: Re: Usenet Posting Guide? > Some of us prefer to view the forest rather than count the trees down > there > The point was this, back in the days when you had to think before you > posted, those who, to borrow your analogy, visited the forest actually > appreciated it, and contributed to the ecology of the forest, rather than > now, where too many people use the forest as a place to hold their drug and > alcohol parties, while thrashing the forest. It's more like now the forest has been clearcut and paved over to make room for a Wal-Mart. -- Visit my blahg site. http://myblahg.blogspot.com/ === Subject: Re: complex integral.....?? > um...... i think that root of 1+(z^3)+(z^5)}] exist between -1 ~ -0.5 > thus, f(z) is analysis of |z|<1/2 > thus.......i think that answer is 0 > it is right?? > All the roots of a_n*z^n + a_{n-1}z^(n-1) + ... + a_1*z + a0 verify > 1/(1 + B/|a_0|) < |z| < 1 +A/|a_n| > Where A = max(|a_0|, |a_1|, ..., |a_{n-1}|) and B = max(|a_1|, ..., > |a_{n-1}|, |a_n|) I think it's easier to notice that if |z| <= 1/2, then |z^3 + z^5| <= ..., hence z^3 + z^5 could not = -1. === Subject: Re: Difficult social problem [...] > Mr. Harris, > even if your work were correct (and it is not), it would be as > awe inspiring as my morning piss after waking up. > And people that's math society. > That's how math people *really* are, so forget the movies. > Keep that image of a math person taking his morning piss. > James Harris James, how soon you forget: : David Ullrich is a ing piece of dog. : : I think it's funny that I can call a professor at Oklahoma State : University a ing piece of dog knowing that he'll keep : replying in my threads. : : You see, he has to keep replying pushing the same old lies. : : He's stuck. He's trapped in something that he can't get out of, : so it doesn't matter what I call him, or what I say about him, : he has to come back. : : You see I'm the person who has the correct math argument, so : posters like David Ullrich or Arturo Magidin are *compelled* : to reply out of fear that if they go away, then I'll get some : people who'll pay attention to the truth. : : So David Ullrich, the math professor at Oklahoma State University, : is demeaned by me as the piece of ing dog he is, and he : *has* to keep coming back. : : : James Harris === Subject: Re: Modelling the movement of an electro mechanical device. > I have a servo system, which moves through a certain angle, depending on the > amount of voltage applied to it's input. > The response of the movement from when the control signal is input, until > the armature actually reaching the desired position varies as the device > operates through different angles. This can be measured over a range of > angles and frequencies. > Does anyone have any suggestions as to what route I should be looking to > take if I wanted to produce a mathematical model of the device, so that it > is possible to predict the movement of the device? Well, plot (angular) movement versus electrical input as one graph and plot movement versus time as another plot. If both graphs are straight lines then calculating movement from electrical input is just a proportion. Then calculating time from movement is just another proportion. If the graphs are not straight lines you might begin with a statistical curve fit... === Subject: Re: Math dependency logic REVISED >> [...] >Of course, you don't deny that if one starts with something true and >apply correct reasoning, then the conclusion must be true. > Well that's progress - for some reason you've decided not to be > entirely stupid today. > So of course I didn't deny that. So that example is totally > irrelevant regarding your (_repeated_) statement that I was > _lying_ when I said nobody was denying that. Which brings > us back to the beginning: If I was lying when I said nobody > was denying that you should produce an example where > someone _has_ denied that. Or admit that you were being > either stupid or dishonest when you said I was lying. > Of course neither of those is going to happen, because > you have more concern for exposing evildoers than for > telling the truth. > But >coming from you these are empty words, for you have demonstrated (on more >than one occasion!) your inability to reason from C1-C4 to Ex~(x=x). > Oops. Back to the irrelvancies... When (with no evidence) *you* accuse someone of faulty reasoning, you expect to be taken seriously. But when glaring instances of your own imperviousness to logic are cited and referenced, you dismiss these as irrelevant! Are you Rush Limbaugh? Do you do prescription drugs? --John C1 AxAy[x=y -> Az(z in x <-> z in y)] C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)Classification C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] (Weak Extensionality) Exhibit of proof of Ex~(x=x) from C1-C4 and someone will point out the error. === Subject: Re: 3 x 3 matrix / eigenvalue bojdsa$c4rt$1@netnews.upenn.edu... > If A is a nonsingular 3 x 3 matrix with nonnegative entries, then why must A > have a positive real eigenvalue? > Mike Perron-Frobenius theorem. === Subject: Re: Uncle Al is Sadistic . The LaRouche Show is a weekly audio talk show, broadcast live on the Internet every Saturday, featuring interviews with Lyndon LaRouche, his associates, and special guests. Hosted by Michele Steinberg, Counterintelligence co-director of Executive Intelligence Review, and by Marcia Merry Baker, EIR Economics Intelligence director. Live Broadcast TODAY 3:00-4:00 p.m. Eastern Standard Time (2000-2100 UTC) Speaker: William Wertz on Schiller: Poet of Freedom. High-speed audio: Stream Low-speed audio: Stream During the live broadcast, you can ask questions by calling one of the following numbers: > been practiced all over the place. Utopia would be at hand. --ils duces d'Enron! http://larouchepub.com/radio/index.html === Subject: Factorization dispute, again Notice, (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where you see that the constant terms match as now you have 7(7)(22) = 1078, which is the constant term of the polynomial 49(300125 x^3 - 18375 x^2 - 360 x + 22). Various people have debated me about what happens when you divide off 49, where for some odd reason, some of them seem to believe that you can have w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and (5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 where the w's vary as x varies, which is a rather naive notion. That's because you can multiply *everything* out, and simplify to get (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 which should be simple enough for all of you. Now those of you who usually work in the field of complex numbers may think that it's not a big deal, as you may think it doesn't matter if w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but you see, as 22 is coprime to 7 in the ring of algebraic integers, if w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring. You know, it's like how in integers 1/2 doesn't exist. It's not an integer, so it's not in the ring. So you see, my argument is correct and simple, and mathematicians are indeed running from a little gut check in their field. They're pussies too scared to handle the truth. But you should also understand, some people will be able to see that, which is part of my plan. I can let mathematicians destroy themselves proving they can't be trusted based on what they *see*, while they forget what they can't see: the wearing down of the mathematician mystique. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: Uncle Al is Sadistic . > Speaker: William Wertz on Schiller: Poet of Freedom. Schiller, poet of Romantic Dreck. Have your read the words to -Ode an die Freude-? Bob Kolker === Subject: Re: Proof of Loan Amortization Formula In sci.math, Jay I know you guys get tired of guys like me asking for proofs but, this > will be the last time I ask. I think seeing the proof actually helps > me visualize and learn how to apply it in real life. When I look at > the loan amortization formula, I just dont see that link right away. > Could you take some time to explain or recommend a book that has the > P.S Here is the demon below: > L t(1+t)^n > p = _________________ > [ (1+t)^n - 1 ] The simplest way of proving it would be to posit the problem in this fashion, perhaps. Assume one has applied for loan of a principal L, paid to him at the start of the loan. He is to pay back this loan in n months, at a constant (t * 100) % interest rate (per month). [*] Basically, one pays interest on any outstanding principal at the start of the month. (We assume that his first payment is a month after the loan, as well. Note that this is a *compound interest* loan, as opposed to a *simple interest* loan, which is rarely used nowadays. Another variant is a continuously compounded interest loan, which computes the interest on the outstanding loan amount using a formula such as L * exp(k * ln(1+t)). Still other variants are possible, such as ARMs, which posit a variable t; usually the variations are such that one can use a table lookup, as they are adjusted every 6 months or so.) At month 0 his balance sheet (relative to the loan company) looks like: B(0) = L At month 1, we assume the payment is promptly credited and one has interest on the loan: B(1) = L * (1+t) - P At month 2: B(2) = (L * (1+t) - P) * (1+t) - P At month 3: B(3) = ((L * (1+t) - P) * (1+t) - P) * (1+t) - P At this point you might want to gather terms, as I for one smell a possible induction hypothesis here: B(3) = L * (1+t)^3 - P * ( (1+t)^2 + (1+t) + 1) = L * (1+t)^3 - P * ((1+t)^3 - 1) / ((1+t) - 1) Let's see if this works for B(4): B(4) = L * (1+t)^4 - P * (1+t) * ((1+t)^3 - 1) / ((1+t) - 1) - P = L * (1+t)^4 - P * ((1+t)^4 - (1+t) + (1+t) - 1) / ((1+t) - 1) = L * (1+t)^4 - P * ((1+t)^4 - 1) / ((1+t) - 1) Oooh! Now let's set up a more formal hypothesis. We assume B(k) = L * (1+t)^k - P * ((1+t)^k - 1) / ((1+t) - 1) for an integer k, and prove that B(k+1) = L * (1+t)^(k+1) - P * ((1+t)^(k+1) - 1) / ((1+t) - 1) using similar algebraic manipulation (which I leave to the interested reader, as it's very similar to my B(3)=>B(4) transition above). We also trivially verify that B(0) = L * (1+t)^0 - P * ((1+t)^0 - 1) / ((1+t) - 1) = L so weak induction follows; we now have a closed form for the balance after the k'th month. The terms of the loan are that after n months the loan is paid off, so B(n) = 0, and therefore B(n) = 0 = L * (1+t)^n - P * ((1+t)^n - 1) / ((1+t) - 1) or L * (1+t)^n = P * ((1+t)^n - 1) / ((1+t) - 1) or P = L * t * (1+t)^n /((1+t)^n - 1). QED In practice, the equality is not quite exact because of rounding of P to the nearest penny, but the amount of variation is at most 0.01 * ((1+t)^n - 1) / ((1+t) - 1) regardless of the value of L. If we substitute n = 360 (a 30 year house mortgage) and t = 0.005 (about a 6% per annum rate), we get $10.045 ..., which is miniscule compared to most house loans nowadays; the last payment is increased by at most this amount, or perhaps the mortgage company remits a check. [*] just to confuse things: the payments in many loans are *per month*, but the rate is quoted as an interest rate *per year*, which may complicate the analysis in real life, even for this relatively simple loan. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: naive geometry questions > 2) Besides planes and spheres, is there any other surface S such that a > piece of S can be moved around adlibitum while each of its points remains > in contact with S? A surface with the property that I think you're describing is called homogeneous. I think that planes and spheres are the only homogeneous surfaces that can be constructed in three-dimension space. There are others but they can't be constructed in three-dimensional space. For example there are the flat tori and hyperbolic spaces. In three dimensions, a torus has to have a bulging outer area and a caving inner area but in four-dimensions, you can make a torus is flat, that is, it doesn't bulge or cave anywhere. I don't know how to describe hyperbolic space. Maybe someone else in this group can. Since I'm on the topic, does anyone know how many dimensions you need for a Euclidean space in which you can hyperbolic space? Have a tolerable existence. Eli === Subject: Re: Math dependency logic REVISED >> Ain't it touchin' the lengths to which Logikoi will go to bail >> one another out? See Camaraderie of the Experts >I did go to that link. Here's how it begins. >Lonely, are you? > If not for the way _he_ tends to speculate on people's personal lives > when he can't refute their arguments I'd say this wasn't very nice, > pointing out what a pathetic character he must be, forced to talk to > himself like this in public where everyone can see. >Anyway, the lengths the Logikoi will go to bail each other out of >*what*? Was there something threatening in this thread? > I know _I've_ been terrified of the possible consequences. I mean > of course everything he's said here has been nonsense, but > regardless, what if someone at OSU found out that there was > someone on the internet saying bad things about me? > (Giggle. Worse yet: You may not have noticed, but he often > quotes me saying wild things like everything is equal to > itself. What if someone at OSU found out I was promulgating > that sort of heresy? I can just picture it, once that post-tenure > review he mentioned elsewhere is implemented: There's a > committee meeting in Whitehurst. Ullrich says everything is > equal to itself? Off with his head.) Of more concern to this committee might be your UseNet stalking of JSH. Searched Groups for JSH OR James OR Harris author:ullrich@math.okstate.edu. Results 1 - 10 of about 2,700. Search took 0.70 seconds. Then again, you could always plead insanity... http://www.fetchfido.co.uk/sound_files/giggle.wav --John === Subject: f(n+1)/f(n)=g(n)? Hi if f(n+1)/f(n)=g(n), is it possible to write the function f(n) on a non-recursive form? That is to say, on the form f(x)=...... for some functions g(n) it is easy, like for f(n+1)/f(n)=3, which gives f=3^x*C, But are there any general formula for any given g(n)? === Subject: Re: Greek Alphebet > John Savard >> But in other contexts, the letters of the Greek alphabet are fraught >> with meaning! >> Thus, alpha comes from aleph, which means Ox. >References please? The _name_ alpha might be a degenerate form of the >Hebrew aleph, but the letter itself started AS the symbol for an ox, in >Greek. It actually comes from the upside down capital alpha, which was the >actual symbol for the head of an ox. (Just picture an upside down A). >Who says it's not the other way around and aleph is not a degenerate form >of alpha? With all due respect to the Greeks, the Semitic alphabet was well established before the Greeks considered using an alphabet; they started with it, and the Semitic alphabet already had many versions. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: derivate of x^x Ok, I've probably asked this question before, but I've forgot the answer... What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ? I guess asking for the integral of these functions would be foolish? === Subject: Poisson Distribution In the poission distribution like this: the monthly average number of airplane crashes is 2.2, can I imply that the average number of airplane crashes is 4.4 in 2 months? If I use that assumption, I will lead to a different solution to the problem: the probability of 5 crashes in 2 months, from using Poisson process for 1 month and binomial distribution. -- Khoa Tran === Subject: Re: Factorization dispute binomial factorizations, a fruitful field of endeavor; what's that one method, Partial Quotients, used in calculus?... I never really got the hang of it, but it was interesting. the funny thing about the Army (and all of the services, and the NSA etc.) is that they have lots of applied mathematicians ... and generals can ask them to look at stuff ... perhaps, they already have looked at monsieur Harris and his ordnances! > Let's take 'em out. > http://mathforprofit.blogspot.com/ --ils duces d'Enron! http://larouchepub.com/radio/index.html === Subject: Re: Math dependency logic REVISED > Ain't it touchin' the lengths to which Logikoi will go to bail > one another out? See Camaraderie of the Experts > I did go to that link. Here's how it begins. > Lonely, are you? Not at all. But don't fail to read Camaraderie of the Experts at It explains why you Boyz go to such great lengths to fend off assaults on one or another's 'expertise'. But you'll find nothing there about bum-fuzzling. Sorry. > Anyway, the lengths the Logikoi will go to bail each other out of > *what*? Was there something threatening in this thread? Oooh, were > my mortgage[1] and all, we can't have that. --John === Subject: Re: Key Core Error Argument [snip long boring discussion, look upthread if interested] > You keep writing a ratio because apparently you think a ratio is more > powerful or mysterious, capable of doing something that it can't. > Now then, if you admit that a_1(x)/w(x) is an algebraic function, it > can be replaced by f(x), and if you admit that 7/w(x) is an algebraic > function, it can be replaced by g(x), so then you have f(x) + g(x). > But the constant term of f(x) + g(x) is 1, so let h(x) + 1 = f(x) + > g(x), to isolate constant terms as before. Which line contains the error? 1. The constant term of (h(x) + 1) is 1 2. (h(x) + 1) = (f(x) + g(x)) (Noting that f(x) = a_1(x)/w(x), g(x) = 7/w(x) 3. (h(x) + 1) = (a_1(x)/w(x) + 7/w(x)) 4. The constant term of a_1(x)/w(x) + 7/w(x) is 1 -William Hughes === Subject: Re: Routine technique, analysis >> Last I remembered setting a variable to 0 to clear it out, knowing >> that pulled out terms independent of it was routine in analysis, which >> is why a lot of this is funny, ironic, and very, very sad. >I must say, it certainly brings a tear to MY eye. I knew when programming one must be careful using uninitialized variables in case they contain unexpected values, but didn't realize the same applied in pure mathematics. Maybe this explains why my algebra sometimes goes horribly wrong; I've reused x or theta from the previous calculation and forgotten to clear it out afterwards. Yeah, that's the ticket. I didn't make a mistake, it was the variables that somebody else had just used that were out of whack. Honest! === Subject: My research, publication announcement There's more to my work than just arguing on Usenet, so I'd like to point out that my paper Advanced Polynomial Factorization is slated to be published: See http://www.megasociety.net/NoesisHighlights.html The Mega Foundation is an organization of high IQ people, and I'm glad to be associated with them. To learn further about the organization you can use Google, or see: Why a group like the Mega Foundation? http://www.ultrahiq.org/Mega/WhyMega.htm I hope at least some of you will appreciate that often the most important ideas in history have to get past people limited by their lack of imagination and their prejudices, who act against scientific progress. What I want you to see is that there's more to me than Usenet, so that you can begin to understand that the revolution I'm giving you a chance to be a part of is bigger than the small-minded people who continually throughout history work to halt progress. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: Key Core Error Argument wow, how embarrassing for you. please, don't tell me how many that I'm on -- I'm trying to get out of here, anyway! > I am now at 2710, but I am only around in this newsgroup since > january 1988. --ils duces d'Enron! http://larouchepub.com/radio/index.html === Subject: Re: Key Core Error Argument there is also a proof of the isomorphism of deductive proofs with inductive ones, which may perhaps be amenable to combining the two forms into a tautology, or necklace. > before the final link (you missed the green, dood !-) --ils duces d'Enron! http://www.wlym.com/covers/7101contents.png === Subject: Re: Ex(~x=x), counterpart theory, and contingent identity > Well, when you claim that you can define scope so that self-identity >is always implicit, you must deal with a Kantian > possibility--... > The logical determination of a concept by reason > is based upon a disjunctive syllogism, in which the > major premiss contains a logical division (the division > of the sphere of a universal concept), the minor > premiss limiting this sphere to a certain part, and > the conclusion determining the concept by means > of this part. > --Immanuel Kant > Critique of Pure Reason A577/605 I'm not sure how this cashes out where the logic of identity is concerned. One might construe the classical Ax[x=x <-> Ey(x=y)] as characterizing by exclusion the relation between self-identity and identity-with-something: no self-identical is an identical-with-nothing, and no identical-with-something is a non-self-identical. In respect of the foregoing, ~AxEy(x=y) might be the minor premise and ~Ax(x=x) the conclusion. > ...--namely, the non-self-identical. > Of course, the concept is still a fiction in Kantian epistemology. > However, it is also the reason for the apparent > complexity of his ideas--he does not trivially assert > self-identity as self-evident. --John === Subject: Re: derivate of x^x > Ok, I've probably asked this question before, but I've forgot the answer... > What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ? > I guess asking for the integral of these functions would be foolish? x^x = e^(x*log(x)) Now use the chain and product rules. === Subject: Re: Uncle Al is Sadistic . > The LaRouche Show is a weekly audio talk show, broadcast live on the > Internet every Saturday, featuring interviews with Lyndon LaRouche, > his associates, and special guests. Hosted by Michele Steinberg, > Counterintelligence co-director of Executive Intelligence Review, and > by Marcia Merry Baker, EIR Economics Intelligence director. > Live Broadcast TODAY > 3:00-4:00 p.m. Eastern Standard Time (2000-2100 UTC) > Speaker: William Wertz on Schiller: Poet of Freedom. > High-speed audio: Stream > Low-speed audio: Stream > During the live broadcast, you can ask questions by calling one of the > following numbers: > been practiced all over the place. Utopia would be at hand. > --ils duces d'Enron! > http://larouchepub.com/radio/index.html === Subject: Re: My research, publication announcement > There's more to my work than just arguing on Usenet Yes, but that's your crowning achievement. , so I'd like to > point out that my paper Advanced Polynomial Factorization is slated > to be published: > See http://www.megasociety.net/NoesisHighlights.html How nice that the cranks have gotten together and put up a website. > The Mega Foundation is an organization of high IQ people, and I'm glad > to be associated with them. To learn further about the organization > you can use Google, or see: > Why a group like the Mega Foundation? > http://www.ultrahiq.org/Mega/WhyMega.htm > I hope at least some of you will appreciate that often the most > important ideas in history have to get past people limited by their > lack of imagination and their prejudices, who act against scientific > progress. Not any more, now that we have the Internet . . . > What I want you to see is that there's more to me than Usenet, so that > you can begin to understand that the revolution I'm giving you a > chance to be a part of is bigger than the small-minded people who > continually throughout history work to halt progress. Is there a publication party? Free drinks? Are we all invited? as you didn't label your binomial factorization (binomials being polynomials, and monomials can also be considered that, I guess, but I dygress), what is the import of transforming a coefficient of a linear variable, x, into a function? I also had New Math in the 3rd grade, so I can empathize with trite Bourbaki-isms! as for rings where 7 is not a factor of 22, isn't it true that in anything where it is a factor, it dyssolves anything of interest? Bud, I know I ain't the first top ask that! > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2 > as long as you're in a ring where 7 is not a factor of 22. > I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. > You may see people who reply claiming that the word polynomial has > some significance, as if it's a mystical thing which refutes basic > logic, so if something isn't polynomial it no longer behaves > logically. --ils duces d'Enron! http://larouchepub.com/radio/index.html === Subject: Re: f(n+1)/f(n)=g(n)? primefinder grava .88 la saucisse et au marteau: > Hi > if f(n+1)/f(n)=g(n), is it possible to write the function f(n) on a > non-recursive form? That is to say, on the form f(x)=...... > for some functions g(n) it is easy, like for f(n+1)/f(n)=3, which > gives f=3^x*C, But are there any general formula for any given g(n)? f(n+1) = g(n)g(n-1)....g(0)f(0) That's the best you can do in the general case. -- Nicolas === Subject: Re: Modelling the movement of an electro mechanical device. > I have a servo system, which moves through a certain angle, depending on the > amount of voltage applied to it's input. > The response of the movement from when the control signal is input, until > the armature actually reaching the desired position varies as the device > operates through different angles. This can be measured over a range of > angles and frequencies. > Does anyone have any suggestions as to what route I should be looking to > take if I wanted to produce a mathematical model of the device, so that it > is possible to predict the movement of the device? > JD Is the device a spring/mass device with torque proportional to voltage? Is it something else? I have in my lab an instrument that is perfect for this whatever it is if you have an angle transducer (the SRS785 two channel dynamic analyzer). The down side of this instrument is the $11,000 price tag. Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net === Subject: Re: Ex(~x=x), counterpart theory, and contingent identity > (1) states necessary and sufficient conditions for the necessity of > (material) identity: (1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y))) If identicals x and y are necessarily self-identical, then--and > only then--is their identity a necessary one. Beyond this, > (1) states nothing more: from (1) it neither follows that > John Correy is necessarily self-identical nor that John > Correy is contingently self-identical--or indeed that John > Correy is self-identical at all. (1) does not say which of > the foregoing is the case. We can sit around and argue about whether you or I are necessarily > self-identical. However, although logicians do argue about such > matters--Who else would bother?--it is not as logicians that they > argue but as metaphysicians, or as pataphysicians, or as what have > you? So, when I claim that (e.g.) Benjamin Franklin is necessarily > self-identical while the inventor of bifocals is not, my main > warrant for this claim is that if Benjamin Franklin is necessarily > self-identical but the inventor of bifocals is not, then Benjamin > Franklin and the inventor of bifocals are not necessarily identical > (although they are identical). In other words, I take the necessary > self-identity of the former and the contingent self-identity of the > latter to constitute, together with (1), an *explanation* for the > contingent identity of Benjamin Franklin and the inventor of bifocals. To this you might object that these would be also be contingently > identical if both Benjamin Franklin and the inventor of bifocals were > contingently self-identical. To which I would respond that identities > involving what linguistically oriented analytical philosophers refer > to as rigid designators, are identities whose terms are necessarily > self-identical; whereas identities involving what such philosophers > refer to as non-rigid designators, are identities whose terms are > contingently self-identical. Therefore, granted that I take rigid and > non-rigid designation as the linguistic marks of necessary and > contingent self-identity--putting the cart back behind the horse, > rather than approaching the matter bass ackwards as it is > fashionable to do these days--and granted that I take > Benjamin Franklin and John Correy to be 'rigid' designators > and 'the inventor of bifocals' to be 'non-rigid', I conclude > that the contingent identity of Benjamin Franklin and the inventor > of bifocals has as its basis the contingent self-identity of the > inventor of bifocals, while Benjamin Franklin is necessarily > self-identical. As to whether physical or mathematical objects are contingently > self-identical or necessarily so, some sort of metaphysical argument > (rather than a logical one) warranting one or the other of these > conclusions would have to be made. I suspect that mathematical > properties are both essential in, and necessary to, mathematical > objects--but this is an intuition and nothing more. John PS It won't surprise me if Paul Holbach or G. Frege bring in talk > about 'scope', which I think is only peripherally relevant to > discussions of necessary and contingent identity. > Let me bring in a quote: > One must distinguish between the claim that identity sentences are > contingent and the claim that the identity relation itself is > contingent. For the relation to be contingent, there need to be things > between which it holds merely contingent. For it to be necessary, it > has to be that if the relation obtains between things, it obtains > between those very things of necessity. [...] One can consistently say > that there are contingent identity sentences, though the relation > itself is necessary. Thus one could say that The first > Postmaster-General of the US was the inventor of bifocal lenses. is > contingent and is an identity sentence, but that if we consider the > object, x, which is in fact referred to by the first > Postmaster-General of the US and the object, y, which is in fact > referred to by the inventor of bifocal lenses, it is necessary that > x is identical to y. > [Sainsbury, M. (1995). Philosophical Logic. In A. C. Grayling (Ed.), > /Philosophy. A Guide Through the Subject/ (pp. 61-122). Oxford: Oxford > University Press. (p. 93)] > PH This sounds so much like what Kripke says, either in Identity and Necessity or in _Naming and Necessity_, that I hope Sainsbury cited him. Of course, what Kripsbury says represents the Party Line on contingent identity: There are *statements* of contingent identity but no instances of contingent identity itself. Oops! Before I forget, let me forestall the inevitable bUllrich-ism: Duh. You're not saying anything *new* you know. 'AxAy(x=y -> (N(x=x & y=y) <-> N(x=y)))' is a theorem of standard quantified modal logic with identity. (Giggle) --John === Subject: Re: My research, publication announcement > There's more to my work than just arguing on Usenet, so I'd like to > point out that my paper Advanced Polynomial Factorization is slated > to be published: > See http://www.megasociety.net/NoesisHighlights.html > The Mega Foundation is an organization of high IQ people, and I'm glad > to be associated with them. To learn further about the organization > you can use Google, or see: > Why a group like the Mega Foundation? > http://www.ultrahiq.org/Mega/WhyMega.htm > I hope at least some of you will appreciate that often the most > important ideas in history have to get past people limited by their > lack of imagination and their prejudices, who act against scientific > progress. > What I want you to see is that there's more to me than Usenet, so that > you can begin to understand that the revolution I'm giving you a > chance to be a part of is bigger than the small-minded people who > continually throughout history work to halt progress. > James Harris > http://mathforprofit.blogspot.com/ Crank Information http://www.crank.net/harris.html http://www.crank.net/usenet.html === Subject: [JSH] [Lunatic-Crank-Reposted Nonsense] Re: Factorization dispute, again >> So you see, my argument is correct and simple, and mathematicians are > indeed running from a little gut check in their field. They're > pussies too scared to handle the truth. Mr. Harris, still thinking that pissing out this flawed argument will ever make it true. You are so sad and your so-called math work is even sadder! Happy pissing! === Subject: Oneness of a number I am trying to build fractals generated by the principal of the oneness of a number. For the definition of the oneness of a number, view the internet, or below.* I have been unseccesful in creating a pattern around the oneness although I have struck on some intriguing details. These mainly resolve around the issue of, if we take the oneness of the oneness of a number, and keep doing that, how long does it take 'til we reach 1? (note that for 5 this never happens) I think I might be able to find a pattern in that since, if I call this the recursive oneness loop length (if anybody can think a better name, tell me), then, the recursive oneness loop length of a number like 20000000 is still only 15. I am trying to make a large-scale 2d map of it, but so far it is still pure chaos, even if the range is so limited. Does anybody know any other numerical iterations like the oneness that create a chaotic result? I think this is very interesting.. * For the oneness of a number 'n' : if(n is even) n = n/2; if(n is odd) n = 3*n + 1; Keep doing this until n = 1. The amount of necessary steps for this is called the oneness of n. -- Quaternion === Subject: Re: JSH: Difficult social problem congratulation! > 7/x = 1 requires that x=7. > It's an intriguing problem. > http://mathforprofit.blogspot.com/ --ils duces d'Enron! http://www.movisol.org/ http://members.tripod.com/~american_almanac/ === Subject: Re: JSH: Difficult social problem >And believe me, if you don't speak any French, the >waiters in France won't bring you soup. Have you considered the possibility that the reason waiters don't bring you soup might *not* be that you don't speak French? -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === Subject: Re: derivate of x^x > Ok, I've probably asked this question before, but I've forgot the answer... > What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ? > I guess asking for the integral of these functions would be foolish? Use logarithmicdrivative: f'(x) = x^x(1 + Ln(x)) For the other, aply that twice. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: NOVA strings and branes it's hearsay, about the US casualty estimates. now, the Japanese were negotiating through the VAtican Special Office and the OSS, and MacArthur's air & sea blockade was very successful; the fact is that he was not consulted! the tie by the parliament was a matter of saving face; damn right, it was broken by the emperor. you point is also taken that it was no worse than the German situation circa '45, because they were also victim of the British terror-bombings. thus was the Nuclear Age begun. The next day, a political and diplomatic storm erupted. Truman, who had previewed the speech and approved it on Sept. 11, lied and told the press that Wallace never showed him the speech. Secretary of State Byrnes and the press went ballistic, and on Sept. 20, Truman asked for Wallace's resignation and got it. Truman promptly appointed Averell Harriman in Wallace's place. For the next two decades, Wallace continued to battle for national policy direction as he saw it. That is a story for another telling. References > Now consider that, if anything, the Japanese had a (well earned) > reputation of keeping fighting even in situations where the Germans > surrendered and draw your own conclusions. > The last large scale military operation in the Pacific preceding > Hiroshima was the battle of Okinawa. Okinawa was defended by about > 100000 Japanese troops, with practically no heavy weaponry and no > close air or naval support. The island is small enough so that every > part of could've been covered by direct fire from US navy ships (and, > of course, air support from the carriers). Yet, it still took nearly > 3 months of fighting and close to 50000 US casualties (about 12000 > Japanese forces, less than 10% were taken prisoner. In addition, some > 50000-100000 civilians died. > That was a rather small island. In Japan itself, the Japanese > military had ten times as many troops (granted, poorly equipped, but > so were those in Okinawa) and many millions of civilians who were > ready to fight to the end. Based on the Okinawan experience you can > begin to estimate the cost, in casualties and destruction that would > result from landing in Japan. --ils duces d'Enron! === Subject: Re: Factorization dispute It turns out that I can isolate the current dispute easily enough by > focusing on the factorizations: Consider [snipped] > Crank Information > http://www.crank.net/harris.html > http://www.crank.net/usenet.html > Readers should please check out *all* the links Sam the Worm listed. > As for the math, notice that with > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > no other factorization works as long as 7 is not a factor of 22. You have only demonstrated anything for the case x=0, not 'x' in general. -- 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: Oneness of a number === >Subject: Oneness of a number >Message-id: For the definition of the oneness of a number, view the internet, or below.* It's more commonly called the Collatz Conjecture or the 3x+1 problem. >I have been unseccesful in creating a pattern around the oneness although I >have struck on some intriguing details. These mainly resolve around the >issue of, if we take the oneness of the oneness of a number, and keep doing >that, how long does it take 'til we reach 1? Depends on how many binary digits the number has (which is related to the magnitude of the number) AND ALSO on the pattern of 1s and 0s in the binary representaion (which is NOT related to the magniyude of the number). >(note that for 5 this never happens) It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1 >I think I might be able to find a pattern in that since, if I call >this the recursive oneness loop length (if anybody can think a better >name, tell me), It's sometimes refered to as the stopping time. Others make up their own terminology. > then, the recursive oneness loop length of a number like >20000000 is still only 15. I am trying to make a large-scale 2d map of it, How are you plotting this in 2D? >but so far it is still pure chaos, even if the range is so limited. Maybe you just need to look at the chaos from the right perspective, say by looking at the forest instead of the trees: http://members.aol.com/mensanator666/collatz/bang0010.htm >Does anybody know any other numerical iterations like the oneness that >create a chaotic result? I think this is very interesting.. >* For the oneness of a number 'n' : >if(n is even) n = n/2; >if(n is odd) n = 3*n + 1; >Keep doing this until n = 1. The amount of necessary steps for this is >called the oneness of n. >-- >Quaternion -- Mensanator Ace of Clubs === Subject: Re: JSH: Difficult social problem >And believe me, if you don't speak any French, the >waiters in France won't bring you soup. > Have you considered the possibility that the reason waiters don't > bring you soup might *not* be that you don't speak French? This is unthinkable if you have ever been to Quebec or France. The only way I could get a cup of coffee in Quebec City was for my wife to order it for me. :-) Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net === Subject: Re: NOVA strings and branes > than the German situation circa '45, because > they were also victim of the British terror-bombings. > thus was the Nuclear Age begun. Terror bombings? The Germans started bombing cities first. They simply were getting as good (or bad) as they gave. Arthur Bomber Harris said something to the effect that the Germans had sown the wind, and in due course would reap the whirlwind. So true. Bob Kolker === Subject: Re: Factorization dispute > Let's take 'em out. James Harris > Take 'em out? What do you mean by that? Are you back to your previous > threat of calling out the military or the CIA and the FBI, or are you > advocating the formation of a private militia? > I think James wants his vast silent audience to treat mathematicians to > dinner and a movie. I suspect that there's an ulterior motive. Physics students are a cool crowd and I know because I was once a physics undergrad, and I'd get into all kinds of fascinating discussions, as physics is that kind of field. When you deal with crackpots and cranks, which mathematicians are now, who have been backed into a logical corner, you go ahead and deliver the coup de grace of the irrefutable facts. That is, you take 'em out. Now then, the crackpots may still believe their nonsense, but civilized society is once again protected by the vigillance of active and highly intelligent minds, as the bulk of society can move on with the truth. I've isolated mathematicians by a logical argument, having made a major discovery--that's what physics people do--in what they probably think of as their area alone, and caught them trying to dodge the result and its implications. Remember mathematicians have claimed that pure math was for the benefit of humanity, but I've shown them trying to hide one of the most astounding results in math history (one of three, mind you). So it's time for the physics people to look at the facts, stand for logic over social needs, and take 'em out. James Harris === Subject: Re: Oneness of a number === >>Subject: Oneness of a number >>Message-id: I am trying to build fractals generated by the principal of the oneness of > number. >>For the definition of the oneness of a number, view the internet, or >>below.* > It's more commonly called the Collatz Conjecture or the 3x+1 problem. >>I have been unseccesful in creating a pattern around the oneness although >>I have struck on some intriguing details. These mainly resolve around the >>issue of, if we take the oneness of the oneness of a number, and keep >>doing that, how long does it take 'til we reach 1? > Depends on how many binary digits the number has (which is related to the > magnitude of the number) AND ALSO on the pattern of 1s and 0s in the > binary representaion (which is NOT related to the magniyude of the > number). >>(note that for 5 this never happens) > It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1 No, you didn't listen well, I take the oneness of the oneness for the recursive oneness looplength. The oneness of 5 is 5. the oneness of 5 is 5, etc.. >>I think I might be able to find a pattern in that since, if I call >>this the recursive oneness loop length (if anybody can think a better >>name, tell me), > It's sometimes refered to as the stopping time. Others make up their own > terminology. >> then, the recursive oneness loop length of a number like >>20000000 is still only 15. I am trying to make a large-scale 2d map of it, > How are you plotting this in 2D? 1. Using complex number representation and messing around with that. 2. Using y as a period of x (y = x/WIDTH, so to speak) -- Quaternion >If you saw >(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) >with the c's algebraic integers, I think few of you would have a >problem realizing that only two of the c's have 7 as a factor. No - the problem here is that this polynomial does not factor at all in the algebraic integers in the form you have shown. Its factorization is necessarily of the form 49*(x - r1)*(x - r2)*(x - r3), where r1, r2, and r3 are nonunit *algebraic integers* which are the roots of x^3 + 5*x^2 + 3*x + 2 = 0, and are such that -r1*r2*r3 = 2. Knowing this, you can distribute the 49 among the factors in many different ways. However, *none* of those ways will end up having constant terms of 7, 7, and 2. Why? Because 7 and 2 are coprime in the algebraic integers. It is not possible to write 2, which you want as the constant term of the third factor, as a product of a nonunit factor of 7 and a nonunit factor of 2. Andrzej. >But, of course, you're looking at *functions* of x, as you have >f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, >so I could also write it as >(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2). >Notice that dividing both sides by 49 gives >(f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2 >as long as you're in a ring where 7 is not a factor of 22. >I want to emphasize that point as notice there's only *one* way to >divide through by 49 if 7 is not a factor of 22. >Usually you can *see* the other factors of 7, but I want you to >abstract, and generalize. >Please pay careful attention to that example. >You may see people who reply claiming that the word polynomial has >some significance, as if it's a mystical thing which refutes basic >logic, so if something isn't polynomial it no longer behaves >logically. >Now then, in my advanced factorization work, I just use functions of x >that are a lot more complicated than f_1(x) = c_1 x, and unfortunately >there are people who can use an unfamiliar leap in complexity to >confuse others. >Some of you have learned various advanced math topics, now imagine if >in your classrooms there were some hecklers who continually hollered >out at your teacher, or otherwise disrupted the class? >What if when there were difficult concepts those hecklers would try to >confuse everyone as they sought to discredit the mathematics? >If you find that hard to imagine, imagine me in your class with you >questioning the professor and calling him names. >How much would you have learned? >I need those of you interested in mathematics to focus on the basics, >so that you can understand the advanced. >James Harris >http://mathforprofit.blogspot.com/ === Subject: Re: derivate of x^x >Ok, I've probably asked this question before, but I've forgot the answer... >What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ? Logarithmic differentiation. Doug === Subject: Re: Factorization dispute, again > Notice, http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.net Stupid is as stupid does. There is no fixing stupid because it is not broken. Harris would be much happier as a priest proclaiming god amdist heretics and buggering altar boys to His greater glory. Hey Harris, your village called: Its idiot is missing. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Usenet Posting Guide? With all respect, dear man, you strike me as that type more enamored > with process rather than function. Most of us out here have far better > things to do than screw around for weeks configuring newsreaders and > the like, even if we were so inclined, which most of us clearly are > not. I suppose we could memorize the phone book, too, but would that > help us communicate our ideas any better? > Some of us prefer to view the forest rather than count the trees down > there. Better view, too. Well, I find the technical aspects of how USENET functions much more > interesting than most of the discussions that take place on it. By the > same token, I am much more interested in the hardware and operating system > software of the systems I administer than in any of the applications > for which they are used. Once I hoped that the growing popularity of personal computers meant > that nearly everyone would learn to think like programmers. It never > occurred to me that, sadly, the opposite would happen: That computers > would be designed to be used by people who *can't* program them. > thinking here a little more closely. You're seeing the world from a > very narrow vantage point. Technology and inventive genius exist to > serve those who use and take advantage of them, not merely those who > design and invent, otherwise you might just as easily make the > argument that the world was a far better place when mostly car > builders and those who tinkered all day with Model T Fords were out > there driving on the roads. Certainly the roads were safer then, and > that those driving them generally knew their cars inside and out, but > that ignores the fundamental purpose behind that of automobiles, which > is to facilitate transportation. > The same exact argument can be made for telephones, television, > airplanes and a whole host of other modern conveniences we now take > for granted and which represent fundamental shifts in the way we > communicate and get around. They were all once the domain of a select > few tinkerers and inventors who had absolutely no idea of what they > were about to unleash on the world. > With all respect, sir, I can guess your politics right across the > board. Just for openers, you're a Sierra Club, save-the-whales type. > Gotcha, huh? I think you're entrenching him. He said it was cooler when programmers used 'em. You said it's cooler when people who knew vehicles inside out were the ones driving them. You guys are agreeing. === Subject: Re: Skeptickal Inquirer UFO I hope you're not speaking poorly about my Venus lizard folk, Cathars and all. This may be getting some folks a wee bit off topic from group focus, or that of my intended agenda upon Venus life (lizard folk and all), but according to some fairly recent feedback, I've learned a thing or two about our nasty moon, as a place that I believe we've long needed to establish a lunar space elevator (LSE) in order to be efficiently getting ourselves off to visiting the wizard of Oz at Venus L2, as well as for reaching out to those thoroughly irradiated to death snowman/snowwoman situated on Mars: http://guthvenus.tripod.com/gv-cm-ccm-01.htm Here's a little typical feedback of supposed facts from: Jay Windley (webmaster@clavius.org) High-energy cosmic rays do not come from the sun. They come from outside the solar system, and our sun is the primary defense against lower energy and thus their secondary effects in the ambient are minimal. My thoughts: http://guthvenus.tripod.com/gv-moon-radiation.htm Unfortunately, I may have incorrectly utilized the terminology of cosmic rays, though fortunately, I never specified upon any specific spectrum of high-energy cosmic rays, just pointing out that our sun is certainly capable of tossing out its fair share of far worse things than visible photons plus IR worth of BTUs and of those nasty UVs. Obviously a supernovae is worth a thousand fold in terms of being nasty, thereby from the far off generated galactic influx must offer a measurable degree of such events, and of the secondary radiation given off by all that infamous clumping lunar dirt should become a fairly darn good as well as unobstructed indicator of whatever cosmic/galactic influx. Seems fairly odd that shuch measurements aren't common place, as where's the justification for not otherwise providing this level of information? The assertions or premise offered by the likes of Jay Windley, that of our moon not only lacking an atmosphere but also without a Van Allen buffer zone is not such a bad thing if you're out and about on the lunar surface, seems somewhat risky if not downright lethal. I might have come into that understanding if we're referring to an earthshine illuminated lunar surface, but I'm not going so far if that's of any fully solar illuminated environment while wearing a mostly synthetic moon suit because, we're not talking about avoiding a 270 nm UV sun burn. Sorry about all my make-due reverse engineering logic, or lack thereof. I was simply trying to establish upon the amount of solar radiation that becomes hard X-Ray class. So exactly how much is it on a typical lunar day, or how about on a good day as well as on a truly High-energy cosmic rays do not come from the sun ??? Do we suppose that happens to include the likes of the last couple of weeks of horrific solar flak? Seems there should be some specific knowledge (excluding Apollo) of what's what pertaining to the solar illuminated surface, as opposed to the absolute lunar nighttime environment and, of something specific pertaining to whatever earthshine contributes. This task of obtaining knowledge is somewhat like my getting a grasp upon the applied energy (thrust or torque) involved in accelerating something the size and mass of the moon, so that it accelerates and thereby recedes form Earth at 38 mm/year. As worthy feedback provided from: Ami Silberman (silber@mitre.org) The mechanisms for the lunar recession have been well understood for decades. In a nutshell, tides cause friction between the oceans and the ocean floors, which transfers energy from the solid part of the earth to the oceans. One of the effects of this friction is that the tidal bulge is off-center, and is located eastward of the moon. (So the high tide actually occurs when the moon is west of overhead.) The result of the tidal bulge being off center is that there is a torgue effect placed on the moon, and this in turn transfers energy from the earth to the moon. The earth's spin rate slows, the moon is speeded in its orbit and therefor moves further away from the earth. (This transfer of energy is essentially a transfer of angular momentum, which is a conserved quantity.) The historical (over geological eras) rate of recession has varied due to varying amounts of tidal friction due to shallower or deeper oceans, and the positions of the continents. For the benefit of all my loyal critics, I've conceded that there's a darn good chance that the likes of Tim Thompson has more than a few valid points as to his version of what's what. This following page is just another example of my learning from the pros, of accepting other input, which may even including the likes of what Ami Silberman just presented, that I'd not be calling flak, as there actually seems to be some considerable worth to at least Tim's version of the lunar recession, if I don't say so myself. http://guthvenus.tripod.com/earth-moon-energy.htm === Subject: Re: Skeptical Inquirer UFO Unfortunately Scott, this sting/ruse is way so much bigger than even you can imagine. I've posted this folowing in at least a half dozen topics that I believe are related to obtaining the truth. This may be getting some folks a wee bit off topic from group focus, or that of my intended agenda upon Venus life (lizard folk and all), but according to some fairly recent feedback, I've learned a thing or two about our nasty moon, as a place that I believe we've long needed to establish a lunar space elevator (LSE) in order to be efficiently getting ourselves off to visiting the wizard of Oz at Venus L2, as well as for reaching out to those thoroughly irradiated to death snowman/snowwoman situated on Mars: http://guthvenus.tripod.com/gv-cm-ccm-01.htm Here's a little typical feedback of supposed facts from: Jay Windley (webmaster@clavius.org) High-energy cosmic rays do not come from the sun. They come from outside the solar system, and our sun is the primary defense against lower energy and thus their secondary effects in the ambient are minimal. My thoughts: http://guthvenus.tripod.com/gv-moon-radiation.htm Unfortunately, I may have incorrectly utilized the terminology of cosmic rays, though fortunately, I never specified upon any specific spectrum of high-energy cosmic rays, just pointing out that our sun is certainly capable of tossing out its fair share of far worse things than visible photons plus IR worth of BTUs and of those nasty UVs. Obviously a supernovae is worth a thousand fold in terms of being nasty, thereby from the far off generated galactic influx must offer a measurable degree of such events, and of the secondary radiation given off by all that infamous clumping lunar dirt should become a fairly darn good as well as unobstructed indicator of whatever cosmic/galactic influx. Seems fairly odd that shuch measurements aren't common place, as where's the justification for not otherwise providing this level of information? The assertions or premise offered by the likes of Jay Windley, that of our moon not only lacking an atmosphere but also without a Van Allen buffer zone is not such a bad thing if you're out and about on the lunar surface, seems somewhat risky if not downright lethal. I might have come into that understanding if we're referring to an earthshine illuminated lunar surface, but I'm not going so far if that's of any fully solar illuminated environment while wearing a mostly synthetic moon suit because, we're not talking about avoiding a 270 nm UV sun burn. Sorry about all my make-due reverse engineering logic, or lack thereof. I was simply trying to establish upon the amount of solar radiation that becomes hard X-Ray class. So exactly how much is it on a typical lunar day, or how about on a good day as well as on a truly High-energy cosmic rays do not come from the sun ??? Do we suppose that happens to include the likes of the last couple of weeks of horrific solar flak? Seems there should be some specific knowledge (excluding Apollo) of what's what pertaining to the solar illuminated surface, as opposed to the absolute lunar nighttime environment and, of something specific pertaining to whatever earthshine contributes. This task of obtaining knowledge is somewhat like my getting a grasp upon the applied energy (thrust or torque) involved in accelerating something the size and mass of the moon, so that it accelerates and thereby recedes form Earth at 38 mm/year. As worthy feedback provided from: Ami Silberman (silber@mitre.org) The mechanisms for the lunar recession have been well understood for decades. In a nutshell, tides cause friction between the oceans and the ocean floors, which transfers energy from the solid part of the earth to the oceans. One of the effects of this friction is that the tidal bulge is off-center, and is located eastward of the moon. (So the high tide actually occurs when the moon is west of overhead.) The result of the tidal bulge being off center is that there is a torgue effect placed on the moon, and this in turn transfers energy from the earth to the moon. The earth's spin rate slows, the moon is speeded in its orbit and therefor moves further away from the earth. (This transfer of energy is essentially a transfer of angular momentum, which is a conserved quantity.) The historical (over geological eras) rate of recession has varied due to varying amounts of tidal friction due to shallower or deeper oceans, and the positions of the continents. For the benefit of all my loyal critics, I've conceded that there's a darn good chance that the likes of Tim Thompson has more than a few valid points as to his version of what's what. This following page is just another example of my learning from the pros, of accepting other input, which may even including the likes of what Ami Silberman just presented, that I'd not be calling flak, as there actually seems to be some considerable worth to at least Tim's version of the lunar recession, if I don't say so myself. http://guthvenus.tripod.com/earth-moon-energy.htm === Subject: Re: Vito Rizzuto - February 21st 1946 > But DSK regularly changes his email address thus frustrating > those of us who try to killfile him. As you've no doubt noticed though Robin, his posts (to sci.math anyway) do have an Achilles heel - a four-digit year at the end of the title. If your newsreader supports wildcard titles in kill rules, all one needs is a filter on titles of the form /(19|20)dd$/ (in Perl wildcard syntax). --------------------------------------------------------------------------- John R Ramsden (jr@adslate.com) --------------------------------------------------------------------------- Eternity is a long time, especially towards the end. Woody Allen === Subject: Re: Usenet Posting Guide? >But that's always been the case. We made a lot of money doing >work that our customers didn't want or have to do. What we >didn't do (when we weren't ing idiots) was hide all the >warts under pretty pictures. I'm sorry, WHAT did you say you were doing to the idiots?! === Subject: Re: associative one-way function >Does anyone know if there is any potential one-way function >which is associative other than modular exponentiation? How is this associative? Is (a^b)^c = a^(b^c)? I think not! (And therefor I am not?) >That is, given a function f(x,y) (which is easy to compute), >- given z and x, it is difficult to find y such that z = f(x,y) >- f(f(x, y1), y2) = f(f(x, y2), y1) That doesn't look like the associative law, either. Setting the words aside, it sort of looks like what you want occurs in any group: given a group element x and an integer n we compute the power x^n in the group; certainly (x^n)^m = (x^m)^n, which is what you appear to request in your last line. And yes, it's often difficult to take an element in a cyclic group and decide what power of the generator it is. Your initial example is of this type, where the group is (Z/kZ)^* , the multiplicative group mod k (where k is any fixed integer). Another example commonly used for one-way functions is to use the group of points in an elliptic curve mod p. dave === Subject: Re: division by general integer using register shifts > Hi all, > As most here are already aware (but just in case some aren't), > if > you shift the digits of a number to the left or right one time then it's > like multiplying or dividing the number by its radix. So if you perform a > binary left shift on say: > 00011011 (27 base 10) > you get: > 00110110 (54 base 10 or 27 * 2) > So you can multiply a number by any power of two very quickly (for a > computer, register shifts are much faster than mult/div. instructions) by > shifting. What about numbers that are not powers of two? They can be done > by distributing the operation across numbers that are powers of two and sum > to the number you're trying to multiply by. Say you want to multiply by > 800. 800 is not a power of two, but 32, 256, and 512 are (32+256+512=800). > So: > n*800 = (n*32)+(n*256)+(n*512) = (n<<5)+(n<<8)+(n<<9) > The same is true for division. Binary right shifting is the same as > dividing by two. I can't seem to work out the algebra though. Can someone > tell me how (or is it possible) to do division by a general integer using > shifts? > -- > Best wishes, > Allen You could try not working binary but convert to the divisor as radix instead, and then reconvert to binary. Does this help? === Subject: differentiable...problem... if f is differentialbe on (0, infinite) and lim [f(x) +f'(x)] = L (x->infinite) show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite) ------------------------------------ it seems to be trivial. but i no touch.....oops........ help.....me......please...... In sci.math, James Harris (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) > with the c's algebraic integers, I think few of you would have a > problem realizing that only two of the c's have 7 as a factor. Not clear at all without knowing precisely what the roots are. Since two of the roots are complex in this case I might have a chance of not using trig, but it would take some work. I can tell you that the roots, as reported by Pari/GP, are approximately: -4.424076847035404679966063971 -0.2879615764822976600169680144 - 0.6075770270241740255965589263*I -0.2879615764822976600169680144 + 0.6075770270241740255965589263*I but this doesn't help much as algebraic units are provably dense around (0 + 0i) -- sqrt(n+1) - sqrt(n) is a unit. > But, of course, you're looking at *functions* of x, as you have > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > so I could also write it as > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2 > as long as you're in a ring where 7 is not a factor of 22. Only an issue modulo 3 or 5, AFAICT.... :-) I can posit a simpler example. Let's assume for giggles the following equation, similar to yours: (d_1 * x + 7) * (d_2 * x + 7) * (d_3 * x + 2) = 49 * (x^3 + 2) and carry through your logic. I know the roots to this equation. :-) We know that x^3 + 2 = (x + 2^(1/3)) * (x + 2^(1/3) * (-1/2 + sqrt(3) * I/2)) * (x + 2^(1/3) * (-1/2 - sqrt(3) * I/2)) (GP/Pari verifies this nicely, as it turns out). Therefore, one can pick d_1 = 7 / (2^(1/3)), d_2 = 7 / (2^(1/3) * (-1/2 + sqrt(3) * I/2)), d_3 = 2 / (2^(1/3) * (-1/2 - sqrt(3) * I/2)). Since -1/2 + sqrt(3) * I/2 and -1/2 - sqrt(3) * I/2 are both units (as is easily verified by multiplying them together), they factor little in the analysis. I can just as easily posit d_1' = d_2' = 7 / (2^(1/3)) d_3' = 2 / (2^(1/3)) = 2^(2/3) Now d_3' turns out to be an algebraic integer, solving the equation y^3 - 4 = 0. (d_3 solves this equation too.) d_1' ? Well, lessee; the most obvious equation is 2 * y^3 - 343 = 0 which is clearly not going to lead to anything resembling an algebraic integer. Since d_1' requires a cube root no equation of lesser degree will do. But I may need those roots to your equation, and not just approximations, either. If one posits N(x) = x^3 + 5*x^2 + 3*x + 2 then, if one substitutes x = y-5/3, one can rewrite this as: N(y-5/3) = y^3 - 16/3 * x + 169/27 Now substitute y = w + 16/(9*w) (Vi.8fta's substitution, with p = -16/3) and get N(w+16/(9*w)-5/3) = (729*w^6 + 4563*w^3 + 4096)/(729*w^3) Surprise: w^3 = (-4563 ± sqrt(4563^2-4*729*4096)) / (2 * 729) = (-4563 ± sqrt(8877033)) / (2 * 729) , by our old friend, the quadratic formula. Since 8877033 = 3^9*11*41 and 4563 = 3^3 * 13^2, we can divide by 3^3/3^3 (or sqrt(3^6)/3^3), resulting in w^3 = (-169 ± sqrt(12177)) / (2 * 27) or w^3 = (-169 ± 3*sqrt(1353)) / (2 * 27) I'm going to take the + sign; the - sign will yield slightly identical results, as it turns out. Because GP/Pari does odd (but somewhat logical) things with (-1)^3, I'm going to take the sign out now. The results can be written: x1 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) - 16/(3*((169 - 3*sqrt(1353)) / 2)^(1/3)) x2 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 - sqrt(3) * I/2) - 16/(3*((169 - 3*sqrt(1353)) / 2)^(1/3)) * (-1/2 + sqrt(3) * I/2) x3 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 + sqrt(3) * I/2) - 16/(3*((169 - 3*sqrt(1353)) / 2)^(1/3)) * (-1/2 - sqrt(3) * I/2) It's worth noting that (169 - 3*sqrt(1353)) * (169 + 3*sqrt(1353)) = 16384 = 2^14, so one can rewrite the above as x1 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) - 1/3 * ((169+3*sqrt(1353))/2)^(1/3) x2 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 - sqrt(3) * I/2) - 1/3 * ((169+3*sqrt(1353))/2)^(1/3) * (-1/2 + sqrt(3) * I/2) x3 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 + sqrt(3) * I/2) - 1/3 * ((169+3*sqrt(1353))/2)^(1/3) * (-1/2 - sqrt(3) * I/2) If you look carefully you'll see that using the alternate solution w^3 = (-169 - 3*sqrt(1353)) / (2 * 27) will merely swap x2 and x3 in the final list. In this form, figuring out whether xi is an algebraic integer is a Lost Cause(tm), although we know the answer is yes since they all solve the equation x^3 + 5*x^2 + 3*x + 2 = 0, as one can verify if one multiplies (x-x1)*(x-x2)*(x-x3) in GP/Pari. However, there's a different route, making the exact roots entirely irrelevant. (Sorry, guys -- but it *was* an interesting solution using Vi.8fta's substitution. :-) ) We are hypothesizing (c_1*x + 7)(c_2*x + 7)( c_3*x + 2) = 49(x^3 + 5 x^2 + 3x + 2). for some algebraic numbers ci. This hypothesis turns out to be quite valid, as it turns out -- but what are the ci? Since we know x^3 + 5x^2 + 3x + 2 = (x-x1)*(x-x2)*(x-x3) we can deduce that c1 = -7/x1, c2 = -7/x2, and c3 = -2/x3, for some rotation of x1, x2, x3. Are these algebraic integers? We note that, if xi solves x^3 + 5 x^2 + 3x + 2 = 0, then 1/xi has to solve 2*z^3 + 3*z^2 + 5*z + 1 = 0, 7/xi has to solve 2*z^3 + 3*7*z^2 + 5*7^2*z + 7^3 = 0, and -7/xi has to solve 2*z^3 - 3*7*z^2 + 5*7^2*z - 7^3 = 0. -7/xi is clearly *not* an algebraic integer (as the above is not a reducible polynomial over Q). Fortunately, -2/xi is: 2*z^3 - 3*2*z^2 + 5*2^2*z - 2^3 = 0 or z^3 - 3*z^2 + 10*z - 4 = 0 Well, 1 out of 3 ain't bad. This counter-proof generalizes quite nicely for most cubic polynomials x^3 + a_2 * x^2 + a_1 * x + a_0 with non-unit a_0 not divisible by 7. I'm afraid you may be out of luck, James. :-) [rest snipped] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Uncle Al is Sadistic . In sci.math, Nicolas Le Roux : > amanda grava .88 la saucisse et au marteau: > [SNIP] > Could you *please* the lines you are not responding to. Erm...are you sure you didn't inadvertantly snip the word snip? Pleasing the lines sounds impossible; I've never quite figured out whether chocolates or flowers will make the sentence This statement is false happy. :-) ;-) :-) > Otherwise, this > is very hard and unpleasant to read and, besides, it takes more time to > download (and it's significative with a low-speed modem). I can relate to that to some extent; I've a 56k myself. Fortunately, I also use leafnode (fetchnews) -- though that doesn't do a thing for having to pay by the connection minute. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: torque T = r x F and basic tensors > Q( a x b ) = 1/(detQ) ( Qa x Qb ) > where Q is the matrix of the change of basis, but I'm not quite seeing > how that matches the A' = Q^{-1} A Q form. > As you say, under the change of orthonormal basis, the vectors a and b > go to a' = Qa and b' = Qb. In component form, > a_i = (sum over m) Q_(im) a_m and b_i = (sum over n) Q_(in) b_n > I've used different summation indices in order that the substitutions > below makes sense. > Because both the new and old bases are orthonormal, Q is an orthogonal > matrix, i.e., Q^T = Q^(-1), i.e if M = Q^(-1), M_(ij) = Q_(ji). Thus, > det Q = +/- 1. If in addition, right-handed orthonormal bases are > taken to right-handed orthonormal bases, then det Q = 1, and Q is called > a special orthogonal matrix. The set of all special orthogonal matrices > forms the group SO(3), where the notation is fairly self-explanatory - > S for special (det +1), O for orthogonal (inverse = transpose), 3 for > 3 by 3. > Define matrix A by A_(ij) = a_i b_j - a_j b_i. Under a transformation, > A'_ij = a'_i b'_j - a'_j b'_i > = (sum over m,n) [Q_(im) a_m Q_(jn) b_n > - Q_(jn) a_n Q_(im) b_m] > = (sum over m,n) [Q_(im) a_m b_n Q_(jn) > - Q_(im) a_n b_m Q_(jn)] > = (sum over m,n) [Q_(im) (a_m b_n - a_n b_m) Q_jn] > = (sum over m,n) [Q_(im) A_(mn) Q^(-1)_nj] > A = Q A Q^(-1) > This is a little different than A = Q A Q^(-1) because, the way Feynman > defines things, a'_i = (sum over n) Q^(-1)_(in) a_n. It's a good > exercise to show this. sure what you were meaning to show here, if you will note the last 4 lines of your posting. that is inconsistent with how I specified the change of basis, it should have been A' = Q A Q^{-1} and that is what you have. I also note that Feynman's matrix object T_ij = x_i F_j - x_j F_i does indeed transform as T' = Q T Q^{-1}, as you have shown. I am still mystified as to what real value there is in taking a nice axial vector object like Torque and making the anti-symmetric matrix out of it... === Subject: Re: Uncle Al is Sadistic . grava .88 la saucisse et au marteau: > Could you *please* the lines you are not responding to. > Erm...are you sure you didn't inadvertantly snip the word snip? Actually, as far as I remember, I think it was the word delete. -- Nicolas Wow, THAT'S an interessant discussion === Subject: Re: differentiable...problem... > if f is differentialbe on (0, infinite) > and lim [f(x) +f'(x)] = L (x->infinite) > show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite) Well. Perhaps L'Hopital's Rule has some use, after all. --Ron Bruck === Subject: question on finding an isogeny for two curves Can the direct isogeny be found between these two elliptic curves? It is known that they are in a set of 8 isogenous curves, but what is the exact transformation that takes x <--> X, i.e. points on (1) to points on (2)? (1) y^2 = (x + 366)*(x^2 - 366*x + 2625489) (2) Y^2 = (X - 1551)*(X^2 + 1551*X + 497214) Both curves have conductor 210. More generally, how might one find the isogeny between y^2 = x^3+ax+b and Y^2 = X^3+AX+B, when the curves have the same conductor? (a,b,A,B are known) === Subject: Re: Uncle Al is Sadistic . > amanda grava .88 la saucisse et au marteau: > [SNIP] > Could you *please* the lines you are not responding to. Otherwise, this > is very hard and unpleasant to read and, besides, it takes more time to > download (and it's significative with a low-speed modem). Well..when I used to do that (not in this group though), I was accused of leaving out the info to mislead the reader, etc. Anyway, I am very sorry. I will do that from now on. === Subject: Re: Continuum fluently presume at least that you are presumably fluent in what passes for English in the state that gave the world Hollywood . It's sheer fun to see how all those quick to stomp into dirt some kid asking for homework help, wouldn't find a pick better then notational for my babbling. Know ye, philistines, that Z stands for Ze Any Set With Cardinality of Ze Continuum. amateur > Besides, to comment on you > awareness, what languages do you fluently read in except english ? > You presume that I am fluent in English? === Subject: Re: Continuum fluently presume at least that you are presumably fluent in what passes for English in the state that gave the world Hollywood . It's sheer fun to see how all those quick to stomp into dirt some kid asking for homework help, wouldn't find a pick better then notational for my babbling. Know ye, philistines, that Z stands for Ze Any Set With Cardinality of Ze Continuum. amateur > Besides, to comment on you > awareness, what languages do you fluently read in except english ? > You presume that I am fluent in English? === Subject: [JSH] Re: My [Crank] research [Yeah Right], publication [in Crank Net] announcement [Joke] > There's more to my work than just arguing on Usenet, so I'd like to > point out that my paper Advanced Polynomial Factorization is slated > to be published: > See http://www.megasociety.net/NoesisHighlights.html > The Mega Foundation is an organization of high IQ people, and I'm glad > to be associated with them. To learn further about the organization > you can use Google, or see: > Why a group like the Mega Foundation? > http://www.ultrahiq.org/Mega/WhyMega.htm Mr. Harris, to me, this seems equivalent to the copyright you had on the web page you created for your famed Proof of FLT. Now, you are once again hell-bent on crying, cursing, cheating and trying to do anything to propagate your flawed lies. You also have some sort of greatness disorder that makes you think you are above every person you come in contact with. IIRC, this is a Narcissistic Personality Disorder. Here is excerpt from the web site: http://www.suite101.com/welcome.cfm/npd Narcissists attract abuse. Haughty, exploitative, demanding, insensitive, and quarrelsome - they tend to draw opprobrium and provoke anger and even hatred. Sorely lacking in interpersonal skills, devoid of empathy, and steeped in irksome grandiose fantasies - they invariably fail to mitigate the irritation and revolt that they induce in others. Here is a clinical definition: http://www.behavenet.com/capsules/disorders/narcissisticpd.htm *** Diagnostic criteria for 301.81 Narcissistic Personality Disorder (cautionary statement) A pervasive pattern of grandiosity (in fantasy or behavior), need for admiration, and lack of empathy, beginning by early adulthood and present in a variety of contexts, as indicated by five (or more) of the following: (1) has a grandiose sense of self-importance (e.g., exaggerates achievements and talents, expects to be recognized as superior without commensurate achievements) (2) is preoccupied with fantasies of unlimited success, power, brilliance, beauty, or ideal love (3) believes that he or she is special and unique and can only be understood by, or should associate with, other special or high-status people (or institutions) (4) requires excessive admiration (5) has a sense of entitlement, i.e., unreasonable expectations of especially favorable treatment or automatic compliance with his or her expectations (6) is interpersonally exploitative, i.e., takes advantage of others to achieve his or her own ends (7) lacks empathy: is unwilling to recognize or identify with the feelings and needs of others (8) is often envious of others or believes that others are envious of him or her (9) shows arrogant, haughty behaviors or attitudes Reprinted with permission from the Diagnostic and Statistical Manual of Mental Disorders, fourth Edition. Copyright 1994 American Psychiatric Association *** Perhaps instead of spewing your useless nonsense, you should study this disorder, go see a doctor and then drug and drink yourself to death. Think about, eh, let's wack you! === Subject: Re: torque T = r x F and basic tensors at 05:01 PM, jjensen14@hotmail.com (J Jensen) said: >1. Why is there a cross product in the definition of torque T = r x >F ? Because you're doing Physics in 3 dimensions, you can map skew-symmetric tensors into vectors. For more than 3 dimensions, you have to explicitly treat the torque as a skew symmetric tensor or a 2-form. >2. How is torque a tensor? Because the definition of cross product makes it transform like one. If you don't want to treat it as a tensor or a 2-form, than you need to treat it as a vector and pay close attention to your weights. Either way, you need to be careful to distinguish covariant indices from contravariant. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Hard tensor question (kst). <3fa668f5$9$fuzhry+tra$mr2ice@news.patriot.net> at 11:23 PM, dynamics@vianet.on.ca (Ken S. Tucker) said: >As I understand, the indices forming the relatively >defined tensor K are meaningless after the definition, >but I have some uncertainty. I'm not sure what you're trying to say, but change to polar coordinates and observe what happens. >The relative tensor K_uv is actually a 4th rank >(counting weight) like the Riemann-Christoffel >Curvature tensor. The weight has nothing to do with the rank. >It is very evident the Kronecker delta (d^u_v) >serves as an orthogonal metric, No it doesn't. >relativity of K_uv ? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: [JSH] [LUNATIC FRINGE] Re: My [Flawed] research, [JOKE] publication announcement Mr. Harris, Psychopathology ------------------------ Narcissism: A pattern of traits and behaviors which signify infatuation and obsession with one's self to the exclusion of all others and the egotistic and ruthless pursuit of one's gratification, dominance and ambition. Perhaps research in to your delusional fits of grandeur would be a more appropriate vocation for you? Still pissing out your research? === Subject: Re: Why is math so difficult for some people? >> Similarly, Cramer's proof of the Levy-Cramer theorem that >> the sum of two independent random variables is not normal >> unless they both are is probably the only easy proof, but >> it likewise obscures the ideas. Any time characteristic >> functions are used to prove a probability theorem, the >> concepts are not even present. >Do you happen to know of a probabilistic proof of the Cramr-Lvy >theorem? > There are entropy proofs. Any published? -- A. === Subject: Re: Why is math so difficult for some people? >> Many of the so-called elegant proofs completely [sic] hide the concepts. >Elegance is characterized by the propensity toward dispensing with >extraneous inessentials. >> Proving the Central Limit Theorem for sums of >> independent identical distributions is certainly the most >> elegant, but it [sic] hides virtually everything; it has no >> probability in it at all. But the characteristic function proof includes extraneous inessentials. The Lindeberg proof and the Skorokhod embedding use different strictly probability approaches, although they are not completely probability arguments; I do not believe there can be a completely probability argument, as versions of the Berry-Esseen bounds just cannot be done strictly probabilistically; the bounds on the difference of sums of iid random variables with a finite third moment and sums of iid normal random variables with the same variance, but the sequences dependent, does not increase with the number. Both of these proofs go over almost verbatim for martingales. The characteristic function proof does go over, but not as easily. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 > If you saw > (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2) > with the c's algebraic integers, I think few of you would have a > problem realizing that only two of the c's have 7 as a factor. > But, of course, you're looking at *functions* of x, as you have > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > so I could also write it as > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2 > as long as you're in a ring where 7 is not a factor of 22. > I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. > Usually you can *see* the other factors of 7, but I want you to > abstract, and generalize. > Please pay careful attention to that example. > You may see people who reply claiming that the word polynomial has > some significance, as if it's a mystical thing which refutes basic > logic, so if something isn't polynomial it no longer behaves > logically. All functions that I know of, all of them behave logically when examined from the view point of calculus. Well, I know that a derivative measures change and I also know that when the instantaneous change is 0, there is some sort of a critical point (determined by the 2nd derivative). Let's say we want to describe the graph of f(x) in words. Begin by finding f'(x). f'(x)=cos(x) We know that a function has a critical point at a point where the derivative is 0. cos(x)=0 at x=pi/2 and 3pi/2 Ok, we know where the critical points are, but we don't know what happens at those points. So take the 2nd derivative. f''(x)=-sin(x) When x=pi/2, f''(x)=-1. When x=3pi/2 f''(x)=1 Therefore I can now analyze my results. When x=pi/2, the 2nd derivative is negative, which means it's decreasing so pi/2 must be a maximum. When x=3pi/2, the 2nd derivative is positive so 3pi/2 must be a minimum. In fact, if I graph this, my predictions are correct. My point is that all elementary functions that I know of behave logically if you examine it from a calculus viewpoint, which may not be a bad idea for you to do. David Moran > Now then, in my advanced factorization work, I just use functions of x > that are a lot more complicated than f_1(x) = c_1 x, and unfortunately > there are people who can use an unfamiliar leap in complexity to > confuse others. > Some of you have learned various advanced math topics, now imagine if > in your classrooms there were some hecklers who continually hollered > out at your teacher, or otherwise disrupted the class? > What if when there were difficult concepts those hecklers would try to > confuse everyone as they sought to discredit the mathematics? > If you find that hard to imagine, imagine me in your class with you > questioning the professor and calling him names. > How much would you have learned? > I need those of you interested in mathematics to focus on the basics, > so that you can understand the advanced. > James Harris > http://mathforprofit.blogspot.com/ > You may see people who reply claiming that the word polynomial has > some significance, as if it's a mystical thing which refutes basic > logic, so if something isn't polynomial it no longer behaves > logically. No both polynomial and non-polynomial functions behave logically. It is just that polynomials have some properties that are not true in general. Let P(x) by a polynomial divisible by p, and let g_1(x) and g_2(x) be polynomials such that P(x) = g_1(x) * g_2(x) Then p divides at least one of g_1(x) and g_2(x). If g_1(x) and g_2(x) are not polynomials, this is no longer true. Counterexample. Let P(x) = 15-3x. Then P(x) is divisible by 3. Let g_1(x)=4-sqrt(1+3x) and g_2(x)=4+sqrt(1+3x). Then P(x) = g1(x)*g2(x) list a few values P( 0) = 15, g_1( 0) = 3, g_2( 0) = 5 P( 1) = 12, g_1( 1) = 2, g_2( 1) = 6 P( 5) = 0, g_1( 5) = 0. g_2( 5) = 8 P( 8) = -9, g_1( 8) = -1. g_2( 8) = 9 P(16) = -33, g_1(16) = -3. g_2(16) = 11 Note how the factor of 3 switches back and forth between g_1 and g_2 depending on the value of x. Thus neither g_1(x) nor g_2(x) is divisible by 3 for all x. In particular the fact that g_1(0) is divisible by 3 does not mean that g_1(1) is divisible by 3. - William Hughes > If you saw > (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) > with the c's algebraic integers, I think few of you would have a > problem realizing that only two of the c's have 7 as a factor. How do you know that this is possible with algebraic integer c1 to c3? I do not think this is a valid factorisation in the algebraic integers. Could you provide the monic polynomial with integer coefficients of which the c's are the roots? More to the point, x^3 + 5x^3 + 3x + 2 factors as (x - r1)(x - r2)(x - r3) in the algebraic integers. Where the r's are the roots of the polynomial. We can multiply the factors by 7/r1, 7/r2 and 2/r3 to get your factorisation. So we get: (7/r1 x + 7)(7/r2 x + 7)(2/r3 x + 2). So c1 = 7/r1, c2 = 7/r2, c3 = 2/r3. Of these only c3 is an algebraic integer (r3 is a divisor of 2). Neither 7/r1, nor 7/r2 are algebraic integers. Unless you claim that r1 and r2 are units (they are divisors of 2). But they are not, because the constant term of a polynomial (see the correct usage here, James) must be 1 for a root to be a unit. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: differentiable...problem... > if f is differentiable on (0, infinite) > and lim [f(x) + f'(x)] = L (x->infinite) > show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite) Hint: L'Hopital Note: this classic problem is well-known due to the fact that it appeared in Hardy's classic calculus textbook with a solution more awkward than the elegant L'Hopital approach. For further details and references on the L'Hospital-based approach see my prior post [1], and [2]. -Bill Dubuque [2] Martin D. Landau; William R. Jones. A Hardy Old Problem. Math. Magazine 56 (1983) 230-232. === Subject: Re: differentiable...problem... >if f is differentialbe on (0, infinite) >and lim [f(x) +f'(x)] = L (x->infinite) >show that lim f(x) = L (x->infinite) and >lim f'(x) = 0 (x->infinite) Bill Dubuque (wgd@berne.ai.mit.edu) > f e^x (f + f') e^x > lim f + f' = L => lim f = lim ----- = lim ------------ = L >x->oo x->oo x->oo e^x x->oo e^x Provided f e^x -> +-oo, in which case 0 = L - L = lim f+f' - lim f = lim f' However when f e^x -> k; then f = fe^x e^-x -> 0 f e^-x = f e^x e^-2x -> 0 L = lim f+f' - lim 2f = lim f'-f 0 = lim f = lim f e^-x / e^-x = (f' - f)e^-x / -e^-x = lim f-f' = -L 0 = lim f+f' - lim f = lim f' So if lim f = k and lim f' exists, then lim(x->oo) f+f' = k + lim f' k = lim(x->oo) f = k + lim f'; lim f' = 0 If lim f' doesn't exist, then for n >= 2 f_n(x) = sin(x^n)/x -> 0 (f_n)'(x) = nx^(n-1) cos (x^n)/x - (sin x^n)/x^2 -> oscillation ---- === Subject: Re: differentiable...problem... > if f is differentialbe on (0, infinite) > and lim [f(x) +f'(x)] = L (x->infinite) > show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite) Let's suppose L=0 (substitute f by f-L). We must prove that if f(x)+f'(x)->0 (x->inf), then f(x)->0 (x->inf) f(x)+f'(x)=o(1) (x->inf) multiply by e^(x) : e^(x)[f(x)+f'(x)]=o(e(x)) (x->inf) integrate from 0 to t : (the derivative of e^(x)f(x) is e^(x)[f(x)+f'(x)]) e^(t)f(t)-f(0)=o(e(t)) (t->inf) Divide by e^t : f(t)=o(1) which means that f(t)->0 (t->inf) (excuse my english) === Subject: if x has normally distributed digits, does 1/x? Is there a simple proof, or counterexample, to the proposition that if we consider the expansion of the fractional part of some (irrational) number x in some base, and know that it is normal, then it follows that the expansion of 1/x is also normal? I'd appreciate a pointer to any literature in this area. Rob Shaw === Subject: Re: if x has normally distributed digits, does 1/x? >Is there a simple proof, or counterexample, to the proposition >that if we consider the expansion of the fractional part >of some (irrational) number x in some base, and know that it is normal, >then it follows that the expansion of 1/x is also normal? >I'd appreciate a pointer to any literature in this area. What do you mean by normally distributed digits? The normal distribution is a continuous one. Nor would it make sense to speak of x - floor(x) being normally distributed, since the normal dsitrubtion is supported by the entire real line. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: if x has normally distributed digits, does 1/x? >Is there a simple proof, or counterexample, to the proposition >that if we consider the expansion of the fractional part >of some (irrational) number x in some base, and know that it is normal, >then it follows that the expansion of 1/x is also normal? >I'd appreciate a pointer to any literature in this area. > What do you mean by normally distributed digits? The normal > distribution is a continuous one. > Nor would it make sense to speak of x - floor(x) being normally > distributed, since the normal dsitrubtion is supported by the entire > real line. A real number is said to be normal (to the base 10) if for all n every string of digits of length n appears with frequency 10^(-n) in its decimal expansion. This has nothing to do with the (Gaussian) normal distribution that you are thinking of. As others have noted, it seems unlikely that x normal implies 1/x normal, but I don't know what's in the literature. Let y be normal to base 2, and let x be the number you get by making believe y is a decimal expansion. Then x isn't normal - it has only the digits 0 and 1 - but I'd expect 1/x to be normal (though I wouldn't be able to prove it). -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: if x has normally distributed digits, does 1/x? >Is there a simple proof, or counterexample, to the proposition >that if we consider the expansion of the fractional part >of some (irrational) number x in some base, and know that it is normal, >then it follows that the expansion of 1/x is also normal? I seriously doubt it. Here's an indication of why: Let's talk about base 2. Start with x = .01010101... . Of course x is not normal, but it does contain the right number of 0's and 1's; it's weakly normal in base 2, in a sense. But x = 1/3, so 1/x = three = 11.000... , which is far from normal. I wouldn't be surprised if you could actually construct a counterexample along these lines - you'd need to verify a few things... >I'd appreciate a pointer to any literature in this area. >Rob Shaw David C. Ullrich === Subject: Re: if x has normally distributed digits, does 1/x? >Is there a simple proof, or counterexample, to the proposition >that if we consider the expansion of the fractional part >of some (irrational) number x in some base, and know that it is normal, >then it follows that the expansion of 1/x is also normal? > I seriously doubt it. Here's an indication of why: > Let's talk about base 2. Start with x = .01010101... . > Of course x is not normal, but it does contain the right > number of 0's and 1's; it's weakly normal in base 2, in > a sense. But x = 1/3, so 1/x = three = 11.000... , which is > far from normal. You're in the land of rationals. Rationals and normality don't mix. (Take that out of context, man!) Phil -- Unpatched IE vulnerability: window.open search injection Description: cross-domain scripting, cookie/data/identity theft, command execution Reference: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-Content.HTM Exploit: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htm === Subject: Re: if x has normally distributed digits, does 1/x? >>Is there a simple proof, or counterexample, to the proposition >>that if we consider the expansion of the fractional part >>of some (irrational) number x in some base, and know that it is normal, >>then it follows that the expansion of 1/x is also normal? >> I seriously doubt it. Here's an indication of why: >> Let's talk about base 2. Start with x = .01010101... . >> Of course x is not normal, but it does contain the right >> number of 0's and 1's; it's weakly normal in base 2, in >> a sense. But x = 1/3, so 1/x = three = 11.000... , which is >> far from normal. >You're in the land of rationals. >Rationals and normality don't mix. >(Take that out of context, man!) Well, that's amusing enough to make it seem possible that you were just trying to be funny. In case you were also trying to make a serious point: I certainly didn't mean that this was a counterexample, or something that would lead to a counterexample with no new ideas required. But what I had in mind was something _vaguely_ like this: Suppose we could find x_n such that each n-bit sequence in the binary expansion of x_n occurs with the right frequency, but 1/x_n has a terminating binary expansion. Then let x consist of a long stretch of the bits of x_1, followed by a much longer stretch of the bits of x_2, etc. For suitable values of long and longer x will be normal, but it doesn't seem so unlikely that 1/x would turn out to be abnormal. >Phil >-- >Unpatched IE vulnerability: window.open search injection >Description: cross-domain scripting, cookie/data/identity > theft, command execution >Reference: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-Content.HTM >Exploit: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htm David C. Ullrich === Subject: Image of a straight line in a convex cylindrical mirror What is the reflection in a cylindrical mirror x^2+y^2 = a^2 of a straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c) ? Is it a catenary ? If not, what is it? === Subject: Re: Image of a straight line in a convex cylindrical mirror When is the homework due? > What is the reflection in a cylindrical mirror x^2+y^2 = a^2 of a > straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c) > ? Is it a catenary ? If not, what is it? === Subject: Re: Image of a straight line in a convex cylindrical mirror > When is the homework due? :) this is my curiosity about curvarture of lines when the virtual image is distorted in reflection. I observed clear images of square lines of floor mosaic in a polished cylindrical flowerpot in Twelve Oak Mall, Detroit. What is commonly known is the optics: image-object distances are connected by standard formula (u-f)(v-f)=f^2 where f=a/2 or focal length, u,v are object & image distances etc. in reflection. I did not do much of image / ray tracing in 3D before . I believed that cos(slope angle) in the x-z projection is inversely proportional to z-coordinate, as it is for a catenary z=A cosh(x/A). I asked this question expecting that there could be elementary geometrical operations with straight lines, cylinders, reflection &c. My imagination surrounded vague thoughts about geodesic deviation, but I right now I would thank in advance for any ensuing pointers and clarifications. === Subject: Re: Image of a straight line in a convex cylindrical mirror > What is the reflection in a cylindrical mirror x^2+y^2 = a^2 of a > straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c) > ? Is it a catenary ? If not, what is it? What have you done so far? === Subject: another idea about Harris This guy is totally out of touch with reality. So what would be the harm in simply ignoring him from this day forward and forever, and permitting him to retire in the triumphant knowledge (as far as he's concerned) that he is absolutely and totally correct in everything he's ever said? He would be quite happy, and then you all could have a lot more fun communicating with real mathematicians. You all must know by now that you are wasting your time. I mean the ones of you who are responding in a thoughtful reasoned way to James. The others, who are simply calling him a horse's ass, should join another group, maybe the mutualasskicking.math group. === Subject: Re: another idea about Harris > This guy is totally out of touch with reality. So what > would be the harm in simply ignoring him from this > day forward and forever, and permitting him to retire > in the triumphant knowledge (as far as he's concerned) > that he is absolutely and totally correct in everything > he's ever said? he wouldn't retire. he is a world-class crackpot and troll. have you seen his picture? > He would be quite happy, and then you > all could have a lot more fun communicating with real > mathematicians. You all must know by now that you > are wasting your time. I mean the ones of you who > are responding in a thoughtful reasoned way to James. > The others, who are simply calling him a horse's ass, > should join another group, maybe the > mutualasskicking.math group. === Subject: Re: another idea about Harris > This guy is totally out of touch with reality. So what > would be the harm in simply ignoring him from this > day forward and forever, and permitting him to retire > in the triumphant knowledge (as far as he's concerned) > that he is absolutely and totally correct in everything > he's ever said? He would be quite happy, and then you > all could have a lot more fun communicating with real > mathematicians. You all must know by now that you > are wasting your time. I mean the ones of you who > are responding in a thoughtful reasoned way to James. > The others, who are simply calling him a horse's ass, > should join another group, maybe the > mutualasskicking.math group. Besides being a crank, he is a master troll. His posts are irresistable. You yourself have added yet another JSH-related post to this newsgroup. === Subject: Re: another idea about Harris >>This guy is totally out of touch with reality. So what >>would be the harm in simply ignoring him from this >>day forward and forever, and permitting him to retire >>in the triumphant knowledge (as far as he's concerned) >>that he is absolutely and totally correct in everything >>he's ever said? He would be quite happy, and then you >>all could have a lot more fun communicating with real >>mathematicians. You all must know by now that you >>are wasting your time. I mean the ones of you who >>are responding in a thoughtful reasoned way to James. >>The others, who are simply calling him a horse's ass, >>should join another group, maybe the >>mutualasskicking.math group. > Besides being a crank, he is a master troll. His posts are > irresistable. You yourself have added yet another JSH-related post to > this newsgroup. And you another ... and me too. Gib === Subject: Re: another idea about Harris ( [JSH] for killfiles) Francis Harrington > This guy is totally out of touch with reality. So what > would be the harm in simply ignoring him from this > day forward and forever, and permitting him to retire > in the triumphant knowledge (as far as he's concerned) > that he is absolutely and totally correct in everything > he's ever said? He would be quite happy, and then you > all could have a lot more fun communicating with real > mathematicians. You all must know by now that you > are wasting your time. I mean the ones of you who > are responding in a thoughtful reasoned way to James. > The others, who are simply calling him a horse's ass, > should join another group, maybe the > mutualasskicking.math group. Talking to Harris _about mathematics_ is a waste of time, yes. After eight years of claiming to have a proof of FLT, he doesn't know an algebraic integer is, or what a ring extension is. Conclusion? He doesn't _care_ whether his statements are true or false. Even if he had no audience at sci.math, would he stop posting? It costs him nothing to cross-post to other rooms from which he would still get his arse kicked. The JSH thing is tedious most of the time, and morbid all the time, but still good for the occasional laugh. Try Google with these keywords: guffaw idiot polynomial LH === Subject: Analysis problem... I'm having a hard time with the following problem: Let f be a continuous function on [1,oo) such that f^4 is Lebesgue integrable. Prove that if a > 4 then |integral of f(x^a)| < oo where the integral is also over [1,oo). Also, give an example where the integral of f(x^4) is infinite. Any hints or input on this would be greatly appreciated. Hugh === Subject: Re: Analysis problem... > I'm having a hard time with the following problem: > Let f be a continuous function on [1,oo) such that > f^4 is Lebesgue integrable. > Prove that if a > 4 then |integral of f(x^a)| < oo where > the integral is also over [1,oo). > Also, give an example where the integral of f(x^4) is infinite. > Any hints or input on this would be greatly appreciated. Can you write the integral of f(x^a) differently? Do you know Hoelders inequality? HTH, Michael. -- &&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&& Dr. Michael Ulm FB Mathematik, Universitaet Rostock michael.ulm@mathematik.uni-rostock.de === Subject: Re: Analysis problem... [[ This message was both posted and mailed: see the To, Cc, and Newsgroups headers for details. ]] > I'm having a hard time with the following problem: > Let f be a continuous function on [1,oo) such that > f^4 is Lebesgue integrable. > Prove that if a > 4 then |integral of f(x^a)| < oo where > the integral is also over [1,oo). > Also, give an example where the integral of f(x^4) is infinite. I'll give you a hint for a = 5: consider a change of variable in int_1^infty f(x^5)^4 x^4 dx, then think Holder. --Ron Bruck === Subject: Re: JSH: Difficult social problem === >Subject: Re: JSH: Difficult social problem >Message-id: And believe me, if you don't speak any French, the >waiters in France won't bring you soup. Have you considered the possibility that the reason waiters don't > bring you soup might *not* be that you don't speak French? >I think that when I order soup in France (in English), I would get soup. >As the French have the word soupe which is some form of soup. E.g. >bouillabaisse is a soupe. But in that case it will be the main >course. >A better example would be ordering cold water in Italy. Do not be >surprised when you get hot water. I ordered two lattes at an AutoGrill in Italy, and got exactly (and only) what I ordered...two cups of milk. Marvin Sebourn osugeography@aol.com >dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, >+31205924131 >home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Difficult social problem >And believe me, if you don't speak any French, the >waiters in France won't bring you soup. > Have you considered the possibility that the reason waiters don't > bring you soup might *not* be that you don't speak French? I think that when I order soup in France (in English), I would get soup. As the French have the word soupe which is some form of soup. E.g. bouillabaisse is a soupe. But in that case it will be the main course. A better example would be ordering cold water in Italy. Do not be surprised when you get hot water. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: help me, please! > William Elliot ha scritto nel messaggio >Is there a fast way to determine X and Y both integer such that: >X^2 - Y^2 = A >>Of course the answer to your question is yes. >>Indeed, you phrased your question incorrectly. >>Factor x^2 - y^2 = (x-y)(x+y) = a >>Thus x-y and x+y are factors of a >>For each factorization of a = nm you have >>x-y = n; x+y = m >>solve for x and y. Beware, >>if the solution turns out not be an integer, discard it. >>Do this for all factorizations of a. >>Collect the solutions you find. >>To make sure you understand the process >>solve x^2 - y^2 = 12 for integers, x,y. >>First list all the ways of factoring 12. >>Let's see what you try. >>What are your answers? >>x^2 - y^2 = 66 has no solution. Why? > bacause it has tre factors. Think of the form of the factorization of the left side: x^2 - y^2 = (x - y)(x + y) Now, let A = x - y, and B = x + y. What can you say about the relationship between A and B? What sorts of numbers can occur as A and B? For instance, you can probably see that B - A = 2y. What does that say about A and B (thought of together, not just one at a time)? Having decided what numbers A and B that can appear in this factorization, now look at all the factorizations of 66. What sorts of numbers do you get from those factorizations? If this isn't enough of a hint, you're trying too hard. The answer is really pretty simple. > Thanx > marco Dale. === Subject: Re: help me, please! ha scritto nel messaggio > Is there a fast way to determine X and Y both integer such that: > X^2 - Y^2 = A > Factor x^2 - y^2 = (x-y)(x+y) = a > Thus x-y and x+y are factors of a > For each factorization of a = nm you have > x-y = n; x+y = m > solve for x and y. Beware, > if the solution turns out not be an integer, discard it. > Do this for all factorizations of a. > Collect the solutions you find. > To make sure you understand the process > solve x^2 - y^2 = 12 for integers, x,y. Clearly you skipped this step because of the question below you asked. So show us some work and what answers you get. If you just guess instead of following the process then to learn what's to be know, solve for integers x^2 - y^2 = 4972 > First list all the ways of factoring 12. > Let's see what you try. > What are your answers? > x^2 - y^2 = 66 has no solution. Why? > bacause it has tre factors. No. The number of factors has nothing to do with it. === Subject: Re: sum of random variables > Saying that one random variable is > equal to another one means that the value of the first is, with > probability 1, equal to the value of the second. > Huh? I think you're understating. Saying that one random variable > is equal to another is stronger than what you're stating -- it means > that *every instance* of the value of the first variable is equal to > the value of the second one. (yes, what I just said implies that > the probability is one -- but probability one does not imply what I > said). > Please correct me if I'm mistaken I won't insist on this point. I think it depends on your point of view. E.g. in analysis L^1 functions are not really functions, but rather equivalence classes for equality almost everywhere. 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: sum of random variables > Saying that one random variable is > equal to another one means that the value of the first is, with > probability 1, equal to the value of the second. Huh? I think you're understating. Saying that one random variable is equal to another is stronger than what you're stating -- it means that *every instance* of the value of the first variable is equal to the value of the second one. (yes, what I just said implies that the probability is one -- but probability one does not imply what I said). Please correct me if I'm mistaken Carlos -- === Subject: Re: Solution to an inequation >Here is an important equation I am encountering. >b <= x * ( (1-x)^a ) > Inequality, not equation. > I accept. >Solve for x in terms of a and b. > Unlikely to have a closed-form solution in general. > Okay, for my purposes, a is an even integer and b is a ratioanl number less > than one. > I am looking for a solution in the range (0,1). > Particularly, I am interested in the smallest value of x which will satisfy > the inequality. If you can't get me the smallest value, can you get me > something as small as possible in a closed form solution. Right now the > smallest I have is the place where x * ( (1-x)^a ) is maximised. ... that place being x_0 = 1/(1+a). I assume b < f(x_0) = a^a/(1+a)^(1+a), where f(x) = x (1-x)^a. It seems that f'''(x) > 0 for 0 < x < 1/(1+a), so if x_1 is any point with 0 <= x_1 <= x_0 with f(x_1) < b, f(x) > f(x_1) + f'(x_1) (x-x_1) + 1/2 f''(x_1) (x - x_1)^2. So if x_1 is not itself a solution, solve the quadratic f(x_1) + f'(x_1) (x - x_1) + 1/2 f''(x_1) (x - x_1)^2 = b; any solution x_2 of this with x_1 < x_2 < x_0 will have f(x_2) > b. In particular, with x_1 = 0 you get x_2 = (1 - sqrt(1-4ab))/(2a) if b<1/(4a). 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: Solution to an inequation >Here is an important equation I am encountering. >b <= x * ( (1-x)^a ) > Inequality, not equation. I accept. >Solve for x in terms of a and b. > Unlikely to have a closed-form solution in general. Okay, for my purposes, a is an even integer and b is a ratioanl number less than one. I am looking for a solution in the range (0,1). Particularly, I am interested in the smallest value of x which will satisfy the inequality. If you can't get me the smallest value, can you get me something as small as possible in a closed form solution. Right now the smallest I have is the place where x * ( (1-x)^a ) is maximised. === Subject: Re: Who contributed most to mathematics? >but the Greeks started it. > Just about every thing the Greeks did was rediscovered by others > as it was needed in their time. Much of Archemedies work was for War > efforts as they were needed. Sure they have being first name > recognition but played no more role than India and other countries > to thought today. > I have heard that Euclid's Elements is the book with more editions > published than any other except the Bible. But wasn't his school in > Alexandria? So perhaps he should be assigned to Egypt (even if he was > Greek by culture and ancestry). And in fact the MacTutor site does assign Euclid to Egypt, although there's no reliable information about his birthplace or ancestry. 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: Who contributed most to mathematics? >>but the Greeks started it. >> Just about every thing the Greeks did was rediscovered by others >> as it was needed in their time. Much of Archemedies work was for War >> efforts as they were needed. Sure they have being first name >> recognition but played no more role than India and other countries >> to thought today. >> I have heard that Euclid's Elements is the book with more editions >> published than any other except the Bible. But wasn't his school in >> Alexandria? So perhaps he should be assigned to Egypt (even if he was >> Greek by culture and ancestry). >And in fact the MacTutor site does assign Euclid to Egypt, although >there's no reliable information about his birthplace or ancestry. The Greeks in Alexandria seem to have been mainly of Greek ancestry, or to be from those of other groups, mainly not Egyptians, who joined the Greeks. The Ptolemies made no attempt to submerge the other cultures, and the Greeks were sufficiently xenophobic (barbarian was the Greek for those who did not speak Greek, basically) that they made no attempt to do so. When the Seleucids did, they ran into open revolts. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Who contributed most to mathematics? >and the Greeks were >sufficiently xenophobic (barbarian was the Greek for those >who did not speak Greek, basically) Use of the term barbarian might suggest xenophobia in a modern English speaker, but Herodotus for example seems to use it in a quite neutral sense to mean non-Greeks. -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === Subject: Re: Who contributed most to mathematics? Ì Herman Rubin ó.8d.98.87.8b.8c .97.99.95 .92.86.94.9d.92.87 > , and the Greeks were > sufficiently xenophobic (barbarian was the Greek for those > who did not speak Greek, basically) Right, but the word THEN, did not have a derogative meaning. To native Greeks of the time, foreign languages sounded very much like nonsensical babling similar to var-var-var noises, thus the adjective. (b in Greek is pronounced as v in English). > -- > This address is for information only. I do not claim that these views > are those of the Statistics Department or of Purdue University. > Herman Rubin, Department of Statistics, Purdue University > hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: Who contributed most to mathematics? >I was just trying to see who in fact borned the most mathematicians. >My SWAG list was not intended to be complete but just a start. Ok, I >blew it after France at least according to MacTutor link below. >However I was close with Italy, Germany and former Russia but totally >underestimated England and the U.S. and way over estimated India and >Greece. This is a quantative analysis. The next step is qualitative >analysis which is why I asked you guys the question. I doubt that anybody would claim that being in the MacTutor list is all that significant as an indication of mathematical importance: it's mainly a question of which people somebody has bothered to write a biography of. >>There sure have been a lot of important French mathematicians; >>Fourier, Galois, d'Alembert... >Fermat - but his last theorem was published after his death. >So during his life time it did not play a major role in mathematical >developement. Fermat's importance in mathematics does not rest on his Last Theorem. He made many contributions in various areas. Although he didn't publish much, he corresponded with many important mathematicians, so he was quite influential in his lifetime. > The same can be said of Da Vinci and other Greats whose >certain works were suppressed during their lifetime. Although da Vinci did study a lot of mathematics, I'm not aware of any really significant contributions he made to mathematics. >>but the Greeks started it. >Just about every thing the Greeks did was rediscovered by others >as it was needed in their time. Much of Archemedies work was for War >efforts as they were needed. Sure they have being first name >recognition but played no more role than India and other countries >to thought today. This is just silliness. The main contribution of the Greeks was to discover the idea of mathematical proof. Nearly all Western mathematicians until very recent times got their first taste of mathematics from studying Euclid. 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: Who contributed most to mathematics? >Please forgive me my ignorance, I missed a few classes in college >elaborate on this list especially regarding Czechoslovakia and >Taiwan? > Others have addressed some of the other countries. My recollection is > that Hamilton was from Ireland and that Chern was from Taiwan. Shiingshen Chern. Born 26 Oct 1911 in Chia-hsing (or Jiaxing), Chekiang province(now Zhejiang), China. You, too, need some well-hopped Pilsner to refresh your memory ... === Subject: Re: Who contributed most to mathematics? >but the Greeks started it. > Just about every thing the Greeks did was rediscovered by others > as it was needed in their time. Much of Archemedies work was for War > efforts as they were needed. Sure they have being first name > recognition but played no more role than India and other countries > to thought today. I have heard that Euclid's Elements is the book with more editions published than any other except the Bible. But wasn't his school in Alexandria? So perhaps he should be assigned to Egypt (even if he was Greek by culture and ancestry). === Subject: Re: [Probability] Convergence in probability metrizable? > There are lots of metrics. In fact, if d is any bounded > metric, and there are lots of bounded metrics equivalent > to any one, take D(f,g) = int d(f(x),g(x)) dP(x). This > is equivalent to the topology you specified. Noel. === Subject: Re: [Probability] Convergence in probability metrizable? >I am given a probability space (W,F,P) and a seperable metric space E. >I consider the set M(E) of all maps f:W->E which are Borel measurable. >I consider on M(E) the topology generated by the base {B(f,a,e): f in >M(E), a,e>0} where: >B(f,a,e)={g in M(E): P(d(f,g)>=a)(fn) in M(E) and f in M(E), the convergence fn->f with respect to this >topology is equivalent to the convergence in probability, i.e.: >For all a>0, P(d(fn,f)>=a)->0 as n->+oo >I was trying to find a metric for this topology and am starting to >suspect it may not be metrizable. Any hint is appreciated. There are lots of metrics. In fact, if d is any bounded metric, and there are lots of bounded metrics equivalent to any one, take D(f,g) = int d(f(x),g(x)) dP(x). This is equivalent to the topology you specified. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Question on 7d geometry -Steven > Hello list, > This question may be too simple, but it is important to me. In 7D space, I have two hyper planes: > (*1): > (x1 x2 x3 x4 x5 x6 x7)*(a1 a2 a3 a4 a5 a6 a7)T = 0 (*2): > (x1 x2 x3 x4 x5 x6 x7)*(b1 b2 b3 b4 b5 b6 b7)T = 0 where T is the transpose sign, * is the matrix multiplication sign. Now I first get the subspace(sub-hyperplane) S1 which is the > intersection of (*1) and (*2). Then I combine the subspace S1 and a > vector > v1 = (a1 a2 a3 a4 a5 a6 a7)T to form a hyper plane (*3), so that this > new hyperplane is the span of S1 and v1, and S1, v1 are on the > hyperplane (*3). The questions what is the formula of this new > hyperplane? Let u be a vector such that the induced hyperplane x*(u^T)=0 (like in (*1) > and (*2)) contains S1. Then u is a linear combination of a and b. > Proof: Every linear combination of a and b clearly has this property, this > gives a 2D vector space. On the other hand, the orthogonal complement of S! > (that is the set of all such u) only is 2D because S1 is 5D and 2+5=7. qed. > So we make the ansatz u=s*a+t*b with unknown coefficients s and t. Because > v1 also lies in the hyperplane (*3) we have v1*(u^T)=0 which gives a linear > condition on s and t. The result is 1D solution space for s and t. It is 1D > because every multiple of any solution u also is a solution. Just pick any > value where not both s and t are zero and calculate u. === Subject: Re: Question on 7d geometry > Hello list, > This question may be too simple, but it is important to me. > In 7D space, I have two hyper planes: > (*1): > (x1 x2 x3 x4 x5 x6 x7)*(a1 a2 a3 a4 a5 a6 a7)T = 0 > (*2): > (x1 x2 x3 x4 x5 x6 x7)*(b1 b2 b3 b4 b5 b6 b7)T = 0 > where T is the transpose sign, * is the matrix multiplication sign. > Now I first get the subspace(sub-hyperplane) S1 which is the > intersection of (*1) and (*2). Then I combine the subspace S1 and a > vector > v1 = (a1 a2 a3 a4 a5 a6 a7)T to form a hyper plane (*3), so that this > new hyperplane is the span of S1 and v1, and S1, v1 are on the > hyperplane (*3). The questions what is the formula of this new > hyperplane? Let u be a vector such that the induced hyperplane x*(u^T)=0 (like in (*1) and (*2)) contains S1. Then u is a linear combination of a and b. Proof: Every linear combination of a and b clearly has this property, this gives a 2D vector space. On the other hand, the orthogonal complement of S! (that is the set of all such u) only is 2D because S1 is 5D and 2+5=7. qed. So we make the ansatz u=s*a+t*b with unknown coefficients s and t. Because v1 also lies in the hyperplane (*3) we have v1*(u^T)=0 which gives a linear condition on s and t. The result is 1D solution space for s and t. It is 1D because every multiple of any solution u also is a solution. Just pick any value where not both s and t are zero and calculate u. -- reverse my forename for mail! === Subject: Re: Maximizing and minimizing in optimization >Is it possible to convert a problem where we are trying to maximize a >quantity (profit) into a problem where we are trying to minimize a >(different) quantity (loss)? Maximizing f is the same as minimizing -f. Or perhaps what you have in > mind is the equivalence in Linear Programming between solving the primal > problem (a max problem) and the dual problem (a min problem). > Yes. The equivalence in Linear programming was what I was thinking > about. If yes, can we do the transform in linear >time? In particular, is it possible to transform a Network flow >problem where we try to maximize the flow between the source and sink >into a shortest path problem where we are trying to find the shortest >between two points (say source and sink)? Maximum flow is equivalent to minimal cut. If it's on a planar graph, > and the source and sink are on the same face, then this is equivalent > to a shortest-path problem in the dual graph. Here, I want to know if the Maximum flow in a network can be > formulated as a Linear programming problem. Also, can you please > elaborate as to why the Maximal flow is equivalent to the > shortest-path problem _only_ for planar graphs? Certainly maximum-flow (with source s and sink t) is a linear programming problem: maximize f subject to sum_j (x_{sj} - x_{js}) = f sum_j (x_{tj) - x_{jt)) = -f sum_j (x_{ij) - x_{ji}) = 0 for all other i 0 <= x_{ij} <= c_{ij} for all i,j where c_{ij} is the capacity of the arc i -> j, and x_{ij} the flow in that arc (make x_{ij} = 0 if there is no arc i -> j). 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: Maximizing and minimizing in optimization >Is it possible to convert a problem where we are trying to maximize a >quantity (profit) into a problem where we are trying to minimize a >(different) quantity (loss)? Maximizing f is the same as minimizing -f. Or perhaps what you have in > mind is the equivalence in Linear Programming between solving the primal > problem (a max problem) and the dual problem (a min problem). Yes. The equivalence in Linear programming was what I was thinking > about. If yes, can we do the transform in linear >time? In particular, is it possible to transform a Network flow >problem where we try to maximize the flow between the source and sink >into a shortest path problem where we are trying to find the shortest >between two points (say source and sink)? Maximum flow is equivalent to minimal cut. If it's on a planar graph, > and the source and sink are on the same face, then this is equivalent > to a shortest-path problem in the dual graph. Here, I want to know if the Maximum flow in a network can be > formulated as a Linear programming problem. Also, can you please > elaborate as to why the Maximal flow is equivalent to the > shortest-path problem _only_ for planar graphs? > Certainly maximum-flow (with source s and sink t) is a linear programming problem: > maximize f > subject to > sum_j (x_{sj} - x_{js}) = f > sum_j (x_{tj) - x_{jt)) = -f > sum_j (x_{ij) - x_{ji}) = 0 for all other i > 0 <= x_{ij} <= c_{ij} for all i,j > > where c_{ij} is the capacity of the arc i -> j, and x_{ij} the flow in > that arc (make x_{ij} = 0 if there is no arc i -> j). Pradip === Subject: Re: Maximizing and minimizing in optimization >Is it possible to convert a problem where we are trying to maximize a >quantity (profit) into a problem where we are trying to minimize a >(different) quantity (loss)? > Maximizing f is the same as minimizing -f. Or perhaps what you have in > mind is the equivalence in Linear Programming between solving the primal > problem (a max problem) and the dual problem (a min problem). Yes. The equivalence in Linear programming was what I was thinking about. > If yes, can we do the transform in linear >time? In particular, is it possible to transform a Network flow >problem where we try to maximize the flow between the source and sink >into a shortest path problem where we are trying to find the shortest >between two points (say source and sink)? > Maximum flow is equivalent to minimal cut. If it's on a planar graph, > and the source and sink are on the same face, then this is equivalent > to a shortest-path problem in the dual graph. Here, I want to know if the Maximum flow in a network can be formulated as a Linear programming problem. Also, can you please elaborate as to why the Maximal flow is equivalent to the shortest-path problem _only_ for planar graphs? Pradip. === Subject: Re: Oneness of a number === >>Subject: Re: Oneness of a number >>Message-id: === >>Subject: Oneness of a number >>Message-id: I am trying to build fractals generated by the principal of the oneness >>of a > number. >>For the definition of the oneness of a number, view the internet, or >>below.* It's more commonly called the Collatz Conjecture or the 3x+1 problem. >>I have been unseccesful in creating a pattern around the oneness >>although I have struck on some intriguing details. These mainly resolve >>around the issue of, if we take the oneness of the oneness of a number, >>and keep doing that, how long does it take 'til we reach 1? Depends on how many binary digits the number has (which is related to > the magnitude of the number) AND ALSO on the pattern of 1s and 0s in the > binary representaion (which is NOT related to the magniyude of the > number). >(note that for 5 this never happens) It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1 >>No, you didn't listen well, I take the oneness of the oneness for the >>recursive oneness looplength. > That is a bit hard to follow. >>The oneness of 5 is 5. >>the oneness of 5 is 5, etc.. > Let me see if I got this straight. If S(n) is the stopping time, you are > then finding the stopping time of the stopping time? > S(n) > S(S(n)) > S(S(S(n))) > etc. > until that reaches 1? Yes, and the amount of recursive calls needed for that I call the recursive oneness loop length. Let's call this ROLL(n) > So if the stopping time of n is n (as with 5) you get a loop? Yeah, > I haven't considered that before, any other loops other than 5? Anything > else interesting about doing that? 5 is not the only one that loops, any number of which the oneness is 5 will cause a loop. The oneness of 32 is 5. Very few numbers have a oneness of 5. I'm not sure but something tells me the only next one of which ROLL(n) becomes infinite is 2^32. (which causes the trap 2^32->32->5->5->5->... of course) And very few numbers n cause ROLL(n) to become infinite. Again like I said, it takes a long while for ROLL(n) to become larger than 16. This only happens regularily when n exceeds 2000000000 (2 billion) You just gave me a cool idea btw. Suppose you have a number 's' and you call N(s) the amount of numbers for which the oneness is 's'. I'll try that later. There must be, for each 's', at least one even and at least one odd number that reaches them? Could those two systems form two two-dimensional axises? Also, I made another recursive loop that does S(n) S(S(n)) S(S(S(n))) Until this reaches 'n', not one. Let's call this looplength the recursive oneness identity loop length, ROILL. Note that this only happens for 1,2,3,4,5,7, and 16 Also note that in my definition ROLL(1) = ROILL(1) = 6. Since S(1) = 3 S(3) = 7 S(7) = 16 S(16) = 4 S(4) = 2 S(2) = 1 I think this set of numbers might be interesting, in intuitive terms 3 and 7 might be seen as the 'prime' counterparts of the 2-powers 16 and 4, and 2 as the main axis, being as well prime as the smallest power of 2. Well perhaps you are into intuitive stuff and care to think about it. Again note that this is my definition, some definitions would say S(1) = 0. I'm not a believer in that :) >>I think I might be able to find a pattern in that since, if I call >>this the recursive oneness loop length (if anybody can think a better >>name, tell me), It's sometimes refered to as the stopping time. Others make up their own > terminology. > then, the recursive oneness loop length of a number like >>20000000 is still only 15. I am trying to make a large-scale 2d map of >>it, How are you plotting this in 2D? >>1. Using complex number representation and messing around with that. >>2. Using y as a period of x (y = x/WIDTH, so to speak) > That doesn't help much. Can you be more specific? That's because called brainstorm coding and it's a bit hard to express with written language. I did say what is necessary : I've tried to visualize any possible patterns in the one-dimensional series, let's leave it at that because I've tried many things. As for the complex numbers, I tried dividing by 2+2i when the complex and real parts are even, and multiplying with 3+3i and adding 1+1i when they are odd. I'm still experimenting with that.. -- Quaternion === Subject: Re: Oneness of a number Does two have more oneness than one? === Subject: Re: Oneness of a number > Does two have more oneness than one? In my defintion, the oneness of one is 3. 1->4->2->1 takes three iterations It could be that some definitions say that the oneness of 1 is 0, since it starts at the stopping time. It doesn't matter that much, and I prefer it this way, where S(1)=3. And thus larger than S(2) -- Quaternion === Subject: Re: Oneness of a number > Does two have more oneness than one? > In my defintion, the oneness of one is 3. > 1->4->2->1 > takes three iterations > It could be that some definitions say that the oneness of 1 is 0, since it > starts at the stopping time. It doesn't matter that much, and I prefer it > this way, where S(1)=3. And thus larger than S(2) > -- > Quaternion If you use an index to represent the collatz-transformations, like C_0 = 1 C_1 = 1*2^1 C_2 = 1*2^2 C_a = 1*2^a C_2,0 = (C_2 -1)/3 C_a,0 = (C_a -1)/3 C_a,b = C_a,0 * 2^b with the periodic property that C_0 = C_2,0 = C_2,2,0 = C_2,2,2,2,0 = C_2,...2,0 = 1 (which reflects the loop 1-4-2-1....) and so on, then this indicates in a very natural way the oneness and stopping time of a number: it's just the sum of all indexes plus the number of them: C_4,0 = 5 Stopping-time = (4+0) + (1) = 5 C_4,1,0 = 3 Stopping-time = (4+1+0) + (2) = 7 Because of the loop at 1 it is also C_4,0 = C_2,4,0 = C_2,2,....,2,4,0 = 5 so the stopping time is not exactly unique. With the additional restriction of allowing only the smallest number of indexes the definition is unique then. For 1 the stopping time is then also unique, namely 0, since the representation with the shortest index is C_0 = 1 Gottfried Helms === Subject: Re: Oneness of a number >> Does two have more oneness than one? >> In my defintion, the oneness of one is 3. >> 1->4->2->1 >> takes three iterations >> It could be that some definitions say that the oneness of 1 is 0, since >> it starts at the stopping time. It doesn't matter that much, and I prefer >> it this way, where S(1)=3. And thus larger than S(2) >> -- >> Quaternion > If you use an index to represent the collatz-transformations, > like > C_0 = 1 > C_1 = 1*2^1 > C_2 = 1*2^2 > C_a = 1*2^a > C_2,0 = (C_2 -1)/3 > C_a,0 = (C_a -1)/3 > C_a,b = C_a,0 * 2^b > with the periodic property that > > C_0 = C_2,0 = C_2,2,0 = C_2,2,2,2,0 = C_2,...2,0 = 1 > (which reflects the loop 1-4-2-1....) > and so on, then this indicates in a very natural way the oneness and > stopping time of a number: it's just the sum of all indexes plus the > number of them: > C_4,0 = 5 Stopping-time = (4+0) + (1) = 5 > C_4,1,0 = 3 Stopping-time = (4+1+0) + (2) = 7 > Because of the loop at 1 it is also > C_4,0 = C_2,4,0 = C_2,2,....,2,4,0 = 5 > so the stopping time is not exactly unique. With the additional > restriction of allowing only the smallest number of indexes the > definition is unique then. For 1 the stopping time is then also unique, > namely 0, since the representation with the shortest index is > C_0 = 1 Damn let me take that in tonight, it's quite complex for me to understand what you mean to represent with 'C_a1,a2,a3..,an' right now and how it can fight the logic that if you follow the iteration scheme you get a bijection between n and the amount of iterations that are needed to transform it to 1.. -- Quaternion === Subject: Re: Oneness of a number > Damn let me take that in tonight, :-)) So I wish, that the nightmares won't catch you... Gottfried (P.S. I had them already, just some weeks ago :-) === Subject: Re: Oneness of a number >Does two have more oneness than one? That seems vague, and depends on context. Is this ONENESS part of some higher aspect of Mathematics beyond the typical undergraduate study of Calculus? At a simpler level, it seems that you could treat any single group of a given quantity as ONE WHOLE; so a grouped unit is one whole set. This allows us to find a fraction of any number as well as find the fraction of 1. G C === Subject: Re: Oneness of a number === >Subject: Re: Oneness of a number >Message-id: === >Subject: Oneness of a number >Message-id: a >> number. >For the definition of the oneness of a number, view the internet, or >below.* >> It's more commonly called the Collatz Conjecture or the 3x+1 problem. >I have been unseccesful in creating a pattern around the oneness although >I have struck on some intriguing details. These mainly resolve around the >issue of, if we take the oneness of the oneness of a number, and keep >doing that, how long does it take 'til we reach 1? >> Depends on how many binary digits the number has (which is related to the >> magnitude of the number) AND ALSO on the pattern of 1s and 0s in the >> binary representaion (which is NOT related to the magniyude of the >> number). >(note that for 5 this never happens) >> It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1 >No, you didn't listen well, I take the oneness of the oneness for the >recursive oneness looplength. That is a bit hard to follow. >The oneness of 5 is 5. >the oneness of 5 is 5, etc.. Let me see if I got this straight. If S(n) is the stopping time, you are then finding the stopping time of the stopping time? S(n) S(S(n)) S(S(S(n))) etc. until that reaches 1? So if the stopping time of n is n (as with 5) you get a loop? I haven't considered that before, any other loops other than 5? Anything else interesting about doing that? >I think I might be able to find a pattern in that since, if I call >this the recursive oneness loop length (if anybody can think a better >name, tell me), >> It's sometimes refered to as the stopping time. Others make up their own >> terminology. > then, the recursive oneness loop length of a number like >20000000 is still only 15. I am trying to make a large-scale 2d map of it, >> How are you plotting this in 2D? >1. Using complex number representation and messing around with that. >2. Using y as a period of x (y = x/WIDTH, so to speak) That doesn't help much. Can you be more specific? >-- >Quaternion -- Mensanator Ace of Clubs === Subject: existence of envelope singular solutions What conditions should be satisfied so that a singular solution envelope for F(x,y,c) = 0 exists for parametric variation of c ? It is known that singular solution should satisfy the above F and del F/ del c = 0 ; perhaps it is a condition using second order partial derivative. In the crank - connecting rod example, we have a crank of radius R centered on origin, angle c with x-axis. The end [R cos(c), R sin(c)] connects to a point on x-axis (reciprocating piston) with connecting rod length L ; It is found that an envelope exists in c limits + - ArcTan(L/R), and there is no envelope outside these limits. === Subject: Re: parallelizability of manifolds >>so every continuous image of it also is simply connected. > No! Take a balloon (S^2), deflate it to a disk, roll it up like a crepe > or something to get an interval, and then join the ends together > to get a circle, which is not simply connected. What do you mean with roll it up like a crepe? And joining the ends together is no continuous operation!? > Sort of at the basis of the theory of covering spaces is the > observation that the continuous image of a simply-connected space is > not necessarily simply connected. Yes, I see that I was wrong. But given a simply connected top. space X and a top. space Y together with an epimorphism f:X->Y, can we say that Y is simply connected? My idea was that any continuous curve s:[0,1]->Y can be pulled back to a continuous curve r:[0,1]->X such that f(r)=s so that we can prove that Y has to be simply connected. Can we turn this idea into a proof? According to your post, the answer seems to be no. So can you give an example where this pull-back does not work? In the present case, we would have X=S^2, Y=R^2{(0,0)} and f any continuous map X->Y. Of course it need not be an epimorphism, so the conclusion was wrong even if the proposition mentioned above was correct. I'll think about it further, but this is it for now. Tobias -- reverse my forename for mail! === Subject: Re: parallelizability of manifolds > so every continuous image of it also is simply connected. >> No! Take a balloon (S^2), deflate it to a disk, roll it up like a crepe >> or something to get an interval, and then join the ends together >> to get a circle, which is not simply connected. > What do you mean with roll it up like a crepe? Sorry, too many math for poets lectures. That really just means to take a projection to an interval. Everything to this point is just the projection pi : R^3 --> R^1, restricted to S^2. (That's continuous.) >And joining the ends together is no continuous operation!? Sure it is; it's the map f : t |--> ( cos(2pi t), sin(2pi t) ). Perhaps you're thinking of continuous maps _with continuous inverses_, which this isn't. If that's what you mean by continuous image, then the right phrase is homeomorphic image, and then yes, the homeomorphic image of a simply-connected space is also simply-connected. >> Sort of at the basis of the theory of covering spaces is the >> observation that the continuous image of a simply-connected space is >> not necessarily simply connected. > Yes, I see that I was wrong. But given a simply connected top. space X and a > top. space Y together with an epimorphism f:X->Y, can we say that Y is > simply connected? No, that is still exactly the setting for covering spaces. The usual example is the same f I gave above, where X is the real line and Y is the circle. > My idea was that any continuous curve s:[0,1]->Y can be > pulled back to a continuous curve r:[0,1]->X such that f(r)=s That's called the path-lifting property; it's a consequence of f being a covering map. (It's also possessed by other maps, such as fibrations.) > so that we can prove that Y has to be simply connected. No! First of all, the fact that you can lift paths does not mean that you can lift loops; that is, if s is a loop in Y, then it's also a path, and so it lifts to a path r if f is a covering (say); but this lifted path r need not be a loop, i.e., there's no reason you must have r(0)=r(1). If you can't see this with formulas, draw a picture of a helix (a spring) X, casting a circular shadow. The shadow, Y is a circle and the projection is the map f. Draw a loop around Y, and then lift it back to X and you'll see you don't get a loop. Also you want to be a little careful : the fact that you can lift paths does not by itself guarantee that you can lift other maps such as homotopies (maps from [0,1]x[0,1] into Y). If you _are_ in the context of covering spaces, there is a pretty characterization of what can be lifted: a map r : Z --> Y induces a homomorphism r* : pi_1(Z) --> pi_1(Y), from which you can show that a _necessary_ condition for liftability is that the subgroup r*(pi_1(Z)) must be contained in the image f*(pi_1(X)). If f is a covering, then this condition is also sufficient. In particular, any map from a contractible space Z into a space Y will lift to a map from Z into any cover X of Y. I know you never said you interested only in covering maps, but these things put your confusion into stark relief. You might want to think about some of the classical examples of these. Start with any simply-connected space X and a discrete group G of self-homeomorphisms of X and then let Y be the quotient X/G . You could take (X,G) to be (R, Z), (R^2, Z^2), (S^2, Z/2Z), (complex half-plane, modular group), (S^3, binary icosahedral group), etc. dave === Subject: Re: parallelizability of manifolds > Sorry, too many math for poets lectures. That really just means to take > a projection to an interval. Everything to this point is just the > projection pi : R^3 --> R^1, restricted to S^2. (That's continuous.) > Sure it is; it's the map f : t |--> ( cos(2pi t), sin(2pi t) ). Ah, now your example is clear. Nice! > Perhaps you're thinking of continuous maps _with continuous inverses_, > which this isn't. If that's what you mean by continuous image, then > the right phrase is homeomorphic image, and then yes, the > homeomorphic image of a simply-connected space is also simply-connected. >> Yes, I see that I was wrong. But given a simply connected top. space X >> and a top. space Y together with an epimorphism f:X->Y, can we say that Y >> is simply connected? In the meantime, I could figure this out. A morphism (here: a continuous map) f:X->Y is called an epimorphism iff there is a morphism g:Y->X such that fg=id_Y. Using this definition, and any closed curve s:[0,1]->Y, we can easily lift it to gs:[0,1]->X and still it is closed: gs(0)=gs(1). Because X is simply connected, there is continuous c:[0,1]^2->X with c(0,t)=gs(t) and c(1,t)=x for some x in X. Applying f gives us fc:[0,1]^2->Y with fc(0,t)=fgs(t)=s(t) and fc(1,t)=f(x). So Y is simply connected. But what does it mean for f to be an epimorphism? fg=id_Y is equivalent to: for every S subset Y we have g-(f-(S))=S (where the - sign denotes pre-images; this is just set-theoretic). Because of continuity reasons, a necessary condition is S open <=> f-(S) open (*) For surjective f with the property (*), you can take any set-theoretic g:Y->X with fg=id_Y to prove the other direction of the following theorem: f is an epimorphism in the category of topological spaces, iff it is surjective and has property (*). This is equivalent to that f be a quotient map. So being an epimorphism is pretty special, so I agree with you that continuous images of simply-connected spaces need not be simply-connected. Tobias -- reverse my forename for mail! === Subject: Re: parallelizability of manifolds >> Yes, I see that I was wrong. But given a simply connected top. space X >> and a top. space Y together with an epimorphism f:X->Y, can we say that Y >> is simply connected? > In the meantime, I could figure this out. A morphism (here: a continuous > map) f:X->Y is called an epimorphism iff there is a morphism g:Y->X such > that fg=id_Y. No - that's _not_ the usual (category-theoretic) definition of epimorphism. Instead, f: X --> Y is an epimorphism iff whenever g_1,g_2 : Y --> Z are morphisms such that g_1 o f =g_2 o f, it follows that g_1 = g_2. (Your notion might be called a split epimorphism ... ) > Using this definition, and any closed curve s:[0,1]->Y, we > can easily lift it to gs:[0,1]->X and still it is closed: gs(0)=gs(1). > Because X is simply connected, there is continuous c:[0,1]^2->X with > c(0,t)=gs(t) and c(1,t)=x for some x in X. Applying f gives us > fc:[0,1]^2->Y with fc(0,t)=fgs(t)=s(t) and fc(1,t)=f(x). So Y is simply > connected. In the context of covering spaces (where Dave Rusin was explaininf stuff), the only covering maps f: X --> Y with X simply connected and f an epimorphism in your sense are _homeomorphisms_ ... so Y _would_ be simply connected in this (fairly uninteresting) case. > But what does it mean for f to be an epimorphism? fg=id_Y is equivalent to: > for every S subset Y we have g-(f-(S))=S (where the - sign denotes > pre-images; this is just set-theoretic). Because of continuity reasons, a > necessary condition is > S open <=> f-(S) open (*) > For surjective f with the property (*), you can take any set-theoretic > g:Y->X with fg=id_Y to prove the other direction of the following theorem: > f is an epimorphism in the category of topological spaces, iff it is > surjective and has property (*). > This is equivalent to that f be a quotient map. ... Yep ... > So being an epimorphism is > pretty special, so I agree with you that continuous images of > simply-connected spaces need not be simply-connected. > Tobias === Subject: Re: parallelizability of manifolds > No - that's _not_ the usual (category-theoretic) definition > of epimorphism. Instead, f: X --> Y is an epimorphism iff > whenever g_1,g_2 : Y --> Z are morphisms such that g_1 o f =g_2 o f, > it follows that g_1 = g_2. > (Your notion might be called a split epimorphism ... ) Sorry for this one, the only reference I have for category theory is Lang's Algebra, and he defines epimorphism and monomorphism only for abelian categories. My notion is much stronger; it implies the other, but if we take a set X and let X_1 be X with the discrete topology, X_2 with the indiscrete topology, and define f:X_1->X_2 as the identity, then f is continuous. It is an epimorphism, but does not have a right inverse. > In the context of covering spaces (where Dave Rusin was explaininf > stuff), the only covering maps f: X --> Y with X simply connected and > f an epimorphism in your sense are _homeomorphisms_ ... so Y _would_ > be simply connected in this (fairly uninteresting) case. Maybe, but since I am not familiar with covering maps, I did not consider this case. -- reverse my forename for mail! Tobias Fritz Student of Mathematics and Physics University of Heidelberg, Germany === Subject: Re: parallelizability of manifolds >Indeed, algebraic topology seems to be an extremely powerful tool: Yes! >S^2 is simply connected, Yes! >so every continuous image of it also is No! Take a balloon (S^2), deflate it to a disk, roll it up like a crepe or something to get an interval, and then join the ends together to get a circle, which is not simply connected. Sort of at the basis of the theory of covering spaces is the observation that the continuous image of a simply-connected space is not necessarily simply connected. dave === Subject: v sp over field {0,1} I learned all of my linear algebra over the reals or complexes, so this seemingly simply problem has baffled me. X is a vector space over the field F_2 = {0,1} of finite dimension. K is a subspace and K^perp is its orthogonal complement, K^perp := {x in X with =0 for all k in K}. The equation I want to prove is dim K + dim (K^perp) = dim X. I've tried on my own, and noticed many discouraging things. Normally (in R^n, say), we have that K and K^perp only intersect at zero. But this is no longer necessarily true. I don't think that you can extend any orthonormal basis of a subspace to an orthonormal basis of the entire space. For one thing, Gram-Schmidt certainly doesn't work any more. Many of these troubles probably stem from the fact that there are many nonzero vectors v for which =0. -Tyler === Subject: Re: v sp over field {0,1} I think I understand now. The equality dim(Null(A)) + dim(Range(A)) = dim(X) is still true, where A is a linear transformation in the finite- dimensional space X. I say still true because I was not certain that it would hold for vector spaces over arbitrary fields. (I reviewed the proof to confirm it still works.) We want to check that dim(F) + dim(F^perp) = dim(X). Choose a basis of F, and let A have these elements as rows. Then Null(A) = F^perp; and the row rank of A is clearly dim(F). Since row rank = col rank = dim(Range(A)), we are done. === Subject: Re: v sp over field {0,1} > I think I understand now. The equality > dim(Null(A)) + dim(Range(A)) = dim(X) > is still true, > where A is a linear transformation in the finite- > dimensional space X. I say still true because > I was not certain that it would hold for vector > spaces over arbitrary fields. (I reviewed the > proof to confirm it still works.) > We want to check that dim(F) + dim(F^perp) = dim(X). > Choose a basis of F, and let A have these elements > as rows. Then Null(A) = F^perp; and the row rank > of A is clearly dim(F). Since row rank = col rank > = dim(Range(A)), we are done. You can't assume Null(A) = F^perp without knowing more about the inner product (sesquilinear form) you're using. As others have pointed out, you really need to take into account whether the form is degenerate. For example, if X has a non-zero radical R then dim(R) + dim(R^perp) = dim(R) + dim(X) > dim(X). -- Jim Heckman === Subject: Re: v sp over field {0,1} Adjunct Assistant Professor at the University of Montana. >I learned all of my linear algebra over the reals or complexes, >so this seemingly simply problem has baffled me. >X is a vector space over the field F_2 = {0,1} of >finite dimension. K is a subspace and K^perp is its >orthogonal complement, What is the inner product for the vector space, though? There are canonical inner products in R^n and in C^n, but I don't know if there is one for vector spaces over F_2... >K^perp := {x in X with =0 for all k in K}. >The equation I want to prove is >dim K + dim (K^perp) = dim X. >I've tried on my own, and noticed many discouraging things. >Normally (in R^n, say), we have that K and K^perp only >intersect at zero. But this is no longer necessarily true. >I don't think that you can extend any orthonormal basis of >a subspace to an orthonormal basis of the entire space. For >one thing, Gram-Schmidt certainly doesn't work any more. >Many of these troubles probably stem from the fact that there >are many nonzero vectors v for which =0. What is your definition of inner product? What are the axioms you require of an inner product? In some books, an inner product is required to be positive definite (i.e., =0, and =0 if and only if v = 0)... It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: v sp over field {0,1} >X is a vector space over the field F_2 = {0,1} of >finite dimension. K is a subspace and K^perp is its >orthogonal complement, >K^perp := {x in X with =0 for all k in K}. >The equation I want to prove is >dim K + dim (K^perp) = dim X. As you noted, inner products don't always behave like the standard one in R^n. But this result still holds if, for example, is still defined as the dot product. In that case, you can describe K^perp as the solution to a system of linear equations. Do you know how to compute the dimension of a nullspace (kernel)? Note that the result is _false_ if for example =0 for all u and v. You'll need to look to see what properties you're assuming about the inner product before you try to prove something about it! (Nondegeneracy, for starters.) dave === Subject: Re: v sp over field {0,1} > I learned all of my linear algebra over the reals or complexes, > so this seemingly simply problem has baffled me. > X is a vector space over the field F_2 = {0,1} of > finite dimension. K is a subspace and K^perp is its > orthogonal complement, > K^perp := {x in X with =0 for all k in K}. > The equation I want to prove is > dim K + dim (K^perp) = dim X. > I've tried on my own, and noticed many discouraging things. > Normally (in R^n, say), we have that K and K^perp only > intersect at zero. But this is no longer necessarily true. > I don't think that you can extend any orthonormal basis of > a subspace to an orthonormal basis of the entire space. For > one thing, Gram-Schmidt certainly doesn't work any more. > Many of these troubles probably stem from the fact that there > are many nonzero vectors v for which =0. > -Tyler It is best to think of K^perp not as a subspace of X but as a subspace of the dual space X*. (Now via your pairing , it is true that X* is identified with X, but that just confuses things.) -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: v sp over field {0,1} > I learned all of my linear algebra over the reals or complexes, > so this seemingly simply problem has baffled me. > X is a vector space over the field F_2 = {0,1} of > finite dimension. K is a subspace and K^perp is its > orthogonal complement, > K^perp := {x in X with =0 for all k in K}. > The equation I want to prove is > dim K + dim (K^perp) = dim X. What do you mean by ? You have not defined it. Jose Carlos Santos === Subject: Re: v sp over field {0,1} Sorry. Fix a basis of X (suppose dim X = m). Then let the coordinates/coefficients of x and k be x1...xm and k1...km respectively. Now we can define :=x1k1 + x2k2 + ... + xmkm, in a manner analogous to the standard inner product. It is linear and symmetric, but does not really give a norm. === Subject: Re: v sp over field {0,1} Adjunct Assistant Professor at the University of Montana. >Sorry. >Fix a basis of X (suppose dim X = m). Then let the >coordinates/coefficients of x and k be >x1...xm and k1...km respectively. Now we can >define :=x1k1 + x2k2 + ... + xmkm, >in a manner analogous to the standard inner >product. It is linear and symmetric, but does >not really give a norm. So? Not all inner products are positive definite. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Extremely bizarre number theoretical property > Well, I discovered something rather very odd (at least to me) today. > Let n be a positive integer, and let s(n) denote the sum of all the > divisors of n. So, for example, s(6) = 1+2+3+6 = 12. > It turns out there is a class of numbers with the property that > a) n is a perfect square > b) s(n) is a perfect square This one is in Sloane's online database of offbeat sequences, number A008847. Go to http://www.research.att.com/~njas/sequences/ and enter 1, 9, 20, 180, 1306 and click Search. The search will cough up more terms of the table, and some references in the lit. Larry === Subject: Re: Extremely bizarre number theoretical property > Well, I discovered something rather very odd (at least to me) today. > Let n be a positive integer, and let s(n) denote the sum of all the > divisors of n. So, for example, s(6) = 1+2+3+6 = 12. > It turns out there is a class of numbers with the property that > a) n is a perfect square > b) s(n) is a perfect square > Here's some examples: annotated, with factors of sqrt(n): > s(9^2) = 11^2 3 3 > s(20^2) = 31^2 2 2 5 > s(180^2) = 341^2 [non-primitive - 9*20] > s(1306^2) = 1729^2 2 653 > s(1910^2) = 2821^2 2 5 191 > s(11754^2) = 19019^2 [non-primitive - 9*1306] > s(17190^2) = 31031^2 [non-primitive - 9*1910] > s(32486^2) = 43617^2 2 37 439 > s(38423^2) = 43491^2 7 11 499 > s(47576^2) = 68961^2 2 2 2 19 313 > s(48202^2) = 72219^2 2 7 11 313 > s(50920^2) = 82677^2 2 2 2 5 19 67 > s(51590^2) = 86583^2 2 5 7 11 67 > s(83884^2) = 117831^2 2 2 67 313 104855 117831 5 67 313 132682 187131 2 11 37 163 198534 347529 2 3 7 29 163 247863 347529 3 7 11 29 37 292374 479787 [non-primitive - 9*32486] 300876 503347 2 2 3 25073 312374 414309 2 313 499 313929 436107 3 3 3 7 11 151 334330 496713 2 5 67 499 345807 478401 [non-primitive - 9*38423] 376095 503347 3 5 25073 428184 758571 [non-primitive - 9*47576] 433818 794409 [non-primitive - 9*48202] 458280 909447 [non-primitive - 9*50920] 464310 952413 [non-primitive - 9*51590] 469623 658749 3 7 11 19 107 498892 696787 2 2 191 653 contains 1306=2*653 623615 696787 5 191 653 754956 1296141 [non-primitive - 9*83884] 768460 1348221 [non-primitive - 20*38423] 787127 828723 11 163 439 943695 1296141 [non-primitive - 9*104855] 985369 1125579 7 11 67 191 1194138 2058441 [non-primitive - 9*132682] 1276880 2106853 2 2 2 2 5 11 1451 1378608 2650557 2 2 2 2 3 7 11 11 373 1547052 2728713 2 2 3 13 47 211 1606281 1993719 3 29 37 499 1676904 2854579 2 2 2 3 107 653 contains 1306=2*653 1754039 1903629 7 83 3091 1933815 2728713 3 5 13 47 211 2015094 3310671 2 3 29 37 313 2034423 2501877 3 3 3 151 499 2156730 3969147 2 3 5 29 37 67 2181409 2322957 19 29 37 107 2452440 4657471 2 2 2 3 5 107 191 2552202 4154493 2 3 3 3 151 313 2731590 4980801 2 3 3 3 5 67 151 2811366 4557399 [non-primitive - 9*312374] 3008970 5463843 [non-primitive - 9*334330] 3043401 3779139 3 19 107 499 3817974 6275451 2 3 19 107 313 4086330 7523607 2 3 5 19 67 107 4490028 7664657 [non-primitive - 9*498892] 4957260 10773399 [non-primitive - 20*247863] 5147241 6540807 3 11 61 2557 5545617 7536711 3 7 11 24007 5570344 8390001 2 2 2 13 19 2819 5612535 7664657 [non-primitive - 9*623615] 5643638 8786379 2 7 11 13 2819 5805336 10287381 2 2 2 3 19 29 439 5881722 10773399 2 3 7 11 29 439 6278580 13519317 [non-primitive - 20*313929] 6385703 6457269 67 191 499 6403664 9346701 2 2 2 2 29 37 373 6416984 10185273 2 2 2 7 19 37 163 6916140 14830431 [non-primitive - 9*20*38423] 7084143 9115953 [non-primitive - 9*787127 7542184 11178921 2 2 2 13 47 1543 7642712 12516231 2 2 2 7 11 19 653 contains 1306=652*2 8010922 10722621 2 67 191 313 8868321 12381369 [non-primitive - 9*985369] 8934096 15205827 2 2 2 2 3 373 499 9392460 204212197 [non-primitive - 20*469623] 9821396 14335671 2 2 13 67 2819 Some prime, such as 37, 67, 107, 191, 313, 373, 439, 499, 653 are a factor of n quite frequently? I wonder why. Their squares' sigma contributions are as follows: 107 7 13 127 191 7 31 13^2 373 3 13 73 7^2 439 3 67 31^2 499 3 7 109^2 And some other primes: (2 gives 7 more easily than 653, and 5 gives 31 more easily than 67.) I guess that when looked at in the simplest possible terms you can just create a binary matrix of non-square terms in sigma(p^2) for prime p, and find combinations that sum to zero in exactly the same way as you do in QS (in fact, the similarity with QS is massive, as you'd want to sieve over a quadratic expression to form the vectors in the first place). I'd conjecture that there is an infinite quantity of such numbers, and that it's not hard to construct one with pretty massive numbers of factors. I'll leave that for someone else. Phil -- Unpatched IE vulnerability: Basic Authentication URL spoofing Description: Spoofing the URL displayed in the Address bar Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/15.html === Subject: Re: Algebraic Closure > |There's an interesting description in Conway's On Numbers and Games. > |(You take nim addition/multiplication on the ordinals: the ordinals > |below a certain one form algebraic closure of F_2.) > a somewhat related question: > from fooling around with sloane's encyclopedia of integer sequences, > it appears that the number of: > field structures on the vector space [f_2]^[2^[n-1]] for which the > frobenius automorphism is the obvious rotation operator > is equal to the number of: > circular words of circumference 2^n in the alphabet {0,1} > containing every word of length n as a contiguous sub-word. Such circular words are called de Bruijn sequences. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Algebraic Closure |There's an interesting description in Conway's On Numbers and Games. |(You take nim addition/multiplication on the ordinals: the ordinals |below a certain one form algebraic closure of F_2.) a somewhat related question: from fooling around with sloane's encyclopedia of integer sequences, it appears that the number of: field structures on the vector space [f_2]^[2^[n-1]] for which the frobenius automorphism is the obvious rotation operator is equal to the number of: circular words of circumference 2^n in the alphabet {0,1} containing every word of length n as a contiguous sub-word. does this equality actually always hold, and is there a pleasant bijective proof of it? also, i tried the obvious greedy algorithm for creating such circular words (start with the all-zeros word of length n and then start appending letters to it, preferably 1 but if that causes a repeated word then 0 instead), and it seemed to work; for example 00000111110111001101011000101001. does this greedy algorithm actually always work, and if so then why? also, are there are any particular nice generalizations of any or all of this? (by the way, just in case any of these questions are answered in on numbers and games, i did try to get it out of the library but it was in circulation.) -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Two questions in propositional logic >> >Anyway, as I said in the other post I think I've got this sorted out now >(although I'll want to write my own proof of it before I'm fully happy >with it). >If you're interested, John's proof went roughly as follows: >Define an equivalence relation on C by p ~ q if there is a bijection f >from the set of primitive propositions to itself, such that q = p with >all the p_i in p replaced with f(p_i). >He then showed that there can be at most one element of each equivalence >class in C, and that there are countably many equivalence classes >greater than p (because you can take some countable subset of the whole >set of primitive propositions and represent every formula in C as a >formula using only this countable subset and the primitives that appear >in p). >He then went on to show there were only countably many elements less >than p in a similar manner. I pointed out that this was rather easier, >as you could just reverse the chain by negation and use the previous >result. >> I found this the most natural approach. However (and IBL will probably >> kill me for saying anything at all about this sheet in public) you >> should also be aware there is a one word answer to this question. That >> is, there is one particular word (familiar to everyone) which >> instantly makes this question a triviality. I'll leave you to guess >> what it is. >Hmm. >... >Oh hell. Is it the one-word that you always use to prove obvious >statements? It is, isn't it... ::sketches proof in his head:: Excuse me >while I go sulk. >(To anyone reading this who doesn't understand that line, it's a >Leaderism. :) Come on. For those of us who aren't getting any of this, and who don't even know what a Leaderism is: what the heck is the one word? >> By the way you didn't ask about the last question (if the set of >> primitive propositions is allowed to be uncountable is it true given a >> set S of propositions that you can find an independent set of >> propositions equivalent to S); does that mean you've solved it? I >> still don't have it, one year after taking the course. No hints >> please! >Well... yes and no. I thought I had a solution. There was a tiny problem >in it contained right at the end which I think I can fix (but it may be >a way bigger problem than I thought it was. :) >That being said, my solution to the last question was only one line. My >solution to the previous question did not assume countability. I was >lazy and didn't feel like proving it twice... >David David C. Ullrich === Subject: Re: Two questions in propositional logic >Anyway, as I said in the other post I think I've got this sorted out now >>(although I'll want to write my own proof of it before I'm fully happy >>with it). >>If you're interested, John's proof went roughly as follows: >>Define an equivalence relation on C by p ~ q if there is a bijection f >>from the set of primitive propositions to itself, such that q = p with >>all the p_i in p replaced with f(p_i). >>He then showed that there can be at most one element of each equivalence >>class in C, and that there are countably many equivalence classes >>greater than p (because you can take some countable subset of the whole >>set of primitive propositions and represent every formula in C as a >>formula using only this countable subset and the primitives that appear >>in p). >>He then went on to show there were only countably many elements less >>than p in a similar manner. I pointed out that this was rather easier, >>as you could just reverse the chain by negation and use the previous >>result. >I found this the most natural approach. However (and IBL will probably >kill me for saying anything at all about this sheet in public) you >should also be aware there is a one word answer to this question. That >is, there is one particular word (familiar to everyone) which >instantly makes this question a triviality. I'll leave you to guess >what it is. >>Hmm. >>... >>Oh hell. Is it the one-word that you always use to prove obvious >>statements? It is, isn't it... ::sketches proof in his head:: Excuse me >>while I go sulk. >>(To anyone reading this who doesn't understand that line, it's a >>Leaderism. :) > Come on. For those of us who aren't getting any of this, and who > don't even know what a Leaderism is: what the heck is the one word? Well, the one word that I thought it was is induction. I thought I saw a really simple proof by induction on the length of a proof of q from p. Unfortunately it didn't work. :) This happens to me a lot with Dr Leader's example sheets (Dr Leader is the man who lectures the course lecturing method, which are referred to as leaderisms.) >By the way you didn't ask about the last question (if the set of >primitive propositions is allowed to be uncountable is it true given a >set S of propositions that you can find an independent set of >propositions equivalent to S); does that mean you've solved it? I >still don't have it, one year after taking the course. No hints >please! >>Well... yes and no. I thought I had a solution. There was a tiny problem >>in it contained right at the end which I think I can fix (but it may be >>a way bigger problem than I thought it was. :) >>That being said, my solution to the last question was only one line. My >>solution to the previous question did not assume countability. I was >>lazy and didn't feel like proving it twice... >>David > David C. Ullrich === Subject: Re: Two questions in propositional logic >Anyway, as I said in the other post I think I've got this sorted out now >(although I'll want to write my own proof of it before I'm fully happy >with it). If you're interested, John's proof went roughly as follows: Define an equivalence relation on C by p ~ q if there is a bijection f >from the set of primitive propositions to itself, such that q = p with >all the p_i in p replaced with f(p_i). He then showed that there can be at most one element of each equivalence >class in C, and that there are countably many equivalence classes >greater than p (because you can take some countable subset of the whole >set of primitive propositions and represent every formula in C as a >formula using only this countable subset and the primitives that appear >in p). He then went on to show there were only countably many elements less >than p in a similar manner. I pointed out that this was rather easier, >as you could just reverse the chain by negation and use the previous >result. >>I found this the most natural approach. However (and IBL will probably >>kill me for saying anything at all about this sheet in public) you >>should also be aware there is a one word answer to this question. That >>is, there is one particular word (familiar to everyone) which >>instantly makes this question a triviality. I'll leave you to guess >>what it is. >Hmm. >... >Oh hell. Is it the one-word that you always use to prove obvious >statements? It is, isn't it... ::sketches proof in his head:: Excuse me >while I go sulk. >(To anyone reading this who doesn't understand that line, it's a >Leaderism. :) >> Come on. For those of us who aren't getting any of this, and who >> don't even know what a Leaderism is: what the heck is the one word? >Well, the one word that I thought it was is induction. Ah, then it's a good thing I presuaded him to state the Word explicitly. It's a word I really would not have thought of applying here, and it really does give a one-line solution. (I imagine you've already seen the word in his reply to me. Incredibly keen argument.) >I thought I saw a >really simple proof by induction on the length of a proof of q from p. >Unfortunately it didn't work. :) This happens to me a lot with Dr >Leader's example sheets (Dr Leader is the man who lectures the course >lecturing method, which are referred to as leaderisms.) >>By the way you didn't ask about the last question (if the set of >>primitive propositions is allowed to be uncountable is it true given a >>set S of propositions that you can find an independent set of >>propositions equivalent to S); does that mean you've solved it? I >>still don't have it, one year after taking the course. No hints >>please! >Well... yes and no. I thought I had a solution. There was a tiny problem >in it contained right at the end which I think I can fix (but it may be >a way bigger problem than I thought it was. :) >That being said, my solution to the last question was only one line. My >solution to the previous question did not assume countability. I was >lazy and didn't feel like proving it twice... >David >> David C. Ullrich David C. Ullrich === Subject: Re: Two questions in propositional logic >> >Anyway, as I said in the other post I think I've got this sorted out now >(although I'll want to write my own proof of it before I'm fully happy >with it). If you're interested, John's proof went roughly as follows: Define an equivalence relation on C by p ~ q if there is a bijection f >from the set of primitive propositions to itself, such that q = p with >all the p_i in p replaced with f(p_i). He then showed that there can be at most one element of each equivalence >class in C, and that there are countably many equivalence classes >greater than p (because you can take some countable subset of the whole >set of primitive propositions and represent every formula in C as a >formula using only this countable subset and the primitives that appear >in p). He then went on to show there were only countably many elements less >than p in a similar manner. I pointed out that this was rather easier, >as you could just reverse the chain by negation and use the previous >result. >> I found this the most natural approach. However (and IBL will probably >> kill me for saying anything at all about this sheet in public) you >> should also be aware there is a one word answer to this question. That >> is, there is one particular word (familiar to everyone) which >> instantly makes this question a triviality. I'll leave you to guess >> what it is. >Hmm. >... >Oh hell. Is it the one-word that you always use to prove obvious >statements? It is, isn't it... ::sketches proof in his head:: Excuse me >while I go sulk. >(To anyone reading this who doesn't understand that line, it's a >Leaderism. :) > Come on. For those of us who aren't getting any of this, and who > don't even know what a Leaderism is: what the heck is the one word? I really should have just e-mailled my clue. I didn't want to post the one word because it is a fun problem and it will likely be set in future years as well. However as I've given away the fact there is a one word solution I suppose it is unfair on people who won't find out otherwise. So I'll post it and hope to goodness that no-one else doing the course happens to read this group. The one word is . . . . . . . . . . . . . . . . . . . . . . . PROBABILITY Michael === Subject: Re: Two questions in propositional logic > >Anyway, as I said in the other post I think I've got this sorted out now >>(although I'll want to write my own proof of it before I'm fully happy >>with it). >>If you're interested, John's proof went roughly as follows: >>Define an equivalence relation on C by p ~ q if there is a bijection f >>from the set of primitive propositions to itself, such that q = p with >>all the p_i in p replaced with f(p_i). >>He then showed that there can be at most one element of each equivalence >>class in C, and that there are countably many equivalence classes >>greater than p (because you can take some countable subset of the whole >>set of primitive propositions and represent every formula in C as a >>formula using only this countable subset and the primitives that appear >>in p). >>He then went on to show there were only countably many elements less >>than p in a similar manner. I pointed out that this was rather easier, >>as you could just reverse the chain by negation and use the previous >>result. > I found this the most natural approach. However (and IBL will probably > kill me for saying anything at all about this sheet in public) you > should also be aware there is a one word answer to this question. That > is, there is one particular word (familiar to everyone) which > instantly makes this question a triviality. I'll leave you to guess > what it is. >>Hmm. >>... >>Oh hell. Is it the one-word that you always use to prove obvious >>statements? It is, isn't it... ::sketches proof in his head:: Excuse me >>while I go sulk. >>(To anyone reading this who doesn't understand that line, it's a >>Leaderism. :) >> Come on. For those of us who aren't getting any of this, and who >> don't even know what a Leaderism is: what the heck is the one word? >I really should have just e-mailled my clue. I didn't want to post the >one word because it is a fun problem and it will likely be set in >future years as well. However as I've given away the fact there is a >one word solution I suppose it is unfair on people who won't find out >otherwise. So I'll post it and hope to goodness that no-one else doing >the course happens to read this group. >The one word is >[one word snipped] Took only another minute to realize I wasn't sure why the solution made sense, since there _are_ uncountable chains of reals, for example. Took another minute to see the actual solution: there are no uncountable chains of _rationals_. Very clever - I _really_ never woulda thought of applying that word in a problem about logic. >Michael David C. Ullrich === Subject: Re: Two questions in propositional logic >this that I'm not seeing (which is very possible, on account of it being >Early here and I just got up. :), that needn't be true. >For example take the following sequence: >p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. >> Pointed out a few minutes ago that this is not a counterexample. >> Of course what I said is wrong - a counterexample is >> p_1 & (p_2), p_1 & (p_2 v p_3), ... p_1. >Oops. Should have read this post before my last one. That's exactly the >counterexample I had in mind when I made the (wrong) counterexample this >morning, I was just half asleep and misremembering. :) >> (Realized I wanted to show that the intersection of a strictly >> decreasing sequence of open subsets of the Cantor set had >> empty interior. Realized that that was false. Now, I bet >> that there's no strictly decreasing omega_1-sequence of >> subsets of the Cantor set, but never mind, the proof you >> already have is simpler.) >Are you sure that the cantor set approach works? Also, do you mean >{0,1}^X for some possibly uncountable X rather than the cantor set? When I was thinking along these lines I was trying to prove that false fact, in which only countably many formulas appear... >(which is homeomorphic when X is countable). Because I tried that, and I >couldn't come to the conclusion that you could get the association to >work both ways - you could get a clopen subset of {0, 1}^X for every >proposition, but I wasn't sure you could go the other way... it looked >like if you chose some sufficiently nasty clopen set it wouldn't >correspond to a proposition. (I didn't prove that you couldn't, but it >looked like it was going to go badly wrong if I tried to prove that you >could, so I gave up on that approach). No, not every clopen set corresponds to a proposition, at least I don't think it does. That doesn't matter to the false thing I was trying to prove. (Regardless, it was thinking about open subsets of the middle-thirds Cantor set that led me to realize what I was trying to prove was obviously false.) >David David C. Ullrich === Subject: Re: Two questions in propositional logic > Well... yes and no. I thought I had a solution. There was a tiny problem > in it contained right at the end which I think I can fix (but it may be > a way bigger problem than I thought it was. :) Hmm... Yes. It does indeed appear to be a bigger problem than I initially anticipated. There's a surprise. ::grumble grumble evil frigging example sheets grumble:: David === Subject: Re: Greek Alphebet >
  In my studies on triginometry, and calculus, I am comming across 
many
>> symbols which appear to be the Greek Alphebet, as there was a chart of
>> the Alphebet in the front of the book. I am wondering if someone could
>> tell me what they all mean, or point me to a site where I could find
>> out.
>> Tom
>You need this rule: The symbols mean whatever the author says they mean.
>Take each one in the context where it is discussed. Somewhere else, the
>same symbol is very likely to mean something different.
>Even the venerable PI does not necessarily mean the constant 3.1415...,
>although that is probably the meaning in the context of the subjects you
>mentioned.
>Lynn Killingbeck
>
As has been discussed here, the Greek alphabet is well used by mathematics. I have sometimes wondered why further alphabets have not been borrowed. Aleph from the Hebrew is used and seems to have a meaning even more fixed than pi (but not so well known). But it appears to stop there. I have not seen beth, gimmel used in maths. The Russian alphabet would blend in well but I have not seen it used. Many letters would be hard to distinguish from Latin and Greek but many other would be usable. And of course there are many, many more alphabets available. Even more extreme we could use Chinese and have thousands of symbols available. I have also wondered what mathematics in Greek and Russian look like. Do the Greeks use Latin and Greek letters in a similar way to us Latin alphabet users? How about the Russians, do they spurn their own alphabet and use Latin and Greek letters? J === Subject: Re: Greek Alphebet >Do the Greeks use Latin and Greek letters in a similar way to us Latin >alphabet users? How about the Russians, do they spurn their own >alphabet and use Latin and Greek letters? Others have answered your question about the Russians, and I can assure you that Greeks use Greek letters the same way everybody else does: they are not Greek to them :-) [Curious thread, this one, and I have probably missed some good stuff :-) I liked the comment about an inverted A looking like an ox head -- good point, except that the contemporary Greek word for cow (AGELADA, from AGELI = herd) first shows up in Greek around the 10th century *AD* ... whereas the ancient Greek word for ox starts with, alas, a B (BOUS*, modern Greek BODI)...] *Speaking of oxen, there is an apocryphal story, attributed to Apollodorus, about Pythagoras having sacrificed one hundred of them upon discovering the eponymous theorem; the story survives anywhere from Diogenes Laertius (4th century AD) to Veniamin of Lesbos (early 19th century AD) -- who disputes it on the grounds of Pythagoras having been a metempsychosis believer :-) [I recently read this in http://195.134.75.8/0709011144070000/main.htm (Veniamin of Lesbos' rendering of The Elements into modern Greek), p. 6] baloglouAToswego.edu === Subject: Re: Greek Alphebet >>
 symbols which appear to be the Greek Alphebet, as there was a chart 
of
> the Alphebet in the front of the book. I am wondering if someone 
could
> tell me what they all mean, or point me to a site where I could find
> out.
> Tom
>>You need this rule: The symbols mean whatever the author says they 
mean.
>>Take each one in the context where it is discussed. Somewhere else, the
>>same symbol is very likely to mean something different.
>>Even the venerable PI does not necessarily mean the constant 3.1415...,
>>although that is probably the meaning in the context of the subjects 
you
>>mentioned.
>>Lynn Killingbeck
>>mathematics.  I have sometimes wondered why further alphabets have not
>been borrowed.
>Aleph from the Hebrew is used and seems to have a meaning even more
>fixed than pi (but not so well known).  But it appears to stop there. 
>I have not seen beth, gimmel used in maths.
Cantor introduced aleph, beth, gimel, and daleth as the
cardinality of the integers, the reals, functions on the
reals, and functions on that.  These were constants, not
variables.  At this time, aleph_z is the z-th smallest
transfinite cardinal of a well-ordered set, where z is an
ordinal number, starting from 0, and omega_z is the
corresponding initial ordinal.  At least with the Axiom
of Choice, beth_z is the z-th least power transfinite
cardinal; with the GCH, beth_z is the cardinality of the
power set of a set with cardinality aleph_z.  
I have seen zayin introduced as the opposite of aleph in
models without choice in one paper.
>The Russian alphabet would blend in well but I have not seen it used. 
>Many letters would be hard to distinguish from Latin and Greek but
>many other would be usable.
I have not seen Russian letters used for mathematical 
constants or variables in any Russian mathematics papers.
>And of course there are many, many more alphabets available.  
>Even more extreme we could use Chinese and have thousands of symbols
>available.
>I have also wondered what mathematics in Greek and Russian look like. 
>Do the Greeks use Latin and Greek letters in a similar way to us Latin
>alphabet users?  How about the Russians, do they spurn their own
>alphabet and use Latin and Greek letters?
I have not seen Greek papers, but I have seen papers in 
Russian and Chinese and other languages.  Almost without
exception, they use Latin and Greek letters as we do.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558
===
Subject: Re: Greek Alphebet
>Even more extreme we could use Chinese and have thousands of symbols
>available.
Tohoku Journal, in which he used a Japanese character for the name
of a field. I cannot recall ever seeing an arabic character used
(apart from Arabic numerals, of course), nor a Norse rune, nor any
of the south Asian alphabets (e.g. Thai) nor the old English thorn, 
nor an accented Latin character from one of the modern European 
alphabets, though Cech (co)homology is usually indicated by the 
use of that diacritical mark (it's a hachek or something like that)
on the H. The German Fraktur lettering is a font, really, not an
alphabet, but those gylphs have been used heavily in the past,
especially in algebraic number theory. There are zillions of other
symbols used, some of which appear to be letters, but I don't think
they really are -- nabla and pe come to mind. There are some ancient
Greek letters which I don't think I have ever seen used in math
(including qoppa or something like that; I think the letters were
never used in post-Homeric Greek). There are of course many more
alphabets from which one can draw; take a look at the full Unicode suite.
And that doesn't include Klingon! (although it WAS proposed!)
>I have also wondered what mathematics in Greek and Russian look like. 
>Do the Greeks use Latin and Greek letters in a similar way to us Latin
>alphabet users?  How about the Russians, do they spurn their own
>alphabet and use Latin and Greek letters?
Well, research mathematics is usually communicated in one of just a few
languages, so people whose native language is, say, Cherokee, don't usually
write much high-level mathematics in that language -- not if they want
their results to get much exposure. Before the (first) October revolution,
that included Russia; the preferred language for the St Petersburg society
was French. But during the 20th century there was quite a bit of
research mathematics published in Russian, and indeed the text would be
in Russian (using Cyrillic, of course) and the mathematics using exactly
the same notation used in the west. I don't recall even seeing a point
in a geometric diagram ever labelled with a Cyrillic character.
(The fact that the typesetting of mathematics was done using Latin and
Greek characters must have driven the Soviet printers crazy, but it was
a boon to the translators into western languages, who would typically
just cut and paste the displays. This was occasionally confusing since,
for example, the names of the trig functions are not the same in all
western languages.)
There is not always a clear distinction between research monograph and
graduate text, so university materials followed the same conventions.
I am not sure whether the same is true in elementary-school texts;
obviously the bulk of the text would be in the native language. 
If I had to write the primary-grade materials for them, I guess I would
opt for using letters which appear identical in both alphabets, at
least for a couple of years until the youngsters start the study of
European languages in earnest.
The same situation occurs in the Chinese texts I have seen.
This helps a lot when I have to assess the courses an international
student has taken in the past. They think I'm kidding when I say,
show me the book you used, but it's pretty easy to recognize what
was in a course from the equations which appear.
Just imagine the joys of writing mathematics in, say, Hebrew, where
the bulk of the writing is done right-to-left but the equations are
written left-to-right. Back in the typewriter days, that meant having
to guess how much space would be required, jumping ahead (i.e. left),
and proceeding to type (to the right).
Please don't ask me what typewriter means.
dave
===
Subject: Re: Greek Alphebet
> Aleph from the Hebrew is used and seems to have a meaning even more
> fixed than pi (but not so well known).  But it appears to stop there. 
> I have not seen beth, gimmel used in maths.
Beth is used in set theory for cardinal arithmetic.
Russian letter Sha is used for the Shafarevich group.
-- 
G. A. Edgar                               
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Greek Alphebet
> John Savard  But in other contexts, 
the letters of the Greek alphabet are fraught
>> with meaning!
>> Thus, alpha comes from aleph, which means Ox.
>References please? The _name_  alpha might be a degenerate form of 
the
>Hebrew aleph, but the letter itself started AS the symbol for an ox, 
in
>Greek. It actually comes from the upside down capital alpha, which 
was
the
>actual symbol for the head of an ox. (Just picture an upside down 
A).
>Who says it's not the other way around and aleph is not a degenerate
form
>of  alpha?
> With all due respect to the Greeks, the Semitic alphabet
> was well established before the Greeks considered using
> an alphabet; they started with it, and the Semitic alphabet
> already had many versions.
And just when do you think the Greeks were using the Cadmian alphabet.
1438 BC is when. This is 50 years before Abraham.
There were already 5 varianets of the Cadmian script by 900 BC.
===
Subject: Re: Greek Alphebet
>> With all due respect to the Greeks, the Semitic alphabet
>> was well established before the Greeks considered using
>> an alphabet[...]
>And just when do you think the Greeks were using the Cadmian alphabet.
>1438 BC is when. This is 50 years before Abraham.
The Greeks couldn't have invented an alphabet.  The language has the
wrong structure.  They would have [or: already had [1]] invented a 
syllabry.
Only a Semitic language has the vowel-consonant factorization in its
morphology to enable an illiterate person to surmise the segmentation
of morphemes at the phonemic level.  Without the syntactic redundancy
of vowels an illiterate person isn't going to see below the syllabic
level.  Most newly developed writing systems in the world were pictographic
or syllabic.  In fact, I'm not even sure there ARE any alphabetic systems
outside the Greek/Roman/Phoenician/Arabic/Hebrew lineage.
The sequencing and names are best understood as probably arising from
the names and constellations of a lunar zodiac, which were once prevalent
throughout the old world.  It's easy to see a situation where the
constellations in a Semitic language are, as a mnemonic device, are
effectively labelled by consonants in a Abe, Baker, Charlie, Daniel, 
etc.
type system.  In later times, the icons that stand for the constellations
come to also denote the consonants themselves.
There's a passage in Psalms that might be a latter-day survival of that
tradition, in which the representations and names of the alphabet symbols
are embedded in mnemonic verses.
Note:
[1] I believe Linear B (or was it A?) was syllabic, as a case in point.
===
Subject: Re: Greek Alphebet
> With all due respect to the Greeks, the Semitic alphabet
> was well established before the Greeks considered using
> an alphabet[...]
>>And just when do you think the Greeks were using the Cadmian alphabet.
>>1438 BC is when. This is 50 years before Abraham.
>The Greeks couldn't have invented an alphabet.  The language has the
>wrong structure.  They would have [or: already had [1]] invented a 
syllabry.
>Only a Semitic language has the vowel-consonant factorization in its
>morphology to enable an illiterate person to surmise the segmentation
>of morphemes at the phonemic level.  Without the syntactic redundancy
>of vowels an illiterate person isn't going to see below the syllabic
>level.  Most newly developed writing systems in the world were 
pictographic
>or syllabic.  In fact, I'm not even sure there ARE any alphabetic systems
>outside the Greek/Roman/Phoenician/Arabic/Hebrew lineage.
There are; the languages of India use an alphabet partly 
derived from the ancient Semitic alphabet.  Also, ancient
Persian used an alphabet not sufficiently close to the
Semitic alphabet, but they knew about it.
The Hangul alphabet for Korean is not derived from the
Semitic, although it is probable that the existence of an
alphabet was known to the designers.  A Korean character
has two to four alphabetic characters.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558
===
Subject: Re: Greek Alphebet
>> With all due respect to the Greeks, the Semitic alphabet
>> was well established before the Greeks considered using
>> an alphabet[...]
>And just when do you think the Greeks were using the Cadmian alphabet.
>1438 BC is when. This is 50 years before Abraham.
[ZAP!] Lies and *CHEAP* Propaganda
WHAT A BUNCH OF IDIOCIES!!!!
You are out of your tiny little mind bloke!! No need to comment
*ANYTHING* of your CHEAP UNBELIEVABLE lies and propaganda!!!! Yor lies
are *ENORMOUS*, you little bwoy. Is Goebbels your spiritual
father??? The bigger the lie it is, the more believable it is. ??
But I can explain your Ignorance AND your cause if you are Jewish.
Are you Jewish bwoy? That could explain a lot!
propagandist.
-Hades
===
Subject: Re: Greek Alphebet
>> With all due respect to the Greeks, the Semitic alphabet
>> was well established before the Greeks considered using
>> an alphabet[...]
>And just when do you think the Greeks were using the Cadmian alphabet.
>1438 BC is when. This is 50 years before Abraham.
> The Greeks couldn't have invented an alphabet.  The language has the
> wrong structure.  They would have [or: already had [1]] invented a
syllabry.
> Only a Semitic language has the vowel-consonant factorization in its
> morphology to enable an illiterate person to surmise the segmentation
> of morphemes at the phonemic level.  Without the syntactic redundancy
> of vowels an illiterate person isn't going to see below the syllabic
> level.  Most newly developed writing systems in the world were
pictographic
> or syllabic.  In fact, I'm not even sure there ARE any alphabetic systems
> outside the Greek/Roman/Phoenician/Arabic/Hebrew lineage.
TWADDLE.
The Greeks were already using 4 scripts before the arrival of Cadmus, Greek
Linear A (Aeolic and Ionic), Cypriot Linear A (Mycenaean), Linaer B
(Mycenaean), and Phaistos Script (Ionic). The Akkadian-Phoenician scribes
knew absolutely nothing about vowels. It was the Greeks who invented vowels
both in Linear A and B and in Cadmian script. Up until 1400 BC which is the
date of the Ras Sharma tablets the Phoenicians were still using cuneiform 
on
clay.
Cadmus arrival in Greece in 1438 indicates that it was in Greece that the
Cadmian script was developed. According to Phoenician historical records
translated by Philo of Byblis the so-called 3 letters were invented by
Eisirius (Celix) who was a foreigner and the brother of Phoenix (Chna) and
by implication the brother of Cadmus as well.
The divergence between Cadmian script and Canaanite script took place
immediately on Cadmus arrival in Greece. The Greeks added vowels to the
alphabet but the Phoenicians did not.
> The sequencing and names are best understood as probably arising from
> the names and constellations of a lunar zodiac, which were once prevalent
> throughout the old world.  It's easy to see a situation where the
> constellations in a Semitic language are, as a mnemonic device, are
> effectively labelled by consonants in a Abe, Baker, Charlie, Daniel,
etc.
> type system.  In later times, the icons that stand for the constellations
> come to also denote the consonants themselves.
PREPOSTEROUS. The names of the Phoenician constellations are known and they
do not go A, B, C. Besides which both Greek and Hebrew start of as A, B, G
and there are no zodiacal constellations in that order either, nor enough
for an alphablet.
An upside down Alpha looks more like a horses had than that of an ox so it
is more likely that is was derived from the Greek Alogo.
> There's a passage in Psalms that might be a latter-day survival of that
> tradition, in which the representations and names of the alphabet symbols
> are embedded in mnemonic verses.
> Note:
> [1] I believe Linear B (or was it A?) was syllabic, as a case in point.
Wrong. Linear A and B had separate individual symbols for VOWELS as well as
bi-syllabic consonant-vowel and tri-syllabic combinations.
===
Subject: Re: Greek Alphebet
>And just when do you think the Greeks were using the Cadmian alphabet.
>1438 BC is when. This is 50 years before Abraham.
>There were already 5 varianets of the Cadmian script by 900 BC.
And just where do you get these dates from?  If you believe the Bible,
your date for Abraham is too late.  If you don't believe the Bible,
why do you think there was an Abraham at all?  As for the Greek alphabet,
various dates are suggested for it, but I'm not aware of any 
archaeological evidence for it before the mid 8th century BCE.
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: Greek Alphebet
>And just when do you think the Greeks were using the Cadmian alphabet.
>1438 BC is when. This is 50 years before Abraham.
>There were already 5 varianets of the Cadmian script by 900 BC.
> And just where do you get these dates from?  If you believe the Bible,
The Traditional Chronology as recorded by Diodorus, Castor and Apollonius 
et
al, and used by Tatian, Eusebius and Clement. The siege of Troy began 1193
BC so add up the figures and generations.
> your date for Abraham is too late.  If you don't believe the Bible,
If you knew how to do some simple arithmetic and statistics you would know
that the numbers in the bible are all the numerators of fractions.
http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm
(the above research is copyrighted)
> why do you think there was an Abraham at all?  As for the Greek alphabet,
> various dates are suggested for it, but I'm not aware of any
> archaeological evidence for it before the mid 8th century BCE.
Then you know nothing about the subject. 5 different scripts are known to
have existed in 900 BC. The only way there could have been so many is if 
all
of them derived from a common root long before the Ionian mitigation of 
1071
BC. 1200 BC is the latest possible date for the use of Cadmian script by 
the
Greeks and the start of its divergence into 5 independent Greek scripts.
Herodotus records an inscription made in the time of Amphytrion and Laius
which firmly dates to Amhytrions victory over the Teleboans which is dated
to 1286 BC by the occurrence of a total solar eclipse over the Aegean which
occurred on February 10 and is confirmed by the traditional chronology.
> 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: Greek Alphebet
>If you knew how to do some simple arithmetic and statistics you would know
>that the numbers in the bible are all the numerators of fractions.
Yes, all the integers used in the bible are the numerators of fractions.
As are all the integers that are not used in the bible.  So?
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: Greek Alphebet
>If you knew how to do some simple arithmetic and statistics you would
know
>that the numbers in the bible are all the numerators of fractions.
> Yes, all the integers used in the bible are the numerators of fractions.
> As are all the integers that are not used in the bible.  So?
Do the maths.
http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm
> 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: Greek Alphebet
> Yes, all the integers used in the bible are the numerators of 
fractions.
> As are all the integers that are not used in the bible.  So?
> Do the maths.
Agamemnon,
I think you may want to be a little more courteous with Robert. He can
certainly do the maths much easier than anyone here, so you'll probably 
be
very lucky if you can convince him to read your webpage.
And you never know....If he does, you may become famous :*)
> http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm
> Robert Israel                                israel@math.ubc.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
--
Ioannis Galidakis
http://users.forthnet.gr/ath/jgal/
------------------------------------------
Eventually, _everything_ is understandable
===
Subject: Re: Greek Alphebet
> Yes, all the integers used in the bible are the numerators of
fractions.
> As are all the integers that are not used in the bible.  So?
> Do the maths.
> Agamemnon,
> I think you may want to be a little more courteous with Robert. He can
> certainly do the maths much easier than anyone here, so you'll 
probably
be
> very lucky if you can convince him to read your webpage.
If he can do the maths then he should have no problem in confirming that I
am right.
> And you never know....If he does, you may become famous :*)
> http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm
 Robert Israel                                israel@math.ubc.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
> --
> Ioannis Galidakis
> http://users.forthnet.gr/ath/jgal/
> ------------------------------------------
> Eventually, _everything_ is understandable
===
Subject: Re: Greek Alphebet
Agamamnonas is a person who is a man of all trades but master of none. He 
is
infamously famous, notoriously famous already. He has absolutely NO need of
mere professors of maths!
Why, Agamamnonas the linguist has already told us that Dolmades is from
the Greek word gemistes or gamistes or something like that and not 
from
the Turkish root word dol/dolma meaning
fill/stuff/filling/stuffing/filled/stuffed.
Why don't we all get together to tell Agamemnonas to go, get stuffed.
Pigenai na gemithis, re Agamemnona!
Gia sou re Agamemnona;-) Ean esei mou stelleis loukanika, kai egw na sou
stellw Ntpolmades? 8a kanomai allagei/trampa, re. Entaksi?
Entaksi, Mparataksi!
-- 
choro-nik
********
> Yes, all the integers used in the bible are the numerators of
fractions.
> As are all the integers that are not used in the bible.  So?
> Do the maths.
> Agamemnon,
> I think you may want to be a little more courteous with Robert. He can
> certainly do the maths much easier than anyone here, so you'll 
probably
be
> very lucky if you can convince him to read your webpage.
> And you never know....If he does, you may become famous :*)
> http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm
 Robert Israel                                israel@math.ubc.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
> --
> Ioannis Galidakis
> http://users.forthnet.gr/ath/jgal/
> ------------------------------------------
> Eventually, _everything_ is understandable
===
Subject: Re: Greek Alphebet
> Agamamnonas is a person who is a man of all trades but master of none. He
is
> infamously famous, notoriously famous already. He has absolutely NO need
of
> mere professors of maths!
> Why, Agamamnonas the linguist has already told us that Dolmades is 
from
> the Greek word gemistes or gamistes or something like that and 
not
from
> the Turkish root word dol/dolma meaning
> fill/stuff/filling/stuffing/filled/stuffed.
I said it was from the Latin word Tomaculum meaning sausage and Toomentum
meaning stuffing. There is no Altaic root for Dolma.
> Why don't we all get together to tell Agamemnonas to go, get stuffed.
> Pigenai na gemithis, re Agamemnona!
> Gia sou re Agamemnona;-) Ean esei mou stelleis loukanika, kai egw na sou
> stellw Ntpolmades? 8a kanomai allagei/trampa, re. Entaksi?
Mpikse angkouri mes to kolo sou.
> Entaksi, Mparataksi!
> --
> choro-nik
> ********
> Yes, all the integers used in the bible are the numerators of
> fractions.
> As are all the integers that are not used in the bible.  So?
 Do the maths.
> Agamemnon,
> I think you may want to be a little more courteous with Robert. He can
> certainly do the maths much easier than anyone here, so you'll
probably
> be
> very lucky if you can convince him to read your webpage.
> And you never know....If he does, you may become famous :*)
> http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm
> Robert Israel                                israel@math.ubc.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
> --
> Ioannis Galidakis
> http://users.forthnet.gr/ath/jgal/
> ------------------------------------------
> Eventually, _everything_ is understandable
>
===
Subject: Re: Greek Alphebet
> Agamamnonas is a person who is a man of all trades but master of none.
He
> is
> infamously famous, notoriously famous already. He has absolutely NO 
need
> of
> mere professors of maths!
> Why, Agamamnonas the linguist has already told us that Dolmades is
from
> the Greek word gemistes or gamistes or something like that and 
not
> from
> the Turkish root word dol/dolma meaning
> fill/stuff/filling/stuffing/filled/stuffed.
> I said it was from the Latin word Tomaculum meaning sausage and Toomentum
> meaning stuffing. There is no Altaic root for Dolma.
Absolutely brilliant! What's wrong with you, Agamemnona? Can't even find a
word in a dictionary? Oh, I know why. It is because it is Turkish. The root
word is dol with the verb being doldurmak meaning to fill. Dolma 
is
that which has been stuffed. Dolgu is a tooth filling. Doldur is 
what
you say at the petrol pump if you want to say Fill her up.
Tomaculum and Toomentum my foot!!!!
-- 
choro-nik
********
> Why don't we all get together to tell Agamemnonas to go, get 
stuffed.
> Pigenai na gemithis, re Agamemnona!
> Gia sou re Agamemnona;-) Ean esei mou stelleis loukanika, kai egw na 
sou
> stellw Ntpolmades? 8a kanomai allagei/trampa, re. Entaksi?
> Mpikse angkouri mes to kolo sou.
> Entaksi, Mparataksi!
> --
> choro-nik
> ********
> Yes, all the integers used in the bible are the numerators of
> fractions.
> As are all the integers that are not used in the bible.  So?
 Do the maths.
 Agamemnon,
 I think you may want to be a little more courteous with Robert. He 
can
> certainly do the maths much easier than anyone here, so you'll
> probably
> be
> very lucky if you can convince him to read your webpage.
 And you never know....If he does, you may become famous :*)
 http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm
> Robert Israel                                israel@math.ubc.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
 --
> Ioannis Galidakis
> http://users.forthnet.gr/ath/jgal/
> ------------------------------------------
> Eventually, _everything_ is understandable
===
Subject: Re: Greek Alphebet
> Agamamnonas is a person who is a man of all trades but master of 
none.
> He
> is
> infamously famous, notoriously famous already. He has absolutely NO
need
> of
> mere professors of maths!
 Why, Agamamnonas the linguist has already told us that Dolmades 
is
> from
> the Greek word gemistes or gamistes or something like that and 
not
> from
> the Turkish root word dol/dolma meaning
> fill/stuff/filling/stuffing/filled/stuffed.
> I said it was from the Latin word Tomaculum meaning sausage and
Toomentum
Tomentum rather.
> meaning stuffing. There is no Altaic root for Dolma.
> Absolutely brilliant! What's wrong with you, Agamemnona? Can't even find 
a
> word in a dictionary? Oh, I know why. It is because it is Turkish. The
root
> word is dol with the verb being doldurmak meaning to fill. 
Dolma is
> that which has been stuffed. Dolgu is a tooth filling. Doldur is 
what
> you say at the petrol pump if you want to say Fill her up.
POPPYCOCK.
The Turks got the word from the Greeks. The Turkish word Dol is a 
corruption
of the Latin Tom.
> Tomaculum and Toomentum my foot!!!!
> --
> choro-nik
> ********
 Why don't we all get together to tell Agamemnonas to go, get
stuffed.
> Pigenai na gemithis, re Agamemnona!
 Gia sou re Agamemnona;-) Ean esei mou stelleis loukanika, kai egw na
sou
> stellw Ntpolmades? 8a kanomai allagei/trampa, re. Entaksi?
> Mpikse angkouri mes to kolo sou.
 Entaksi, Mparataksi!
> --
> choro-nik
> ********
> Yes, all the integers used in the bible are the numerators of
> fractions.
> As are all the integers that are not used in the bible.  So?
 Do the maths.
 Agamemnon,
 I think you may want to be a little more courteous with Robert. He
can
> certainly do the maths much easier than anyone here, so you'll
> probably
> be
> very lucky if you can convince him to read your webpage.
 And you never know....If he does, you may become famous :*)
 http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm
> Robert Israel                                israel@math.ubc.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
 --
> Ioannis Galidakis
> http://users.forthnet.gr/ath/jgal/
> ------------------------------------------
> Eventually, _everything_ is understandable
>
===
Subject: Re: Greek Alphebet
No, Agamemnona, the Turkish word Dol is a corruption of the Greek Ta
Vgaleis Ap'ton Kolosou!
Ma nomizeis oth monon esei esai Karpasitis?
-- 
choro-nik
********
 Agamamnonas is a person who is a man of all trades but master of
none.
> He
> is
> infamously famous, notoriously famous already. He has absolutely NO
> need
> of
> mere professors of maths!
 Why, Agamamnonas the linguist has already told us that Dolmades 
is
> from
> the Greek word gemistes or gamistes or something like that 
and
not
> from
> the Turkish root word dol/dolma meaning
> fill/stuff/filling/stuffing/filled/stuffed.
 I said it was from the Latin word Tomaculum meaning sausage and
> Toomentum
> Tomentum rather.
> meaning stuffing. There is no Altaic root for Dolma.
> Absolutely brilliant! What's wrong with you, Agamemnona? Can't even 
find
a
> word in a dictionary? Oh, I know why. It is because it is Turkish. The
> root
> word is dol with the verb being doldurmak meaning to fill. 
Dolma
is
> that which has been stuffed. Dolgu is a tooth filling. Doldur 
is
what
> you say at the petrol pump if you want to say Fill her up.
> POPPYCOCK.
> The Turks got the word from the Greeks. The Turkish word Dol is a
corruption
> of the Latin Tom.
> Tomaculum and Toomentum my foot!!!!
> --
> choro-nik
> ********
> Why don't we all get together to tell Agamemnonas to go, get
> stuffed.
> Pigenai na gemithis, re Agamemnona!
 Gia sou re Agamemnona;-) Ean esei mou stelleis loukanika, kai egw 
na
> sou
> stellw Ntpolmades? 8a kanomai allagei/trampa, re. Entaksi?
 Mpikse angkouri mes to kolo sou.
> Entaksi, Mparataksi!
> --
> choro-nik
> ********
> Yes, all the integers used in the bible are the numerators of
> fractions.
> As are all the integers that are not used in the bible.  So?
 Do the maths.
 Agamemnon,
 I think you may want to be a little more courteous with Robert. 
He
> can
> certainly do the maths much easier than anyone here, so 
you'll
> probably
> be
> very lucky if you can convince him to read your webpage.
 And you never know....If he does, you may become famous :*)
 http://www.enthymia.co.uk/myths/bible/EstablishChronology.htm
> Robert Israel
israel@math.ubc.ca
> Department of Mathematics
http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
 --
> Ioannis Galidakis
> http://users.forthnet.gr/ath/jgal/
> ------------------------------------------
> Eventually, _everything_ is understandable
>
===
Subject: Re: Greek Alphebet
>>  > Thus, alpha comes from aleph, which means Ox.
>>  > References please?
>> Alpha does not come from aleph, they simply have a common source.
>Punic (Phonecian) alphabet. Hebrew, Aramaic and Arabic are all 
>descendants of the Urguritic -language-. I am not sure what the original 
>scripts were (assuming there was any scripts).
>Bob Kolker
The Ugaritic alphabet is a 32 character cuneiform 
modification of the early Semitic alphabets; there
were lots of them.  The earliest known Semitic
alphabetic writing is from about 1800 BCE, found
on a cliff wall in Egypt.
Hebrew and Ugaritic are almost identical as languages;
Aramaic is similar, but different, and these probably
diverged before alphabetic writing; Arabic is from
a different branch of the Semitic family.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558
===
Subject: Re: The crazy counterfeit coin problem strikes again!
> Could some1 plz explain to me how to do the following problem? I have my 
reputation at stake! The deadline for me to come up with the answer is 
tomorrow! I got the weighings done to 14 but I can't get it to eleven:
> You have 20 blocks of the same size and appearance.  Some are 
> aluminum, and some are duraluminum, which is heavier.  Using at most 
> 11 weighing on a pan balance, how can you determine how many blocks 
> are aluminum?
Can I check what you want to find out?  
If you want to find which are the heavier blocks then I do not think
that it can be done.  There 2 ^ 20 = 1048576 possible answers.  Each
weighing could give 3 answers: left pan is heavier; right pan is
heavier; the pans are equal.  So with 11 weighings, you can
distinguish at most 3 ^ 11 = 177147 cases.  You would need at least 13
weighing.  Even then it is not certainly possible but only might be.
Or is it enough to be able to say N blocks are heavier without
indicating which those blocks are?  Since then only 21 answers are
possible, we can hope to achieve in much fewer weighings.  The type of
analysis above would suggest a minimum of 3 weighings but I rather
doubt you would get anywhere near that minimum.
Finally in ether case, you have a problem if all the blocks are the
same.  So in case 2 you would have to say: 0 or 20.
J
===
Subject: Re: The crazy counterfeit coin problem strikes again!
> What if one doesn't need to find which blocks are Ds? How many less 
weighings will it take?
number between 0 and 20, which 2-3 tribits (i.e. 2 Saw it yesterday.
> In German.
> Uhm, did the synchro botch everything up or is it
> a lot of senseless mathbabble already in the original? :-)
The basic idea was cool but everything else was terrible.  And then there's 

the sequel Hypercube which was even stupider.
Have a tolerable existence.  Eli
===
Subject: Re: Differentiation
===
Subject: Re: Differentiation
 >>Is differentiation a one way function?
 >>Is differentiation a one way functional?
 >Assuming you mean is solving the analytical form for the
 >anti-derivative hard, I think this depends on your function
 >space. For example, in P_n, the polynomial function space of
 >degree n, it's clearly not one-way since you can easily
 >differentiate and integrate any polynomial.
Functions in an extended notion of closed or elementary form.
Here's another one way function,
        likely more fruitful for including the first.
Is construction of differential equations from
        a set of equations a one way function?
----
===
Subject: Re: Differentiation
>Is differentiation a one way function?
>More exactly
>Is differentiation a one way functional?
Assuming you mean is solving the analytical form for the
anti-derivative hard, I think this depends on your function space.
For example, in P_n, the polynomial function space of degree n, it's
clearly not one-way since you can easily differentiate and integrate
any polynomial.
===
Subject: Re: Does compact+continuum connected+locally connected==>pathwise
 connected?Countable cofinite space counter example
Does compact+continuum connected+locally connected==>pathwise connected?
Continuum connected means that any two points of
the space lie in a continuum (= compact connected set).
 > Usually continuum means compact connected Hausdorff.
 >> Yes.  But I posted quickly, and neglected to mention I was thinking
 >> in a METRIC SPACE.
Ddddaah. Quick posting = GIGO.
 >Pssht. Metric spaces are for wimps. :)
Bbbblht. Making title first sentence is for dimps.
 >At this point it would be useful to know what you mean by locally
 >connected. Some people say a property holds locally if for each x
 >it holds on a neighbourhood of x, and some say it holds locally if
 >for each x and U a neighbourhood of x there is a neighbourhood V of
 >x such that it holds on V and V is a subset of U.
Steen says the latter.  I've not see the former.
 >If you mean the former, then the closed topologists sine curve is a
 >counterexample. If you mean the latter... I'm working on it. Steen
 >and Seebach doesn't seem to have an example of such a space, which
 >means it's probably a fair bet that there is no such space. However
 >it can't hurt to try both sides...
It's out of their book's realm.
Random thought of the day:  Hahn-Mazurkiewicz theorem
Any compact, metrizable, connected, locally connected space
        is a continuous image of [0,1].
Is continuous image [0,1] path connected?  Yes
Continuum connected S ==> S connected
        pick p in S.  For all s in S, some connected Ks with p,s in Ks
        /{ Ks | s in S } nonnul;  S = /{ Ks | s in S } connected
continuum connected overkill.
S subspace connected when
        for all a,b in S some connected K with a,b in K
subspace connected S ==> S connected
Thus as problem space is connected.  Apply H-M theorem.  QED
----
===
Subject: Re: Does compact+continuum connected+locally connected==>pathwise
 connected?Countable cofinite space counter example
>>Continuum connected means that any two points of
>>the space lie in a continuum (= compact connected set).
>> Usually continuum means compact connected Hausdorff.
> Yes.  But I posted quickly, and neglected to mention I was thinking in
> a METRIC SPACE.
> --Ron Bruck
Pssht. Metric spaces are for wimps. :)
At this point it would be useful to know what you mean by locally 
connected. Some people say a property holds locally if for each x it 
holds on a neighbourhood of x, and some say it holds locally if for each 
x and U a neighbourhood of x there is a neighbourhood V of x such that 
it holds on V and V is a subset of U.
If you mean the former, then the closed topologists sine curve is a 
counterexample. If you mean the latter... I'm working on it. Steen and 
Seebach doesn't seem to have an example of such a space, which means 
it's probably a fair bet that there is no such space. However it can't 
hurt to try both sides...
David
===
Subject: Re: Does compact+continuum connected+locally 
connected==>pathwiseconnected?
 Use the Hahn-Mazurkiewicz theorem any compact, metrizable, connected, 
locally
 connected space is a continuous image of [0,1].
>  >Continuum connected means that any two points of
>  >the space lie in a continuum (= compact connected set).
>  Usually continuum means compact connected Hausdorff.
Yes.  But I posted quickly, and neglected to mention I was thinking in
a METRIC SPACE.
----
===
Subject: Re: topology.....problem...
===
Subject: topology.....problem...
 >X,Y is topology space.
 >function  f : X -> Y
 >any B <= Y (symbol <= is inclusion relation),
 >cls{f^(-1) (B)} <= f^(-1) cls(B) satisfy   (symbol cls = closure)
 >show that f is continuous.
 >i use that
 >any closed A <= Y, f^(-1) (A) is closed set. <=> f : conti
 >i would show that f^(-1) cls(B) is closed set.
 >cls{f^(-1) cls(B)} <= f^(-1) cls(cls(B)) = f^(-1) (cls(B))
 >thus f^(-1) cls(B) is closed.
 >it is right??
Almost.
You forgot to start with f^-1(cl B) subset cl f^-1(cl B)
hypothesis and closed B ==> f^-1(B) closed
        f^-1(B) subset cl f^-1(B) subset f^-1(cl B) = f^-1(B)
        f^-1(B) = cl f^-1(B)
Extend exercise unto theorem
f continuous  iff  for all A, f(cl A) subset cl f(A)
                iff  for all A, cl f^-1(A) subset f^-1(cl A)
                iff  for all A, bd f^-1(A) subset f^-1(bd A)
----
===
Subject: Re: topology.....problem...
bok3np$ocd$1@news.hananet.net...
> X,Y is topology space.
> function  f : X -> Y
> any B <= Y (symbol <= is inclusion relation),
> cls{f^(-1) (B)} <= f^(-1) cls(B) satisfy   (symbol cls = closure)
> show that f is continuous.
Take B a closed subset of Y; you have cl(f^(-1)(B)) C f^(-1)(cl(B)) =
f^(-1)(B).
Hence cl(f^(-1)(B)) = f^(-1)(B), and f^(-1)(B) is closed.
--
Julien Santini,
Etudiant en licence au CMI de Ch.89teau-Gombert,
FRANCE.
===
Subject: Re: topology.....problem...
Julien Santini  grava .88 la saucisse et au 
marteau:
>  Julien Santini,
>  Etudiant en licence au CMI de Ch.89teau-Gombert,
>  FRANCE.
Est-ce vraiment n.8ecessaire? :)
-- 
Nicolas
===
Subject: Re: topology.....problem...
Nicolas Le Roux  scribbled the following:
> Julien Santini  grava .88 la saucisse et au 
marteau:
>>  Julien Santini,
>>  Etudiant en licence au CMI de Ch.89teau-Gombert,
>>  FRANCE.
> Est-ce vraiment n.8ecessaire? :)
Que est-ce que est vraiment n.8ecessaire?
-- 
/-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland --------
-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
Ice cream sales somehow cause drownings: both happen in summer.
   - Antti Voipio & Arto Wikla
===
Subject: Re: topology.....problem...
> X,Y is topology space.
> function  f : X -> Y
> any B <= Y (symbol <= is inclusion relation),
> cls{f^(-1) (B)} <= f^(-1) cls(B) satisfy   (symbol cls = closure)
> show that f is continuous.
> --------------------------------
> um......i think......
> i use that any closed A <= Y, f^(-1) (A) is closed set. <=> f : 
conti
> i would show that f^(-1) cls(B) is closed set.
> cls{f^(-1) cls(B)} <= f^(-1) cls(cls(B)) = f^(-1) (cls(B))
> thus f^(-1) cls(B) is closed.
> it is right??
No. You said that you wanted to proved that a certain condition implies
that f is continuous, but what you did prove was the reverse
implication.
Jose Carlos Santos
===
Subject: topology.....problem...
X,Y is topology space.
function  f : X -> Y
any B <= Y (symbol <= is inclusion relation),
cls{f^(-1) (B)} <= f^(-1) cls(B) satisfy   (symbol cls = closure)
show that f is continuous.
--------------------------------
um......i think......
i use that any closed A <= Y, f^(-1) (A) is closed set. <=> f : 
conti
i would show that f^(-1) cls(B) is closed set.
cls{f^(-1) cls(B)} <= f^(-1) cls(cls(B)) = f^(-1) (cls(B))
thus f^(-1) cls(B) is closed.
it is right??
please......advice....for me....thank you
===
Subject: Apocalypse NOW!
Action Device to generate unidirectional force.
http://www.geocities.com/actiondevice
-Abhi.
===
Subject: Re: Apocalypse NOW!
Abhi,
you posted this before and it was quickly pointed out that your math is
faulty and precisely where your mistake is. Do you want me to post such an
analysis again, or would you prefer to actually go and build your device 
and
tell us all whether it worked or not?
Krill
> Action Device to generate unidirectional force.
> http://www.geocities.com/actiondevice
> -Abhi.
===
Subject: Re: Apocalypse NOW!
> Action Device to generate unidirectional force.
ah, thank you for resurfacing abhi, now my new killfile is complete.
===
Subject: Re: Apocalypse NOW!
> Action Device to generate unidirectional force.
> ah, thank you for resurfacing abhi, now my new killfile is complete.
Then start from zero and recreate new one.
Let me try to explain how unidirectional force is generated through
this action device. Please take printout, read and look at figuire
carefully.
This device is in vertical position i.e. upper end of bolt 5 pointing
towards sky. Attach UPPER end of bolt 5 to centre horizontal circular
plate, diameter of which is greater than length AC. This circular
plate is in your hand. Let mass of whole device be 10 Kg. So downward
gravitational force acting on this device is 10g Newton. Due to spring
forces in spring 1 and 2, let resultant force generated at point D 9
be 20g Newton. And direction of this resultant force at point D 9 is
UPWARD. So it is transmitted through bolt 5 in UPWARD direction to
circular plate in your hand.
What will happen next? Will you feel weight of Action Device in
downward direction or upward force pushing the circular plate and your
hand in upward direction?
If you feel upward force, take this circular plate near the roof of
your room. This circular plate will stick to roof of your room like
magnet. It will not fall.
In space, this 20g force generated at circular plate will push /
accelerate any mass in only one direction. If this circular plate is
in hands of Astronaut, that Astronaut will be accelerated only in one
direction.
And if some rope is tied between round plate and Astronaut, the
accelerating Astronaut will pull the action device in the direction of
his acceleration. So now the action device and astronaut are moving in
space and no reaction mass is expelled.
Needless to add, the force generated at point D 9 depends on magnitude
of force in spring 1 and 2.
Please keep, the angle ABC as small as possible. Page No. 42 of Sixth
edition of Fundamentals of Physics by halliday states that c =
ab[sin(ABC)]
Just wonder, if this device is so simple, then why don't I make it
myself?
God has His own ways to create problems for everyone. I am finding
myself alone and I just don't know where to go. But let me leave this
RCM cyber cafe in Matunga and try to find someone in this Bombay.
Like Jodie Foster said in Contact..
OK, GO..
-Abhi.
===
Subject: Re: Apocalypse NOW!
Abhi,
your math/physics is faulty. The device will not float to the ceiling and
stay there, unless you nail it up. It will accelerate towards the floor at
approximately 9.8 meters per second squared.
Your error is in assuming that the forces from the springs acting at your
call each of those F here, but you can call them what you want) must be
summed as a vector. You can use many methods to do this, but I'll use your
cartesian coordinate system. I'll call the angles ABD/CBD, BDA/BDC and
DAB/DCB alpha, beta and gamma respectively (but again you can call them 
what
you want). In that case, the two forces acting at point B from the left and
right hand springs can be written as:
{-Fsin(alpha),-Fcos(alpha)} and
{Fsin(alpha),-Fcos(alpha)} respectively
net force acting on point B from the springs is therefore
{ 0, -2Fcos(alpha) }
component acting along your y axis does not. There is a net force acting on
point B towards point D.
If you do the same calculation for point D, you'll find a net force acting
on that point of equal magnitude but of opposite sign. In other words, 
there
is no net force acting on the rod BD. No unitary force.
Sorry, but it doesn't work
Krill
> Action Device to generate unidirectional force.
 ah, thank you for resurfacing abhi, now my new killfile is complete.
> Then start from zero and recreate new one.
> Let me try to explain how unidirectional force is generated through
> this action device. Please take printout, read and look at figuire
> carefully.
> This device is in vertical position i.e. upper end of bolt 5 pointing
> towards sky. Attach UPPER end of bolt 5 to centre horizontal circular
> plate, diameter of which is greater than length AC. This circular
> plate is in your hand. Let mass of whole device be 10 Kg. So downward
> gravitational force acting on this device is 10g Newton. Due to spring
> forces in spring 1 and 2, let resultant force generated at point D 9
> be 20g Newton. And direction of this resultant force at point D 9 is
> UPWARD. So it is transmitted through bolt 5 in UPWARD direction to
> circular plate in your hand.
> What will happen next? Will you feel weight of Action Device in
> downward direction or upward force pushing the circular plate and your
> hand in upward direction?
> If you feel upward force, take this circular plate near the roof of
> your room. This circular plate will stick to roof of your room like
> magnet. It will not fall.
> In space, this 20g force generated at circular plate will push /
> accelerate any mass in only one direction. If this circular plate is
> in hands of Astronaut, that Astronaut will be accelerated only in one
> direction.
> And if some rope is tied between round plate and Astronaut, the
> accelerating Astronaut will pull the action device in the direction of
> his acceleration. So now the action device and astronaut are moving in
> space and no reaction mass is expelled.
> Needless to add, the force generated at point D 9 depends on magnitude
> of force in spring 1 and 2.
> Please keep, the angle ABC as small as possible. Page No. 42 of Sixth
> edition of Fundamentals of Physics by halliday states that c =
> ab[sin(ABC)]
> Just wonder, if this device is so simple, then why don't I make it
> myself?
> God has His own ways to create problems for everyone. I am finding
> myself alone and I just don't know where to go. But let me leave this
> RCM cyber cafe in Matunga and try to find someone in this Bombay.
> Like Jodie Foster said in Contact..
> OK, GO..
> -Abhi.
===
Subject: Factorization example, end of argument
Here's an example that shows the problem with the ring of algebraic
integers.
Consider
(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
                 
                          49(x^3 + 5x^2 + 3x + 2)
where the c's *should* be algebraic integers.  Notice that only two of
the c's can have 7 as a factor.
Now then, you have as a zero of the factorization x = -7/c_1, so let
x= -7/c_1, so you have
49(-7^3/c_1^3 + 5(49)/c_1^2  - 21/c_1 + 2) = 0
which is
2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.
But that is a non-monic primitive irreducible over Q, so c_1 and by
symmetry c_2 cannot be algebraic integers.  However they must be
algebraic numbers, and it can be shown that any algebraic number can
be written as the ratio of algebraic integers.
So then there must exist f, such that fc_1 is an algebraic integer,
and letting g = fc_1 and multiplying both sides by f, I have
(gx + 7f)(c_2 x + 7)( c_3 x + 2) = 
                 
                          49f(x^3 + 5x^2 + 3x + 2)
so a zero is now x = -7f/g, which gives
49f(-7^3f^3/g^3 + 5(49)f^2/g^2  - 21f/g + 2) = 0
which is
2g^3 - 21f g^2 + 245f^2 g - 343 f^3 = 0
which proves that f must have 2 itself as a factor for g to be an
algebraic integer.
But looking back again at
(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
                 
                          49(x^3 + 5x^2 + 3x + 2)
that would mean that c_3 has a factor that is 2, which can distribute
to c_1 x + 7, but that leaves c_2 x + 7, which also needs a factor of
2 by symmetry.
But you see, you only have just that one 2.
Now then, are mathematicians and math groupies rational, or are you
cranks?
Will you give up when you see the truth, or will you keep fighting as
if mathematics is just a joke to you, as if you never really cared
whether or not it was true, as long as everyone you cared about agreed
with you?
What's important to you sci.math newsgroup?
James Harris
http://mathforprofit.blogspot.com/
===
Subject: Re: Factorization example, end of argument
> Here's an example that shows the problem with the ring of algebraic
> integers.
> Consider
> (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 2)
> where the c's *should* be algebraic integers.  
You have never given any reason why the c's *should* be algebraic 
integers.  You go on to prove that c_1 and c_2 cannot be algebraic
integers.  Don't you think it is time to drop the *should*.
> Notice that only two of the c's can have 7 as a factor.
> Now then, you have as a zero of the factorization x = -7/c_1, so let
> x= -7/c_1, so you have
> 49(-7^3/c_1^3 + 5(49)/c_1^2  - 21/c_1 + 2) = 0
> which is
> 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.
> But that is a non-monic primitive irreducible over Q, so c_1 and by
> symmetry c_2 cannot be algebraic integers.  However they must be
> algebraic numbers, and it can be shown that any algebraic number can
> be written as the ratio of algebraic integers.
> So then there must exist f, such that fc_1 is an algebraic integer,
(note that f can be chosen so that f and fc_1 are coprime)
> and letting g = fc_1 and multiplying both sides by f, I have
> (gx + 7f)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49f(x^3 + 5x^2 + 3x + 2)
> so a zero is now x = -7f/g, which gives
> 49f(-7^3f^3/g^3 + 5(49)f^2/g^2  - 21f/g + 2) = 0
> which is
> 2g^3 - 21f g^2 + 245f^2 g - 343 f^3 = 0
> which proves that f must have 2 itself as a factor for g to be an
> algebraic integer.
I assume that you are saying that the above polynomial is
irreducible over the algebraic integers and thus f must be divisible
by 2 or g satifies a non-monic irreducible polynomial.  
You need to prove irreducibility of course, but let that
slide for a moment.
> But looking back again at
> (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 2)
> that would mean that c_3 has a factor that is 2, which can distribute
> to c_1 x + 7, but that leaves c_2 x + 7, which also needs a factor of
> 2 by symmetry.
Less obscurely:
You can write
(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
((g_1/f_1) x + 7) ((g_2/f_2) x + 7) (c_3 x + 2)
= 49(x^3 + 5x^2 + 3x + 2)
Matching coefficients of x^3 gives us 
           (g_1/f_1) (g_2/f_2) c_3 = 49
and we conclude c_3 has a factor of 4. 
> But you see, you only have just that one 2.
Why?  c_3 can be divisible by 4.  (c_3 x + 2) cannot be divisible by
4 but there is no reason for it to be.
                       - William Hughes
===
Subject: Re: Factorization example, end of argument
> Here's an example that shows the problem with the ring of algebraic
> integers.
 Consider
 (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 2)
 where the c's *should* be algebraic integers.  
> You have never given any reason why the c's *should* be algebraic 
> integers.  You go on to prove that c_1 and c_2 cannot be algebraic
> integers.  Don't you think it is time to drop the *should*.
I don't see a problem at this time with dropping the should.
> Notice that only two of the c's can have 7 as a factor.
 Now then, you have as a zero of the factorization x = -7/c_1, so let
> x= -7/c_1, so you have
 49(-7^3/c_1^3 + 5(49)/c_1^2  - 21/c_1 + 2) = 0
 which is
 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.
 But that is a non-monic primitive irreducible over Q, so c_1 and by
> symmetry c_2 cannot be algebraic integers.  However they must be
> algebraic numbers, and it can be shown that any algebraic number can
> be written as the ratio of algebraic integers.
 So then there must exist f, such that fc_1 is an algebraic integer,
> (note that f can be chosen so that f and fc_1 are coprime)
Why?
James Harris
http://mathforprofit.blogspot.com/
===
Subject: Re: Factorization example, end of argument
> Here's an example that shows the problem with the ring of algebraic
> integers.
 Consider
 (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 2)
 where the c's *should* be algebraic integers.  
 You have never given any reason why the c's *should* be algebraic 
> integers.  You go on to prove that c_1 and c_2 cannot be algebraic
> integers.  Don't you think it is time to drop the *should*.
> I don't see a problem at this time with dropping the should.
> Notice that only two of the c's can have 7 as a factor.
 Now then, you have as a zero of the factorization x = -7/c_1, so let
> x= -7/c_1, so you have
 49(-7^3/c_1^3 + 5(49)/c_1^2  - 21/c_1 + 2) = 0
 which is
 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.
 But that is a non-monic primitive irreducible over Q, so c_1 and by
> symmetry c_2 cannot be algebraic integers.  However they must be
> algebraic numbers, and it can be shown that any algebraic number can
> be written as the ratio of algebraic integers.
 So then there must exist f, such that fc_1 is an algebraic integer,
 (note that f can be chosen so that f and fc_1 are coprime)
> Why?
The choice of f is not unique.  Some of your divisibility results
needed g=fc_1 coprime to 2.  The issue is
moot since it has been shown that f is not divisible by 2.
                         - William Hughes
===
Subject: Re: Factorization example, end of argument
> Here's an example that shows the problem with the ring of algebraic
> integers.
 Consider
 (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 2)
 where the c's *should* be algebraic integers.  
 You have never given any reason why the c's *should* be algebraic 
> integers.  You go on to prove that c_1 and c_2 cannot be algebraic
> integers.  Don't you think it is time to drop the *should*.
 I don't see a problem at this time with dropping the should.
 Notice that only two of the c's can have 7 as a factor.
 Now then, you have as a zero of the factorization x = -7/c_1, so 
let
> x= -7/c_1, so you have
 49(-7^3/c_1^3 + 5(49)/c_1^2  - 21/c_1 + 2) = 0
 which is
 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.
 But that is a non-monic primitive irreducible over Q, so c_1 and by
> symmetry c_2 cannot be algebraic integers.  However they must be
> algebraic numbers, and it can be shown that any algebraic number 
can
> be written as the ratio of algebraic integers.
 So then there must exist f, such that fc_1 is an algebraic integer,
 (note that f can be chosen so that f and fc_1 are coprime)
 Why?
> The choice of f is not unique.  Some of your divisibility results
> needed g=fc_1 coprime to 2.  The issue is
> moot since it has been shown that f is not divisible by 2.
>                          - William Hughes
Oh, make a claim, get called on it, and then just shift to another claim, 
eh?
Crank.
James Harris
===
Subject: Re: Factorization example, end of argument
> Here's an example that shows the problem with the ring of algebraic
> integers.
 Consider
 (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 2)
 where the c's *should* be algebraic integers.  
> You have never given any reason why the c's *should* be algebraic 
> integers.  You go on to prove that c_1 and c_2 cannot be algebraic
> integers.  Don't you think it is time to drop the *should*.
> Notice that only two of the c's can have 7 as a factor.
 Now then, you have as a zero of the factorization x = -7/c_1, so let
> x= -7/c_1, so you have
 49(-7^3/c_1^3 + 5(49)/c_1^2  - 21/c_1 + 2) = 0
 which is
 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.
 But that is a non-monic primitive irreducible over Q, so c_1 and by
> symmetry c_2 cannot be algebraic integers.  However they must be
> algebraic numbers, and it can be shown that any algebraic number can
> be written as the ratio of algebraic integers.
 So then there must exist f, such that fc_1 is an algebraic integer,
> (note that f can be chosen so that f and fc_1 are coprime)
> and letting g = fc_1 and multiplying both sides by f, I have
 
> (gx + 7f)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49f(x^3 + 5x^2 + 3x + 2)
 so a zero is now x = -7f/g, which gives
 49f(-7^3f^3/g^3 + 5(49)f^2/g^2  - 21f/g + 2) = 0
 which is
 2g^3 - 21f g^2 + 245f^2 g - 343 f^3 = 0
 which proves that f must have 2 itself as a factor for g to be an
> algebraic integer.
 I assume that you are saying that the above polynomial is
> irreducible over the algebraic integers and thus f must be divisible
> by 2 or g satifies a non-monic irreducible polynomial.  
> You need to prove irreducibility of course, but let that
> slide for a moment.
> But looking back again at
 (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 2)
 that would mean that c_3 has a factor that is 2, which can distribute
> to c_1 x + 7, but that leaves c_2 x + 7, which also needs a factor of
> 2 by symmetry.
> Less obscurely:
> You can write
> (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
> ((g_1/f_1) x + 7) ((g_2/f_2) x + 7) (c_3 x + 2)
> = 49(x^3 + 5x^2 + 3x + 2)
> Matching coefficients of x^3 gives us 
>            (g_1/f_1) (g_2/f_2) c_3 = 49
> and we conclude c_3 has a factor of 4. 
 But you see, you only have just that one 2.
> Why?  c_3 can be divisible by 4.  (c_3 x + 2) cannot be divisible by
> 4 but there is no reason for it to be.
>                        - William Hughes
Scratch that.  Set x = 1 to get 
((g_1/f_1) + 7) ((g_2/f_2)  + 7) (c_3 + 2) = (49)(11)
but unless (c_3 +2) is divisible by 4 the LHS cannot be an integer.
So if James can prove that 
   2g^3 - 21f g^2 + 245f^2 g - 343 f^3
is irreducible [1] the entire structure of math will fall in ruins.
                      - William Hughes
[1]  This might be difficult in light of a recent post by Dik Winter
===
Subject: Re: Factorization example, end of argument
 > Consider
 > (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
 >                  
 >                           49(x^3 + 5x^2 + 3x + 2)
 > where the c's *should* be algebraic integers.  Notice that only two of
 > the c's can have 7 as a factor.
Why should they be algebraic integers?  What do you mean with that?
Any ring where such a factorisation exists has 7 or 2 as unit, which
means that 1/7 or 1/2 is also in that ring.
 > Now then, you have as a zero of the factorization x = -7/c_1, so let
 > x= -7/c_1, so you have
 > 49(-7^3/c_1^3 + 5(49)/c_1^2  - 21/c_1 + 2) = 0
 > which is
 > 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.
 > But that is a non-monic primitive irreducible over Q, so c_1 and by
 > symmetry c_2 cannot be algebraic integers.  However they must be
 > algebraic numbers, and it can be shown that any algebraic number can
 > be written as the ratio of algebraic integers.
(It is even stronger, every algebraic number can be written as the
quotient of an algebriac integer and a rational (normal) integer.)
Yes, they are algebraic numbers.  Given (r1 to r3 are the negatives
of the roots):
  49(x^3 + 5x^2 + 3x + 2) = 49.(x + r1)(x + r2)(x + r3)
we find (to get the constant terms 7, 7 and 2:
  (7.x/r1 + 7)(7.x/r2 + 7)(r1.r2.x + 2)
So
  c1 = 7/r1,
  c2 = 7/r2 and
  c3 = r1.r2
(for some permutation of the roots).  As the base-polynomial in x is
irreducible and the constant term is 2, none of the roots is a unit.
And because r1.r2.r3 = 2, r1, r2 and r3 are non-unit divisors of
2.
 > So then there must exist f, such that fc_1 is an algebraic integer,
 > and letting g = fc_1 and multiplying both sides by f, I have
f = r1, g = 7, for instance (because c1 = 7/r1).
 > (gx + 7f)(c_2 x + 7)( c_3 x + 2) = 
 >                  
 >                           49f(x^3 + 5x^2 + 3x + 2)
 > so a zero is now x = -7f/g, which gives
x = -r1.  Yup, just one of the orginal roots...
 > 49f(-7^3f^3/g^3 + 5(49)f^2/g^2  - 21f/g + 2) = 0
 > which is
 > 2g^3 - 21f g^2 + 245f^2 g - 343 f^3 = 0
 > which proves that f must have 2 itself as a factor for g to be an
 > algebraic integer.
No, that is not shown here.  Set g = 7 (an algebraic integer), we get:
   686 - 1029 f + 1715 f^2 - 343 f^3 = 0
but that can be divided through by 343 and we get:
   2 - 3 f + 5 f^2 - f^3 = 0
Looks familiar.  Setting f = 2h we get:
   2 - 6 h + 20 h^2 - 8 h^3 = 0
dividing by 2 we get:
   1 - 3 h + 10 h^2 - 4 h^3 = 0,
so h is *not* an algebraic integer, and so f does *not* have 2 as a
factor.
Remainder of argument deleted because it is based on a false premissa.
I will try to explain your error.  Given:
 > 2g^3 - 21f g^2 + 245f^2 g - 343 f^3 = 0
and requiring that g be an algebraic integer, you know that g should be
the root of a monic polynomial with integer coefficients.  So you think
that the polynomial should be divisible by 2, and so f should have a
factor of 2.
Your error is two-fold:
1.  Algebraic integers *can* be roots of non-monic polynomials, where not
    all coefficients are divisible by the leading coefficient.  E.g. 7
    is (in the integers) a root of 2 x^2 - 13 x - 7.  The point is that
    algebraic integers *can not* be roots of non-monic *irreducible*
    polynomials where not all coefficients are divisible by the leading
    coefficient.  Leading to:
2.  I get:
        2 g^3 - 21 f g^2 + 245 f^2 g - 343 f^3 =
        (g - 7)[2 g^2 + (41 - 21 f) g + (245 f^2 - 147 f + 98)] -
              343(f^3 - 5 f^2 + 3 f - 2).
    As the last term is 0 when f is r1, r2 or r3, the polynomial is
    *not* irreducible when f has one of these values.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
===
Subject: Re: Factorization example, end of argument
> Here's an example that shows the problem with the ring of algebraic
> integers.
> Consider
> (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 2)
> where the c's *should* be algebraic integers.  
  It is simply false that there is a factorization of the form 
you give above with c_1, c_2, and c_3 algebraic integers, and you 
yourself have given a proof of that below.  The only mystery is 
why you say *should*.
> Notice that only two of
> the c's can have 7 as a factor.
> Now then, you have as a zero of the factorization x = -7/c_1, so let
> x= -7/c_1, so you have
> 49(-7^3/c_1^3 + 5(49)/c_1^2  - 21/c_1 + 2) = 0
> which is
> 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.
> But that is a non-monic primitive irreducible over Q, so c_1 and by
> symmetry c_2 cannot be algebraic integers.  However they must be
> algebraic numbers, and it can be shown that any algebraic number can
> be written as the ratio of algebraic integers.
  Seems to me you should stop right here.  You have just proved that
for a factorization of the form you started with, the coefficients
cannot all be algebraic integers.  There is no factorization of the
form
     (c1*x + 7)*(c2*x + 7)*(c3*x + 2)
where c1, c2, and c3 are algebraic integers.  I have no reason to 
think there should be, and I have no disagreement or problem with that 
conclusion.
  So the question in the remainder of what you are doing here is:
what are you trying to prove?  
> So then there must exist f, such that fc_1 is an algebraic integer,
> and letting g = fc_1 and multiplying both sides by f, I have
> (gx + 7f)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49f(x^3 + 5x^2 + 3x + 2)
> so a zero is now x = -7f/g, which gives
> 49f(-7^3f^3/g^3 + 5(49)f^2/g^2  - 21f/g + 2) = 0
> which is
> 2g^3 - 21f g^2 + 245f^2 g - 343 f^3 = 0
> which proves that f must have 2 itself as a factor for g to be an
> algebraic integer.
  I don't see this. Certainly you can write
      f*(21*g^2 - 245*f*g + 343*f^2) = 2*g^3,
but how do you know that 21*g^2 - 245*f*g + 343*f^2 is not
a multiple of 2 ?  Note that it is not necessarily an odd
integer, because it is not necessarily an integer at all.
It is a sum of three algebraic integers.
> But looking back again at
> (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 
>                  
>                           49(x^3 + 5x^2 + 3x + 2)
> that would mean that c_3 has a factor that is 2, which can distribute
> to c_1 x + 7, but that leaves c_2 x + 7, which also needs a factor of
> 2 by symmetry.
  See above.
> But you see, you only have just that one 2.
> Now then, are mathematicians and math groupies rational, or are you
> cranks?
  Again, I cannot figure out what you are trying to prove here.
If you want to claim that there is no factorization of the
form you started with, you succeeded with that halfway through, but 
then you continued with the stuff about divisibility by 2, and you never
really stated a conclusion.  I see no reason to think there would 
be such a factorization of the form you assumed in the first 
place.  What are your reasons behind the word should ?
  But mostly, what are you trying to prove here?
> Will you give up when you see the truth, or will you keep fighting as
> if mathematics is just a joke to you, as if you never really cared
> whether or not it was true, as long as everyone you cared about agreed
> with you?
  In this case we might even agree with you if we knew what you were 
asserting.  But you have not stated a conclusion.
  Nora B.
> What's important to you sci.math newsgroup?
> James Harris
> http://mathforprofit.blogspot.com/