mm-1028 === Subject: Re: Your baby was damned to death, but who made the choice? > Perhaps we could compromise and agree to just abort you? As long as you take Bloggs... -- Keith === Subject: Re: Your baby was damned to death, but who made the choice? >>Perhaps we could compromise and agree to just abort you? > As long as you take Bloggs... Youre one we could do without. How does it feel to be getting your ass kicked, you big bad warmonger, hot air bag, blowhard and illiterate? === Subject: Re: stupid memes >>You cant prove a negative is one of the most annoying and stupid ones >>that gets thrown in my face a lot. > I remember hearing it a lot on the old TV series Project UFO. The > investigators were apologizing for the fact that they were able to show > that the UFO featured in this particular episode was, or could have > been, something other than an alien spaceship. > And it makes sense. It isnt mathematical *proof* that somewhere, > somehow, there really is an alien spaceship landing on the Earth; you > would have to investigate _every_ reported UFO sighting. > The standard example is that one cannot prove there are no black swans, > unless one has seen every swan in the world. But perhaps today DNA > analysis could prove that no conceivable mutation would give a swan > black feathers without being lethal: although I do not expect this to be > possible. There is a black swan which is native to Australia. There is also a black-necked swan found in Chile. > The mathematical equivalent would be not being able to prove that no > natural number has property X, if one needed to prove that a natural > number does not have property X by a completely separate method for each > natural number tried. > Of course, you do not *need* to prove that no alien spaceship has ever > landed on the planet Earth to convincingly demonstrate that looking for > them is a waste of time and effort. > John Savard > http://home.ecn.ab.ca/~jsavard/index.html -- -- Geo. Michael Henry No! Bad dog! I said sit! anonymous === Subject: Re: stupid memes >> The standard example is that one cannot prove there are no >> black swans, unless one has seen every swan in the world. But >> perhaps today DNA analysis could prove that no conceivable >> mutation would give a swan black feathers without being lethal: >> although I do not expect this to be possible. >In fact, there are black swans in Western Australia: >http://www.dpc.wa.gov.au/emblems/swan.html Black swans are common throughout Australia. For myself, it is difficult to think of white swans. David ----- === Subject: One world or many? http://www.arxiv.org/abs/astro-ph/0501189 Consensus of chaotic inßation e.g. see Max Tegmarks website of an infinity of parallel universes is being challenged by a wide angle anomaly in the WMAP data. New data that was supposed to be published has apparently been held back? The above paper makes the case that there is only one relatively small universe - smaller than the observable universe! Its close to a virtual reality cosmic pong game. There are walls of a cosmic dodecahedron you pass through a wall and instantly come out on an antipodal wall billions of light years away. There are multiple images of the same objects in the celestial sphere in this model. What is really true? We will not know for awhile. The Shape of Space after WMAP data Jean-Pierre Luminet Laboratoire Univers et ThÇeories, CNRS-UMR 8102, Observatoire de Paris, F[CapitalEth]92195 Meudon cÇedex, France. Abstract What is the shape of space is a long-standing question in cosmology. In this talk I review recent advances in cosmic topology since it has entered a new era of experimental tests. High redshift surveys of astronomical sources and accurate maps of the Cosmic Microwave Background radiation (CMB) are beginning to hint at the shape of the universe, or at least to limit the wide range of possibilities. Among those possibilites are surprising ñwrap aroundî universe models in which space, whatever its curvature, may be smaller than the observable universe and generate topological lensing effects on a detectable cosmic scale. In particular, the recent analysis of CMB data provided by the WMAP satellite suggest a finite universe with the topology of the PoincarÇe dodecahedral spherical space. Such a model of a ñsmall universeî, the volume of which would represent only about 80 % the volume of the observable universe, offers an observational signature in the form of a predictable topological lens effect on one hand, and raises new issues on the early universe physics on the other hand. === Subject: Re: POLL: Some finite coin sequences are not computable <344013F45nfv5U1@individual.net> <344ggqF46mj3aU1@individual.net> <345uvdF469qimU1@individual.net> <3460t6F450klgU1@individual.net> <346f8jF480pn6U1@individual.net> <34je9gF4bbmslU1@individual.net> <34jgbiF49o0jlU1@individual.net> <34rkutF4dgl9rU1@individual.net> posting-account=sAS5-AwAAABlKnmtMjBbYHvhxI6W0cAg -------------------------------------s-o-s------------------- --------------- -- > Herc > > Yes, but the entire sequence is not computable. > > Every finite segment of the sequence is computable. That is, for each > n, there exists a general recursive function f, such that for all k > with 1<=k<=n, f(k) is the k-th digit of the sequence. > > But the entire sequence is not computable. That is, there does not > exist a general recursive function f, such that for all k, f(k) is > the > k-th digit of the sequence. > > Understand? > of course he doesnt. this guy is a phenomenal 9 on the crank scale. > better to give him the silent treatment and let him throw his loony > fits. > I understand the CLAIM, but theres a step missing in your deduction. > > all digits of the entire sequence are computable up to that digit > To prove your CLAIM : Yes, but the entire sequence is not computable > You have to construct a new sequence of digits. > That is going to be a mighty feat when **oo digits of all sequences** already appear in order. > Lets pull a function out of the air : D(x) = Comp(x, x) + 1 mod 9 > Obviously there is no D in Comp(d, ..) so what does that mean? > Put D back where it came from. D doesnt make a new sequence of digits > like you claim it does. > REMEMBER THIS IS YOUR REASONING > 123 > 456 > 789 > the diagonal is 159 > modify each digit 260 > NOW 260 does not appear on the list! QED > Now consider an entirely different beast, the infinite list. > Modify the diagonal > OOPS THAT SEQUENCE IS COMPUTED TO OO DIGITS TOO > Herc If I construct the diagonal number for the list of all computable reals, it will not be computable, because the halting problem is insoluble. === Subject: Abucus--Do Companies Make Them Anymore? posting-account=eiD2oQwAAACDx7NI-hagGyM0s6bsA2Ac Looking for one. But not the kind for a child. Does anyone know of any web sites? Kara === Subject: Re: Abucus--Do Companies Make Them Anymore? > Looking for one. But not the kind for a child. Does anyone know of any > web sites? If you search for abacus (the correct spelling in English and in other languages), you will have better luck. Abacuses are still very much for sale in east Asia, and there is a barely sufficient literature in English about how to use abacuses. What does Google turn up when you search with the correct spelling? -- Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 Learn in Freedom (TM) http://learninfreedom.org/ remove .de to email === Subject: Re: Abucus--Do Companies Make Them Anymore? > Looking for one. But not the kind for a child. Does anyone know of any > web sites? > If you search for abacus (the correct spelling in English and in other > languages), you will have better luck. Abacuses are still very much for > sale in east Asia, and there is a barely sufficient literature in > English about how to use abacuses. What does Google turn up when you > search with the correct spelling? You might also Google soroban which is the name of the variety of abacus used in Japan, and possibly Korea. === Subject: Re: Question about Presidents Social Security plan > Why is it that people who are too stupid to manage their own affairs are > smart enough to elected officials who are smart enough to manage things for > them? Why is it that presidents who are too stupid to manage the government are smart enough to appoint officials who are smart enough to manage it for them? Why is that people with no electrical skill are capable of hiring good electricians? Youre a smart guy, Jim, but I bet even _you_ hire people smarter than yourself. -- Tony P. === Subject: Re: Question about Presidents Social Security plan > tonyp says... >> Private Learjets are of course _much_ better for the economy :-) >Actually, thats true. > Actually, its false. Actually, it was ironic. >Private investment is much better for the economy than >military investment. As Timothy Little said, military expenditure can be considered a form of insurance. But insurance premiums are an expense, not an investment. > For one thing, that is a _general_ rule that has many exceptions, and > for another, private Learjets are more an investment in conspicuous > consumption than in productive capital. Roy, if you _travel_ by Learjet you are indeed making an investment in extravagance. When you buy a Learjet to rent out to conspicuous consumers, Id say you are making an actual investment. > The most expensive military of all is one that is > not strong enough to deter aggression. The above sentence applies literally to the US military: in September 2001 it was (as it still is) the most expensive military in the world, but it demonstrably was too weak to deter aggression. Now, back to Learjets. Or big-screen TVs, or cosmetic surgery, or live opera. They all give pleasure to their consumers, and employment to their producers. None of them may be to your taste or mine, but the joy of the free market is that it caters to all appetites. Well, not all: I value quiet conversation with congenial friends more than I value opera, but having a nice chat adds nothing to GDP, whereas buying an opera ticket _does_. Life is more than the economy. Still, employing people to produce stuff that other people are willing to pay for is what the economy is all about. CEOs and movie stars are willing to pay for Learjets, taxpayers are willing to pay for F16s. Producing either plane employs people and thus adds to GDP. Buying either plane can seem like a misguided investment or a frivolous expense, to some people. Its a matter of taste: if taxpayers as a group are willing to pay for F16s, I cannot condemn their taste any more than I can condemn that of CEOs and rock stars. Our current problem is along a different axis: we have a president who thinks taxpayers should not _pay_ for F16s, but _finance_ them. Borrow, dont tax, is the Bush line. Borrowing money from rich people, rather than taxing rich people, is of course very good for the rich people. Youd think the same would apply to not-rich people. Social Security (a pool of not-rich peoples money) has been lending to the Treasury for about 20 years. The government has been borrowing from the not-rich, but Bush now wants to convince us that _those_ government IOUs are worthless, and our only hope for retirement security is private accounts. (As if _that_ will somehow change the 2:1 worker:retiree ratio.) Of course, the _rich_ must never be allowed to think that _their_ government IOUs are worthless. So far Bushs tightrope act is working. As Bush stirs the SS pot, however, the not-rich might wake up and see the double standard hes trying to pass off as the natural order of things. Or they might not. You never know, in a democracy. -- Tony P. === Subject: Re: Am I thick?--- Aptitude Tests posting-account=eiD2oQwAAACDx7NI-hagGyM0s6bsA2Ac I agree with Tim. Dont pay any attention. When I was in elem. school the powers that be decided that all children needed an IQ test. I scored so low that I was labeled as severely mentally retarded. I was put into classes for others like myself. My parents fought to have me removed. It was later discovered that I needed glasses!! I wasnt in those classes long, but long enough for the teachers to tell me what I could not achieve. *** On a side note, my Dr. (an MD & PHD--former Microbiology teacher at George Washington) scored so low that he was told not to attend college--go to a tech school and learn a trade instead! === Subject: Re: Am I thick?--- Aptitude Tests > psychometric test which I presume will include maths, spatial awareness > etc... > Anyway, having only done one beofe around 15 years ago -- which I did > badly > in -- I thought Id have a practice. > So anyway, I didnt get below average, I managed to get LOW in ALL the > tests. Which was quite dissapointing to say the least. > Anyway, I dont consider myself to be a genius, but I do possess a BSc > First Class Honours in Computing. Im also a reasonable programmer - not > brilliant, but not too bad - hell I even get paid to do for a living. > Now my question is, why am I so hopeless at these tests? Can I improve my > scores? Or should I give my career to become a road-sweeper? > Am I just a very lucky idiot who ßuked a First (75% average) in very a > technical degree at a decent university? > Whats wrong with me? Nothings wrong with you. The test simply shows that you dont perform well at those tests - nothing to do with your intelligence, ability to reason or solve problems or to generally be useful in a work situation. I dont care what psychologists or whoever prepared the test try to tell you. The skills that got you through tertiary studies probably did not consist of a lot of psychometric tests which is why you performed better there. This doesnt mean you cant get better at these tests by practice if you feel that it is seriously hindering your ability to find better work. I rarely do well at those sort of tests either. Meanwhile I continue to do real work and problem solving in a real work environment, working with real people. At my place of work I tend to get the best projects involving new technologies that are up against the hardest challenges? I get these jobs owing to the quality of work Ive delivered in past projects. If my company had hired me on the basis of a psychometric test Im sure I wouldnt be working there now. Despite discussion with psychologists on the subject I still fail to see how a stressful environment like a job interview and and a battery of esoteric, contrived and alien questions can lead to any sensible conclusions as to how well a prospective employee will perform in most jobs. I have several unresolved issues with most of the interview techniques / recruitment tests that are in use today and those at a management level that tend to do most of the interviewing. It is astonishing the number of technical skilled staff that are employed without a single technical staff representative at the interview. Theres no doubt that hiring suitable staff based on a brief interview is a difficult task and employers are still not doing a very good job at it. Tim === Subject: Re: More analytic functions questions > Let G be an open subset of C. Show that if f : G---> C is continuous and > f^2 is analytic on G, then f is analytic on G. > I have no idea how to prove this. I tried expanding f^2 as a power series > and then from that I couldnt find anything. If f^2 took values only in the > plane minus 0 and minus a line from 0 to infinity, this would be true. > Is it easier to prove the statement if f has no zeroes? Using the fact that bounded isolated singularities are removable as Jose suggested, we need only show f is analytic where f is nonzero. David gave one proof, heres another: Suppose f(a) is nonzero. Then for z near a, (f(z) - f(a))/(z-a) = [(f(z)^2 - f(a)^2)/(z-a)]/(f(z) + f(a)). Therefore f(a) = (f^2)(a)/(2f(a)). === Subject: Re: More analytic functions questions >>Let G be an open subset of C. Show that if f : G---> C is continuous and >>f^2 is analytic on G, then f is analytic on G. >>I have no idea how to prove this. I tried expanding f^2 as a power series >>and then from that I couldnt find anything. If f^2 took values only in >>the >>plane minus 0 and minus a line from 0 to infinity, this would be true. >>Is it easier to prove the statement if f has no zeroes? > Yes. Lets say g = f^2 to prevent some confusion. If g has no > zero in a disk then you know that g has an analytic square root > h in that disk. Now you must have h(z) = plus or minus f(z) > for every z in the disk; since f is continuous the same > choice of plus or minus works at every point of the disk, > so f is analytic in the disk. What is h(z)? I also am not sure I understand your statement about f and its continuity. > So no you just have to show that f is analytic near p, if f(p) = 0. > Say g has a zero of order n at p. Then g(z) = (z-p)^n s(z) near > p, where s is non-zero near p. If n is even then you can use the > same argument as before. So you just have to show n is even. > But s has an analytic square root near p; if n were odd that > would mean there was a function (z-p)^(n/2) continuous near > p, and theres no such function. >>A side question (I apologize if this is obvious and I should know this) : >>Are all power series analytic? > Whether its obvious or not yes, its certainly something you should > know. Look for material on power series in the book. >>If so, why? If not, why not? >>Tony > ************************ > David C. Ullrich === Subject: Re: Variance multiplication >how do I multiply 2 different distributions? >Adding variances are possible, but Im sure >it wont work on multiplication! >Does somebody has any idea what will happen to >the variance? I think what youre talking about is: if X and Y are independent random variables, Var(X + Y) = Var(X) + Var(Y). If so, again assuming X and Y are independent, Var(XY) = E[(XY)^2] - E[XY]^2 = E[X^2] E[Y^2] - E[X]^2 E[Y]^2 = E[X^2] E[Y^2] - E[X]^2 E[Y^2] + E[X]^2 E[Y^2] - E[X]^2 E[Y]^2 = Var(X) E[Y^2] + E[X]^2 Var(Y) = Var(X) Var(Y) + Var(X) E[Y]^2 + E[X]^2 Var(Y) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: You pedantic ignoramuses In sci.logic, |-|erc <34ttcoF3tibk3U1@individual.net>: >> lets pull a function out of the air, R(x) = Comp(random, x) + 1 mod 10 >> >> according to you, this LIARS statement F(x) = ~F(x) is >> a new random sequence and proves the set of all programs >> Comp cant spit out any random string at all. >> Herc >> Without knowing (or finding earlier implementations of) Comp(x,y), I >> cant process this statement. I can of course assume that >> Comp(x,y) is machine x running on input y, in which case the correct >> equality >> R(x) = (Comp(x,x) + 1) mod 9 >> and that is indeed a new sequence, not in the Comp() list. Of >> course, it is not a computable sequence, though it is well-defined. > Is R computable? If youre asking whether every r in R is computable, the answer is no. Chaitins Constant is a nice counterexample. Besides, the set of computable reals is of measure 0, since the machines computing said reals can be mapped 1-1 with the integers. Since R is obviously not measure 0, there are uncomputable reals. (Presumably, a real is computable if there exists a TM that can spit out its digits. There are admittedly a few technical issues, such as which part of the tape can be used for digit-spitting, and which for scratchwork. In a pinch, one can use two tapes; TM2 can be mapped to TM by mangling the states and changing the alphabet, so one merely simplifies the problem thereby; the number of computable reals is unchanged.) Of course, there are also computable reals: sqrt(2), pi, and 1/3 are perfectly computable. One can also get into complexity space, asking questions such as how much time and storage might be needed to compute sqrt(2), pi, and 1/3. But this is a detail. And then theres TX_10 = {k/10^n: n >= 0, n, k in J}. This is a perfectly computable and denumerable set, consisting of all finite prefixes (base 10). It is dense, but does not cover R or Q. > Herc -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: You pedantic ignoramuses <34oiivF4a82laU1@individual.net> posting-account=sAS5-AwAAABlKnmtMjBbYHvhxI6W0cAg Actually, aleph_1 is the cardinality of the set of all isomorphism classes of well-orderings of N. It may or may not be equal to the cardinality of the set of all subsets of N, that depends on the continuum hypothesis. Likewise, aleph_(n+1) is the cardinality of the set of all isomorphism classes of well-orderings of a set of cardinality aleph_n. It may or may not be equal to the cardinality of the set of all subsets of a set whose cardinality is aleph_n, that depends on the generalized continuum hypothesis. === Subject: Density of C(T) in L^(T), T=|z|=1 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0G4tH610011; Could someone please help me prove that C(T), the set of continuous functions on the circle is dense in the set of L^2 functions on the circle? . I have been trying to use the map g(x)=e^ix :R->T to try to prove this, without === Subject: Re: Density of C(T) in L^(T), T=|z|=1 posting-account=Kb0T_QwAAACc9B9LpxfLjH0hHHYjPxft > Could someone please help me prove that C(T), the set > of continuous functions on the circle is dense in the set > of L^2 functions on the circle? . I have been trying to use > the map g(x)=e^ix :R->T to try to prove this, without i think this is a simple consequence of the theory of fourier series in L^2. any L^2 function can be approximated arbitarily accurately in the L^2 norm by a truncated fourier series, and each truncated fourier series is continuous since it is a sum of continuous functions. hence C(T) is dense in L^2. === Subject: How to compute a derivate of complex function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id j0G4r5n09801; The complex function is f(z)=sin(arctan(zi/zr))/z, where z=zr+zr*i. Do the first and second derivate of above function exist. If so, how to compute them? === Subject: Re: How to compute a derivate of complex function > The complex function is f(z)=sin(arctan(zi/zr))/z, where z=zr+zr*i. Do the first and second derivate of above function exist. If so, how to compute them? it seems that this function may not be analytic. to check simply rewrite f(z) as F(zr,zi)=u(zr,zi)+iv(zr,zi) where u, v are real valued functions and represent the real and imaginary part of F(zr+izi). then check the Cauchy-Riemann conditions are satisfied: u_x=v_y, u_y=-v_x. If this fails then your function is not analytic and none of the derivatives will exist. you can work out the rest if this is your homework. === Subject: Re: Slippery number theory lemma Bart Goddard > Larry Hammick > With difficulty I found a messy proof of this simple-looking lemma. > See any tidy proof? > For any two relatively prime odd numbers p,q >2, let f(p,q) be the > number of pairs (x,y) of elements of {1,2,...,(pq-1)/2} such that: > 1) x = (meaning is congruent to) some element of {1,2,...,(p-1)/2} mod > p 2) x = some element of {1,2,...,(q-1)/2} mod q > 3) y = x mod p > 4) y = -x mod q. > Then f(p,q) = f(q,p). > Fix x. By the Chinese remainder theorem, there is a unique > y satisfying (3) and (4). Then -y satisfies (3) and (4) > (which are (3) and (4) with the roles of p and q reversed.) > Since 1leq y leq (pq-1)/2 iff (pq+1)/2 leq -y leq pq-1 > and vice versa, we have exactly as many solutions to > (3) and (4) as we do to (3) and (4), with the correspondence > given by y <--> -y. === Subject: Re: Differential problem >>Alan and Bob have a keg of L litres filled with wine. >Its supposed to be ALICE and Bob : > http://en.wikipedia.org/wiki/Characters_in_cryptography Alice got tired of being an enabler for Bobs drinking, so she tossed him out, and he moved in with Alan. Robert (not Bob) Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada