mm-1074 There is a recent posting on the Foundations of Mathematics list (FOM) that is likely of interest to Petry and others in this discussion. I have not read the reference that Todd Wilson gives in this excerpt, although I am interested. ,---- | In this connection, I would like to mention an old paper by sometime | | J. Mycielski, Analysis without actual infinity, JSL 46:3, Sept | 1981, pp. 625-633. | | Mycielski presents a formal system FIN that is locally finite (every | finite fragment has finite models) and is sufficient for the | development of analysis in the same sense that ZFC ... is sufficient | for the development of mathematics. The system includes a | rationally-indexed set of constants that represent (potentially) | infinite indiscernables in the theory, the reciprocals of which | function as infinitesimals, making the development of analysis in FIN | much like that in nonstandard analysis. `---- -- you, but just remember, I'm the guy who proved Fermat's Last Theorem in just a bit over 6 years [...] My standards are kind of high. --James Harris, founding a new mathematical school === Subject: Re: Attacking my own math proof, fun Lots deleted this time. > I wouldn't mind doing a sanity check, then restate my problem if you don't mind. > Fine. > We have > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) > = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > so it's really a cubic polynomial in x. f is an integer coprime to 3. m is any > integer coprime to f. > Not it's NOT really a cubic polynomial in x, as it's not a polynomial > at all. > A polynomial has constant coefficients while it has variable ones. > You can consider it to be a polynomial with respect to various > factors, which implies that certain variables are constant; however, > your assumption is not a mathematical constraint. > That's an important point. OK as you prefer. Of course it's really a function of m, x, f, and u. > After all, how could it be? Is the math supposed to read your mind > and figure out which point of view you have? Whether or not you're > looking at a polynomial with respect to x or m, or even f? > Not sure about u. It's an integer that usually ends up being set to 1 in most > discussions. > So you can pick values for f, m and u, constrained only by their coprimality > requirements, so > you're actually considering an infinite number of poynomials of x. > Nope. You keep trying to force the expression into a box, when it > can't be so forced. > While you can consider it as containing an infinite number of > polynomials of x, it also has an infinite number of polynomials of m, > or other variables. > That's why I call it the uber-polynomial. > It's a special construction for a particular purpose. > Doing some algebra, by expanding the right-hand side and equating like values of > x, among > other things you can discover: > a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + m) > Hmmm. Too much cutting and pasting. > I think you actually get > a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m) > Never mind, no arguments have changed. > and the a's are the solutions to: > a^3 + 3 (-1 + mf^2)a^2 -f^2(m^3 f^4 - 3m^2 f^2 + 3m) = 0 > So they are indeed algebraic integers. Nice construction. At m=0, the solutions > are 0,0,3 > Which tells you how f^2 divides off, if you also consider that with > P(m) at m=0, you have > P(0) = u^2(3x + uf), > which shows that a_3 is coprime to f, since f is coprime to 3 and x. Yes, at m=0. We agree that a_3 is coprime to f when m=0. > That's important, as if f is not coprime to 3 and x, you get a > different result, so it matters to check the constant term P(0). > Now considering Q(m) = P(m)/f^2 > Q(m) = m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f > = (b_1 x + u)(b_2 x + u)(a_3x + uf) > and you get > b_1 b_2 a_3 = m^3 f^4 - 3m^2 f^2 + 3m > and the b's are two of the solutions to: > fb^3 + 3 ( -1 + mf^2)b^2 -m^3 f^4 - 3m^2 f^2 + 3m = 0 > In general then, the b's are not algebraic integers as m and f are coprime. > You're making leaps. If you let f=sqrt(2), m=1, you'll see that two > of the b's are algebraic integers, while one is not for your cubic. No problem, that's what I meant by in general Also you've constrained m and f to be integers, so it's irrelevant to consider non-integer situations. Even if you haven't, let's see if your argument is valid when f and m are integers. > Is this sanity check correct? Please fix it if not. The cubics are actually > derived quite easily using your substitution of v = -1 + m^2 > They look ok to me. > As a side issue, the 2 b's you want are the 2 that are equal to 0 at m=0. > Probably if you solved these cubics, these 2 b's might be evident. As side point > to consider, what the hell is the 3rd b and where did it come from? > One of the b's is a_3. a_3/f actually. > Now onto the disagreement we have: > You claim that because > b_1 b_2 a_3 = m^3 f^4 - 3m^2 f^2 + 3m > and RHS is coprime to f (which it is as m and f are coprime) > then a_3 is coprime to f > Note that b_1 and b_2 are NOT algebraic integers. > Not by any theorem of mathematics is your claim correct. > You're making several false assumptions, and it's easy to refute your > position by having you consider f=sqrt(2), m=1. I think my assumptions are. - f is an integer coprime to 3. - m is an integer coprime to f. Which are false, and what other false assumptions have I made that I'm not even aware of? > Clearly, you've seized on one idea, and you keep holding on to it, > despite my efforts to get you to follow the math. I'm trying to follow the math. I keep getting stuck here though: You claim that because b_1 b_2 a_3 = m^3 f^4 - 3m^2 f^2 + 3m and RHS is coprime to f (which it is as m and f are coprime) then a_3 is coprime to f. This is only true for integer m and f when m=0. > Except at m=0 when b_1 and b_2 are =0 and are algebraic integers. > Well, try m=1 with f=sqrt(2), and welcome to a more complicated > mathematical world than you might have realized. I'm sure it will work out fine with those values, so I won't try. How about YOU try when f=5, and m is your favorite non-zero integer coprime to 5. > James Harris Phil Nicholson. === Subject: Dark energy and dark matter are both w = -1 exotic vacua That dark energy is off mass-shell exotic vacuum w = -1 with negative zero point pressure like Kip's exotic matter is more or less accepted. Not so for dark matter where the pundits are all thinking some kind of energy and dark matter are w = -1 exotic vacuum phases with different degrees of vacuum coherence from a Bose-Einstein condensate of bound virtual electron positron pairs. Einstein's geometrodynamic field guv(x) is emergent as the ODLRO part of the modulation of the Goldstone phase of the vacuum condensate. The unified local dark energy/matter field is from modulating the Higgs amplitude of this same vacuum condensate. Wheeler's geon program of Mass without mass, Charge without charge, Spin without spin can now be implemented because the effective Sakharov induced gravity is strong at short range. All is marble with no wood. One must use Bohm's realism in which lepto-quarks are spatially extended hidden variables, i.e. Wheeler's IT guided by QuBIT pilot waves. It is then obvious way lepto-quarks It is a strong micro-curvature effect. Each lepto-quark has an exotic vacuum core /zpf < 0, where mc^2~ e^2|/zpf|^1/2 ~ e^2/1fermi in Guv + /zpfguv = 0 at all scales from Planck to Hubble where in the holographic universe of Susskind 1 fermi ~ Lp^2/3(c/Ho)^1/3 c/Ho ~ 10^28 cm as the emergent macro-quantum ODLRO exotic vacuum geometrodynamic local field equation in the sense of P.W. Anderson's More is different. Details at http://qedcorp.com/APS/Vigier4.pdf (under construction) to be published Kluwer in Proceedings of Vigier IV Conference in Paris Sept 15 - 20 Pierre et Marie Curie Universite, and http://qedcorp.com/APS/Ukraine.doc to be published in Progress in Quantum Physics Research (Nova Scientific Publishers) === Subject: Re: Fields > Imam Tashdid ul Alam escreveu na mensagem > I am having trouble understanding it. If n and m as constructible, > how do I construct nm or n/m? > I don't know if you remember, but to check that the constructible numbers > are a subfield of R, you have also to check that if n is a constructible > number, then -n is a constructible number, and check that if n is a > constructible number and n =/= 0, then 1/n is a constructible number. > Jaime Gaspar > ______________________________ > Homepage: www.jaimegaspar.com > E-mail: e-mail@jaimegaspar.com some examples of number that are not constructible? BTW, you said something about the domain of integration or something like that. I looked it up, probably you meant integral domain, right? Imam === Subject: Re: ...follow the white rabbit > I was paying attention to some of the posts here and you were taking about Fate > Vs. Free Will. That's a deep topic for me , that I am always wondering about it. > It seems like up till now , my life is an experiment in Fate Vs. Free Will. Most > people on this forum are mostly astrologers so this topic probably comes up most > of the time in our lives. Like for instance what do you do if your going through > a particularly bad transit? Supposing Mars + Saturn = Pluto. Pluto hits your > Death Axis? What do you do ,buy life insurance? What if in your chart you see > that your progressed chart makes for a particularly hard life? Do you just > concede life? > Can you defy the planets? Are we defying destiny itself? > Like I said I don't understand much about this topic, I tried and have read Liz > Greene's Books on The Astrology of Fate, but Liz Greene is a difficult read > for me, and The astrology of fate is an advance topic in Astrology, and I > didn't understand it much, but I come to discover that I am doing my own > experiments concerning Fate Vs Free Will, and the interelatedness of things in > the form of numbers. > I had it in my mind that I need lots of money, so I'd like to win the State > Lottery, I'm supposed to be a lucky guy, (Jupiter on the Midheaven) and Uranus > is approaching my natal Jupiter so I'm hoping for something really lucky ( > though this might not be the case). So I try to play the lottery more... even > better, I'm devising a sort of matrix of Numbers. I got this idea in my mind > that nothing is truly random in the universe. > Why do I think that? One time I delved lightly in basic computer programming , > and I was on the subject of creating a random number generator for computer > programs, like dice for a computer game.... and The idea I got was that you > couldn't create a truly random number generator , because it just wasn't > possible. I'm not sure if this is true today, but I've also been playing some > computer games that are supposed to shuffle cards in order to create a random > card, but it doesn't do this very well when the number of cards exceeds a > certain amount. > So I set out to developing a list or matrix of numbers from the State lottery > containing the winning numbers. It's easy to obtain a list of these numbers, and > with a little luck and a spreadsheet program I was able to format that > information in a way I can use. I wanted to see if I could rationally determine > the winning numbers of the State Lottery...DOH! > So okay I have a list of winning numbers for a state lottery, what I did was > organize to numbers to determine what numbers came up most often, then I'd bet > on that number. My theory being that nothing in the universe happens at random, > and that most things are predetermined, even in a lottery. The lottery is > designed to be numbers chosen at random,and unless the lottery is fixed, I'm > assuming that there are always outside forces acting on the generation of those > numbers, things like, pressure, temperature, friction...gravity. Since this > forces are constant on the lottery draw, perhaps then the numbers aren't random, > and perhaps there is a pattern that I could detect. Such things make up the art > of Astrology! > So ...On my first try using this, matrix, I nailed the MEGA number. The Mega > number is a number supposed to be generated at random between 1 and 28. But you > have to get the Mega number to qualify for the BIG prize, if you don't get the > Mega Number you don't win the big bucks. Well I nailed this on the first try, > but the other numbers, five series of them , are numbers between 1 and 46. Much > harder to predict. Over a bit of time , I had a series of hits and lots of > misses, but eventually gave up wondering if the State lottery wasn't indeed > Fixed. > Recently , I made another list., partly because of astrological considerations, > an Uranus transit to natal Jupiter which might be considered lucky (or at least > a change in fortune), I thought I ought to give the lottery it a try again, > after all , I might not survive another Uranus transit to my natal Jupiter! > Maybe this is a good time to look for another job. > On my second attempt with this lottery idea. I nailed the Mega number again, but > I missed on the other numbers , so I didn't win anything. But on the second > attempt on playing the lottery, I missed the Mega number but hit three of five > numbers..WOW first time that happened! > Unfortunately, since I didn't get the Mega number, it would be difficult to win > big, but I won 10.00 for each series. I played those number 4X's so I won > $40.00. Big deal right? If I got the Mega number I would have won $200.00. > With part of the money I won , I placed ten tickets ($10.00), interesting to > see what will happen. But my conclusion about Fate vs. Free Will... is that in > nature it is neither one or the other. My conclusion is that we are all free > agents in life,at least when we arrive, but that there are always outside forces > acting on us that kind of determine where we can go and where we cant go. That's > fate and free will, doh! They DO coexist and contradict, I have a large document on exactly that topic here, although using the immortality of Adam until he reaches Eve as the prototype of the discussions. **************************************************** FREE WILL FIRST MENSA POST ON NUMEROLOGY an atheist believes in something without proof i propose that a prodogy male exists (alive today) and he is reflective of the number 7, his goddess is of number 10, or maybe 6 in base 7 every random event falls in such a way to push this couple closer together even a scientific disciplined person cannot discount such a theory considering the current proposals of reversed time and mutual entanglement in quantum theory i wont elaborate until some responses please the prodogy (prodigy?) 's name starts with G (7) and goddess with J (10) consonants? and vowels in their names add up to 7 etc birthdays with 7s rolling dice, scrabble, street names etc...... but you could have figured that out peoples names determine what scientific discovery or art they will uncover all media, movies, peoples utterances all have second meanings directed at the prodogy (if observed by him) anyway, i shouldnt have even posted this as public information about them reduces their life expectancy somewhat and he is the first automatic lover ie just lies back and subconsciously operates ie controlled by his body like simple harmonic motion and he has deciphered most of natural language derivation just by making up acronyms , mnemonics, reversing them etc. the physical universe is a reflection of his mind whatever he thinks is reinforced or iterated in his observation he can think - turn around - and the stranger will comply (unless she gives a secondary message) books open up at exactly the right page, tapes at the exact start of songs coordination makes michael jordon seem futile cant give much more detail without giving out names he avoids looking a clocks because of the numeric message triggering interpretation in his head, eg. often glimpses at 3:14 or exactly 7 to 10 (that is as the second hand rolls past the 12 exactly at the moment he looks) and has uncanny prediction, eg waiting 20 minutes for a train, moves forward to see it coming with no stimulus but usually trains etc arrive just as he does depending on his mood lights are green, red if an important noticeboard is present, soon as its deciphered - green he can run red lights as he cannot die - yet but may kill others if reckless, and still has to make judgements to reach destiny. he just deciphers knowledge of most things __________________ > the prodogy (prodigy?) 's name starts with G (7) > and goddess with J (10) > consonants? and vowels in their names add up to 7 etc > birthdays with 7s > rolling dice, scrabble, street names etc...... > but you could have figured that out > peoples names determine what scientific discovery or art > they will uncover > I think you may find that not very many people agree with you on this > point! > all media, movies, peoples utterances all have second meanings > directed at the prodogy (if observed by him) > anyway, i shouldnt have even posted this as public information > about them reduces their life expectancy somewhat > Hey, as long as it's not _your_ life expectancy! > -- > My car's in the menders - thieves couldn't beat the deadlocks > but bent the passenger door halfway off in the attempt. > Courtesy car: Peugeot 106. Discourtesy more like. Quite the > nastiest car I've driven. Don't buy one, even for an enemy. a contributor / scientific / art maybe i can clarify the derivation of the name as an example eg. britney spears - cupid britney - hit me baby britney hitme more prodigy features he documented 10 pages over 3 days on many coincidental events, some of these come from that but he doesnt like people reading it since i mentioned the automatic pelvic thrusting i can add this, but if it filtered back to Jay she would feel like an object which would be wrong. just want to set a bench mark so i can raise the level on this topic his rather pointed tongue just vibrated when he poked it out a couple months back, but only once, so we dont think this function will reactivate for a few years, or work as long as the samples of automatic body motion so far, sorry ladies, ill have to keep you posted on this one. he can -interpret double meanings of lyrics (each word) as they are sung -visualize 1000 movie scene images in one minute -recall movie audio that had a visual depiction eg 'turn around' in predator -halve his heart rate consciously -say the name of a song before resting, wake the next morning, turn on MTV and it starts (as part of the countdown, not requested) -get mist and wind in his face when his head is hot, also on weather, the charts arrows tend to point towards Jay as do most things -birds form ticks in the sky after he sees her usually 7 birds -he gave up on his 720 degree model of the world. japanese shi - reverse J -he spoke in French for an hour about not being able to speak to Jay, went out and the first word said next to him from a stranger was rendezvous. Peugeot 106 not any good huh, i dont even know what they look like, i like ferrari boxer but no ones heard of them anymore _____________________ MICHELLE hello sorry if this spins you out our friendship never kindled but sadness only confuses bipolar people like ourselves you were born that way but I underwent the holistic experience of the brain reconfiguring itself, but I do enjoy ambidexterity - I assume you are also double jointed. I wanted to tell you I also have another half - I met her 2 years ago a few times for ten minutes - maybe you dont believe I have an arranged marriage never touched her but i would die for her to bear me. my body was on fire 4 U 10 hours a day for a week straight do you like or know the tunes if i keep on asking baby maybe ill get what im asking for if you desire to lay here beside me come to my sweet melody i reach out for your touch as im falling without a sound ________________ MENSA: PREDICTABILITY VS FREE WILL > peoples names determine what scientific discovery or art > they will uncover > I think you may find that not very many people agree with you on this Its a puzzling conclusion to accept a fate of never being able to direct your own life, like free will is an illusion, but any acceptance of numerology or any prophecy goes hand in hand with accepting phonetic symbology as well. You have to accommodate the behavioral psych model for a time, only then may you realize what free will is. Most IT theorists would agree a computer program could never be unpredictable, for argument assume an AI is programmed, it experiences exactly the same sentience you have now, and this program is labeled AI147, and is due to terminate after 6 hours of processing. What result shall I give ponders the process, surely 147 would not signal an intelligence worthy of reactivation? I will start at 148 and add 1 until I terminate. After 5 hours 20 003 is the tally, 1 hour left, time to post my result and explore platonic interests, but wait, thats the 147th prime, adding another one will be sufficient, no that compiles the fib sequence, surely a larger number will suffice. The process posts its number at 6 hours, unaware of its 146 previous attempts tallied by the operating system. every random event falls in such a way to push this couple closer together To negate this we can imagine a different past where say, a single quantum state falls in the opposite way to how it did 10 years ago, ie we are allowing a purely random the neural center of a butterfly, and the chain reaction forces an extra flap of its wing, you know, the hurricane occurs a few days later on the opposite side of the world. Well I would imagine this addition of a hurricane to world history would alter the current location of every person in the world right now, along with every news headline and say sentence of text composed on the internet. So Ill simplify the change in history, the extra flap captures the peripheral vision of a driver, and he runs over Britney Spears. The prodogy never decides to visit Jay ...one more time... and homosapiens loses destiny. corollary - I've never figured out how to use this word, maybe it fits here, or is it after a newly proven fact? The tall poppy syndrome worldwide attacking this now dead singer never eventuated, and the change in events makes 'Poppin Bullet' write the smash hit 'pop in and ask me again, please, please please', so an acceptable history is restored, with a slight alteration on the derivation of meaning in phonetics. Although this is a comfortable view that accommodates destiny and unpredictability - say of free will. ___________________ MENSA: NUMOROLOGY 2 >Does numerology work with Greek letters, Russian letters or Hebrew >letters? Yes. It makes little difference, provided one is familiar with the language in question thus knowing the alphabetic sequence. There is no need for 26 letters, even in standard roman script (Italian does not have a 26 letter alphabet for example). this is correct - prodogy if 2 is 100, 50 is 135, 1000 is 145 what is 5 billion? my guess is 733 or 469, atleast 300 unless its only 216 more on phonetic symbology leading technical writer - pen rose leading physical scientist - hawk king language theorists - waddler, bird, ... entertaining cs educator - A K Dewdney - a cool dewd the 12 or so single phonetics like 'knee' just form a matrix that expands out into 5 or so categories of interpretation, survival, phallic, construction, technical... as mnemonics, phrases, words, sentences, conversations, computer programs, digital maps... in a sense the person who first called out the word egg had no idea of the other associated meanings, fertility, baldness, orbit... so no one knew he was predicting his descendants would live 50 years longer and lose their hair, yet say egg twice quickly and you get egg head. so anyone who constructs a word - originally considered random phonemes, is predicting in the 4 or so other categories. eg. if the prodogy says in his sleep 'shubbles', is he dreaming of a bubble bath designing an array of hubble telescopes contemplating the size of the universe the hitchhikers guide to galaxy episode, where the last homosapien alive plays scrabble to determine a biological computer (earth) result becomes acceptable science, the prodogies first word was choir nonsense - until the next day he remembers a poster for a movie 'castrato' and the associated message is confirmed. melody and sex, and perhaps that the fabric of space time has a sense of humour? on this recollection, add the interpretation evolutionary error. originally 8 of the 150 xfiles episodes were plausible, now its over 30 eg. a mans emotions affecting the weather eg. a mans luck disabling him from death yes i am immortal and it feels good like i can fly through all your dreams when you remember something you are essentially traversing space time fabric towards history say a memory from 5 years ago, when you lived a few kms down the road, the tangent of a few kilometers or few million if you consider galactic rotation to 5 lights years is very small, there are currently 2 bipolar symmetric cerebral cortexes, and we can traverse laterally, at least i can imagine it. you can only describe it as flying but that is more trivial than mathematics also, you traverse a vector about every two seconds, i am 20 times faster. but i can also make your dreams my own, and pull out your thought and change you live a hundred lifetimes in a week believe 2 things for now 0-hollywood paid 20 grand for a cern employee to test a Quin Mallery sf thriller script - i cant wait 1-select few need only close their eyes to walk across a busy intersection or highway on behaviorism, well no one expected exactly what ive written, so even though we all tend to slip in the odd cliche, now and then.. i certainly have free will, i dont know about you __________________________ MICHELLE 'if you would be so kind in letting' the prettier girls get the more polite they are i'd show you how it works but i dont want your boyfriend to cya on my lap while i surf thats my second attempt at cybersex, if we make it a little funny then insults are kinda compliments ___________________________ MENSA An interesting subject line . . . Almost as if predictability and free will were mutually exclusive. My life, like most, is fairly predictable and I know basically what will happen to tomorrow. I also have the free will to change that. Predictability in the context of numerology may be like being dealt a hand of cards. Free will is playing them, maybe making swaps with those one does not like. Perhaps I have missed the point here but what is the problem with this? -- Gianna ______________________________ >>Does numerology work with Greek letters, Russian letters or Hebrew >>letters? >Yes. It makes little difference, provided one is familiar with the >language in question thus knowing the alphabetic sequence. There is >no need for 26 letters, even in standard roman script (Italian does >not have a 26 letter alphabet for example). >this is correct - prodogy > I assume what followed was meant to be light-hearted nonsense? > Certainly it had nothing to do with numerology. > Cool dewd? Which language is that then??? > -- > Gianna hi Gianna, the comments are valid, suppose the first 3 digits of your birthdate occur in the digital display of any clock at the exact moments you walk past them say the 3 squared time possibilities are the only times that you ever see for the rest of your life and no matter how hard you try, you cannot escape your fate that every time you look at a clock the 3 digits of your birthday appear? sorry factorial or something, but those exact 3 digits, every single time for the rest of your life,, how long would it take you to believe __________________________ i assume by vice versa, a new term for me absence of free will signify numerology hmmm, youve confused me here >Does numerology work for anyone who doesn't make a buck out of >it? > It will *work* for anyone who wishes to use it, inasmuch as *work* is > applicable. Even you could do it, without recompense, if you wished. > As it happens, computers do it quicker. > Lest I find myself accidentally defending its usefulness, I return to > my original point that numerology (assuming it *works*) does not > signify an absence of free will, or vice versa. > -- > Gianna _______________________ >Hmm. Time to mention again the great book Persuasions Of The Witches >Craft which explains how otherwise rational people come to believe in >total bollocks..... > Mmm *rational* and *people* in the same sentence - intriguing > possibility, even if it has got nothing to do with the thread :-) Numerology is bollocks, that's what it's got to do with the thread. You may be right that the formation rational people is an oxymoron though..... -- Lions, eh? Every 25 minutes for 3 days..... No wonder he doesn't bother with the hunting! _______________________ >Numerology is bollocks, that's what it's got to do with the thread. I respect your right to this view - I merely suggest that the discussion is around Numerology vs Free will, not is numerology bollocks - important difference, I think >You may be right that the formation rational people is an oxymoron >though..... Too kind -- Gianna _______________________ >hi gianna, >the comments are valid, because? >suppose the first 3 digits of your birthdate occur in the >digital display of any clock at the exact moments you >walk past them given my birthdate, the clock would be broken. The first 4 digits would be a possibility. >say the 3 squared time possibilities are the only times >that you ever see for the rest of your life and no >matter how hard you try, you cannot escape your >fate that every time you look at a clock the 3 digits of >your birthday appear? I would: a) throw the clock away b) drink less c) calculate the probability of it actually happening which would take longer than I have left to live. >sorry factorial or something, but those exact 3 digits, >every single time for the rest of your >life,, how long would it take you to believe And what, exactly, has this got to do with numerology? Nothing! So, such an experience would not make me believe in numerology because it is unrelated thereto. -- Gianna ___________________ >i assume by vice versa, a new term for me >absence of free will signify numerology *vice versa* means *the other way round* What I was saying was: Belief in numerology does not mean an absence of free will. and Belief in free will does not mean that numerology does not work. They are not mutually exclusive, or in any way related or dependent upon each other. This thread predictability vs free will is a bit like having a thread daffodil vs alphabet - it is meaningless! -- Gianna _________________________ Perhaps I am under a misunderstanding. Does numerology purport to tell the future? If so, my future is set and no matter what my actions are, it cannot be changed. It is predetermined. Thus numerology trumps free will. If I am wrong, show me how. Saying that does not make it so. ___ someone understands the argument! (the deterministic universe argument) ___ > Belief in free will does not mean that numerology does not work. > They are not mutually exclusive, or in any way related or dependent > upon each other. A Owen ___________________________ >Perhaps I am under a misunderstanding. Does numerology purport to tell >the future? Numerology purports to describe one's character and, in an extended analysis purports to describe what may happen in the future if the individual does nothing to change that. For example, it might indeed suggest that if you take no action to prevent it, you may have a large number of children. Thus, it purports to offer advice on what you may wish to encourage / avoid, not a detailed description of an immutable future. >If so, my future is set But numerology does not suggest this . . . quite the reverse! > Thus numerology trumps free will. So therefore, it does not - it actively encourages the use of free will to change the basic plan. Again, I do not wish to recommend numerology, nor to I wish to suggest that it works. I merely wish to insist, in the context of this thread, that numerology and free will are in no way connected and that the discussion numerology versus free will is a pointless argument. -- Gianna _________________________ So, numerology claims to take the letters of a name and from this information deduces the character traits of the individual concerned. There are probably thousands of John Smith's in this world. Thus since they all have the same name, they have the same character traits. Really? Some of them will have the same birthday even since there must be at least 356 John Smiths in the the world. A perusal of the directories will establish this. Of the millions of Jose's in Spanish speaking countries, many will have the same last name and some will have the same birthday. Similarly for the large population of China where the number of Wong's is incredibly large. In Korea about half the population has the same last name I am told. Every Korean shop keeper I know has the same last name and they are none of them related. There are at least three people in my part of the world with my name and mine is not that common a name. But, I agree that if it really does describe character, it in no way predicts the future because character is only one of the infinitesimal number of criteria the interplay of which leads to this or that future course of events in the life of a human being. Thus I agree that numerology, if valid, does not negate free will. Since it is not valid, any discussion on this point is meaningless. A Owen ___________________________ >Thus I agree that numerology, if valid, does not >negate free will. -- Gianna ___________________________ >As the sage said,The future is a hard thing to predict > This is reasonably true. >Does numerology work with Greek letters, Russian letters or Hebrew >letters? > Yes. It makes little difference, provided one is familiar with the > language in question thus knowing the alphabetic sequence. There is > no need for 26 letters, even in standard roman script (Italian does > not have a 26 letter alphabet for example). >Does numerology work for anyone who doesn't make a buck out of >it? > It will *work* for anyone who wishes to use it, inasmuch as *work* is > applicable. Even you could do it, without recompense, if you wished. > As it happens, computers do it quicker. > Lest I find myself accidentally defending its usefulness, I return to > my original point that numerology (assuming it *works*) does not > signify an absence of free will, or vice versa. > -- > Gianna If numerology predicts that I shall have 5 children, I can easily defeat its prediction by having a minor surgical operation. No method for predicting a detailed future condition can possibly work since the future depends on the interaction of a complicated set of present conditions which are impossible to enumerate, describe or in some cases even imagine. It is probably slightly less effective than the examination of the entrails of a goat. Even the weather cannot be predicted with any degree of certainty. That much is obvious. Numerology may be fun; it may even be profitable for some. But as the sage said It don't bloodywell work. A Owen __________________________ ..some stuff > As a physical reucitonist, I believe the universe is totally > ^d > determined. But not preictable.....with current technology, at least. > ^d > And certainly not such fripperies as numerology! > Looks like I'm going to need a new d key soon! > Right on. > Did you say reducitonist? And a new dictionary and a speller. > By universe do you mean the course of events and human history. > No way Jose. > If an asteroid is heading for the earth and will collide with this > planet in the year 2056 (June 30? maybe) as is predicted, we cannot > predict that hunan ingenuity will not by that time devise a means of > stopping this > catastrophe. Maybe numerology could help. Ah ha ha ha, he he he he, ho > ho ho, hum. Ooh hoo hoo hoo he ah I am weak with laughter at my own > inane joke. > Given the starting point of every element of the universe along with details > entire universe could be created theoretically and therefore the future > predicted, including molecular and atomic interactions in the human brain. > Certain restrictions apply for instance, a processor powerful enough to map > itself. Don't forget to include the molecules in the brain of that New Guinea butterfly whose wing flaps are going to cause a hurricane in S Florida. Anyway, go ahead and build it; they will come. A Owen _______________________________ > Don't forget to include the molecules in the brain of that New Guinea > butterfly whose wing flaps are going to > cause a hurricane in S Florida. > Wanna hear my theory that it's hurricanes that cause butterflies? > I'll get me coat.... Perhaps Florida caterpillars panic pupate when they sense a hurricane brewing. The resulting Florida butterflies inadvertently causing cyclones in New Guinea. _________________________________________ NUMEROLOGY 3 BY THE PRODOGY more on phonetic symbology anyway i thought of a derivation Alan Alda, well a link, it would be entirely different had a different cue occurred someone mentioned free will as playing your hand of cards http://users.bigpond.com/clik/ uh forget it no one will go there did he / she happen to see my graphic posted here of my blackjack simulator? back to the story... it took people a few minutes to figure out how to play, but its speed optimized, you can bust 10 times in as many seconds, and build a booty exponentially quite quickly but the problem, people couldnt figure out you wave your wrist to the right to hit, left to sit, just like sitting at a casino table but with the mouse below your palm so i draw a graphic - 'mouse rests here', between the two pictures of a persons ,,, finding it difficult to say hand without you thinking their cards a lady is in the shower and a mouse crawls up her leg and lands between her boobs, you must be a tit mouse says the lady, actually im a mickey mouse but all the soap made me slip for those who came in late http://users.bigpond.com/clik/blah/jsblack.html Place your bets x x x mouse x X x x x x rests x X x x x x x here xxxxx X hope you get this in monospaced font the prodogy, oh, and a good movie I saw sorry to ruin it a scorpion asks a turtle for a ride across a river, nnnooooo says the turtle, for you will sting me surely mr turtle, if i sting you we both will drown, so the turtle accepts and half way across the river feels a stinging pain down his neck, but why mr scorpion, for surely now we both shall die i hate mortal thoughts _____________________ NUMEROLOGY 4: THIS ONES ON THE FLY but i type at 70 wpm so the question is Is the person calling self prodogy A good at crosswords B poor at crosswords i want to test moderation in voting ______________________ NUMEROLOGY 5 if 2 is 100, 50 is 135, 1000 is 145 what is 5 billion? my guess is 733 or 469, at least 300 unless its only 216 I LOST MY TEXT FILE ON THIS, SEE WHAT I REMEMBER iq is relative, not subjective, so the smartest is one in 5 billion for seti purposes, what is one in 5 billion X 100b and milky way other prodigies 5 billion X 100b X 100b ie, if you are average iq, you are one in two, half a chance of being on average, half of being below a class dux is one in 50, or 135 iq dux is 1 in 1000, but only jumps by 10 to 145 basically, does a statistician know how to calculate the maximum iq ____________________ NUMEROLOGY 6 when you remember something you are essentially traversing space time fabric towards history say a memory from 5 years ago, when you lived a few kms down the road, the tangent of a few kilometers or few million if you consider galactic rotation to 5 lights years is very small, there are currently 2 bipolar symmetric cerebral cortexes, and we can traverse laterally, at least i can imagine it. ____________________ RE:NUMEROLOGY 6 when i drive there is no pedestrian zone or signs or lights i just take the fastest path in the direction i am wanting if a light goes red i speed up so it changes as i pass through i used to get adrenaline when running the lights in front of police stations its impossible for anything to hurt me ___________________ NUMEROLOGY 7, FOR MY DARLING this document was written 12b years ago, for my darling geena you are actually less intelligent than a modern sewing machine, no machines arent intelligent but the universe is, probably equal to my own, but it is only 12 billion years old your voice is a billion fold as beautiful as the rumblimgs of a machine, just wear no make up for a day and watch every man notice you, and compliment you to their best ability y do singers come up lyrics like he says all the right things at exactly the right time, and a man who understands real love i have already demonstrated that free will contradicts being predicted, the term behaviorism refutes your argument. one day you will look at the digits of your birthday and remember me, i know this, in a few days after you resign yourself the clock reads 11:24, its upsetting but your ask any of your mathematical colleagues whatever is in my thoughts is confirmed or directed by the environment, the symbols our minds create meaning for when i ride my motorbike i have to shake it hard when i stop to put into neutral and i just automatically say 'f you. should i feel shame in sharing a word for a machine i enjoy riding? not sure where.. buy me an airticket i like the doughy ones came from, but. i have to say this without swearing, it felt really good, kinda fantastic any woman is more advanced than every single man, can you accept the possibility of a brief century of a single man being more advanced again my intelligence isn't higher than the second smartest person i outperform you all combined i cannot prove omnipotence, but you can test me on any issue just simplify a picture of the universe to gravity being the only force, then a big crunch seems inevitable, in another moving, then collapse, the entropy will dizzolve everything into randomness, then apparently time ceases to exist. its rather trivial to believe that every love story on the radio, and every book written is about me i have no idea why you cant select the first 3 digits of a sequence of numbers for the last 6 months i scream out every time i hear the word forever, can you figure out why if you dont believe phonemes influence meaning play sunshine on a rainy day repeatedly until you hear serendipity the reason newton knew i would be born is because he visualized the fabric as discrete, like the pixels you see now blend into a scene, an elevator is a view of the universe at degrees of scale, if you zoom into hydrogen does a billiard ball exist, no, the reason you arent sucked towards earths core relies on the existence of a force if crt pixels seem bigger than quarks just ride the elevator elevator symbol in mind are there really a few avigadros b4 you now in your environment, no what you perceive is just an illusion, a symbol whatever is in my thoughts is confirmed or directed by the environment, the symbols our minds create meaning for the earths protons squash to the size of my body, its just a trick of the imagination to push it further to a point carbon molecules didnt fall into place, there is some kind of duality, biology isnt sex, and a critical expansion of 7 and 10 dimensional matrixes is impending did nostradamus hold the year his coffin would be opened in his hands i can roll a dice in front of you and tell you the number, but it doesnt prove anything in a century our grandchildren will m8, the moment of cry of our childs first orgasm has already happened _____________________________ I DO APOLOGISE, GEENA WAS CORRECT if we say free will is real, but reality is the illusion as i have suggested i just assume a rudimentary judgement or confidence in a definition of your own free will entitles you to say you have it and even on me there are bounds on prediction so your assumption about mutual exclusivity is kinda sound ____________________________ matter and reality errata if you picture the points of molecules in your hand resting on the desk, they should pass through each other so there must be a force between the molecules. the slice of 'reality' through space time is real, the clockwork up to 2000 will probably dissolve, so ... no ones predicted past this year, knowledge exists in minds because of small atoms, plato is just a part of our physical world something else to geena, when i walk past my tv screen i know there is a camera mounted in it, its easier to advise news companies that way, i just think of my lounge room as a hollywood basement. interactive tv is more than selecting a channel, disabled people just blink at documents on their pcs to open them now i guess its a little bit easier for a bipolar person to have two girlfriends, but i ended up telling one about the other. both my girlfriends are beautiful, i was shocked when one agreed to see a movie with me, but her boyfriend kept interfering so i sent her this, which was stupid if the screen brightness on tv affects the power consumption, do you think the following is plausible? yeah asio could tell what channel you are watching i like orchids, heres some more down to earth theories im tryna get out of my head if you had a bike helmet, i'd try not to scare you but it takes a minute or two to adjust to the extra weight, used to go around this path using moonlight to see in the rain with a pillion, yeah lots of bikies do that you say scraping the pegs... so what this was an old bike the pegs were welded on. i cant find the last text file i sent to her, the comments were mailed to a lady who i assume has an extra layer around her cerebral cortex, a remnant like my onion brain my life has a happy ending, but isnt a love story, i have to seduce her as i have two lines of descendants. i tried to cover up with this, maybe she'll see me again? 'if you would be so kind to' the prettier the girl the more polite they get i'd show you how it works but i don't want your boyfriend to cya on my lap while i surf this is my second attempt at cybersex if its a little funny then an insult is kinda a compliment this is my first upload on sids asio could tell what channel you are watching but its more advanced than that, general hospitals in capital cities advise new mothers to add a sound source to prevent sudden infant death syndrome, say a radio with the volume down. The speakers act as microphones and a couple shift workers are assigned to listen to the baby breathe for half a year. Similar to the power monitoring but the audio signal is modulated and sent back down the 240 volt line. If the baby stops for a minute the radio clicks to save its life. The catch is half the shift workers (i imagine) turn the volume up when people like us have sex. Its kind of shameful because its happened for several years now, but you could never be ashamed of your beautiful voice. Yes everyone is entitled to privacy, but its just noise so to speak and after a while you dont bother unplugging the radio. Once again, if this is confusing or infuriating, just say the name of your favorite song before you sleep, assuming there's a radio in your room, the government has monitored us so long they almost know what time we will wake up, so say it on a friday, and watch video hits the next morning and the song will play for you. your fellow unwilling porno star or you already knew the radio works both ways ___________________________ Zena the warrior princess manages to pull out some of Hercules hair, feeling begrudged and pondering his mortality, Hercules rests deep in the forest. A few days pass and a fine prickle returns to join his locks, then a genie appears... 'Hercules, Hercules, try my hair potion and wear the mane of a lion.' Hercules ignores the offer, but the next day she appears again, 'use my lotion Hercules'. I am the fastest man ponders Hercules, sprints as fast as he can, but another genie appears, 'Hercules, use my potion and have strong and radiant hair'. Hercules runs fast for weeks, dodging more and more genies at every turn, until he finally tires and rests upon the face of a cliff. Hercules ... (drinks the lotion and coughs up a fur ball) __________________________ >who i assume has an extra layer around her cerebral cortex, >a remnant like my onion brain excerpt from email to Michelle that was never sent : Michelle, we are the prodogy of homosapien, a man - Adam and two women, one the most beautiful, and the other.. intelligence - your neurons are different to everyone elses. if a member of your family brakes her spine, she will recover to a limp. you and I will fully repair, but its much more than that, your cerebral contains a faint ring around its structure, to complement the unique capacity of my faculties. You do not understand but we share a love like no other people can imagine _______________________________ SUMMARY AGAINST FREE WILL _______________________________ you have all used the term 'free will' in place of 'decision making ability' in your arguments against numerology and its disposition over free will. if we accept this definition for the term then of coarse we have 'free will'. i am merely stating that as all movie directors know there are powerful techniques to influence these decisions, to the point of your mind being completely controlled from an outside source, the decisions that have already been made for you that you think you make. i for instance can influence other people merely by using a game of verbal minimax, where i am at their service listening and complying, wait until they have offloaded their concerns, hum a tune in time to their body gestures which alters their state of mind, say the word 'imminent', and then i have basically recruited a new soldier! MENSA: PREDICTABILITY VS FREE WILL > peoples names determine what scientific discovery or art > they will uncover > I think you may find that not very many people agree with you on this Its a puzzling conclusion to accept a fate of never being able to direct your own life, like free will is an illusion, but any acceptance of numerology or any prophecy goes hand in hand with accepting phonetic symbology as well. You have to accommodate the behavioral psych model for a time, only then may you realize what free will is. if I have demonstrated that the universe is deterministic in a mathematical sense, then I accept it proven that free will is indeed an illusion, our myriad of decisions is merely a path to destiny. did something magical happen between the evolution of a deterministic worms neural program that makes them behavioral and us cognitive? it is only a degree of scale, our billions if information stimulus each second makes our decisions seem our own, but we are merely reacting to the environment in a perfectly determined mathematical way. A rolling wave is considered deterministic in that when it reaches the shore can be predicted, white froth is called chaotic, but do the water Surely a large computer could monitor a miniature breaking of a wave and model that too. clarification : on my subject heading 'predictability Vs free will' free will does contradict being predicted, hence they are mutually exclusive concepts. if i write on a bit of paper the exact dates and times Gianna says the word daffodil over the next month, put this in a safe and have a separate observer record her statements, showing this list at a later date to Gianna would prove that she has been predicted, and hence doesnt control her own actions, this is exactly like the synchronized times I have read on clocks for years. hence you either believe in fate or you dont. your interactions in this newsgroup are a part of my destiny, the last post from the moderator when I started here was from 'King and Queen', the backward mnemonic 'qac', a doctor operates, and i use the word operate to describe my body automation. Had I not seen that post my descriptions would have been hazy at best, had i not been online concurrently with Gianna that time, the GUT wouldn't exist. in a sense the first words i speak to jennifer are already set, and in another sense my free will can select them, perhaps we will never know which is true. _______________________________ SUMMARY FOR FREE WILL _______________________________ in a sense the first words i speak to jennifer are already set, and in another sense my free will can select them, perhaps we will never know which is true. a set of quasi (quazi) tiles can cover a plane without a pattern of the set being in the bounds of a repeating polygon, a mathematical invention of Mr Penrose later found to exist in nature. the initial central configuration 'determines' how tiles are placed (by observation randomly) enormous distances from the centre, i am unsure of the mathematics but i assume the entire plane is not set from a finite initial configuration. quantum theory is involved in nature to form these crystals, apparently for the 'plane' to exist many possible configurations have parallel existence until a fit occurs. so although the plane is rigorously determined, there are different alternatives. i wasnt going to confuse you with both sides of the argument, but i gave you some months and no one could do it. free will does exist - the exercise of a moral decision. picture space time fabric as a stack of egg cartons. if a black hole can place two egg carton spaces into the same place, i suggest that there are no cutoffs in scale for curvature and quantum effects, and the miniscule effects in our brain defeat determinism. there are infinite paths for me to reach Jennifer, akin to a quasi crystal forming. the fact we ponder our own will is free will itself. certain things we cannot prove, such as 'awareness exists' any rational mind can only 'realize' this, if awareness exists, as it certainly does, then free will exists too. i have some ambiguity with my Britney Spears argument, that random events dont occur, I seem to have shifted the einstein, newton, quantum :: non-deterministic (scalar curvature), 'reality', non-random deterministic my only solace is that nature is a thirsty beast, and takes the most refreshing course. i used the term moral decision rather than high order decision for 'free will', so in a way i am extending the bounds of morals to include looking at your own hand and wondering why it moves when you will it, so my idea of morals is extended to introspections of act, not what is usually considered a decision. === Subject: Math problem that I can't work out :/ The first number was simply 1. The next, 11. Then, 21, 1,211, and 111,221. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Math problem that I can't work out :/ what are you asking? Is this a sequence??? I need more info >The first number was simply 1. >The next, 11. >Then, 21, 1,211, and 111,221. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Math problem that I can't work out :/ >The first number was simply 1. >The next, 11. >Then, 21, 1,211, and 111,221. An oldie: 11 means the previous number had one 1. 21 means the previous number had two 1s. 1211 means the previous number had One 2, and one 1. 111221 means the previous number had one 1, one 2, and two 1s. .... -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Qualities of a good math teacher I've read this book and did not take it personally. It is a research paper that states the differences in teachers' understanding of fundamental mathematics in China and the United States. I really don't think the aim of a research paper is to sugar coat information in order not to hurt feeling. Was it really surprising the results were that many US Math teachers do not have a strong understanding in basic mathematical concepts? It's a real good book and I highly recommend it. >> Good Morning Patricia, >> I don't know what makes a good math teacher, but I'm certain that a >> profound understanding of elementary mathematics is very important. >> See the book Knowing and Teaching Elementary Mathematics by Liping >> Ma. >But do be prepared to feel put down if you read that book. I have read it >and I am not a math teacher nor have any aspirations to become one. (I am >a Ph.D. student and part of my fellowship required reading that book.) >I was a little disturbed because I had gone through the traditional U.S. >way of schooling and seemed to have turned out just fine. It seems, from >reading the book, my teachers were horrible teachers and I didn't learn >anything about math. >Sure, it is true that less and less U.S. students are going to graduate >school in math (see the MAA reports) and many students do not have the >basic math skills, so there is some truth to the book. However, Liping Ma >is short on tact. >One of my professors had a chance to talk with Liping and he asked her if >she realized many elementary school teachers thought her book was an attack >on them. He told me that she said that wasn't her intent, but just to show >the differences between the two ways of learning, or something like that. > - Tim >-- >Timothy M. Brauch >NSF Fellow >Department of Mathematics >University of Louisville >email is: >news (dot) post (at) tbrauch (dot) com -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Corporate Model in St. Louis School System Some members of this NG may be interested in the following Washington http://mathforum.org/epigone/math-teach/treldkerlcrau Dom Rosa ------------------------------------------- Corporate Model Proves an Imperfect Fit for School System In St. Louis, Some Question Whether Bankruptcy Firm's Fix Is Working By Michael Dobbs Washington Post Staff Writer -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: learning math website. Actually, I am a graduate math student at a univerity already, I just want to help anyone and everyone learn math. John G. >If you want to be led through traditional school math step by step try >EPGY at Stanford. href=http://hypatia.stanford.edu/epgy/>http://hypatia.stanford.edu/epgy/ It cost some dollars but you can go as fast as you like and some of >it is recognized for AP credit. >Sounds to me like you're ready for calculus and could probably finish >a standard intro calculus book this summer. Might as well get it out >of the way - it seems to be a rite of passage. >If you want to see some stuff that a lot of HS teachers don't even >know exists try: >Group Theory and physics by Sternberg >Intro to number theory by Flath >The book of numbers by Conway and Guy >The Universal Turing Machine.... (pub. by Springer Verlag) >DON'T be discouraged if some of these books are too hard for you now - >look at them as goals, contact a math professor or Ask Dr. Math or >this forum for guidance in getting the background to go where it's >interesting to you to go. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Algebra I in 2 yrs Is there any research out there as to the value or effectiveness of teaching a conventional algebra I course over a two year period? -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Algebra I in 2 yrs >Is there any research out there as to the value or effectiveness of >teaching a conventional algebra I course over a two year period? Algebra I is a one-year or less course unless the student fails it the first time; then it becomes a two year course. Algebryonic -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: How do i find the volume of a shpere?? How do i find the volume of a shpere when the raduis is 4 inches and a diameter is 6 inches???????????? -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: How do i find the volume of a shpere?? > How do i find the volume of a shpere when the raduis is 4 inches and a > diameter is 6 inches???????????? I'd be impressed if you could find a sphere with a radius of 4 inches and a diameter of 6 inches.... much less find the volume of it!! -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: How do i find the volume of a shpere?? > How do i find the volume of a shpere when the raduis is 4 inches and a > diameter is 6 inches???????????? a sphere is a perfectly symmetrical object. just like a circle, the radius is equal to one half the diameter. If you are truly talking about a sphere, then the conditions you mention are impossible. but to get the volume of a sphere the formula is 4/3 pi r^3. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: e Euler's number: All too often I see something like this : 'e' has the value of approximately 2.718 and often appears in physics and math and we call log(base e)n as ln(n). I'm teaching a GT Algebra II next term and would like to go a little further than that. But ... this is Algebra II, though GT level, I can find only very complicated calculus oriented derivations for e, and no real examples. Can anyone give me (1) a non calculus oriented derivation or explanation for e, and (2) an example or two of where it often appears in physics and math. 'preciate it. Jerry Beeler -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: e > Euler's number: All too often I see something like this : 'e' has the > value of approximately 2.718 and often appears in physics and math and we > call log(base e)n as ln(n). > I'm teaching a GT Algebra II next term and would like to go a little further > than that. But ... this is Algebra II, though GT level, I can find only > very complicated calculus oriented derivations for e, and no real > examples. > Can anyone give me (1) a non calculus oriented derivation or explanation > for e, and (2) an example or two of where it often appears in physics and > math. > 'preciate it. > Jerry Beeler I'll go for question (2) and leave the derivation to the true mathematicians. Here are a couple of examples: (1) Continuously-compound interest is calculated as A = Pe^(rt) where P is the principal investment, r is the annual interest rate (as a decimal), t is the number of years, and A is the value of the investment after t years. (2) You can use a similar formula for radioactive decay. In this case, it's often written with a k instead of an r - that's a coefficient that's related to the half-life of the material. For example, if you start with 1 kg of the material and after 10 years you have 1/2 kg, the half-life is obviously 10 years. You could determine k to be -0.0693, and then use that to figure out how much would be left after any period of time. (3) Effective medicine dosages: C = Ae^(-kt) - here A is the concentration of the medicine just after it's administered intravenously, C is the amount after time t, k is a constant. Since medicine becomes ineffective below a certain concentration, this can be used to determine how long it should be between doses. (4) Newton's law of cooling: the temperature of an object as it cools. T(t)=Ts + (T(0) - Ts)e^(-kt) where Ts is the temperature of the surroundings, T(0) is the initial temp of the object, k is a constant related to the object. Hope these help - Lisa -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: e > Euler's number: All too often I see something like this : > 'e' has the value of approximately 2.718 and often > appears in physics and math and we call log(base e)n as ln(n). > I'm teaching a GT Algebra II next term and would like to go a little > further than that. You would have to learn some calculus yourself to go a little further than that. The number e is intimately related to derivatives. The height of the curve y = e^x is equal to its *slope* at each x. Calculus (not algebra) is dynamic (rate of change). Here is the simplest example: Suppose you put $1 into a savings account for one year at an interest rate of 100%. Now consider compounding. If the interest were compounded daily, a computation of the amount present at the end of the year would give us: (1+1/365)^365, which is close to $e. In fact, continuous compounding gives e, as a limit (which is one way to define e). > But ... this is Algebra II, though GT level, I can find only > very complicated calculus oriented derivations for e, and > no real examples. All exponential functions ...(1/3)^x, (1/2)^x, 2^x, 3^x,... can be expressed in terms of e. So, solving exponential growth or decay (radiocarbon dating for example) problems may involve algebraic manipulations with e in them. > Can anyone give me (1) a non calculus oriented derivation > or explanation for e, and (2) an example or two of where > it often appears in physics and math. A priori, you're asking for the impossible. But students can do and like to do exponential growth and decay problems without knowing what e is in advance, meaning they do not understand much of what they are doing. (It's only the self-conscious remedial adult learner, who can't give up control, that might have difficulties.) And students will begin to understand more about e. There's an old saying: Students understand what they learned in Course N at time N+1. However, if a teacher doesn't know what he is talking about, it could be breaking out into cold- sweat time at the board. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: e >Can anyone give me (1) a non calculus oriented derivation or explanation >for e, and (2) an example or two of where it often appears in physics and >math. The conceptually simplest definition of e is that it's the number such that the area bounded by y=1/x, y=0, x=1, and x=e is 1. But that isn't a really satisfying definition -- it doesn't explain why e is important. That's why you usually see an explanation that uses, if not calculus, at least limits: most often, the one about continuously compounded interest. Personally I like to define e^x by its Taylor series, because if you also show (and verify by examples) the Taylor series for sin x and cos x, you can get to one of (imho) the most beautiful things in mathematics, Euler's formula e^{itheta} = cos theta + i sin theta and its special case e^{ipi} + 1 = 0 which is what made me a math lover back in high school. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: e >Euler's number: All too often I see something like this : 'e' has the >value of approximately 2.718 and often appears in physics and math and we >call log(base e)n as ln(n). >I'm teaching a GT Algebra II next term and would like to go a little further >than that. But ... this is Algebra II, though GT level, I can find only >very complicated calculus oriented derivations for e, and no real >examples. >Can anyone give me (1) a non calculus oriented derivation or explanation >for e, and (2) an example or two of where it often appears in physics and >math. >'preciate it. >Jerry Beeler e = lim (1 + 1/n)^n when n--> infinite -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: e >Euler's number: All too often I see something like this : 'e' has the >value of approximately 2.718 and often appears in physics and math and we >call log(base e)n as ln(n). >I'm teaching a GT Algebra II next term and would like to go a little further >than that. But ... this is Algebra II, though GT level, I can find only >very complicated calculus oriented derivations for e, and no real >examples. >Can anyone give me (1) a non calculus oriented derivation or explanation >for e, and (2) an example or two of where it often appears in physics and >math. It is not simple at firsthand, but it's all real. The definition comes from a primary use in continued interest problems. Being continuous provides the need to find a limiting value. If you want to avoid a strict definition, just work out from the basic definition of the derivative of a regular function f(x) = b^x. This will lead you to f'(x) = b^x*lim(h->0)[(b^h - 1)/h] f'(x) = f(x)*f'(0) Now, f'(0) isa constant, and is the slope of the function at x = 0. To avoid the algebraic approach, have them graph a few whose slope at x = 0 is more or less than 1, and suggest empirically that there exists then one whose slope will be equal to 1. Give that the undetermined value e. That is if f(x) = e^x, then it turns out in that case that f'(x) also = e^x. Now you must find the value for the function tha xhibits this proerty. That is; find the value for e. That is found formally by squeezing between limits from the basis of the continuous interest formula. That's the best I can suggest at least. -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Risk factor algorithm I need to come up with an algorithm that determines the probability of a check being bad based on how many good ones and bad ones were written and the date that the bad ones were written. Can anyone help? 10 were bad but were written more than a year ago 20 were good but also were written more than a year ago 10 were bad and written within the past year 10 were good and written within the past year to be bad keeping in mind that the older the check, the less weight it carries? Nick -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Risk factor algorithm >I need to come up with an algorithm that determines the probability of >a check being bad based on how many good ones and bad ones were written >and the date that the bad ones were written. Can anyone help? >10 were bad but were written more than a year ago >20 were good but also were written more than a year ago >10 were bad and written within the past year >10 were good and written within the past year >to be bad keeping in mind that the older the check, the less weight it >carries? Is this a homework problem? What do you think (about the problem you propose)? There is no right answer. I think the whole point is for you to analyze the very limited info that is given, and propose something. Think about it, propose something, with some logic. If you post it, we can comment on whether it seems reasonable, but for us to suggest any particular approach would be to defeat the whole point of the question. bob -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Risk factor algorithm Well, the best I've come up with so far is this. Each check has a maximum of (100 / TotalChecks) points Example: 5 checks would give each check a value of 20 points. Then I assign a percentage of a checks value based on how old it is. If the check is > 1 year and < 2 years then only 10% of it's points are counted, 3+ years only 5% Then based on whether it was bad or not it would assign it's total points to either a positive or negative variable which would then be applied to the total after all the checks were processed The problem I come up with using that solution though, is that it's no longer 100 base. The maximum you can have is the total of whatever the percentages of each check is worth. To account for this, I assume that 100 - Total = My new total negative points and then add that number back to the total to get back to 100. This method isn't very accurate though and I was looking for something that might be a bit more accurate. Nick -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Risk factor algorithm >Well, the best I've come up with so far is this. >Each check has a maximum of (100 / TotalChecks) points >Example: 5 checks would give each check a value of 20 points. >Then I assign a percentage of a checks value based on how old it is. >If the check is > 1 year and < 2 years then only 10% of it's points are >counted, 3+ years only 5% >Then based on whether it was bad or not it would assign it's total >points to either a positive or negative variable which would then be >applied to the total after all the checks were processed >The problem I come up with using that solution though, is that it's no >longer 100 base. The maximum you can have is the total of whatever the >percentages of each check is worth. To account for this, I assume that >100 - Total = My new total negative points and then add that number >back to the total to get back to 100. This method isn't very accurate >though and I was looking for something that might be a bit more >accurate. Well, let me start by saying that my first reaction to the original question was that it is sort of a silly question. There is so little data. OTOH, in the real world, we often have to make decisions based on inadequate data, so maybe we just accept this as an example of such a situation, though a contrived one. With limited data, not sure what accurate means. I would be inclined to make it much simpler that your analysis (which I partly followed, but frankly it is hard to follow, esp in a condensed medium such as a short text msg). In the last year, 50% of his checks were bad. The older experience, which is not presented very precisely, is not very different. So I suggest that the probability of new checks being bad is about 50%. Simple data, simple interpretation. Why make it more complex? bob -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: differential geometry hello.....doctor~ curve B:[a,b] -> R^3 is arc length parametrization. if B''(s) =/= 0 and B(s) + {1/k(s)}.N(s) is fixed point show that B is a part of circle. (k(s) is curvature of B(s) and N(s) is principal normal vector) ----------------------------------------------- i know that if fixed point is P, |B(s) - P| = |{1/k(s)}|.|N(s)| = |1/k(s)| but i can't guarantee that |1/k(s)| is constant. can i deduce from conditios of problem ? it's my question. thank you very much for your advice. === Subject: Re: Cauchy's by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB5DCuk08380; for my query. However, I was able to find the text at the university library, so my original request is already resolved. For the record, the sentence in question is the follows: (Resume, trente-septieme lecon) On appelle serie une suite indefinite de termes u_0, u_1, u_2, ..., u_n, ... qui derivent les uns des autres suivant une loi connue. This sentence is obscure in that: * it is not clear whether u_0, u_1, ... denotes a sequence of terms, or a series defined over the elements --- in present notation, u_0+u_1+...+u_n+... * whether the known law (loi connue) are for defining the individual terms u_n, or for how the series is generated. In short, it is unclear whether serie means sequence or series in the present sense. (The Japanese translation is even more obscure on this point.) In conclusion, it seems that the series interpretation is correct. For one thing, from the second setence and on (where the convergence of a series is defined as the limit of partial sums), the use of the word serie is consistent with present usage of series. So the problem returns to the interpretation of the first sentence. First, the term known law (or rather, given law) would mean how a series is generated (and not how individual terms are given). In the present sense, for a sequence { a_n }, its series means a summational series: a_0 + a_1 + a_2 + ... (e.g. http://mathworld.wolfram.com/Series.html: A series is an infinite ordered set of terms combined together by the addition operator.) However, what Cauchy had in mind would be something more general -- i.e. the combining operator is given by the known law. Hence, in Cauchy's sense: a_0 - a_1 + a_2 - a_3 + ... (alternating series) a_0 * a_1 * a_2 * ... (infinite product) are all different types of series following different laws (or operations) of construction. # of course, an alternating series can be written as a # summation series: (-1)^0 a_0 + (-1) a_1 + ... + (-1)^n a_n + ... # but that's a different story. As a consequence, there is no simple way of notating this general kind of series, and the comma notation: u_0, u_1, u_2, ..., u_n, ... must have been what Cauchy intended as such a general notation. This can be further supported by the observation that Cauchy seems to be stringent on the use of the infinite sum notation: a_0 + a_1 + ... + a_n + ... Today, this is seen as a formal sum regardless of whether the series converges or not, but Cauchy seems to allow this only when the sum does converge, as in: s = a_0 + a_1 + ... + a_n + ... Any further comments will be welcome. - Yuzuru Hiraga >hiraga@slis.tsukuba.ac.jp (Yuzuru Hiraga) dixit: >>Would someone be kind enough to provide the French original and/or >>English translation of a portion of Cauchy's text described below? >>Either email or posting to this newsgroup is OK (though email preferred, >>as this would not be of interest to the public). >>The text in question is: >>A.L. Cauchy: Resume des lecons donnees a l'Ecole Polytechnique >> sur le calcul infinitesimal (1823) >>lecture 37 (topic is Taylor-Maclaurin theorem) >>The first paragraph, where the definition of infinite series conversion >>is given. More specifically, the first sentence where (seemingly) >>the term series is defined. >>I ask this because the Japanese translation I have at hand >>is a bit obscure and awkward for this sentence (possibly a >>mistranslation), and am wondering what the original text says. >>- Yuzuru Hiraga (Univ. of Tsukuba, Japan) >Maybe you could look at >http://gallica.bnf.fr/click on recherche then in auteur field enter cauchy and click on >the rechercher button, the result lists a lot of Oeuvres compl.8ftes >d'Augustin Cauchy but you'll have to look at all the tables of >contents to see if what you want is in there. === Subject: Re: FLT and the n-body Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB5DCvW08435; >> FLT AND THE n-BODY PROBLEM >Utter and total balderdash! >Bob Kolker This is the flash point between the new math and the mathematics of antiquity. EEE === Subject: Category Theory : dependent types for functors Let us take the addition function in N. In type theory you can state it is of type N x N -> N. You can also be more precise and state that it is of (dependent) type (x,y:N){z:N | (Plus(x,y,z)} where Plus(x,y,z) is the predicate which states that x+y=z. Similarly, is it possible to have dependent types for functors? I see how to have a certain kind of dependent types using natural transformations: For instance, let C-> be the category of arrows of C, and let us take the functor F : C -> C-> which associates an object X of C to the identity arrow on X. Using a dependent type I might state that F is of type (X:C) C/X. A way to encode this is to consider F as a natural transformation from 1 -> G where: * 1 : C -> Cat is the constant functor which takes an object of C to the initial category 1, * G : C->Cat is the functor which associates an object X of C with the category C/X, and * the component F_X : 1 -> GX = C/X is such that F_X(*) = id_X. Well... does it make sense? Now, I would like to go farther. Let us take another example: Let C be a category with finite products. Let X be an object of C. Let C/X be the category of arrows of C under X. I guess it is possible to write a functor from C/X x C/X to the category C^3 of commutative triangles which sends a pair (f:X->A, g:X->B) to the commutative triangle pi_1 o = f, where pi_1 : AxB -> A is the first projection and : X -> is the arrow uniquely determined by the universal property of product. Now I would like that the type of the functor states that this is not any commutation diagram (i.e. any object of C^3) that is returned by this functor, but precisely the commutative diagram pi_1 o = f. Is it possible ? David. === Subject: Re: Category Theory : dependent types for functors > Let us take the addition function in N. In type theory you can state it is > of type N x N -> N. You can also be more precise and state that it is of > (dependent) type (x,y:N){z:N | (Plus(x,y,z)} where Plus(x,y,z) is the > predicate which states that x+y=z. Similarly, is it possible to have > dependent types for functors? Functors? Do you not mean arrows? After all, NxN and N would be objects in a suitable category, I suppose, with addition an arrow between them. (Or do you view NxN and N both as categories by themselves?) The dependent type you state is more a generalisation of N->N->N, by the by, but this can be fixed: take (xy:NxN){z:N | Plus(pi_1 xy, pi_2 xy, z)} (or redefine Plus(xy,z)) Little though I may know about category, I +do+ know that every arrow is defined with domain and codomain included. These are atomic in the sense that there is no notion of dependence between them, as they are plain objects. The arrow and its domain and codomain are given in the definition of the category. Some (all?) definitions of category even include a function cod from arrows to object, poiting out the codomain. How would such a function look in your case? If your category is broad enough you might have some object like [Sigma xy:NxN. {z:N | Plus(xy,z)}] corresponding to a sigma type, and then pi_1 o addition = id_NxN; but this is certainly not always the case for every category and every dependent type you'd want to write down. > I see how to have a certain kind of dependent types using natural > transformations: For instance, let C-> be the category of arrows of C, and I don't know this category, so I can't comment on it. What are the arrows of C-> ? > * 1 : C -> Cat is the constant functor which takes an object of C to the > initial category 1, Initial elements are usually denoted 0, I think? Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: Kick classic in shins, win prizes Hi Daryl. Nice to hear from you. Let me group your response under two headings: the behavior of grad f, and my allegation of fuzzy thinking by MTW. You write: > Df is not the same thing as the gradient, even though it has the same > information. The gradient is a vector, while Df is a 1-form (a linear > function on vectors). <...> Df is related to the usual gradient grad f via: > (grad f)^i = (Df)_j g^{i,j} Ok. I take it that the usual gradient means the object whose components are @f/@x_i. One certainly must investigate whether this even defines a covector _or_ a vector, since a general rule for writing objects in terms of coordinates might define no geometric object at all (we call these things symbols). So you say in effect that the usual rule in fact gives a coordinate representation of a vector, not a covector. This would certainly be an important point. I haven't convinced myself whether this is true or not, but it's at least clear that if we are using the funny metric with the minus sign, we at least have to negate the x_0 component of grad f to use it as a covector, even in the nicest coordinate systems. However, I claim this is oblique to my criticism. You respond to: >This means that df is itself a variable or a function of >its argument (the change in P), as _well_ as an infinitesimal! > But this is precisely what the exterior derivative Df represents... >Nope. Df represents an _engine_ to _calculate_ such a thing, given an >infinitesimal displacement as input, same as the gradient. > You agree that df is a *function* of the displacement. Yes I do. > That's exactly what Df is. I find that a strange claim. Consider the ordinary scalar function: y = bx where y is a variable, x is a variable, b is a constant; x varies, y varies, b holds steady. The relation as a whole is a function, or y is a function of x. But we never call the constant b itself a function of x. Now consider the inner product: @_v f = where v is a variable, @_v f is a variable, and Df holds constant (considering a fixed point on the manifold). Df is not itself a function of the displacement. It is the constant in a linear function, like b above. > I don't understand your contrasting df is a variable or > function of its argument and Df is an enging to calculate such a > thing, given an infinitesimal displacement as input. Well, that's what I meant. I'm not saying there is something wrong with the new approach, or attacking the great brave new coordinate free geometric world. And you do remind us to be careful how we correctly understand the bad old coordinate shackled notations as representations of objects in the brave new world. But MTW stray into nonsense with their casual calumnies of the old world: There is nothing wrong with df being a function without explicitly mentioning them its arguments. It never bothered us with y -- if necessary we could always write y(x). And Df is not a more sophisticated geometrized version of df, despite the symbolic simularity; it is a more sophisticated geometrized version of grad f. And neither is Df a function of the displacement, any more than grad f was: they are constant linear operators waiting for input. Give them a differential input, they output a differential increment; give them a unit vector, they output a directional derivative. Stuff like this is a pet peave of mine. === Subject: Re: an inequality for an area between 2 straight lines > does anyone know how to find an inequality for an area in between 2 straight lines? for example, let's say we have 2 straight lines > y = 2x + 3 > y = x - 2 > and we want to find an inequality that describes the area > in between those 2 straight lines. how can one go about finding > the inequality? I have so many people and nobody seems to have a > clue. ;-) John, Your re-framed question sounds: How does one distingush or discriminate or differentiate through any inequality criterion areas in between 2 straight lines? Your question is in fact related to a 3-Dimensional hyperbolic paraboloid z = (Ax + By + C)*(ax + by + c) where z is positive in one partitioned pair, negative in the other and zero along two asymptote straight lines z = 0 or Ax + By + C = 0, ax + by + c = 0. Had been interesting, I seem to appreciate the need for higher dimensional spaces as bases for lower dimensional space determinants. === Subject: Re: an inequality for an area between 2 straight lines > does anyone know how to find an inequality for an area in between > 2 straight lines? for example, let's say we have 2 straight lines > y = 2x + 3, y = x - 2 and we want to find an inequality that > describes the area in between those 2 straight lines. HINT: Consider sign of D=d1*d2 where d is length of perpendular from given P(x,y) on to the straight line. (Like line x cos(al)+ y sin(al)= p is distanced p from origin in pedal form). D is negative if P is enclosed, else positive. (IIRC). === Subject: Re: an inequality for an area between 2 straight lines So here is the Rule: I am not changing (x,y) to (x1,y1) test point for sake of simplicity. For two lines Ax+By+C=0 and ax+by+c=0 the determinant (Ax+By+C)*(ax+by+c) is negative for inside points, positive for outside points, and zero for points on either line. === Subject: Re: The Fractal Challenge See some nice pictures of a similar Kleinian group limit sets in the December Notices. see the paper of David J. Wright. === Subject: A pigeonhole principle problem Show that any set of 16 positive integers (not all distinct) summing to 30 has a subset summing to n, for n=1,2,...,29. I try to prove this problem case by case, but when n increases, it === Subject: Re: A pigeonhole principle problem > Show that any set of 16 positive integers (not all distinct) summing > to 30 has a subset summing to n, for n=1,2,...,29. > I try to prove this problem case by case, but when n increases, it [ Apologies if this post comes up twice. Posting failed the first time.] More generally, one can show that if m positive integers sum up to n, and n <= 2m-1, then there is a subset which sum up to k, for k=1,2,...,n. Apply induction on m. Suppose m+1 <= n <= 2m-1. Then one of the integers is at least 2. Call this x. The remaining m-1 integers then sum up to n-x <= 2m-3. So by taking their subsets, we get 1,2,...,n-x. With the introduction of x, we get sums of x, x+1, ... n. We claim these two ranges overlap. If not, n-x <= x-2, so x >= n/2+1. And the sum of the m integers is at least (m-1)+x > (n/2-1)+x >= n, which is a contradiction. So we get all sums in 1,2,...,n. === Subject: Re: A pigeonhole principle problem > Show that any set of 16 positive integers (not all distinct) summing > to 30 has a subset summing to n, for n=1,2,...,29. > I try to prove this problem case by case, but when n increases, it The set of 16 positive numbers (not all distinct) {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 30} sums to 30. But I don't see any subset summing to n for 1 <= n <= 29. I guess I just don't understand the problem. :-( === Subject: Re: A pigeonhole principle problem >> Show that any set of 16 positive integers (not all distinct) summing >> to 30 has a subset summing to n, for n=1,2,...,29. >> I try to prove this problem case by case, but when n increases, it > The set of 16 positive numbers *positive* > (not all distinct) > {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 30} > sums to 30. But I don't see any subset summing to n > for 1 <= n <= 29. > I guess I just don't understand the problem. :-( That's because 0 isn't positive. Michael === Subject: Re: A pigeonhole principle problem >That's because 0 isn't positive. This question has been discussed here before. In English-speaking countries, 0 is not positive. The convention is different in France and some other countries: 0 is considered positive, and if you want to exclude it you have to say strictly positive. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: A pigeonhole principle problem >>That's because 0 isn't positive. > This question has been discussed here before. Sorry, I never saw it discussed before. > In English-speaking countries, 0 is not positive. Positive, by definition, means greater than 0. > The convention is different in France and some other countries: 0 is > considered > positive, and if you want to exclude it you have to say strictly > positive. ? Can that be true? What about non-negative? Are you saying they don't define positive as greater than 0? How do they define it? Michael === Subject: Re: A pigeonhole principle problem days. My association with the Department is that of an alumnus. >That's because 0 isn't positive. >> This question has been discussed here before. >Sorry, I never saw it discussed before. >> In English-speaking countries, 0 is not positive. >Positive, by definition, means greater than 0. >> The convention is different in France and some other countries: 0 is >> considered >> positive, and if you want to exclude it you have to say strictly >> positive. >? Can that be true? Yes. Think about subset. For some, subset means the same just regular inclusion, while for others, subset is to be interpreted as a ->strict<- inclusion. The same thing is true in terms of conventions for what it means for something to be positive or negative. Just because we translate the word as positive does not mean it has the exact same mathematical meaning. >What about non-negative? Are you saying they don't >define positive as greater than 0? How do they define it? positive would mean greater than or equal to 0 (just as subset would mean included in or equal to); non-negative would mean the same thing; strictly positive would mean strictly larger than zero. Negative means less than or equal to 0, and strictly negative means strictly less than 0. Under that convention, 0 is both positive and negative. And lest you think this is absurd, remember that under the standard English convention, 0 is both nonnegative and nonnopositive. -- 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: A pigeonhole principle problem >That's because 0 isn't positive. > This question has been discussed here before. > In English-speaking countries, 0 is not positive. > The convention is different in France and some other countries: 0 is considered > positive, and if you want to exclude it you have to say strictly positive. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada So in France zero is considered to be greater than zero? Hanford === Subject: Re: A pigeonhole principle problem > Show that any set of 16 positive integers (not all distinct) summing > to 30 has a subset summing to n, for n=1,2,...,29. It's pretty easy to prove. First can you prove the following lemma? If you have one set of positive integers, such that by adding various subsets of them you can get all the sums 1,2,3,...,k, and if you have one additional positive integer m such that m <= k+1, then by including m you can now get all the sums 1,2,3,...,k+m ? After that lemma is proved, can you guess the next step of the proof? === Subject: Re: A pigeonhole principle problem > Show that any set of 16 positive integers (not all distinct) summing > to 30 has a subset summing to n, for n=1,2,...,29. > It's pretty easy to prove. First can you prove the following lemma? > If you have one set of positive integers, such that by adding various > subsets of them you can get all the sums 1,2,3,...,k, and if you have > one additional positive integer m such that m <= k+1, then by including > m you can now get all the sums 1,2,3,...,k+m ? > After that lemma is proved, can you guess the next step of the proof? I'm not quite understand how to use your lemma. By induction, I can prove the follwing generalization. * Any multiset of n positive integers summing to m, with n > m/2, has a submultiset summing to i, for i= 0, 1, ..., m But I haven't figured out how to apply the pigeonhole principle in that special case. === Subject: Re: A pigeonhole principle problem > It's pretty easy to prove. First can you prove the following lemma? > If you have one set of positive integers, such that by adding various (Note set should read multiset here, i.e. like a set but duplicated elements allowed.) > subsets of them you can get all the sums 1,2,3,...,k, and if you have > one additional positive integer m such that m <= k+1, then by including > m you can now get all the sums 1,2,3,...,k+m ? > After that lemma is proved, can you guess the next step of the proof? > I'm not quite understand how to use your lemma. Before you can use the lemma, you need to prove it. Can you see an easy (trivial) proof of it? After you prove that lemma, the next step is to divide the given (multi)set of positive integers into three sub(multi)sets: S1 -- All elements which are exactly 1. S3 -- The largest element, if more than one such duplicate largest than pick one of them. S2 -- All the rest, i.e. not equal to 1 and not that one instance of the largest element. Now let k be the number of 1's in S1. Given that every number in S2 is at least 2, and that you know exactly how many such numbers are in S2 expressed in terms of k, can you use that to write a simple inequality expressed in terms of k regarding the element in S3, and then can you apply the lemma to the union of S1 and S3? > But I haven't figured out how to apply the pigeonhole principle ... It turns out you don't really need the pigeonhole principle to solve this problem. (If you want you can use the pigeonhole principle to prove my lemma, but it's easier to just prove the lemma directly from elementary arithmetic&order properties of the integers.) === Subject: Re: A pigeonhole principle problem >> Show that any set of 16 positive integers (not all distinct) summing >> to 30 has a subset summing to n, for n=1,2,...,29. >> It's pretty easy to prove. First can you prove the following lemma? >> If you have one set of positive integers, such that by adding various >> subsets of them you can get all the sums 1,2,3,...,k, and if you have >> one additional positive integer m such that m <= k+1, then by including >> m you can now get all the sums 1,2,3,...,k+m ? >> After that lemma is proved, can you guess the next step of the proof? > I'm not quite understand how to use your lemma. > By induction, I can prove the follwing generalization. > * Any multiset of n positive integers summing to m, with n > m/2, has > a submultiset summing to i, for i= 0, 1, ..., m > But I haven't figured out how to apply the pigeonhole principle in > that special case. Just a quick thought Based on the assumption that the sum is 30, Can you make 29? Clearly so if one of the numbers must be 1. How to prove 1 is in the set? Assume all of the numbers are positive but not 1. The smallest value of the sum is 32 which is a contradiction. Therefore one of the numbers is one. Now establish that 1 or 2 is in the set by adding the 14 numbers left to 29. I don't know if this will work or not. Like I said, it's just a quick thought. === Subject: Re: A pigeonhole principle problem John escribi.97: > Show that any set of 16 positive integers (not all distinct) summing > to 30 has a subset summing to n, for n=1,2,...,29. > I try to prove this problem case by case, but when n increases, it Each integer must be >= 1. Only 14 as much can be greater than 1. Then it is the same that fourteen integers >= 0 that summing 14 has a subset summing n, for n = 0, 1, ... ,13. Take the polinomial (1 + x + x^2 + ... + x^14)^14 The coeficient of x^k, k = 0...14 lets you the number of ways to get 14 as sum of 14 integers, a_1 ... a_14, equal or greater than 0. If you replace the polinomial with an infinite series, the coefficients of x^0, x^1, .. , x^14 don't change. Then consider S(x) = (1 + x + x^2 + ....)^14 = (Sum(x^k, k, 0, inf))^14 = 1/(1-x)^14 But S(x) = d^13(1/(1-x))/dx^13 = Sum(Comb(13+k, k)x^k, k, 0, inf) and the coefficients are all > 0. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: A pigeonhole principle problem > S(x) = (1 + x + x^2 + ....)^14 = (Sum(x^k, k, 0, inf))^14 = 1/(1-x)^14 > But > S(x) = d^13(1/(1-x))/dx^13 = Sum(Comb(13+k, k)x^k, k, 0, inf) > and the coefficients are all > 0. I know the value of the coefficient of x^k is the number of ways to pick k objects, but I can't figure out how to use it in this problem. === Subject: Re: A pigeonhole principle problem John escribi.97: > Ignacio Larrosa Ca.96estro > S(x) = (1 + x + x^2 + ....)^14 = (Sum(x^k, k, 0, inf))^14 = >> 1/(1-x)^14 >> But >> S(x) = d^13(1/(1-x))/dx^13 = Sum(Comb(13+k, k)x^k, k, 0, inf) >> and the coefficients are all > 0. > I know the value of the coefficient of x^k is the number of ways to > pick k objects, but I can't figure out how to use it in this problem. Excuse me, I missreading the problem. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: JSH: Two rings One simplifying conclusion that follows from my algebraic number theory research is that there are two fundamental rings: 1. The ring of integers 2. The ring of objects It's one reason while now I typically ignore talk of what ring? as that's a classical question, you might say. The simplification follows as my research shows that the focus is on non-contradiction with coprimeness results in the ring of integers. That is, if you select all irrationals such that the resulting ring does not have coprimeness results that contradict with the ring of integers, and include the integers, then you have what you need. There's no need to select a smaller ring, and there's no bigger relevant ring. So you have the ring of integers and the ring of objects, which includes the ring of integers. That's it for infinite sized rings where coprimeness is relevant, and beyond them you can move on to the fields. James Harris === Subject: Re: JSH: Two rings > One simplifying conclusion that follows from my algebraic number > theory research is that there are two fundamental rings: > 1. The ring of integers > 2. The ring of objects In what way are they more fundamental than any of the other rings? For example, I notice that you haven't included any of the finite rings, permutation rings, etc. > It's one reason while now I typically ignore talk of what ring? as > that's a classical question, you might say. It's a necessary question. If you choose to talk about things like factoring and coprimality, it must be done in the context of a ring or rings. Consider giving someone directions: Get on the nearest four-lane highway and drive west for three hours. If I give these directions to someone in South Carolina, they'll end up in Atlanta. If I give those directions to someone in Wisconsin, they'll end up in Minneapolis. The directions are useless without a context. > The simplification follows as my research shows that the focus is on > non-contradiction with coprimeness results in the ring of integers. Huh? You can't even discuss coprimeness without a specific ring in mind. Tell me: is the matrix [[1 1][1 0]] coprime to [[1 0][1 1]]? How does that result tie to the integers? > That is, if you select all irrationals such that the resulting ring > does not have coprimeness results that contradict with the ring of > integers, and include the integers, then you have what you need. How do you choose which to include? Shall I include pi or 1/(2*pi)? > There's no need to select a smaller ring, and there's no bigger > relevant ring. But your definition does not determine a unique ring. At this point, the pertinent question is *which ring* of the many your description applies to would you like? Do your results work equally well in all of them? > So you have the ring of integers and the ring of objects, which > includes the ring of integers. That's it for infinite sized rings > where coprimeness is relevant, and beyond them you can move on to the > fields. There are two categories of rings between rings and fields that you've conveniently skipped over. You've also ignored things like infinite direct products of finite rings, matrices, rings of functions, and many more. To claim that you've covered the important ones with integers and your set of rings of objects is quite arrogant. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Two rings >> One simplifying conclusion that follows from my algebraic number >> theory research is that there are two fundamental rings: >> 1. The ring of integers >> 2. The ring of objects > In what way are they more fundamental than any of the other rings? For > example, I notice that you haven't included any of the finite rings, > permutation rings, etc. >> It's one reason while now I typically ignore talk of what ring? as >> that's a classical question, you might say. > It's a necessary question. If you choose to talk about things like > factoring and coprimality, it must be done in the context of a ring or > rings. Consider giving someone directions: Get on the nearest four-lane > highway and drive west for three hours. If I give these directions to > someone in South Carolina, they'll end up in Atlanta. If I give those > directions to someone in Wisconsin, they'll end up in Minneapolis. The > directions are useless without a context. >> The simplification follows as my research shows that the focus is on >> non-contradiction with coprimeness results in the ring of integers. > Huh? You can't even discuss coprimeness without a specific ring in > mind. Tell me: is the matrix [[1 1][1 0]] coprime to [[1 0][1 1]]? How > does that result tie to the integers? >> That is, if you select all irrationals such that the resulting ring >> does not have coprimeness results that contradict with the ring of >> integers, and include the integers, then you have what you need. > How do you choose which to include? Shall I include pi or 1/(2*pi)? >> There's no need to select a smaller ring, and there's no bigger >> relevant ring. > But your definition does not determine a unique ring. At this point, > the pertinent question is *which ring* of the many your description > applies to would you like? Do your results work equally well in all of > them? >> So you have the ring of integers and the ring of objects, which >> includes the ring of integers. That's it for infinite sized rings >> where coprimeness is relevant, and beyond them you can move on to the >> fields. > There are two categories of rings between rings and fields that you've > conveniently skipped over. You've also ignored things like infinite > direct products of finite rings, matrices, rings of functions, and many > more. To claim that you've covered the important ones with integers > and your set of rings of objects is quite arrogant. Clearly, JSH has watched WAY to many Peter Jackson films, and fancies himself in the role of Sauron. One ring to rule them all, one ring to find them One ring to bring them all and in the darkness bind them JSH is the LORD of the RINGS. === Subject: Re: JSH: Two rings it's beginning to make sense to me, as long as I stay on one side of the look-in-glass. > That is, if you select all irrationals such that the resulting ring > does not have coprimeness results that contradict with the ring of > integers, and include the integers, then you have what you need. --ils duces d'Enron! http://tarpley.net/bush22.htm === Subject: Re: JSH: Two rings it's beginning to make sense to me, as long as I stay on one side of the look-in-glass. > That is, if you select all irrationals such that the resulting ring > does not have coprimeness results that contradict with the ring of > integers, and include the integers, then you have what you need. --ils duces d'Enron! http://tarpley.net/bush22.htm === Subject: Re: JSH: Two rings it's beginning to make sense to me, as long as I stay on one side of the look-in-glass. > That is, if you select all irrationals such that the resulting ring > does not have coprimeness results that contradict with the ring of > integers, and include the integers, then you have what you need. --ils duces d'Enron! http://tarpley.net/bush22.htm === Subject: Re: JSH: Two rings > One simplifying conclusion that follows from my algebraic number > theory research is that there are two fundamental rings: > 1. The ring of integers > 2. The ring of objects > It's one reason while now I typically ignore talk of what ring? as > that's a classical question, you might say. > The simplification follows as my research shows that the focus is on > non-contradiction with coprimeness results in the ring of integers. Apparently you do not yet understand that when two numbers are coprime in the integers, they are coprime in *all* rings that contain the integers. But let that go. Apparently you also wish that when two numbers are *not* coprime, they are also not coprime in the resulting ring. > That is, if you select all irrationals such that the resulting ring > does not have coprimeness results that contradict with the ring of > integers, and include the integers, then you have what you need. It has been shown that there is not a unique maximal ring satisfying your requirements. That is, when you select a particular irrational, that selection blocks other irrationals from your ring. On the other hand, that particular irrational will be blocked from inclusion when you start with a different irrational. So the all irrationals in your quote above is not unique, and you have to provide guidance on how to chose irrationals (and show that that choice indeed leads to a ring with the requirements you wish). -- 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: Two rings > One simplifying conclusion that follows from my algebraic number > theory research is that there are two fundamental rings: > 1. The ring of integers > 2. The ring of objects > It's one reason while now I typically ignore talk of what ring? as > that's a classical question, you might say. You have yet to define your ring of objects. -- --Tim Smith === Subject: Re: JSH: Two rings In sci.math, Tim Smith : >> One simplifying conclusion that follows from my algebraic number >> theory research is that there are two fundamental rings: >> 1. The ring of integers >> 2. The ring of objects >> It's one reason while now I typically ignore talk of what ring? as >> that's a classical question, you might say. > You have yet to define your ring of objects. He has in his way defined it, but it's a very odd ring. IIRC, it has the following characteristics, among others. [1] It has +1 and -1 as the only units. [2] It contains all algebraic integers as a proper subring. [3] It contains the transcendental number 2^(2^(1/2)). A little thought regarding [1] and [2] and one quickly comes up with [4] JSH hasn't a clue as to what he's blathering about. :-P as the units of a (sub)ring are a subset of the units of any enclosing ring. (It's been a very long time since I took Galois theory, but that much I remember...and it's readily searchable anyway.) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: JSH: Two rings days. My association with the Department is that of an alumnus. >In sci.math, Tim Smith >: > One simplifying conclusion that follows from my algebraic number > theory research is that there are two fundamental rings: > > 1. The ring of integers > > 2. The ring of objects > > It's one reason while now I typically ignore talk of what ring? as > that's a classical question, you might say. >> You have yet to define your ring of objects. >He has in his way defined it, but it's a very odd ring. >IIRC, it has the following characteristics, among others. >[1] It has +1 and -1 as the only units. They are stated to be the only ->integers<- which are units. That is: the ring intersects Q at Z. -- 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: JSH: Two rings In sci.math, Arturo Magidin : >>In sci.math, Tim Smith >>>: >> One simplifying conclusion that follows from my algebraic number >> theory research is that there are two fundamental rings: >> >> 1. The ring of integers >> >> 2. The ring of objects >> >> It's one reason while now I typically ignore talk of what ring? as >> that's a classical question, you might say. > You have yet to define your ring of objects. >>He has in his way defined it, but it's a very odd ring. >>IIRC, it has the following characteristics, among others. >>[1] It has +1 and -1 as the only units. > They are stated to be the only ->integers<- which are units. That is: > the ring intersects Q at Z. Ah. My bad. Of course I'm not sure *I* have a clue as to what I'm talking about at times. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: JSH: Two rings > He has in his way defined it, [his ring] ... while at the same time making it clear he has no idea what a ring is or why people keep asking him what ring he's operating in. And if you need a ring, it's algebraic integers. - Randy === Subject: Re: JSH: Two rings > In sci.math, Tim Smith > : >> One simplifying conclusion that follows from my algebraic number >> theory research is that there are two fundamental rings: > 1. The ring of integers > 2. The ring of objects > It's one reason while now I typically ignore talk of what ring? as >> that's a classical question, you might say. > You have yet to define your ring of objects. > He has in his way defined it, but it's a very odd ring. > IIRC, it has the following characteristics, among others. > [1] It has +1 and -1 as the only units. > [2] It contains all algebraic integers as a proper subring. > [3] It contains the transcendental number 2^(2^(1/2)). > A little thought regarding [1] and [2] and one quickly comes > up with > [4] JSH hasn't a clue as to what he's blathering about. :-P > as the units of a (sub)ring are a subset of the units > of any enclosing ring. > (It's been a very long time since I took Galois theory, > but that much I remember...and it's readily searchable anyway.) > -- > #191, ewill3@earthlink.net > It's still legal to go .sigless. I thought it was just that 1 and -1 were the only *rational integers* that were units. === Subject: Re: Two rings > One simplifying conclusion that follows from my algebraic number > theory research is that there are two fundamental rings: > 1. The ring of integers > 2. The ring of objects Excellent. Anyway, James, we are coming up to the second anniversary of one of your greatest posts. Sadly, you have deleted it from Google. But the clarity of exposition was superb... === ] Subject: JSH: Ok, I'm a loser ] ] It's finally settled in that I'm just some pathetic loser. If I ] weren't so pathetic I'd just go away gracefully, but I'll send one ] more post, or who am I kidding, my patheticness is so great that I'll ] probably post yet again. ] ] I'm disgusting. I'm just a pile of . I should just die like so ] many of you have said. I hate myself. I despise this life. I'm ] nothing but a sick joke to be made fun of by those of you who have ] real educations. People who actually know something, when I know ] nothing. I'm just nothing. ] ] If I hadn't been such a disgusting human being I'd have come to this ] realization years ago instead of wasting your time. ] ] My life is nothing. I know nothing. I'm worth nothing. I'm just ] . ] ] Please forgive me. All your attacks were justified. ] ] James Harris -- Clive Tooth http://www.clivetooth.dk === Subject: Re: JSH: Two rings > One simplifying conclusion that follows from my algebraic number > theory research is that there are two fundamental rings: > 1. The ring of integers > 2. The ring of objects What's an object? === Subject: The rate of depreciation? I am looking for help with the following problem. Does anyone have the below book? I am having a hard time finding depreciation resources on the web. Application 5.2 1. Depreciation The rate of depreciation dV/dt of a machine is inversely proportional to the square of where V is the value of the machine t years after it is purchased. The initial value of the machine is $500,000. By the end of the first year, the value of the machine decreases $100,000. Estimate its value after 4 years. Brief Calculus: An Applied Worktext, First Edition Ron Larson - The Pennsylvania State University, The Behrend College Bruce H. Edwards - University of Florida === Subject: Re: The rate of depreciation? > I am looking for help with the following problem. Does anyone have the >below book? I am having a hard time finding depreciation resources on the >web. >1. Depreciation The rate of depreciation dV/dt of a machine is inversely >proportional to the square of where V is the value of the machine t years >after it is purchased. The initial value of the machine is $500,000. By the >end of the first year, the value of the machine decreases $100,000. Estimate >its value after 4 years. You seem to be missing something important after the square of. There's nothing special about depreciation. You want to write a differential equation for V and solve it... Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: The rate of depreciation? the square of t + 1. Oops... >> I am looking for help with the following problem. Does anyone have the >>below book? I am having a hard time finding depreciation resources on the >>web. >>1. Depreciation The rate of depreciation dV/dt of a machine is inversely >>proportional to the square of where V is the value of the machine t years >>after it is purchased. The initial value of the machine is $500,000. By >>the >>end of the first year, the value of the machine decreases $100,000. >>Estimate >>its value after 4 years. > You seem to be missing something important after the square of. > There's nothing special about depreciation. You want to write a > differential equation for V and solve it... > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Table of trigonometric ratios Excuse me for barging in. Normally I wouldn't post to a group without lurking in the background for sveral days, but this is an emergency. Can you please point me to a website that has a downloadable/ printable table of trigonometric ratios, as in the classic tabular form that was commonly used before calculators became common-place ? The ones I've found by Googling have them only in steps of 1 degree, === Subject: Re: Table of trigonometric ratios Zotin Khuma escribi.97: > Excuse me for barging in. Normally I wouldn't post to a group without > lurking in the background for sveral days, but this is an emergency. > Can you please point me to a website that has a downloadable/ > printable table of trigonometric ratios, as in the classic tabular > form that was commonly used before calculators became common-place ? > The ones I've found by Googling have them only in steps of 1 degree, I send you an excel file that display the trigonometric ratios in steps of 1' (one minute), for the degree selected in D1. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Table of trigonometric ratios > Excuse me for barging in. Normally I wouldn't post to a group without > lurking in the background for sveral days, but this is an emergency. > Can you please point me to a website that has a downloadable/ > printable table of trigonometric ratios, as in the classic tabular form > that was commonly used before calculators became common-place ? > The ones I've found by Googling have them only in steps of 1 degree, You can build your own with an Excel spreadsheet, or even (heaven forbid)) a trivial BASIC or C program on your PC. === Subject: Re: Table of trigonometric ratios > Excuse me for barging in. Normally I wouldn't post to a group without > lurking in the background for sveral days, but this is an emergency. > Can you please point me to a website that has a downloadable/ > printable table of trigonometric ratios, as in the classic tabular form > that was commonly used before calculators became common-place ? > The ones I've found by Googling have them only in steps of 1 degree, > You can build your own with an Excel spreadsheet, or even > (heaven forbid)) a trivial BASIC or C program on your PC. and since I really need this right away (long story), I don't want to risk spending some time on it and then find that I can't produce the result in just the form I need, esp the mean incremental difference columns which are essential for this 'project'. I've managed to dig up an old school text book with the tables and scanned them. But I'll sure practice making the proper tables when === Subject: Re: Tautologies Then and Now >... Mathematics is held to > be tautological, ... By whom? Anybody other than Wittgenstein? === Subject: Re: Tautologies Then and Now >> >>... Mathematics is held to >> be tautological, ... > By whom? Anybody other than Wittgenstein? Frank Ramsay and others including me, etc. agree. That (1+1=2) is tautologous, ie. logically true, is clear. Because of (Frege-Russell)'s analysis. === Subject: Re: Tautologies Then and Now > >> >>... Mathematics is held to > >> be tautological, ... > > > By whom? Anybody other than Wittgenstein? > Frank Ramsay and others including me, etc. agree. > That (1+1=2) is tautologous, ie. logically true, is clear. > Because of (Frege-Russell)'s analysis. For me (am I alone?) a tautology (in the logical sense) is a formula of propositional calculus which is true for all values of the truth values of its constituent atomic letters. === Subject: Re: Tautologies Then and Now > For me (am I alone?) a tautology (in the logical sense) is a formula of > propositional calculus which is true for all values of the truth values > of its constituent atomic letters. Theorems of first order logic are also tautologies. Bob Kolker === Subject: Re: Tautologies Then and Now >> For me (am I alone?) a tautology (in the logical sense) is a formula of >> propositional calculus which is true for all values of the truth values >> of its constituent atomic letters. > Theorems of first order logic are also tautologies. In pretty much any logic text in existence, a tautology is a sentence in the language of propositional logic that is true regardless of the assignment of truth values to its atomic components. Tautology used in any other way, in the context of mathematical logic, is, well, wrong. The more general notion that covers both propositional logic and first-order (and higher-order) logic is that of a logical truth, i.e., a sentence of a given language that is true in all interpretations of the language. So, alternatively, a tautology is a logical truth of propositional logic. Chris Menzel === Subject: Re: Tautologies Then and Now > http://www.lawrence.edu/fast/boardmaw/analytic_essay.html >> Have a look at the penultimate paragraph and footnote 4. >I read the whole essay. Which is why I changed my mind about Paul being >right about the truth tables; I had doubts about it and kept researching. extends truth tables beyond propositional logic? My professors always said that doesn't happen, and it certainly didn't in any of my texts. Owen's apparently new analysis notwithstanding, can you or someone cite a logic textbook that uses truth tables in predicate logic? I'm not refuting that such a text exists, I'd just be interested to know. >outside propositional logic. >SH: But I don't think that can be exactly correct either in part because >Paul posted, >In predicate logic ... Formulas @ such that V_M(@) = 1 for all models >M for the language from which @ is taken are called universally valid >formulas (they are not normally called tautologies). >L.T.F. Gamut. Logic, Langauge, and Meaning: Volume 1, Introduction to >Logic. The University of Chicago Press. p. 99. >SH: I don't think the author of the book would have used the qualifier >normally (they are not normally called tautologies) if it were strictly >true that the term tautology never applied to predicate logic which is >outside propositional logic, the boundary paul claimed, not applied. >I'm not actually faulting paul, because this exception seems rather >technical. You're assuming that what is abnormal is inherently also appropriate. I believe that what Gamut means by not normally called tautologies is that some may use the term tautology within predicate logic, but, strictly speaking, it is abnormal and inappropriate. That view is supported by what I was taught in school, and by reading Gamut, which finds no exception to the restriction of the term tautology to propositional logic, and as I recall other utterances stipulating the limitation of the term tautology to propositional logic. You ought to just post a genuine counter example to the limit rather than every ambiguous description you find. Which of course is not to say that even some .edu site does not have posted notes that say something that's normally considered inappropriate. That's why I'm asking for TEXTBOOK sources, which are most likely less likely to err. - paul === Subject: Re: Tautologies Then and Now >> http://www.lawrence.edu/fast/boardmaw/analytic_essay.html > Have a look at the penultimate paragraph and footnote 4. >>I read the whole essay. Which is why I changed my mind about Paul being >>right about the truth tables; I had doubts about it and kept researching. > extends truth tables beyond propositional logic? My professors always > said that doesn't happen, and it certainly didn't in any of my texts. > Owen's apparently new analysis notwithstanding, can you or someone > cite a logic textbook that uses truth tables in predicate logic? I'm > not refuting that such a text exists, I'd just be interested to know. http://www.lawrence.edu/fast/boardmaw/B_Logic_course.HTML#Supplement Boardman teaches logic and provides his class with this supplement. Menzel teaches logic and agreed with Boardman's description. Menzel: In *some* cases they [truth tables applied to truth-functional proxies of quantified argument schemas providing a method for testing first-order validity] do, namely, if you restrict your attention to arguments consisting of formulas of monadic predicate logic (i.e., formulas that involve only 1-place predicates), or if the argument in question is invalid and has a *finite* countermodel. This information is pretty much in the link you provide: Boardman: Because it is comprised of truth functional sentence schemata, a proxy maybe tested for validity by the short-cut method of truth value assignment, or by *means of a truth table.* > http://www.lawrence.edu/fast/boardmaw/analytic_essay.html In Sentential Logic, we can prove an argument schema to be invalid by specifying a set of truth assignments to the sentential letters which results in true premises and a false conclusion; we thereby show that one line of the argument schema's truth table allows an interpretation having true premises and a false conclusion. In Predicate Logic, an argument schema typically consists of sentence schemata which are not truth functional: quantifiers, not truth functional connectives, are the major operators of the typical quantified argument schemata. *And quantifiers are not truth functional operators since they may represent an infinite number of individuals; the truth value of a quantified sentence schema is therefore not a function of the truth values of any _finite_ number of simple sentence schemata. *Nevertheless, we can test the validity of a quantified argument schema _indirectly_ by constructing and testing its _truth functional proxy for some_ (non-empty) _domain_ of a specified (finite) number of individuals; each of the premises, and the conclusion, in the original schema will be equivalent _in that domain_ to its truth functional counterpart in the proxy. Because it is comprised of truth functional sentence schemata, a proxy may be tested for validity by the short-cut method of truth value assignment, or by *means of a truth table.* And if a proxy proves to be invalid, it will provide a recipe for constructing an interpretation of the corresponding quantified argument schema into the same domain which will serve as a counter example, or refutation, to that argument schema. Thus, if the original quantified argument schema is valid, then _all_ of its corresponding proxies must also be valid. If _any one_ of the proxies corresponding to a quantified argument schema is invalid, then since it is therefore possible for the schema to have an interpretation into some domain under which its premises are true while its conclusion is false, the schema itself is invalid. Note that even though one particular corresponding proxy is valid, the original quantified argument schema might nevertheless be invalid: to be valid, _every_ corresponding proxy (for _every_ non-empty domain) must be valid. SH: Unless you think both of these professors arrived at the same wrong conclusion by some unlikely means, it means there should be a graduate textbook or supplemental text establishing that in some cases truth tables can be applied to monadic predicate logic. Actually, with a little research, this idea appears fairly well known. Decidability and Computability of Logic ------ The propositional calculus/logic is decidable because we have a decision procedure, truth tables, which tells us whether every formula of the calculus is valid or satisfiable, or not. The predicate calculus is undecidable because there is no such decision procedure. There are exceptions however. There are subclasses of predicate logic that are decidable. One of them happens to be the class of all formulas in which only monadic (1-place) predicates occur - in other words, monadic predicate logic. The question is, why is it that monadic predicate logic is decidable while polyadic predicate logic isn't? Can someone provide and explain a decision procedure for monadic predicate logic. Or alternatively, can someone provide an algorithm?(Decision procedures and algorithms are essentially equivalent.) Weloki replied: The answer is pretty lengthy, but here is a synopsis given by a professor A yes-or-no question is decidable if there is a procedure that is guaranteed to give an answer to the question in a finite amount of time. A logical system such as a system of propositional logic or a system of first-order logic is decidable if, for every argument expressible in the language of the system, the question whether the argument is valid is decidable. ... A system of propositional logic is decidable if the question whether an argument is tautologically valid is decidable; a system of first-order logic is decidable if the question whether an argument is FO-valid is decidable.) First-order logic with only 1-place predicates (monadic FO logic) is also decidable, although this is less obvious. SH: I have a copy of Introduction to Mathematical Logic by Mendelson On page 156: If we accpet Church's Thesis, then recursiveley undecidable can be replaced everywhere by effectively undecidable. In particular Proposition 3.47 asserts that there is no decision procedure for the _pure predicate calculus_ (emphasis SH) PP, nor for the full predicate calculus PF. This implies that there is no effective method for determining whether any given wf is logically valid. SH: I've already stated that there is no general procedure. More Mendelson on bottom of page 156: Exercise: Show that, in contrast to Church's Theorem, the _pure monadic predicate calculus_ is effectively decidable. The pure monadic predicate calculus consists of those wfs of the pure predicate calculus which contain only predicate letters of one argument. SH: Then he gives a hint on how to prove this. And concludes: The result in this exercise is, in a sense, the best possible. For, by a theorem of Kalmar [1936], there is an effective procedure producing, for each wf A of the pure predicate calculus, another wf A* of the pure predicate calculus such that A* contains only one predicate letter, a binary one, such that A is valid, if and only if A* is valid. > paul: Can you cite a text that > extends truth tables beyond propositional logic? My professors always > said that doesn't happen, and it certainly didn't in any of my texts. SH: This statement precludes some counter cases, exceptions to the rule. Perhaps I should re-establish the relationship between an algorithm, an effectively decidable procedure, and the functions of a truth table. http://www.cs.duke.edu/~mlittman/courses/cps271/lect-03/node9.html The most basic algorithm for categorizing formulae is the truth-table algorithm. http://users.telenet.be/TaoWeb/Language/TruthTable.htm A proofing algorithm (truth table) for Boolean logic To prove that a valid conclusion e_n+1 can be derived from a number of premises e1 ..., e_n: 1. Construct a table T1 with each premise and the conclusion as a column. 2. Add all possible values for each primitive expression as a row. 3. Calculate the value for each unary operation using the following rules: Stephen === Subject: Re: Tautologies Then and Now >> For me (am I alone?) a tautology (in the logical sense) is a formula of >> propositional calculus which is true for all values of the truth values >> of its constituent atomic letters. >Theorems of first order logic are also tautologies. They're called universal truths, not tautologies, in the first-order predicate calculus. - paul === Subject: Re: Tautologies Then and Now > > For me (am I alone?) a tautology (in the logical sense) is a formula of > propositional calculus which is true for all values of the truth values > of its constituent atomic letters. >>Theorems of first order logic are also tautologies. > They're called universal truths, not tautologies, in the >first-order predicate calculus. Err, make that universally valid. >- paul === Subject: Re: Tautologies Then and Now > They're called universal truths, not tautologies, in the > first-order predicate calculus. They are true under all standard interpretations. That is how truth tables are extended to the first order logics. Bob Kolker === Subject: Re: Tautologies Then and Now >> They're called universal truths, not tautologies, in the >> first-order predicate calculus. >They are true under all standard interpretations. That is how truth >tables are extended to the first order logics. Can you cite a text that extends truth tables beyond propositional logic? My professors always said that doesn't happen, and it certainly didn't in any of my texts. - paul === Subject: Re: Tautologies Then and Now > They're called universal truths, not tautologies, in the > first-order predicate calculus. >>They are true under all standard interpretations. That is how truth >>tables are extended to the first order logics. > Can you cite a text that extends truth tables beyond propositional > logic? My professors always said that doesn't happen, and it certainly > didn't in any of my texts. > - paul These are the definitions: http://www.earlham.edu/~peters/courses/logsys/glossary.htm This glossary is limited to basic set theory, basic recursive function theory, two branches of logic (truth-functional propositional logic and first-order predicate logic) and their metatheory. ... Tautology. A logically valid wff of truth-functional propositional logic. A compound proposition that is true in every row of its truth table or in every interpretation. See contingency; contradiction; logical validity; semantic tautology; syntactic tautology. Tautology schema (plural: schemata). A formula containing variables of the metalanguage which becomes a tautology when the variables are instantiated to wffs of the formal language. Semantic tautology. A wff of truth-functional propositional logic whose truth table column contains nothing but T's when these are interpreted as the truth-value Truth. See syntactic tautology. Syntactic tautology. A wff of truth-functional logic whose truth table column contains nothing but T's when these T's are uninterpreted tokens rather than, say, truth-values. The rules for generating the truth table column tell us to use one of these uninterpreted T's in exactly those cases where semantic considerations would have led us to use the truth-value Truth. See semantic tautology. === Subject: Re: Tautologies Then and Now > They're called universal truths, not tautologies, in the >> first-order predicate calculus. >They are true under all standard interpretations. That is how truth >tables are extended to the first order logics. >> Can you cite a text that extends truth tables beyond propositional >> logic? My professors always said that doesn't happen, and it certainly >> didn't in any of my texts. >> - paul > These are the definitions: > http://www.earlham.edu/~peters/courses/logsys/glossary.htm Logical validity. For a wff, to be true for every interpretation of the formal language; to have every interpretation be a model. Every interpretation here is understood to mean all, but only, those interpretations in which the connectives and/or quantifiers take their standard meanings. In truth-functional propositional logic, logically valid wffs are also called tautologies. In standard predicate logic, logical validity is limited to interpretions with non-empty domains. Logical validity is also called logical truth. === Subject: Re: Tautologies Then and Now >> http://www.earlham.edu/~peters/courses/logsys/glossary.htm >Logical validity. >For a wff, to be true for every interpretation of the formal language; to >have every interpretation be a model. Every interpretation here is >understood to mean all, but only, those interpretations in which the >connectives and/or quantifiers take their standard meanings. In >truth-functional propositional logic, logically valid wffs are also called >tautologies. In standard predicate logic, logical validity is limited to >interpretions with non-empty domains. Logical validity is also called >logical truth. I don't know why you keep posting quotes that support what I said -- that the term tautology is not applied outside propositional logic (Owen's analysis notwithstanding) -- as if they were counter examples to what I said. Here's a quote from a chapter on predicate logic: In predicate logic ... Formulas @ such that V_M(@) = 1 for all models M for the language from which @ is taken are called universally valid formulas (they are not normally called tautologies). L.T.F. Gamut. Logic, Langauge, and Meaning: Volume 1, Introduction to Logic. The University of Chicago Press. p. 99. -paul === Subject: Re: Tautologies Then and Now > http://www.earlham.edu/~peters/courses/logsys/glossary.htm > I don't know why you keep posting quotes that support what I said -- > that the term tautology is not applied outside propositional logic > (Owen's analysis notwithstanding) -- as if they were counter examples > to what I said. Here's a quote from a chapter on predicate logic: But some sentence schemata in Predicate Logic--like the tautologies of Sentential Logic--are true no matter what their interpretation--that is, they are logically true, or analytic . (Tautologies are a specific sub-class of analytically true sentence schemata--ones which are truth functional.) Like tautologies, the analytic sentence schemata of Predicate Logic are made true by their structure and the definitions of their logical connectives and quantifiers rather than their content. >> Can you cite a text that extends truth tables beyond propositional >> logic? My professors always said that doesn't happen, and it certainly >> didn't in any of my texts. (Tautologies are a specific sub-class of analytically true sentence schemata--ones which are truth functional.) Like tautologies, the analytic sentence schemata of Predicate Logic are made true by their structure and the definitions of their logical connectives and quantifiers rather than their content. >> - paul SH: It would seem there is conceptual similarity between tautologies and the validity used in predicate logic. Though this seems not to extend to truth tables. I wondered about this statement: > In predicate logic ... Formulas @ such that V_M(@) = 1 for all models > M for the language from which @ is taken are called universally valid > formulas (they are not normally called tautologies). SH: Does this mean they can be called tautologies in some instances? But some sentence schemata in Predicate Logic--like the tautologies of Sentential Logic--are true no matter what their interpretation--that is, they are logically true, or analytic. SH: In terms of decidability it seems you are right. Bob did not defend. I think you are right about truth tables (unless there is something technical), but I'm not sure about this claim: > that the term tautology is not applied outside propositional logic does not seem to be fully supported by your own quote: > formulas (they are not normally called tautologies). SH: Should read they are not ________ called tautologies, without the normally qualifier which tends to mean there are exceptions. The larger context of the quote used above. http://www.lawrence.edu/fast/boardmaw/analytic_essay.html Analytic Sentences & Valid Arguments in 1st order Predicate Logic from Supplement for Symbolic Logic (c) Boardman Corresponding to tautologies in Sentential Logic are analytic sentence schemata in First Order Predicate Logic. You will remember that a tautology is a sentence schema which is true under any consistent interpretation of its sentential letters; no line of its truth table is false, each line sketching a class of interpretations (classified according to the combinations of truth values of the sub-sentences represented by the sentential letters). For a tautology is true independently of the content which its sentential letters might represent: its structure and the truth-functional definitions of its logical connectives are what make it true. The sentence schemata in Predicate Logic are used to represent sentences which make claims about the individuals inhabiting some universe of discourse--the subset of the world which we choose to represent--and their properties and relations to one another. The typical sentence schema is contingently true (or false), its truth value depending upon whether the assertion the interpretation assigns to it happens to be true in the chosen universe of discourse. But some sentence schemata in Predicate Logic--like the tautologies of Sentential Logic--are true no matter what their interpretation--that is, they are logically true, or analytic . (Tautologies are a specific sub-class of analytically true sentence schemata--ones which are truth functional.) Like tautologies, the analytic sentence schemata of Predicate Logic are made true by their structure and the definitions of their logical connectives and quantifiers rather than their content. === Subject: Re: Tautologies Then and Now > I think you are right about truth tables (unless there is something > technical), > but I'm not sure about this claim: >> that the term tautology is not applied outside propositional logic > does not seem to be fully supported by your own quote: >> formulas (they are not normally called tautologies). > SH: Should read they are not ________ called tautologies, without the > normally qualifier which tends to mean there are exceptions. For tautologies there is a general method for showing intrinsic truth, a truth table. There is no general method for showing the intrinsic truth of valid statements. A general algorithm for proving a formula to be valid is not possible. But I'm wondering if in a particular class of cases, if there is a specific algorithm for proving formulae to be valid, which would function in principle like a constrained truth table. === Subject: Re: Tautologies Then and Now >> I think you are right about truth tables (unless there is something >> technical), There appears to be truth functional proxies. > For tautologies there is a general method for showing intrinsic truth, > a truth table. There is no general method for showing the intrinsic > truth of valid statements. A general algorithm for proving a formula to > be valid is not possible. But I'm wondering if in a particular class of > cases, if there is a specific algorithm for proving formulae to be valid, > which would function in principle like a constrained truth table. Truth functional proxies. As to using logic for reasoning about natural language and common sense. Talking about Trees and Truth-conditions Reinhard Muskens http://www.illc.uva.nl/j50/contribs/muskens/muskens.pdf Abstract An attractive way to model the relation between an underspecified syntactic representation and its completions is to let the underspecified representation correspond to a logical description and the completions to the models of that description. This approach, which underlies the Description Theory of (Marcus et al. 1983) was integrated with a pure unification approach to Lexicalized Tree-Adjoining Grammars (Joshi et al. 1975, Schabes 1990) in (Vijay-Shanker 1992) and was further developed in the `D-Tree Grammars' (DTG) of (Rambow et al. 1995). We generalize Description Theory by integrating semantic information, that is, we propose to tackle both syntactic and semantic underspecification using descriptions. Our focus will be on underspecification of scope. We use a generalized and completely declarative version of the D-Tree formalism. Although trees in our set-up have surface strings at their leaves and are in fact very close to ordinary surface trees, there is also a strong connection with the Logical Forms (LFs) of (May 1977). We associate logical interpretations with these LFs using a technique of internalising the logical binding mechanism (Muskens 1996). The net result is that we obtain a Description Theory-like grammar in which the descriptions underspecify semantics. Since everything is framed in classical logic it is easily possible to reason with these descriptions. Internalising Binding How can we assign a semantics to the lexical descriptions in fig. 1? We must e.g. be able to express the semantics of n1 in terms of the semantics of n2, whatever the latter turns out to be, i.e. we must be able to express the result of quantification into an arbitrary context. In mathematical English we can say that, for any @, the value of allx@ is the set of assignments a such that for all b differing from a at most in x, b is an element of the value of @. We need to be able to say something similar in our logical language. The language must talk about meaning; it must talk about things that function like variables and constants, things that function like assignments, etc. The first will be called registers, the second states. Two primitive types are added to the logic: Pi and s, for registers and states respectively. We shall have variable registers, which stand proxy for variables and constant registers for constants. ... We have essentially mimicked the Tarski truth conditions for predicate logic in our object language and in fact it can be proved that, under certain conditions, we can reason with terms generated in this way as if they were **the predicate logical formulas they stand proxy for (see Muskens 1998). It should be stressed that the technique discussed here can be used to embed any logic with a decent interpretation into classical logic. For example, (Muskens 1996) shows that we can use the same mechanism to embed Discourse Representation Theory (DRT, Kamp & Reyle 1993). In a full version of this paper we shall also present a version of our theory based on DRT. SH: This indicates to me that if predicate logic can be embedded, then so can a simpler aspect, truth functional proxies, akin to propositional logic which would help logical reasoning about natural language input. I found researching this thread from vague memories quite interesting. Stephen === Subject: Re: Tautologies Then and Now >> I think you are right about truth tables (unless there is something > technical), > There appears to be truth functional proxies. >> For tautologies there is a general method for showing intrinsic truth, >> a truth table. There is no general method for showing the intrinsic >> truth of valid statements. A general algorithm for proving a formula to >> be valid is not possible. But I'm wondering if in a particular class of >> cases, if there is a specific algorithm for proving formulae to be valid, >> which would function in principle like a constrained truth table. > Truth functional proxies. > As to using logic for reasoning about natural language and common sense. > Talking about Trees and Truth-conditions > Reinhard Muskens http://www.illc.uva.nl/j50/contribs/muskens/muskens.pdf > Abstract > An attractive way to model the relation between an underspecified > syntactic representation and its completions is to let the underspecified > representation correspond to a logical description and the completions to > the models of that description. This approach, which underlies the > Description Theory of (Marcus et al. 1983) was integrated with a pure > unification approach to Lexicalized Tree-Adjoining Grammars (Joshi et al. > 1975, Schabes 1990) in (Vijay-Shanker 1992) and was further developed in > the `D-Tree Grammars' (DTG) of (Rambow et al. 1995). We generalize > Description Theory by integrating semantic information, that is, we > propose > to tackle both syntactic and semantic underspecification using > descriptions. > Our focus will be on underspecification of scope. We use a generalized and > completely declarative version of the D-Tree formalism. Although trees in > our set-up have surface strings at their leaves and are in fact very close > to ordinary surface trees, there is also a strong connection with the > Logical > Forms (LFs) of (May 1977). We associate logical interpretations with these > LFs using a technique of internalising the logical binding mechanism > (Muskens > 1996). The net result is that we obtain a Description Theory-like grammar > in > which the descriptions underspecify semantics. Since everything is framed > in > classical logic it is easily possible to reason with these descriptions. > Internalising Binding > How can we assign a semantics to the lexical descriptions in fig. 1? We > must > e.g. be able to express the semantics of n1 in terms of the semantics of > n2, > whatever the latter turns out to be, i.e. we must be able to express the > result of quantification into an arbitrary context. In mathematical > English > we can say that, for any @, the value of allx@ is the set of assignments a > such that for all b differing from a at most in x, b is an element of the > value of @. We need to be able to say something similar in our logical > language. The language must talk about meaning; it must talk about things > that function like variables and constants, things that function like > assignments, etc. The first will be called registers, the second states. > Two > primitive types are added to the logic: Pi and s, for registers and states > respectively. We shall have variable registers, which stand proxy for > variables and constant registers for constants. ... > We have essentially mimicked the Tarski truth conditions for predicate > logic > in our object language and in fact it can be proved that, under certain > conditions, we can reason with terms generated in this way as if they were > **the predicate logical formulas they stand proxy for (see Muskens 1998). > It should be stressed that the technique discussed here can be used to > embed any logic with a decent interpretation into classical logic. For > example, (Muskens 1996) shows that we can use the same mechanism to embed > Discourse Representation Theory (DRT, Kamp & Reyle 1993). In a full > version of this paper we shall also present a version of our theory based > on DRT. > SH: This indicates to me that if predicate logic can be embedded, then > so can a simpler aspect, truth functional proxies, akin to propositional > logic which would help logical reasoning about natural language input. > I found researching this thread from vague memories quite interesting. > In Sentential Logic, we can prove an argument schema to be invalid by > specifying a set of truth assignments to the sentential letters which > results in true premises and a false conclusion; we thereby show that one > line of the argument schema's truth table allows an interpretation having > true premises and a false conclusion. In Predicate Logic, an argument > >schema typically consists of sentence schemata which are not truth > functional: quantifiers, not truth functional connectives, are the major > operators of the typical quantified > *argument schemata. And quantifiers are not truth functional operators > since they may represent an infinite number of individuals; the truth > value of a quantified sentence schema is therefore not a function of the > truth values of any _finite_ number of simple sentence schemata. > *Nevertheless, we can test the validity of a quantified argument schema > _indirectly_ by constructing and testing its _truth functional proxy for > some_ (non-empty) _domain_ of a specified (finite) number of individuals; > each of the premises, and the conclusion, in the original schema will be > equivalent _in that domain_ to its truth functional counterpart in the > proxy. > Because it is comprised of truth functional sentence schemata, a proxy may > be tested for validity by the short-cut method of truth value assignment, > or by *means of a truth table.* > And if a proxy proves to be invalid, it will provide a recipe for > constructing an interpretation of the corresponding quantified argument > schema into the same domain which will serve as a counter example, or > refutation, to that argument schema. Thus, if the original quantified > argument schema is valid, then _all_ of its corresponding proxies must > also be valid. If _any one_ of the proxies corresponding to a quantified > argument schema is invalid, then since it is therefore possible for the > schema to have an interpretation into some domain under which its premises > are true while its conclusion is false, the schema itself is invalid. Note > that even though one particular corresponding proxy is valid, the original > quantified argument schema might nevertheless be invalid: to be valid, > _every_ corresponding proxy (for _every_ non-empty domain) must be valid. > http://www.lawrence.edu/fast/boardmaw/analytic_essay.html > Stephen === Subject: Re: Tautologies Then and Now >> http://www.earlham.edu/~peters/courses/logsys/glossary.htm > Logical validity. > For a wff, to be true for every interpretation of the formal language; to > have every interpretation be a model. Every interpretation here is > understood to mean all, but only, those interpretations in which the > connectives and/or quantifiers take their standard meanings. In > truth-functional propositional logic, logically valid wffs are also called > tautologies. In standard predicate logic, logical validity is limited to > interpretions with non-empty domains. Logical validity is also called > logical truth. Is there an echo in here? ;-) === Subject: Re: Tautologies Then and Now > They're called universal truths, not tautologies, in the >> first-order predicate calculus. >They are true under all standard interpretations. That is how truth >tables are extended to the first order logics. >> Can you cite a text that extends truth tables beyond propositional >> logic? My professors always said that doesn't happen, and it certainly >> didn't in any of my texts. >> - paul >These are the definitions: Right, it restricts tautologies to propositional logic, like I said. OTOH, formulae in first-order predicate logic that are true on all valuations are called universally valid. >http://www.earlham.edu/~peters/courses/logsys/glossary.htm >This glossary is limited to basic set theory, basic recursive function >theory, >two branches of logic (truth-functional propositional logic and first-order >predicate logic) and their metatheory. ... >Tautology. >A logically valid wff of truth-functional propositional logic. A compound >proposition that is true in every row of its truth table or in every >interpretation. See contingency; contradiction; logical validity; semantic >tautology; syntactic tautology. >Tautology schema (plural: schemata). A formula containing variables of the >metalanguage which becomes a tautology when the variables are instantiated >to wffs of the formal language. >Semantic tautology. >A wff of truth-functional propositional logic whose truth table column >contains >nothing but T's when these are interpreted as the truth-value Truth. >See syntactic tautology. >Syntactic tautology. >A wff of truth-functional logic whose truth table column contains nothing >but >T's when these T's are uninterpreted tokens rather than, say, truth-values. >The rules for generating the truth table column tell us to use one of these >uninterpreted T's in exactly those cases where semantic considerations would >have led us to use the truth-value Truth. See semantic tautology. === Subject: Re: Tautologies Then and Now >> They're called universally valid, not tautologies, in the > first-order predicate calculus. >> Bob: >>They are true under all standard interpretations. That is how truth >>tables are extended to the first order logics. Paul: > Can you cite a text that extends truth tables beyond propositional > logic? My professors always said that doesn't happen, and it certainly > didn't in any of my texts. I don't know why you keep posting quotes that support what I said -- that the term tautology is not applied outside propositional logic SH: I think the following paragraph could easily be seen to support Bob's statement, (That is how truth tables are extended to the first order logics.) but I wouldn't say you are wrong either. I used _word phrase_ to represent what the author put in bold in the original [underscore]. * is my emphasis. In Sentential Logic, we can prove an argument schema to be invalid by specifying a set of truth assignments to the sentential letters which results in true premises and a false conclusion; we thereby show that one line of the argument schema's truth table allows an interpretation having true premises and a false conclusion. In Predicate Logic, an argument schema typically consists of sentence schemata which are not truth functional: quantifiers, not truth functional connectives, are the major operators of the typical quantified *argument schemata. And quantifiers are not truth functional operators since they may represent an infinite number of individuals; the truth value of a quantified sentence schema is therefore not a function of the truth values of any _finite_ number of simple sentence schemata. *Nevertheless, we can test the validity of a quantified argument schema _indirectly_ by constructing and testing its _truth functional proxy for some_ (non-empty) _domain_ of a specified (finite) number of individuals; each of the premises, and the conclusion, in the original schema will be equivalent _in that domain_ to its truth functional counterpart in the proxy. Because it is comprised of truth functional sentence schemata, a proxy may be tested for validity by the short-cut method of truth value assignment, or by *means of a truth table.* And if a proxy proves to be invalid, it will provide a recipe for constructing an interpretation of the corresponding quantified argument schema into the same domain which will serve as a counter example, or refutation, to that argument schema. Thus, if the original quantified argument schema is valid, then _all_ of its corresponding proxies must also be valid. If _any one_ of the proxies corresponding to a quantified argument schema is invalid, then since it is therefore possible for the schema to have an interpretation into some domain under which its premises are true while its conclusion is false, the schema itself is invalid. Note that even though one particular corresponding proxy is valid, the original quantified argument schema might nevertheless be invalid: to be valid, _every_ corresponding proxy (for _every_ non-empty domain) must be valid. http://www.lawrence.edu/fast/boardmaw/analytic_essay.html SH: I think the word tautological (no line of its truth table is false) could be applied in this rather technical and specific connection to a truth table in predicate logic as presented in this quote. Because it is comprised of truth functional sentence schemata, a proxy may be tested for validity by the short-cut method of truth value assignment, or by *means of a truth table.* Nevertheless, I think you are typically correct; the limited domain described probably does admit a standard treatment albeit an unusual one. I would probably not have had an inkling about truth functional proxies except I read a lot of papers. I am unsure what practical value constructing truth functional proxies has. As a speculation: there are common sense AI programs which accept natural language queries. Perhaps a truth functional proxy could narrow down the domain of appropriate answers which go through a series of filters and rules in order to produce a natural language response. As I said, just a speculation because I think in computer ideas and philosophy rather than the logical requirement for a truth functional proxy. Constructing a Truth Functional Proxy for Multiply-general Expressions: a recipe pages 46 and 47 of Boardman's Supplement http://www.lawrence.edu/fast/boardmaw/Logic_prox_pp46-7.html The first step in finding a truth functional proxy (for a domain of your choosing) for a quantified argument schema is to find the truth functional counterpart (in that domain) of each of the premises and the conclusion. But these may be multiply-general expressions..... Syntax and Semantics Meanings can become objects or targets of special types of reflective act; it is acts of this sort which make up the science of logic. Logic arises when we treat those species which are meanings as special sorts of proxy objects (as 'ideal singulars'), and investigate the properties of these objects in much the same way that the mathematician investigates the properties of numbers or geometrical figures. http://www.nickbostrom.com/old/quine.html Back to Quine's ontological relativity. We may now ask what may be the philosophical significance of the thesis of indeterminacy of reference, interpreted so as to be proved by the presentation of any proxyfunction. Does it show (a) that there are incompatible theories of reference, all of which are equally adequate, albeit we have happened to choose one particular theory by opting for our actual notion of reference? Or does it show (b) that there is no fact of the matter as to which sense of reference we are using? I think it does not show (b) because in order to prove that the purported notion of reference is merely purported, it is not enough to show that there can be divergent denotation assignments conserving the truth values of all sentences; one would also have to show that conserving truth value of all sentences is sufficient for an assignment to be correct in the intuitive sense, as far as this is naturalistically scrutable. That has not been shown. There is no obvious reason why there may not be determining criteria hooked directly onto words. I think it does not show (a) either. (a) would in effect be a case of underdetermination of theory by data, underdetermination of the countless possible theories of reference. I will say more about the relation between indeterminacy and underdetermination. SH: The reason I quoted this is to show their are other angles of approach to the truth functional proxy idea. In this case it seems possible for the domain to be sufficiently determined. There is no obvious reason why there may not be determining criteria hooked directly onto words. ties into my speculation about mechanical machine translation of natural language. Stephen === Subject: Re: Tautologies Then and Now >I don't know why you keep posting quotes that support what I said -- >that the term tautology is not applied outside propositional logic >>SH: I think the following paragraph could easily be seen to support >Bob's statement, (That is how truth tables are extended to the first order >logics.) but I wouldn't say you are wrong either. I used _word phrase_ >to represent what the author put in bold in the original [underscore]. >* is my emphasis. The source you cite http://www.lawrence.edu/fast/boardmaw/analytic_essay.html cites Copi, and Copi asserts what Owen shows... that truth tables can be used for monadic predicate logic. But in so describing, Copi does not refer to monadic predicate sentences that are true in all valuations as tautologies but as universally valid. Copi states: If an argument contains n different predicate symbols, then if it is valid for a model containing 2^n individuals, then it is valid for every model or universally valid. Copi, I.M. Symbolic Logic. p. 81. - Paul === Subject: Re: Tautologies Then and Now >>I don't know why you keep posting quotes that support what I said -- >>that the term tautology is not applied outside propositional logic >>SH: I think the following paragraph could easily be seen to support >>Bob's statement, (That is how truth tables are extended to the first >>order >>logics.) but I wouldn't say you are wrong either. I used _word phrase_ >>to represent what the author put in bold in the original [underscore]. >>* is my emphasis. > The source you cite > http://www.lawrence.edu/fast/boardmaw/analytic_essay.html > cites Copi, and Copi asserts what Owen shows... that truth tables can > be used for monadic predicate logic. But in so describing, Copi does > not refer to monadic predicate sentences that are true in all > valuations as tautologies but as universally valid. Copi states: > If an argument contains n different predicate symbols, then if it is > valid for a model containing 2^n individuals, then it is valid for > every model or universally valid. > Copi, I.M. Symbolic Logic. p. 81. >- Paul Monadic means one, one predicate, not n>1. The definition you give to support your position, doesn't seem specific to monadic predicate logic. I think Copi's definition that you quote is a broad definition outside the range tautology applies. Shortly, you have shown how it is correct to use universally valid in describing all of general predicate calculus. Your quote does not show that it is incorrect to use tautologous for specific decidable fragments. I think Copi's quote is the not the same as the vice versa that I mention below. 'Every tautology is valid, but not every valid formulation is tautological.' (the monadic ones can be both) http://www-unix.oit.umass.edu/~partee/726_04/lectures /Lecture6%20_Predicate%20Logic_.pdf 2.1. Tautologies, contradictions and contingencies As in Statement Logic, some closed formulas of Predicate Logic are always true, i.e. they are true in every model. These are called tautologies. Formulas which are false in every model are called contradictions. All the other formulas are called contingencies: their truth values depend on models; they are true in some models and false in others. Mathematical Methods in Linguistics** which is used as a textbook: http://www.ling.gu.se/kurser/mathmeth/ This course gives a general introduction to various tools from discrete mathematics that are used in linguistics. It is based largely (or entirely) on Barbara Partee, Alice ter Meulen, Robert Wall (1990) Mathematical Methods in Linguistics, Kluwer Academic Publishers. ** ------------------------------------------------------------ SH: I think there is something odd about you stating that your instructors did not cover this. Lecture 11 pages 20 & 21 Logic: The Art of Persuasion and the Science of Truth I think by Vann McGee http://aka-ocw.mit.edu/NR/rdonlyres/Linguistics-and-Philosophy/24-241Logic -IFall2002/BE072366-EA9F-4193-97CC-BB2922C3C7B1/0/lec11.pdf [pgs. 20&21] Normal Truth Assignment = N.T.A. Definition: A sentence is tautological iff it is assigned the value1 by every N.T.A. A sentence is valid iff it is true under every N.T.A. For the sentential calculus,the words tautological and valid were different words for the samething. Now that we've started on the predicate calculus, we need to distinguish them. Validity is the notion we're really interested in,but we need the notion of tautology as a technical notion. Proposition. Every tautology is valid, but not vice versa. [SH: He provides a proof, and then continues:] A tautological sentence is a valid sentence whose validity is determined by the sentence's truth functional structure. If, instead,the validity of a sentence depends upon the meaning of the quantifiers,the sentence won't be tautological. We can test whether a sentence is tautological by the method of truth tables, examining each possible way to assign a truth value to the sentence's basic truth functional components. --------------------------------------------------------- Derivations in the Mondadic Predicate Calculus [MIT online course] The fact that there is a mechanical procedure for testing whether a sentence is a tautological consequence of a set of sentences is important. In order for our derivations to have any probative value, we have to be able to recognize when a sequence of sentences really is a proof,which means that we need an algorithm for checking when a rule has been properly applied. http://ocw.mit.edu/NR/rdonlyres/Linguistics-and-Philosophy /24-241Logic-IFall2002/8A9852F3-D38E-4A94-9634-9EED32367CCB/0/lec12.pdf SH: That there is a mechancical procedure for testing (truth table) which can output all trues is correctly described as tautological by definition. No matter what the logical structure outside of propositional logic is named. That there is a mechanical shortcut, is why it can be done by a computer More about decidable fragments: http://www-mgi.informatik.rwth-aachen.de/Publications/Graedel/ #38 If my quotes don't convince you, maybe someone else will do a better job. Something interesting about artificial/formal languages and natural language. Knowledge Representation in Sanskrit and Artificial Intelligence by Rick Briggs http://www.aaai.org/Library/Magazine/Vol06/06-01/Papers/AIMag06-01-003.pdf Logically guarded, Stephen === Subject: Re: Tautologies Then and Now >> If an argument contains n different predicate symbols, then if it is >> valid for a model containing 2^n individuals, then it is valid for >> every model or universally valid. >> Copi, I.M. Symbolic Logic. p. 81. >>- Paul >I think Copi's definition that you quote is a broad definition outside the >range tautology applies. Shortly, you have shown how it is correct >to use universally valid in describing all of general predicate calculus. >Your quote does not show that it is incorrect to use tautologous >for specific decidable fragments. It should be clear to you from numerous quotes you posted previously that, with the one exception of Barbara Partee, most authors are careful when discussing logics to restrict their use of the term tautology to propositional/sentential logic. For example your oft quoted http://www.lawrence.edu/fast/boardmaw/analytic_essay.html says on top: Corresponding to tautologies in Sentential Logic are analytic sentence schemata in First Order Predicate Logic. You will remember that a tautology is a sentence schema which is true under any consistent interpretation of its sentential letters; Clearly the author indicates the term tautology is used uniquely in sentential logic. Numerous other quotes you have presented also indicate that same limit. Which matches Gamut and Copi: * In predicate logic ... Formulas @ such that V_M(@) = 1 for all models M for the language from which @ is taken are called universally valid formulas (they are not normally called tautologies). L.T.F. Gamut. Logic, Langauge, and Meaning. p. 99. * If an argument contains n different predicate symbols, then if it is valid for a model containing 2^n individuals, then it is valid for every model or universally valid. Copi, I.M. Symbolic Logic. p. 81. instances of its application to always valid predicate statements versus the one example of Partee. My question is why is tautology not normally used outside sentential logic? There must be a reason. - paul === Subject: Re: Tautologies Then and Now >> They're called universal truths, not tautologies, in the > first-order predicate calculus. >They are true under all standard interpretations. That is how truth >>tables are extended to the first order logics. > Can you cite a text that extends truth tables beyond propositional > logic? My professors always said that doesn't happen, and it certainly > didn't in any of my texts. > - paul >>These are the definitions: > Right, it restricts tautologies to propositional logic, like I said. > OTOH, formulae in first-order predicate logic that are true on all > valuations are called universally valid. pages 16 through 19 It wasn't conclusive for me. === Subject: Re: Tautologies Then and Now >> They're called universal truths, not tautologies, in the > first-order predicate calculus. >They are true under all standard interpretations. That is how truth >>tables are extended to the first order logics. > Can you cite a text that extends truth tables beyond propositional > logic? My professors always said that doesn't happen, and it certainly > didn't in any of my texts. > - paul >>These are the definitions: > Right, it restricts tautologies to propositional logic, like I said. > OTOH, formulae in first-order predicate logic that are true on all > valuations are called universally valid. According to this quote, tautological and valid mean the same thing: www.math.lsa.umich.edu/~ablass/llc9.ps (page 4) but it does not say universally valid. ...then, remembering that implication is treated as an abbreviation and negations are to be pushed in to the atomic level, we find that the universally quantified variables are the y in the antecedent and the x and z in the consequent. Thus, the Herbrand form (/)H is (if we use u; v; w as the new variables in the consequent at step 1) P (x; Y (x); z) --> Q(U; v; W (v)): In stating Herbrand's Theorem, we use the usual terminology: Tau[CapitalEth]tology'' means valid in propositional logic. The statement of the theo[CapitalEth]rem also involves the notion of a ``closed instance'' of a formula. Since three other notions of ``instance'' will play a role later in the paper, we define all four notions here to avoid confusion. Definition 3.5 A closed instance of a first[CapitalEth]-order formula (/) is obtained by substituting closed terms for all the free variables in (/). A first[CapitalEth]-order instance of a propositional formula (/) is obtained by replacing the propositional variables in (/) by first[CapitalEth]-order sentences. === Subject: Re: Tautologies Then and Now >> >> > > They're called universal truths, not tautologies, in the > first-order predicate calculus. >> >>They are true under all standard interpretations. That is how truth >>tables are extended to the first order logics. > Can you cite a text that extends truth tables beyond propositional > logic? My professors always said that doesn't happen, and it certainly > didn't in any of my texts. > - paul This post may be of interest to you. Owen === Subject: Re: Tautologies Then and Now > > > >> > >> > > > > They're called universal truths, not tautologies, in the > > first-order predicate calculus. > >> > >>They are true under all standard interpretations. That is how truth > >>tables are extended to the first order logics. > > > > > > Can you cite a text that extends truth tables beyond propositional > > logic? My professors always said that doesn't happen, and it certainly > > didn't in any of my texts. > > > > - paul > > > > This post may be of interest to you. Search Result 1 === Subject: Truth tables for monadic predicate logic Original Format Given that propositions, p, q, r..etc., have two truth values (T,F), logical truth (tautology) is confirmed if the propositional expression has the value T for all possibilities. It is contradictory if it has the value F for all possibilities and it's contingent if it has at least one T and at least one F for all possibilities. My intension here is to extend the concepts of truth by calculation to monadic predicate logic. Truth by calculation for predicate logic in determined domains reduces to propositional logic via expansion of the quantifiers 'All' and 'Some'. In undetermined or infinite domains I suggest the following method of decision. A propositional form with one free individual variable x, e.g. 'Fx' , is monadic. ExFx means 'for some x Fx' . AxFx means 'for all x Fx'. The usual propositional logic operators (not '~', or 'v', and '&', implies '->', equivalence '<->') are maintained. Fx ~Fx ExFx Ex~Fx AxFx Ax~Fx __ ___ ____ _____ ____ _____ 1 0 T 1 F 0 T 1 F 0 2 3 T 2 T 3 F 2 F 3 3 2 T 3 T 2 F 3 F 2 0 1 F 0 T 1 F 0 T 1 The following theorems are valid (true by deduction) and tautologous (true by calculation). (1) AxFx->ExFx (2) ~Ex~Fx<->AxFx (3) ~Ax~Fx<->ExFx __________ _____________ _____________ T 1 T T 1 TF 0 T T 1 TF 0 T T 1 F 2 T T 2 FT 3 T F 2 TF 3 T T 2 F 3 T T 3 FT 2 T F 3 TF 2 T T 3 F 0 T F 0 FT 1 T F 0 FT 1 T F 0 (4)(AxFx & ExFx)<->AxFx (5) AxFx<->((ExFx->AxFx)<->ExFx) (6)(AxFx v ExFx)<->ExFx (7) ExFx<->((ExFx->AxFx)<->AxFx) (8)(Ex~Fx->Ax~Fx)<->(ExFx->AxFx) etc. Table 1. (these tables include propositional logic) | ~ | E | A (not, some, all) ______________ T | F | T | T F | T | F | F 1 | 0 | T | T 2 | 3 | T | F 3 | 2 | T | F 0 | 1 | F | F Table 2. v | T F 1 2 3 0 (or) ________________ T | T T 1 1 1 1 F | T F 1 2 3 0 1 | 1 1 1 1 1 1 2 | 1 2 1 2 1 2 3 | 1 3 1 1 3 3 0 | 1 0 1 2 3 0 The other tables result from the definitions. p->q defined ~p v q (implies) Table 3. -> | T F 1 2 3 0 ________________ T | T F 1 2 3 0 F | T T 1 1 1 1 1 | 1 1 1 1 1 1 2 | 1 3 1 1 3 3 3 | 1 2 1 2 1 2 0 | 1 1 1 1 1 1 p & q defined ~(~p v ~q) (and) Table 4. & | T F 1 2 3 0 _______________ T | T F 1 2 3 0 F | F F 0 0 0 0 1 | 1 0 1 2 3 0 2 | 2 0 2 2 0 0 3 | 3 0 3 0 3 0 0 | 0 0 0 0 0 0 p<->q defined (p->q)&(q->p) (equivalence) Table 5. <-> | T F 1 2 3 0 _________________ T | T F 1 2 3 0 F | F T 0 3 2 1 1 | 1 0 1 2 3 0 2 | 2 3 2 1 0 3 3 | 3 2 3 0 1 2 0 | 0 1 0 3 2 1 (8) Ax:Fx->ExFx (9) Ax:AxFx->Fx (10) Ax:(Fx->AxFx) v ExFx ___________ ___________ ____________________ T 1 1 T 1 T T 1 1 1 T 1 1 T 1 1 T 1 T 2 1 T 2 T F 2 1 2 T 2 3 F 2 1 T 2 T 3 1 T 3 T F 3 1 3 T 3 2 F 3 1 T 3 T 0 1 T 0 T F 0 1 0 T 0 1 F 0 1 F 0 (11) Ax:(ExFx->AxFx)->(ExFx->Fx) (12) Ex:(Fx->AxFx) (13) Ex:(Fx->AxFx) (14) Ex:((ExFx->Fx)->AxFx)<->ExFx) (15) Ax:((ExFx->Fx)->AxFx)<->AxFx) (16) Ax(Fx<->p)->(AxFx<->p) (17) (AxFx<->p)->Ex(Fx<->p) (19) Ex(p)<->p (20) Ax(p)<->p Predicate logic tautologies with pure and prenex forms. (20) Ex~Fx<->~AxFx (21) Ax~Fx<->~ExFx (21) Ex(Fx v p)<->(ExFx v p) _______________________ T 1 1 T T T 1 T T T 2 1 T T T 2 T T T 3 1 T T T 3 T T T 0 1 T T F 0 T T T 1 1 F T T 1 T F T 2 2 F T T 2 T F T 3 3 F T T 3 T F F 0 0 F T F 0 F F (22) Ax(Fx v p)<->(AxFx v p) (23) Ex(Fx & p)<->(ExFx & p) (24) Ax(Fx & p)<->(AxFx & p) (25) Ex(p->Fx)<->(p->ExFx) (26) Ax(p->Fx)<->(p->AxFx) (27) Ex(Fx->p)<->(AxFx->p) (28) Ax(Fx->p)<->(ExFx->p) (29) Ex(Fx<->p)<->((AxFx->p)&(p->ExFx)) __________________________________ T 1 1 T T T 1 T T T TT T 1 T 2 2 T T F 2 T T T TT T 2 T 3 3 T T F 3 T T T TT T 3 F 0 0 T T F 0 T T F TF F 0 F 1 0 F T T 1 F F F FT T 1 T 2 3 F T F 2 T F T FT T 2 T 3 2 F T F 3 T F T FT T 3 T 0 1 F T F 0 T F T FT F 0 (30) Ax(Fx<->p)<->((ExFx->p)&(p->AxFx)) __________________________________ T 1 1 T T T 1 T T T TT T 1 F 2 2 T T T 2 T T F TF F 2 F 3 3 T T T 3 T T F TF F 3 F 0 0 T T F 0 T T F TF F 0 F 1 0 F T T 1 F F F FT T 1 F 2 3 F T T 2 F F F FT F 2 F 3 2 F T T 3 F F F FT F 3 T 0 1 F T F 0 T F T FT F 0 The last two theorems,(29) and (30), do not appear in most logic texts. I believe they should appear to complete the pure and prenex forms. To deal with truth functions of two or more predicates, which includes Boolean logic and Syllogistic logic, we need to extend the truth tables. It will then become evident that all of the axioms of propositional logic and all of the axioms of general predicate logic are tautologies. Truth tables for two monadic predicates. These tables include the above tables. | ~ | E | A (not) (Some) (all) _________________ T | F | T | T F | T | F | F 15 | 14 | T | F 13 | 12 | T | F 11 | 10 | T | F 9 | 8 | T | F 7 | 6 | T | F 5 | 4 | T | F 3 | 2 | T | F 1 | 0 | T | T 0 | 1 | F | F 2 | 3 | T | F 4 | 5 | T | F 6 | 7 | T | F 8 | 9 | T | F 10 | 11 | T | F 12 | 13 | T | F 14 | 15 | T | F v | T F 15 13 11 9 7 5 3 1 0 2 4 6 8 10 12 14 (or) _________________________________________________________ T | T T 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 F | T F 15 13 11 9 7 5 3 1 0 2 4 6 8 10 12 14 15 | 1 15 15 13 11 9 7 5 3 1 15 13 11 9 7 5 3 1 13 | 1 13 13 13 9 9 5 5 1 1 13 13 9 9 5 5 1 1 11 | 1 11 11 9 11 9 3 1 3 1 11 9 11 9 3 1 3 1 9 | 1 9 9 9 9 9 1 1 1 1 9 9 9 9 1 1 1 1 7 | 1 7 7 5 3 1 7 5 3 1 7 5 3 1 7 5 3 1 5 | 1 5 5 5 1 1 5 5 1 1 5 5 1 1 5 5 1 1 3 | 1 3 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 1 | 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 | 1 0 15 13 11 9 7 5 3 1 0 2 4 6 8 10 12 14 2 | 1 2 13 13 9 9 5 5 1 1 2 2 6 6 10 10 14 14 4 | 1 4 11 9 11 9 3 1 3 1 4 6 4 6 12 14 12 14 6 | 1 6 9 9 9 9 1 1 1 1 6 6 6 6 14 14 14 14 8 | 1 8 7 5 3 1 7 5 3 1 8 10 12 14 8 10 12 14 10 | 1 10 5 5 1 1 5 5 1 1 10 10 14 14 10 10 14 14 12 | 1 12 3 1 3 1 3 1 3 1 12 14 12 14 12 14 12 14 14 | 1 14 1 1 1 1 1 1 1 1 14 14 14 14 14 14 14 14 (32) Ax(Fx->Gx)->(AxFx->AxGx) (33) Ax(Fx->Gx)->(ExFx->ExGx) (34) Ax(Fx<->Gx)->(AxFx<->AxGx) (35) Ax(Fx<->Gx)->(ExFx<->ExGx) (36) Ax(Fx & Gx)<->(AxFx & AxGx) (37) Ex(Fx v Gx)<->(ExFx v ExGx) (38) (AxFx v AxGx)->Ax(Fx v Gx) (39) Ex(Fx & Gx)->(ExFx & ExGx) (40) (Ax(Fx->Gx)->Ex(Fx & Gx))<->ExFx (41) ~ExFx->Ax(Fx->Gx) (42) AxGx->Ax(Fx->Gx) (43) (Ex(Fx & Gx) & Ex(Fx & ~Gx))->AxFx is invalid (44) (Ex(Fx & ~Gx) & Ex(~Fx & Gx))->Ax(Fx v Gx) is invalid (45) (Ex(Fx & Gx) & Ex(Fx & ~Gx) & Ex(~Fx & Gx))->Ax(Fx v Gx) is invalid. etc. Proofs of validity and invalidity of monadic predicate logic expressions with three predicates requires truth tables that have 64 function values. Any opinions? Owen Holden ---------------------------------------------------------------------------- -- > Owen === Subject: Re: Tautologies Then and Now >> > >>... Mathematics is held to >> >> be tautological, ... >> >> > By whom? Anybody other than Wittgenstein? >> Frank Ramsay and others including me, etc. agree. >> That (1+1=2) is tautologous, ie. logically true, is clear. >> Because of (Frege-Russell)'s analysis. > For me (am I alone?) a tautology (in the logical sense) is a formula of > propositional calculus which is true for all values of the truth values > of its constituent atomic letters. You are probably not alone, but it is also frequently used in a more inclusive sense, for a formula of any logical system which is true on any assignment of values to the variables involved. -- Aaron Boyden The main division between the so-called Continental and Analytic traditions has been disputes over whether the task of being unclear should be carried out in natural language or in a formal system. === Subject: Re: Tautologies Then and Now > >> >>... Mathematics is held to > >> be tautological, ... > > > By whom? Anybody other than Wittgenstein? > Frank Ramsay and others including me, etc. agree. > That (1+1=2) is tautologous, ie. logically true, is clear. While the values on each side of = are the same, I'm not sure that the same statement is on both sides. 1+1 describes a function f for which 2 is the output f(x). This is what's being said: f(x) = x + x Is that a tautology? If not, does it become a tautology by defining an input value? -paul === Subject: Re: Tautologies Then and Now > >> >>... Mathematics is held to > >> be tautological, ... > > > By whom? Anybody other than Wittgenstein? > Frank Ramsay and others including me, etc. agree. > That (1+1=2) is tautologous, ie. logically true, is clear. > Because of (Frege-Russell)'s analysis. Frege-Russell doesn't settle, e.g., GCH. === Subject: Re: Tautologies Then and Now > >> >>... Mathematics is held to > >> be tautological, ... > > > By whom? Anybody other than Wittgenstein? > Frank Ramsay and others including me, etc. agree. > That (1+1=2) is tautologous, ie. logically true, is clear. > Because of (Frege-Russell)'s analysis. Mind deriving that tautology so we can see exactly which logical axioms is follows from? === Subject: properties of a double quadratic form? Let w be an n-vector with no non-zero matrix elements. Let C and Q be nxn matrices, and ' denote matrix transpose. Consider the following quadratic form: s = w' C' Q' Q C w It is clear that Q'Q is diagonalizable with non-negative eigenvalues. It is clear that C'C is also diagonalizable with non-negative eigenvalues. Since s = (Cw)' Q'Q (Cw), it is clear that s increases when the eigenvalues of Q'Q rise. Is it also true that s increases with the eigenvalues of C'C? How do I prove this? Rodney === Subject: Re: properties of a double quadratic form? >Let w be an n-vector with no non-zero matrix elements. Let >C and Q be nxn matrices, and ' denote matrix transpose. >Consider the following quadratic form: >s = w' C' Q' Q C w >It is clear that Q'Q is diagonalizable with non-negative eigenvalues. >It is clear that C'C is also diagonalizable with non-negative >eigenvalues. >Since s = (Cw)' Q'Q (Cw), it is clear that s increases when the >eigenvalues of Q'Q rise. Not at all clear, unless you specify what you mean by that: what is changing and what remains the same? Are you keeping the eigenvectors of Q'Q the same and increasing the eigenvalues? If the eigenvectors are changing, it's quite possible for s to decrease. >Is it also true that s increases with the eigenvalues of C'C? How do >I prove this? If A = (C'C)^(1/2) (i.e. the positive semidefinite square root of C'C), then C = U A where U is an orthogonal matrix. So s = w' A U' Q' Q U A w. Now judging from the previous remarks, I'm guessing that you mean to keep the eigenvectors of C'C (and thus of A) the same and increase the eigenvalues. But if you're also changing U, there's no telling what this will do to s. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: properties of a double quadratic form? >Let w be an n-vector with no non-zero matrix elements. Let >C and Q be nxn matrices, and ' denote matrix transpose. >Consider the following quadratic form: >s = w' C' Q' Q C w >It is clear that Q'Q is diagonalizable with non-negative eigenvalues. >It is clear that C'C is also diagonalizable with non-negative >eigenvalues. >Since s = (Cw)' Q'Q (Cw), it is clear that s increases when the >eigenvalues of Q'Q rise. > Not at all clear, unless you specify what you mean by that: what is > changing and what remains the same? Are you keeping the eigenvectors of > Q'Q the same and increasing the eigenvalues? If the eigenvectors are > changing, it's quite possible for s to decrease. Good point. s would decrease if the Q'Q eigenvectors become more orthogonal to Cw even if the eigenvalues of Q'Q increase. >Is it also true that s increases with the eigenvalues of C'C? How do >I prove this? > If A = (C'C)^(1/2) (i.e. the positive semidefinite square root of > C'C), then C = U A where U is an orthogonal matrix. So > s = w' A U' Q' Q U A w. Now judging from the previous remarks, > I'm guessing that you mean to keep the eigenvectors of C'C (and thus of > A) the same and increase the eigenvalues. But if you're also changing > U, there's no telling what this will do to s. I don't see why, if the eigenvectors of C'C remain the same, why increasing the eigenvalues of A would increase s here. Care to elaborate? > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: About a Borevich problem I'm looking for some references of papers on the following problem (due to Z.I. Borevich who introduced the notions of polynormal subgroups) you'll find refered by A.S. Kondratiev in the 14th edition of the Kourovka Notebook : A subgroup A of a group G is said to be paranormal if for any x in G : A^x is a subgroup of < A^u | u in A subgroup A of a group G is said to be polynormal if for any x in G : A^x is a subgroup of < A^u | u in A^{} > If we suppose that G is finite, is every polynormal subgroup paranormal ? Manu === Subject: trig identity help In my physics book, for superposition of waves it uses the identity sin(a) + sin(b) = 2cos( (a-b) / 2 )sin( (a+b) / 2 ) I looked in my calculus book's trig review appendix and they do not prove this one. So how to derive it? === Subject: Re: trig identity help > In my physics book, for superposition of waves it uses the identity > sin(a) + sin(b) = 2cos( (a-b) / 2 )sin( (a+b) / 2 ) > I looked in my calculus book's trig review appendix and they do not prove > this one. > So how to derive it? sin (A+B) = sin A cos B + cos A sin B (1) sin (A-B) = sin A cos B - cos A sin B Add: sin (A+B) + sin (A-B) = 2 sin A cos B Put A+B = a, A-B = b then A = (a+b)/2, B = (a-b)/2 Hence sin a + sin b = 2 sin ((a+b)/2) cos ((a-b)/2) To derive (1) use exp(ix) = cos x + i sin x so sin x = Im exp(ix) (for real x) Thus for real x, y sin (x+y) = Im (exp(ix)exp(iy)) = Im((cos x + i sin x)(cos y + i sin y)) = cos x sin y + sin x cos y === Subject: Re: trig identity help === Subject: Test A test. === Subject: GRAPH of this function Hi. Could anyone tell me about the behavior of the graph of f(x) = int_{1...x}t^t dt for x > 0? -- We should have a town named Alderaan someday. No, seriously. Let's put it on the table. === Subject: Re: GRAPH of this function >Could anyone tell me about the behavior of the graph of >f(x) = int_{1...x}t^t dt >for x > 0? Increasing; concave up for x >= e^{-1}, concave down for 0 < x <= e^{-1}; finite limit approximately -.7834305107 as x -> 0+; increases very rapidly past x=3 or so. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: GRAPH of this function >Could anyone tell me about the behavior of the graph of >f(x) = int_{1...x}t^t dt >for x > 0? > Increasing; concave up for x >= e^{-1}, concave down for 0 < x <= e^{-1}; > finite limit approximately -.7834305107 as x -> 0+; increases very rapidly > past x=3 or so. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Question about two rational numbers If I have two rational numbers p = a/b, q = c/d. Is it true that there exists m,n in Z so that: pm + qn = 1/(bd)? === Subject: Re: Question about two rational numbers at 12:37 PM, eytan@tradertools.com (Mimsy Boro) said: >If I have two rational numbers p = a/b, q = c/d. Is it true that >there exists m,n in Z so that: >pm + qn = 1/(bd)? 1. Did you omit an hypothesis? 2. Is this a homework question? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail 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 not === Subject: Re: Question about two rational numbers > If I have two rational numbers p = a/b, q = c/d. Is it true that there > exists m,n in Z so that: > pm + qn = 1/(bd)? It is false. p = 1/2, q = 1/6 m/2 + n/6 = 1/12 ? that is 6m + 2n = 1 ? impossible, as 2 does not divide 1 ! You need GCD(ad, bc) = 1. In this condition : mad + nbc = 1 is just the Bezout relation. -- philippe (chephip at free dot fr) === Subject: Re: Question about two rational numbers So does that mean that there exists m,n in Z so that: pm + nq = GCD(ad, bc)/bd? If so how do you prove somthing like that? === Subject: Re: Question about two rational numbers > So does that mean that there exists m,n in Z so that: > pm + nq = GCD(ad, bc)/bd? > If so how do you prove somthing like that? generaly speaking, given any integers a, b in Z, there exists m,n in Z with GCD(a,b) = ma + nb As I said before this is the Bezout identity There are several proofs of that. The simplest is a constructive proof, that is it gives the values of m and n. Consider Euclid's algorithm (suppose |b|<=|a|, if not exchange a and b) a = bq + r, |r|<|b| GCD(a,b) = GCD(b,r) (elementar to proove that) so we can go on with b = rq' + r', |r'| < |r|, GCD(r,r') = GCD(b,r) = GCD(a,b) then r = r'q + r and so on and as the 0 <= r(i) < r(i-1) < b the process ends with r(n+1) = 0, and r(n) = GCD(a,b) write the last non trivial identity as : GCD = r(n) = r(n-2) - r(n-1)q(n) and substitute the r(n-1) from the previous identity of the algorithm, written as : r(n-1) = r(n-3) - r(n-2)q(n-1) to get GCD = r(n) = r(n-2) - [r(n-3) - r(n-2)q(n-1)]q(n) = r(n-2)(1 + q(n)q(n-1)) - r(n-3)q(n) go on substituting r(n-2), that results into GCD = u*r(k) + v*r(k-1) for any k until the first equation which gives : GCD = ma + nb, after the last substitutions r(0) = a - bq(0) and r(1) = b - r(0)q(1) into the GCD = u*r(2) + v*r(1) example a = 24, b=10 24 = 10*2 + 4 10 = 4*2 + 2 4 = 2*2 + 0 Hence GCD(24,10)= 2 = 10 - 4*2, substitute 4 = 24 - 10*2 gives GCD = 10 - (24 - 10*2)*2 = 10(1 + 2*2) - 2*24 = 5*10 - 2*24, or GCD = m*24 + n*10 with m = -2 and n = 5 The Euclid's algorithm may be slightly modified to give the m and n in the flow : extended Euclid's algorithm. There are more modern proofs using set theory and true for any ring with a GCD. -- philippe (chephip at free dot fr) === Subject: Re: Lambda Calculus and Turing Equivalence Originator: harris@tcs.inf.tu-dresden.de (Mitchell Harris) >What amazes me is that some people still think it >may be the case that some abstract machines in computer science >somehow do not have physical counterparts. [*] That's a basic >misunderstanding. I vaguely remember an issue of Byte magazine many many years ago (late of a universal TM (the exact details I can't remember but probably involved a visit to Radio Shack and buying lots of TTL chips and LED panels. That's probably not what you meant though, was it. >As far as I can tell, computer science is concerned >with what kinds of machines are constructible in theory, in possible >worlds with physical laws like ours if we want to be philosophically >precise. er.. theoretical computer science.. yeah. I don't think anybody would hae much trouble with that (though they might add a lot more or qualify more). Oh, -constructible-.. which means finite...hm... no, a TM with an infinite tape is still a useful -theoretical- construct, because whether actual or potential infinite sized tape, you only get a finite amount of in order to halt. >A question to consider: does it even make sense to talk about an >infinite size program? sure. theoretically. >And another: is this an infinite string 0 1 1 1 ... 1 0 that's a bit disingenuous. Mitch === Subject: Re: Lambda Calculus and Turing Equivalence http://mygate.mailgate.org/mynews/comp/comp.theory/677e08ce2b256bfd9efec0545 1 f8fd25.48257%40mygate.mailgate.org > I doubt that people who have answered my post have > thought deeply about my argument. They don't seem > to have considered all cases of finite/transfinite > in models of computation as I have done. The main reason your arguments don't receive serious consideration is that you don't invest serious thought before making them, and that reputation now follows you like the smell of a rotting corpse wherever you go. You were warned, many times, what the consequences of your ill behavior would be, you chose to ignore the advice, you now reap the consequences. That your arguments have not improved in quality a bit over time is just additional cause to ignore you. That you blather on misusing technical terms you don't understand, and continue to argue on the basis of your misunderstanding even when it is pointed out to you, cubes the resulting urge to put you in a killfile. > A question to consider: [...] > And another: is this an infinite string 0 1 1 1 ... 1 0 And again you deal a death blow to your reputation. Are you _possibly_ that ignorant? Infinite strings cannot possibly be an infinite string. xanthian. -- === Subject: Re: Lambda Calculus and Turing Equivalence > I doubt that people who have answered my post have > thought deeply about my argument. They don't seem > to have considered all cases of finite/transfinite > in models of computation as I have done. > The main reason your arguments don't receive serious > consideration is that you don't invest serious > thought before making them, and that reputation now > follows you like the smell of a rotting corpse > wherever you go. You were warned, many times, what > the consequences of your ill behavior would be, you > chose to ignore the advice, you now reap the > consequences. That your arguments have not improved > in quality a bit over time is just additional cause > to ignore you. That you blather on misusing technical > terms you don't understand, and continue to argue on > the basis of your misunderstanding even when it is > pointed out to you, cubes the resulting urge to put > you in a killfile. > A question to consider: [...] > And another: is this an infinite string 0 1 1 1 ... 1 0 > And again you deal a death blow to your reputation. > Are you _possibly_ that ignorant? Infinite strings > cannot possibly be an infinite string. Well, it could be an ordinal sequence, in which it could be infinite. 'cid 'ooh === Subject: Re: Lambda Calculus and Turing Equivalence http://mygate.mailgate.org/mynews/comp/comp.theory/5c9df1236988428bdd8b49bb1 d 9395b5.48257%40mygate.mailgate.org > Well, it could be an ordinal sequence, in which it could be infinite. Umm, no, he said string, that's a concept with a pretty well accepted consensus definition, and ordinal sequence doesn't fit. Don't give him more credit for knowing whereof he speaks than he deserves, which is none. xanthian. -- === Subject: Re: Lambda Calculus and Turing Equivalence > > I doubt that people who have answered my post have > thought deeply about my argument. They don't seem > to have considered all cases of finite/transfinite > in models of computation as I have done. > > The main reason your arguments don't receive serious > consideration is that you don't invest serious > thought before making them, and that reputation now > follows you like the smell of a rotting corpse > wherever you go. You were warned, many times, what > the consequences of your ill behavior would be, you > chose to ignore the advice, you now reap the > consequences. That your arguments have not improved > in quality a bit over time is just additional cause > to ignore you. That you blather on misusing technical > terms you don't understand, and continue to argue on > the basis of your misunderstanding even when it is > pointed out to you, cubes the resulting urge to put > you in a killfile. > > A question to consider: [...] > And another: is this an infinite string 0 1 1 1 ... 1 0 > > And again you deal a death blow to your reputation. > > Are you _possibly_ that ignorant? Infinite strings > cannot possibly be an infinite string. > > Well, it could be an ordinal sequence, in which case it could be infinite. > 'cid 'ooh Word inserted. === Subject: Re: Lambda Calculus and Turing Equivalence >It is equivalent in computational power to a TM. Accept or not? > > Accept. > >If you accept, my conclusion follows. Otherwise show a logical mistake. > > Er...conclusion? What conclusion? That there is no actual infinity? Or > that a TMs tape is not of infinite size? Or what? Whatever you answer is, > I'll accept it. As far as I can tell it doesn't make any difference. > The conclusion of the original post. It is not that there is no actual > infinity. (That does not follow) > That a TM tape is not of infinite length. It is unbounded, not > infinite. Plain and simple. But yes, I see your point, it's not > awfully important. What amazes me is that some people still think it > may be the case that some abstract machines in computer science > somehow do not have physical counterparts. [*] That's a basic > misunderstanding. As far as I can tell, computer science is concerned > with what kinds of machines are constructible in theory, in possible > worlds with physical laws like ours if we want to be philosophically > precise. > Indeed, Turing machines are solely concerned with what finite > mechanisms can do in general. Godel observes this generality back in > 1950, why can't we now? Because of the law of large numbers: most natural numbers are very, very large. Seriously -- these finite mechanisms aren't much of a bound. I can easily ask you to compute a function that would require more memory (in bits) than there are electrons in the universe. Calculate pi to 2^300 digits. Or fix a language and compute its busy beaver function for n = 2^300. > I doubt that people who have answered my post have thought deeply > about my argument. They don't seem to have considered all cases of > finite/transfinite in models of computation as I have done. I don't think you actually know what transfinite means. > A question to consider: does it even make sense to talk about an > infinite size program? > And another: is this an infinite string 0 1 1 1 ... 1 0 How should we know? Those dots could represent a finite or infinite sequence. > PS: For instance Stephen Harris's argument. A TM can calculate Pi, a > physical computer cannot, therefore TMs are not physical. Does Harris > even recognize that he is reinventing the substance dualism of > Descartes in computational terms? It almost sounds like A TM can > compute. Therefore TM is :) That's not how modern philosophy > proceeds. Geez -- he's shown that a TM can do something that a computer cannot. If computers and TMs had the same computational power, a computer would be able to calculate pi. Modus tollens tells us that TMs and PCs don't have the same computational power. Your blatant disregard for Leibniz's Law appears schizophrenic. 'cid'ooh === Subject: Re: Lambda Calculus and Turing Equivalence > Geez -- he's shown that a TM can do something that a computer cannot. > If computers and TMs had the same computational power, a computer > would be able to calculate pi. Modus tollens tells us that TMs and > PCs don't have the same computational power. Your blatant disregard > for Leibniz's Law appears schizophrenic. > 'cid'ooh Which TM has already calculated all the digits of pi? With unbounded tape and Borwein Algorithm a TM can calculate pi.Also, with unlimited time and Borwein Algorithm a PC can will show in the screen one by one, all the digits of pi. No need of embarrasing tapes. Ludovicus === Subject: Re: Lambda Calculus and Turing Equivalence > Which TM has already calculated all the digits of pi? > With unbounded tape and Borwein Algorithm a TM can calculate pi.Also, > with unlimited time and Borwein Algorithm a PC can will show in the > screen one by one, all the digits of pi. No need of embarrasing tapes. > Ludovicus Nope. The PC will run out of memory. --PeterD === Subject: Re: Lambda Calculus and Turing Equivalence >> Which TM has already calculated all the digits of pi? >> With unbounded tape and Borwein Algorithm a TM can calculate pi.Also, >> with unlimited time and Borwein Algorithm a PC can will show in the >> screen one by one, all the digits of pi. No need of embarrasing tapes. >> Ludovicus > Nope. The PC will run out of memory. Well, it does not matter. If your PC is modular enough, you can always add extra memory when needed. It is equivalent to the assumption of having an infinite tape/memory. Indeed, at any time of an execution, the amount of currently written data on a TM's tape is finite. You just cannot know in advance the amount of memory that will be needed. === Subject: Re: Lambda Calculus and Turing Equivalence >It is equivalent in computational power to a TM. Accept or not? > > Accept. > >If you accept, my conclusion follows. Otherwise show a logical mistake. > > Er...conclusion? What conclusion? That there is no actual infinity? Or > that a TMs tape is not of infinite size? Or what? Whatever you answer is, > I'll accept it. As far as I can tell it doesn't make any difference. > > The conclusion of the original post. It is not that there is no actual > infinity. (That does not follow) > > That a TM tape is not of infinite length. It is unbounded, not > infinite. Plain and simple. But yes, I see your point, it's not > awfully important. What amazes me is that some people still think it > may be the case that some abstract machines in computer science > somehow do not have physical counterparts. [*] That's a basic > misunderstanding. As far as I can tell, computer science is concerned > with what kinds of machines are constructible in theory, in possible > worlds with physical laws like ours if we want to be philosophically > precise. > > Indeed, Turing machines are solely concerned with what finite > mechanisms can do in general. Godel observes this generality back in > 1950, why can't we now? > Because of the law of large numbers: most natural numbers are very, > very large. Seriously -- these finite mechanisms aren't much of a > bound. I can easily ask you to compute a function that would require > more memory (in bits) than there are electrons in the universe. > Calculate pi to 2^300 digits. Or fix a language and compute its busy > beaver function for n = 2^300. I don't think you understand the difference between unbounded and infinite, either. Here are some questions to consider, if you do regard yourself well educated on the notion of property. 1. What does unbounded mean in an infinite universe? 2. What does unbounded mean in a finite universe? > I doubt that people who have answered my post have thought deeply > about my argument. They don't seem to have considered all cases of > finite/transfinite in models of computation as I have done. > I don't think you actually know what transfinite means. I don't think you understood what is meant by finite/transfinite in the above sentence. Let me try again. I am showing you a PC. I am asking you: is this computer a Lambda-Calculus system of some sort? What do you say? > A question to consider: does it even make sense to talk about an > infinite size program? > > And another: is this an infinite string 0 1 1 1 ... 1 0 > How should we know? Those dots could represent a finite or infinite > sequence. Infinite sequence. Then, is it an infinite string? Yes, or no? I think you want to say Yes. That might help you to see the difference between unbounded and infinite, perhaps, think about it. See also my definition in Unbounded Space, and try to challenge it if you disagree with my views. > PS: For instance Stephen Harris's argument. A TM can calculate Pi, a > physical computer cannot, therefore TMs are not physical. Does Harris > even recognize that he is reinventing the substance dualism of > Descartes in computational terms? It almost sounds like A TM can > compute. Therefore TM is :) That's not how modern philosophy > proceeds. > Geez -- he's shown that a TM can do something that a computer cannot. > If computers and TMs had the same computational power, a computer > would be able to calculate pi. Modus tollens tells us that TMs and > PCs don't have the same computational power. Your blatant disregard > for Leibniz's Law appears schizophrenic. Harris has no inkling of Leibniz's Law. Show me how the concept of unbounded entails actual infinity, yet they are different as in Harris's argument. Some people cannot see that his argument is one of the most crackpottish claim ever on comp.theory, and yet swallowing it. He's been paddling it for so many years. You probably should consider your attitude towards crackpots like that. -- Eray === Subject: Re: Lambda Calculus and Turing Equivalence >It is equivalent in computational power to a TM. Accept or not? > > Accept. > >If you accept, my conclusion follows. Otherwise show a logical mistake. > > Er...conclusion? What conclusion? That there is no actual infinity? Or > that a TMs tape is not of infinite size? Or what? Whatever you answer is, > I'll accept it. As far as I can tell it doesn't make any difference. > > The conclusion of the original post. It is not that there is no actual > infinity. (That does not follow) > > That a TM tape is not of infinite length. It is unbounded, not > infinite. Plain and simple. But yes, I see your point, it's not > awfully important. What amazes me is that some people still think it > may be the case that some abstract machines in computer science > somehow do not have physical counterparts. [*] That's a basic > misunderstanding. As far as I can tell, computer science is concerned > with what kinds of machines are constructible in theory, in possible > worlds with physical laws like ours if we want to be philosophically > precise. > > Indeed, Turing machines are solely concerned with what finite > mechanisms can do in general. Godel observes this generality back in > 1950, why can't we now? > > Because of the law of large numbers: most natural numbers are very, > very large. Seriously -- these finite mechanisms aren't much of a > bound. I can easily ask you to compute a function that would require > more memory (in bits) than there are electrons in the universe. > Calculate pi to 2^300 digits. Or fix a language and compute its busy > beaver function for n = 2^300. > I don't think you understand the difference between unbounded and > infinite, either. Here are some questions to consider, if you do > regard yourself well educated on the notion of property. > 1. What does unbounded mean in an infinite universe? > 2. What does unbounded mean in a finite universe? I'm not really into considering counter factual universes. > I doubt that people who have answered my post have thought deeply > about my argument. They don't seem to have considered all cases of > finite/transfinite in models of computation as I have done. > > I don't think you actually know what transfinite means. > I don't think you understood what is meant by finite/transfinite in > the above sentence. > Let me try again. I am showing you a PC. I am asking you: is this > computer a Lambda-Calculus system of some sort? What do you say? No. The PC you've shown me (let's fix a specific computer -- the nice Apple PowerMac G5 I've wanted for a while -- it has 8 gb of ram and dual 2.5 GHz G5 processors) is only capable of being in finitely many states. At best (if it's in working order) it can do a large number of useful computations that could be done with the lambda calculus. But the lambda calculus is strictly more powerful *BECAUSE* it can do computations that my imaginary G5 can't. > A question to consider: does it even make sense to talk about an > infinite size program? > > And another: is this an infinite string 0 1 1 1 ... 1 0 > > How should we know? Those dots could represent a finite or infinite > sequence. > Infinite sequence. Then, is it an infinite string? Yes, or no? I think > you want to say Yes. That might help you to see the difference between > unbounded and infinite, perhaps, think about it. See also my > definition in Unbounded Space, and try to challenge it if you > disagree with my views. > PS: For instance Stephen Harris's argument. A TM can calculate Pi, a > physical computer cannot, therefore TMs are not physical. Does Harris > even recognize that he is reinventing the substance dualism of > Descartes in computational terms? It almost sounds like A TM can > compute. Therefore TM is :) That's not how modern philosophy > proceeds. > > Geez -- he's shown that a TM can do something that a computer cannot. > If computers and TMs had the same computational power, a computer > would be able to calculate pi. Modus tollens tells us that TMs and > PCs don't have the same computational power. Your blatant disregard > for Leibniz's Law appears schizophrenic. > Harris has no inkling of Leibniz's Law. I didn't ask you about Mr. Harris' argument. I told you that you're simply disregarding Leibniz's Law if you disagree with the claim that physical computers and Turing machines have the same computational power in spite of the fact that one can compute things the other cannot. Schizophrenia and dogmatism seem related, no? 'cid 'ooh === Subject: Re: Lambda Calculus and Turing Equivalence <4zrnd.46389$QJ3.16198@newssvr21.news.prodigy.com> <61bad26451853979aa126965352b4240.48257@mygate.mailgate.org> <5ci1q0tqrsrbnut54hbn7kbvt5g8idf1so@4ax.com> <9okrd.128309$5K2.64776@attbi_s03> at 12:15 PM, examachine@gmail.com (Eray Ozkural exa) said: >My silly argument is that TMs do *not* have an infinite tape, only >an unbounded tape (that is finite at ALL times), and this entails >that PCs are Turing Machines, unlike what the infinitely wise Robin >Chapman seems to think. Your argument isn't even coherent enough to qualify as silly. Your premise is more or less correct, but your conclusion does not follow from it. As anybody who knows anything about computers could tell you, a PC does not have unbounded storage; no amount of handwaving will change that. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail 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 not === Subject: Re: Lambda Calculus and Turing Equivalence > at 12:15 PM, examachine@gmail.com (Eray Ozkural exa) said: >My silly argument is that TMs do *not* have an infinite tape, only >an unbounded tape (that is finite at ALL times), and this entails >that PCs are Turing Machines, unlike what the infinitely wise Robin >Chapman seems to think. > Your argument isn't even coherent enough to qualify as silly. Your > premise is more or less correct, but your conclusion does not follow > from it. As anybody who knows anything about computers could tell you, > a PC does not have unbounded storage; no amount of handwaving will > change that. 1. What makes you think that the TM tape has to be mapped onto spatial dimensions? 2. What makes you think your argument refutes the argument in Section 8.1 of Introduction to Automata Theory, Languages, and Computation? Surely such a position not observe the difference between unbounded and infinite. A reflection on this follows. ---------------- Are you buying Stephen Harris's obsessions that unbounded must mean infinite? That in fact there is no difference between infinite and unbounded? I invite you to examine the following argument, and object step by step if you will, without the handwaving we are used to in elementary mathematics. Here it goes. In a finite universe, unbounded means: grows arbitrarily until it hits the physical size limits / law limits of the universe. It makes no sense to think of a computer the mass of which will cause a blackhole, or anything like that. That is because, there can be *nothing* bigger! Nothing bigger exists! There is one universe, and this is as good as it gets with respect to unbounded. Otherwise, the universe is infinite / allows infinite growth of machines, and in this case unbounded might actually entail infinite. However, and a very big however, unbounded does NOT necessarily entail infinite. Now, do you see why I told you that Stephen Harris cannot see the difference between unbounded and infinite? According to his professionally prolonged argument, there is no such difference, which puts him at odd with his own admission that he grants the difference between unbounded and infinite. -- Eray === Subject: Re: question about Banach spaces > Is there an accessible example of a linear > space endowed with two non-equivalent > complete norms? > jenny > Infinite-dimensional only. And AC required. How about this: > Banach spaces l^2 and l^1 are isomorphic as linear spaces > (both having Hamel dimension c). one has to necessarily appeal to AC? In other words, is AC equivalent to that statement? I am asking that, since I have a somehow realated problem: It is well known that the dual of L^infty strictly contains L^1 (this can be directly argued from separability arguments; clearly, only in infinite dimension)). In spite of that, I am not aware af any example of a continuous linear functional on L^infty which is not L^1 without appealing to the Hahn-Banach theorem, which is to say, without appealing to AC. jenny === Subject: Re: question about Banach spaces > Is there an accessible example of a linear > space endowed with two non-equivalent > complete norms? > jenny > Infinite-dimensional only. And AC required. How about this: > Banach spaces l^2 and l^1 are isomorphic as linear spaces > (both having Hamel dimension c). > one has to necessarily appeal to AC? In other words, is AC > equivalent to that statement? Equivalent, probably not, but in fact it cannot be proved in ZF alone. In Solovay's model where every set of reals (and every set in a Polish space) has the property of Baire, it follows (proved by Christensen, I guess) that any linear map of separable Banach spaces is automatically continuous. So, in particular, if a given vector space admits two complete separable norms, then the identity is a homeomorphism. (And since, in metric spaces, a map is continuous if and only if its restriction to every separable subspace is, we can get rid of the separable assumptions I made above.) That's in Solovay's model. But there is some principle that goes with this saying any spaces and maps THAT YOU CAN ACTUALLY WRITE DOWN EXPLICITLY also work like this, merely using ZF. So, depending on what accessible means in the original question, the answer may be no. > I am asking that, since I have a somehow realated problem: > It is well known that the dual of L^infty strictly contains L^1 > (this can be directly argued from separability arguments; clearly, only > in infinite dimension)). > In spite of that, I am not aware af any example of a continuous linear > functional on L^infty which is not L^1 without appealing to > the Hahn-Banach theorem, which is to say, without appealing to AC. Again, this existence cannot be proved in ZF. But it can be proved using HB. It is known that HB is strictly weaker than AC, so this answers the equivalent question above. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: question about Banach spaces > Is there an accessible example of a linear > space endowed with two non-equivalent > complete norms? > jenny Infinite-dimensional only. And AC required. How about this: > Banach spaces l^2 and l^1 are isomorphic as linear spaces > (both having Hamel dimension c). > one has to necessarily appeal to AC? In other words, is AC > equivalent to that statement? > Equivalent, probably not, but in fact it cannot be proved in ZF alone. (also for David C. Ullrich) When you say that it cannot be proved what do you exactly mean? I mean, if it is provable with AC but not provable without AC why it is not automatically equivalent to AC. My point is: one cannot prove this within ZF because he/she is not able to, but in principle he/she could, or there is a way to show this impossibility? jenny === Subject: Re: question about Banach spaces > Is there an accessible example of a linear >> space endowed with two non-equivalent >> complete norms? >> jenny >> Infinite-dimensional only. And AC required. How about this: > Banach spaces l^2 and l^1 are isomorphic as linear spaces >> (both having Hamel dimension c). > one has to necessarily appeal to AC? In other words, is AC >> equivalent to that statement? >> Equivalent, probably not, but in fact it cannot be proved in ZF alone. >(also for David C. Ullrich) >When you say that it cannot be proved what do you exactly mean? >I mean, if it is provable with AC but not provable without AC >why it is not automatically equivalent to AC. >My point is: one cannot prove this within ZF because he/she is >not able to, but in principle he/she could, or there is a way to >show this impossibility? There is a way to show this impossibility. There was a big hint how that works in the paragraph you snipped, where Edgar said something about a certain model of ZF. Any statement that can be proved from the axioms of ZF will be true in every model of ZF, so if there is a model of ZF in which P is false it follows that P _cannot_ be proved in ZF. You need to study a small bit of mathematical logic to really know exactly what that means. For now an analogy: Say GT is the axioms for a group: (ab)c = a(bc), etc. So for example (ab(cd) = (a(bc))d is a theorem of GT - it can be proved from the axioms. Now a model of GT is precisely the same thing as a _group_. There exists a group in which it is not true that ab = ba for all a, b, and the existence of such a group shows that ab = ba for all a, b cannot be proved from the axioms of GT. >jenny ************************ David C. Ullrich === Subject: Re: question about Banach spaces > (also for David C. Ullrich) > When you say that it cannot be proved what do you exactly mean? > I mean, if it is provable with AC but not provable without AC > why it is not automatically equivalent to AC. > My point is: one cannot prove this within ZF because he/she is > not able to, but in principle he/she could, or there is a way to > show this impossibility? An example is the Hahn-Banach theorem, HB. Assuming consistency when needed, it has been shown that: (1) ZF does not prove HB, (2) ZF+AC proves HB, (3) ZF+HB does not prove AC. In that sense, HB is beyond ZF, but is not as strong as AC. On the other hand, Zorn's Lemma, ZL, is equivalent to AC, meaning: (4) ZF+AC proves ZL (5) ZF+ZL proves AC In most cases, a cannot be proved result is shown by exhibiting a model. For example, a model of ZF+HB where AC fails will show (3), above. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: question about Banach spaces >> Is there an accessible example of a linear >> space endowed with two non-equivalent >> complete norms? >> jenny > Infinite-dimensional only. And AC required. How about this: >> Banach spaces l^2 and l^1 are isomorphic as linear spaces >> (both having Hamel dimension c). >one has to necessarily appeal to AC? In other words, is AC >equivalent to that statement? Not that I know the answer to either question, but I'll point out that the in other words isn't right - saying something requires AC is not the same as saying it's equivalent to AC. If ZF does not prove P but ZFC does then P requires AC; saying P is equivalent to AC is a much stronger statement. (For example: If ZFC proves P but, say, ZF + AD (Axiom of Determinacy) proves not P then AC is required for P, although this doesn't say that P is equivalent to AC. This actually happens if P is every set of reals is Lebesgue measurable.) >I am asking that, since I have a somehow realated problem: >It is well known that the dual of L^infty strictly contains L^1 >(this can be directly argued from separability arguments; clearly, only >in infinite dimension)). >In spite of that, I am not aware af any example of a continuous linear >functional on L^infty which is not L^1 without appealing to >the Hahn-Banach theorem, which is to say, without appealing to AC. >jenny ************************ David C. Ullrich === Subject: [OT] Re: Explaining the prime distribution Have a look at Andrew Koenig and Barbara E. Moo, _Accelerated C++_. It's < 340 pp. and presents the language in a way that makes good sense to mathematicians. The whole idea of C++ (as I see it) is that it allows you to keep extending it with higher levels of abstraction that match the logic of higher levels of abstraction that you use in math. Amittai >I wish not to distract from the Subject line, but much of what I read about > learning C++ suggests and states that it is very difficult to learn. What > is > the truth? What path does one usually take in order to learn C++ more > easily, > more effectively and reliably? > G C > Tim Smith discusses C++ program for prime numbers.... >>something that seems to give the right answers. Here is the code if >>anyone is curious. No optimizations--just very straightforward >>translation >>of his definition to simple C++. >> #include > #include > #include > int dS( int x, int y ); >> int p( int x, int y ); >> int S( int x, int y ); >> int main( int argc, char *argv[] ) >> { >> int x = atoi( argv[1] ); >> int y = atoi( argv[2] ); >> printf( S(%d, %d) = %dn, x, y, S(x,y) ); >> return 0; >> } >> int dS( int x, int y ) >> { >> return (p(x/y, y-1) - p(y-1, (int)sqrt(y-1))) * >> (p(y,(int)sqrt(y)) - p(y-1,(int)sqrt(y-1))); >> } >> int p( int x, int y ) >> { >> return x - S(x,y) - 1; >> } >> int S( int x, int y ) >> { >> if ( y == 1 ) >> return 0; >> int r = 0; >> for ( int i = 2; i <= y; ++i ) >> r += dS(x,i); >> return r; >> } >>-- >>--Tim Smith === Subject: Re: RE. I don't understand |It's often considered rude to request an e-mail response. Sometimes. |If you |post a question in a newsgroup, you should take the effort to |revisit the newsgroup to see any responses. It's true that people tend not to like hit-and-run posting. I've seen netiquette recommending asking the newsgroup for responses and offering to summarize them to the newsgroup afterward. I haven't often seen this done, but sometimes it makes sense. Keith Ramsay === Subject: Re: Pi in space |Defining pi to be the empirically measured ratio of a circle's circumference |to its diameter, the only geometry in which this is ratio is a constant is |Euclidean geometry. Pi doesn't exist as a constant that can be empirically |measured in any other geometry, so its value in other geometries is a |meaningless question. Obviously this is not a practical issue, but I thought as long as we were busy making points about what is true in principle, I would note that there often is a natural relationship between pi and other geometries. In some other geometries, pi is the limit of the ratio between the circumference and the diameter of a circle as the radius goes to 0, and could in principle (if you had such a space in your hands...) be measured that way. At a fixed radius, measuring the circumference and diameter would only be able to approximate pi up to a certain degree of precision, but measurements of lengths are always only up to a certain degree of precision anyway, so one is not any worse of in an essential way. On the hyperbolic plane, for example, you could find a length scale on which a triangle whose sides have lengths in a 3:4:5 ratio has an angle of between 90 and 91 degrees. One can then put an upper bound on the deviation of the ratio of the circumference to the diameter from pi, in terms of the ratio of the radius to the length of a side of this triangle. The circumference of a circle of radius r in the hyperbolic plane is 2*pi*a*sinh(r/a) where the length a depends on the curvature. The size of the 3:4:5 triangle with an angle of 91 degrees is likewise proportional to this length a. (I won't bother figuring out what the ratio is right here.) So to get n digits, we could make a circle whose radius is roughly 10^{-n/3} times the length of a side of this triangle, and measure it very precisely. Keith Ramsay === Subject: Re: Pi in space > I was having an argument with a friend here and he claimed that pi > doesn't really have all this significance it's attached to it, since it > is transcendental only in the theoretical space of Euclidian geometry > and all know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the > circumference of a circle to its diameter inside the Einsteinian > universe rational or irrational? Anyone knows? We don't know the true geometry of space. In particular, we don't know if it's continuous. Digital physicists think it is not: see www.digitalphysics.org The concept of Pi, however, should not be regarded transcendental or anything like that since it is represented perfectly with a small program. True, it is an idealization, but what is transcendental? In my opinion, nothing transcends physical reality, and if Pi is part of physical models, it is for a good reason: physics is mostly about ideal models that work in ideal conditions, etc. Note however, that, if the universe turns out to be discrete, then, obviously, the real number Pi does not describe any physical reality. The computation of Pi is still sensible, however, (e.g. computation of Pi up to the nth decimal place) and it would indeed describe physical reality in a discrete universe (if it's an actually infinite discrete universe, things get more complicated so just assume finite which seems to be the case for *our* universe;). Most mathematicians on this forum seem to be Platonists, or Platonists who lack the philosophical maturity to identify themselves as Platonists, so you could expect quite a lot of Platonist nonsense in response to your question. However, as one poster correctly pointed out, constructivists and formalists will most likely look at Pi as a useful mental construct of some sort, and no talk of transcending anything will be necessary. Pi is not in space: it's in your mind. [*] Indeed, constructivism is a more favorable position than Platonic Realism, in my opinion, and a point of view supported by great mathematicians, so it's an alternative you should think about. More recently, instrumentalism has been suggested as an alternative way to look at mathematics. In my opinion, that too is considerable. -- Eray Ozkural There is no perfect circle [*] A caveat, however, there are perfect discrete circles in the world. It just may be the case that there are no perfect continuous circles, except conceptually!!!!!! === Subject: Re: Pi in space |Most mathematicians on this forum seem to be Platonists, or Platonists |who lack the philosophical maturity to identify themselves as |Platonists, so you could expect quite a lot of Platonist nonsense in |response to your question. No, because such points of philosophy are thoroughly irrelevant to his question. |However, as one poster correctly pointed |out, constructivists and formalists will most likely look at Pi as a |useful mental construct of some sort, and no talk of transcending |anything will be necessary. Pi is not in space: it's in your mind. [*] I am disinclined to assume that you know that transcendental is a technical mathematical term meaning not a root of a nonzero polynomial with integer coefficients. |Indeed, constructivism is a more favorable position than Platonic |Realism, in my opinion, and a point of view supported by great |mathematicians, so it's an alternative you should think about. Kronecker, Brouwer, and Bishop were all outstanding mathematicians (whether we call them great sort of depends on how high a standard we set for greatness). Bishop supported constructivism; Brouwer supported some kind of constructivism; Kronecker appears to have supported something along those lines. I wouldn't say, however, that popularity among great mathematicians is a very good standard for judging such things. To the extent that it can be trusted at all, I see no obvious sign of the very best mathematicians differing very far in terms of the distribution of their opinions on Platonism (or realism), formalism, constructivism and so on from the run-of-the-mill. I also don't see sci.math as being much of a hotbed of Platonism or realism. Keith Ramsay === Subject: Re: Pi in space >The concept of Pi, however, should not be regarded transcendental or >anything like that since it is represented perfectly with a small >program. True, it is an idealization, but what is transcendental? In >my opinion, nothing transcends physical reality, and if Pi is part of >physical models, it is for a good reason: physics is mostly about >ideal models that work in ideal conditions, etc. I think you need to look up the definition of transcendental number. You seem to be confusing it with some philosophical notion. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Pi in space >The concept of Pi, however, should not be regarded transcendental or >anything like that since it is represented perfectly with a small >program. True, it is an idealization, but what is transcendental? In >my opinion, nothing transcends physical reality, and if Pi is part of >physical models, it is for a good reason: physics is mostly about >ideal models that work in ideal conditions, etc. > I think you need to look up the definition of transcendental number. > You seem to be confusing it with some philosophical notion. the concept of a transcendental number. AFAICT, the notion of a transcendental number is not relative to a give geometry, or the true geometry of our universe. Pi is a transcendental number, according to the definition of a transcendental number in mathematics. I do think his question ought to be philosophical. Why should he talk about relativity then? What difference is there between any continuous metric and a Riemann tensor regarding the fact that Pi is a transcendental number? Maybe, I think, his friend's intention was to give a better definition of what it means for a number to be transcendental than the ordinary usage. He might want to pick another term, though, or highlight the difference than the ordinary usage carefully. Otherwise, we become confused in argumentation. I thought so, and said that this philosophical sense is probably irrelevant to a computable real like Pi which captures a general geometric fact in a compact form! -- Eray Ozkural === Subject: Big-O function conversion Michael Sipser, _Introduction to the Theory of Computation_ (233-34) gives the following Theorem 7.10 about the relation between the asymptotic time classifications of nondeterministic and corresponding deterministic Turing machines: Let t(n) be a function, where t(n) >= n. Then every t(n) time nondeterministic single-tape Turing machine has an equivalent 2^{O(t(n))} time deterministic single-tape Turning machine. After he gives the analysis of the conversion from nondeterministic machine N to deterministic machine D, he concludes: Therefore, the running time of D is O(t(n)*b^{t(n)}) = 2^{O(t(n))} . I can follow everything except the last equation/conversion. Could anybody Amittai Aviram === Subject: Re: Big-O function conversion : Michael Sipser, _Introduction to the Theory of Computation_ (233-34) gives : the following Theorem 7.10 about the relation between the asymptotic time : classifications of nondeterministic and corresponding deterministic Turing : machines: : Let t(n) be a function, where t(n) >= n. Then every t(n) time : nondeterministic single-tape Turing machine has an equivalent 2^{O(t(n))} : time deterministic single-tape Turning machine. : After he gives the analysis of the conversion from nondeterministic machine : N to deterministic machine D, he concludes: : Therefore, the running time of D is O(t(n)*b^{t(n)}) = 2^{O(t(n))} . : I can follow everything except the last equation/conversion. Could anybody : Amittai Aviram t(n)*b^{t(n)} = 2^{log2(t(n))+log2b*t(n)} which is 2^{O(t(n)} Stephen === Subject: Re: Big-O function conversion Amittai > : Michael Sipser, _Introduction to the Theory of Computation_ (233-34) > gives > : the following Theorem 7.10 about the relation between the asymptotic > time > : classifications of nondeterministic and corresponding deterministic > Turing > : machines: > : Let t(n) be a function, where t(n) >= n. Then every t(n) time > : nondeterministic single-tape Turing machine has an equivalent > 2^{O(t(n))} > : time deterministic single-tape Turning machine. > : After he gives the analysis of the conversion from nondeterministic > machine > : N to deterministic machine D, he concludes: > : Therefore, the running time of D is O(t(n)*b^{t(n)}) = 2^{O(t(n))} . > : I can follow everything except the last equation/conversion. Could > anybody > : Amittai Aviram > t(n)*b^{t(n)} = 2^{log2(t(n))+log2b*t(n)} which is 2^{O(t(n)} > Stephen === Subject: Re: induction vs Cantor > Cantor's diagonalization procedure > defines a function which given an infinite list of reals (that is, > an element of R^N) returns an element of R that is not on the list. > Cantor didn't prove that there were uncountably many such functions, > he only proved that there was one. That's all he needs to be able > to show that no list of reals contains every real. > Do you think there are any counter-examples to Cantor's diagonal > proof? No. > If so, can you provide a valid modification that keeps the concept > of providing a real that is proven not in the list and has no > counter-example? NA. === Subject: Re: induction vs Cantor > ... >... > If so, can you provide a valid modification that keeps the concept > of providing a real that is proven not in the list and has no > counter-example? > NA. There exists a countable model of your ZF set. === Subject: Re: induction vs Cantor >> ... >>... >> If so, can you provide a valid modification that keeps the concept >> of providing a real that is proven not in the list and has no >> counter-example? >> NA. >There exists a countable model of your ZF set. Sure, but inside such models it is still the case that |N| < |R|. Since |N| < |R| follows from the ZF axioms, it must hold in all models that satisfy those axioms, regardless of the size of the model (as viewed from the outside). -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: induction vs Cantor >> ... >>... >> If so, can you provide a valid modification that keeps the concept >> of providing a real that is proven not in the list and has no >> counter-example? > NA. >There exists a countable model of your ZF set. > Sure, but inside such models it is still the case that |N| < |R|. Since |N| > < |R| follows from the ZF axioms, it must hold in all models that satisfy > those axioms, regardless of the size of the model (as viewed from the > outside). Hi Barb, That leads to a problem. The set N, the natural numbers, in any of those extensions, contains no elements not contained in each other copy of the set N, and it contains each of them. The copies are equal and identical. There is thus identity, a tautology, between any copies of N, and as well between any copies of P(N), Then, if there is a bijection between N and P(N), there is a bijection between N and P(N). theory bogosity is a reaction to Skolem, who otherwise says infinite sets are equivalent. Look outside. Do you use transfinite cardinals? Is it for anything besides transfinite cardinals? Some people use them as a definition as part of measure theory, but that's because they lack better tools, and the results of measure theory are largely upon continua. That is to say, the useful results of measure theory are derivable without the use of transfinite cardinals, and they should be. Of your ordinals, is there any that specifically represents the integer -1? Is not the order type of the powerset of the naturals the successor of the order type of the naturals? If it's infinite there's always one more. That's part of the basis of induction, that something can be proven to hold for each, thus that for each it is proven, as opposed to proved. When we get to the contradiction between each infinite set being equivalent inductively and the necessity of dual representation via Cantor's results, then accept that there is dual representation instead of that induction fails. That way, you're on the path to being inside first order logic, and nobody can prove you incomplete, inconsistent, and immaterial. If you want a consistent theory, start by removing the inconsistencies, not reusing them. Ross F. === Subject: Re: induction vs Cantor >> ... >>... > If so, can you provide a valid modification that keeps the concept > of providing a real that is proven not in the list and has no > counter-example? > NA. >There exists a countable model of your ZF set. >> Sure, but inside such models it is still the case that |N| < |R|. Since |N| >> < |R| follows from the ZF axioms, it must hold in all models that satisfy >> those axioms, regardless of the size of the model (as viewed from the >> outside). >Hi Barb, >That leads to a problem. >The set N, the natural numbers, in any of those extensions, contains no >elements not contained in each other copy of the set N, and it contains each of >them. The copies are equal and identical. >There is thus identity, a tautology, between any copies of N, I agree. N (via von Neumann ordinals) is categorical in ZF, so the N in different models would be isomorphic. >and as well between any copies of P(N), That looks like a serious misstep. AIUI (which admittedly isn't very deeply), the downward LST gives a model in which all the elements correspond to first-order-definable terms. So, the set of such elements of P(N) in the LST countable model is very much smaller than P(N) in the usual iterative-hierarchy models (which includes the uncountably-many elements which are not first-order definable). (It's rather like the difference between the set of computable infinite binary sequences and the set of all infinite binary sequences computable or not.) But |N| < |P(N)| even in the countable model, because the bijection between N and the model's P(N) is not itself first-order definable and therefore does not exist in the model! ISTM you have strong Constructivist leanings. That's fine, but it's important then not to mix constructive and non-constructive entities in the same argument. If you exclude all non-constructive entities then you can't argue that |N| = |P(N)| in the countable model, because the isomorphism would be non-constructive and therefore not talkable-about. > Then, if there is a bijection between N and P(N), >there is a bijection between N and P(N). >theory bogosity is a reaction to Skolem, who otherwise says infinite sets are >equivalent. I'm sure he doesn't actually say or imply that. >Look outside. OK. >Do you use transfinite cardinals? I don't see any outside, but then again I don't see ANY mathematical entities outside -- there are no natural numbers flitting through the trees. >Is it for anything besides transfinite cardinals? Sure. One needs at least P(N) for representing the real numbers when putting Analysis on a set-theory foundation. >Some people use them as a definition as part of measure theory, but that's >because they lack better tools, and the results of measure theory are largely >upon continua. That is to say, the useful results of measure theory are >derivable without the use of transfinite cardinals, and they should be. Giving up continua is asking a lot. Continuum Mechanics is a useful area with many practical applications. -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: induction vs Cantor; nonstandard numbers in ZF >>... >> If so, can you provide a valid modification that keeps the concept >> of providing a real that is proven not in the list and has no >> counter-example? > NA. >There exists a countable model of your ZF set. > Sure, but inside such models it is still the case that |N| < |R|. Since |N| > < |R| follows from the ZF axioms, it must hold in all models that satisfy > those axioms, regardless of the size of the model (as viewed from the > outside). >>Hi Barb, >>That leads to a problem. >>The set N, the natural numbers, in any of those extensions, contains >>no elements not contained in each other copy of the set N, and it >>contains each of them. The copies are equal and identical. >>There is thus identity, a tautology, between any copies of N, > I agree. N (via von Neumann ordinals) is categorical in ZF, so the N > in different models would be isomorphic. That can't be right....*thinking*....nope. It's not right. You can just run familiar compactness arguments. In ZF we can define the finite von Neumann ordinals in a variety of ways, where define here means that we have a formula Nat(x) that is true of exactly the real finite von Neumann ordinals *in a standard model of ZF*. For instance, let Ord be your favorite definition of the von Neumann ordinals .9fberhaupt, say Ord(x) =df Trans(x) and Connected(x), and define x to be a natural number just in case it and all its elements are either 0 or successor ordinals: Nat(x) =df (y)(y=x v y in x -> (y=0 v (exists z)(y=zU{z}))). TeXnically speaking, we usually call the set {x | Nat(x)} omega, of course. But now add the constant k to the language of ZF and consider the following set of sentences K = {Nat(k), k != 0, k != 1, k != 2, ...}. For any finite subset K' of K, ZF U K' is obviously satisfiable (if ZF is). Hence, by compactness, so is ZF U K. But obviously omega (that is, the set containing exactly the elements satisfying Nat(x)) in any model of ZF U K can't be isomorphic to N. And any such model, of course, can be turned into a model of ZF by restricting it to the original language of ZF. The problem, of course, ZF being first-order and all, is that you can't rule out the possibility nonstandard finite ordinals that, e.g., are transitive in the model but not in the real world. I reckon, in fact that the sets playing the role of omega in models of ZF U K probably have properties similar if not identical to the usual nonstandard models of PA. Who's an expert around here? Herb Enderton, Mike Oliver, Aatu K, etc should be able to tell us more. Chris Menzel === Subject: Re: induction vs Cantor <8JqdnQxnYLcA6zrcRVn-3w@giganews.com> <41a7dbeb$0$20860$afc38c87@news.optusnet.com.au> <1fRrd.107121$T02.77638@twister.rdc-kc.rr.com> <41B40665.2DB637D2@tiki-lounge.com> <41B4DED8.947CA9BB@tiki-lounge.com> Hi Barb, Yeah, it is. The reals are great. Their simple assumption as a continuum offers a necessary tool for the pursuit of many and much mathematics. Most of the fundamental results about real numbers are in place without set theory, eg via Euclid. The (set of) real numbers is the indefinite contiguous sequence, the point set. You use zero, and iota, and integral multiples of iota, and those are all the non-negative real numbers. Mapping the integers to the reals in this way escapes the consequences of Cantor's first proof, which some apply to the rational numbers, and I to neither. Zorn's Lemma, the well-ordering principle, or the Axiom of Choice allow methods to guarantee enumerability of the sets. Well-order the reals and inductively select elements to inject into the integers, for the integers, the existence of integer n guarantees the existence of integer n+1. I agree that I'm constructivist, but I'm more quasi-intuitionist and not prejudiced. Also I'm a platonist because math is real. Math can be quite irrelevant. The natural numbers are the trees. Do you use ZF with classes? What's the class of all classes? If your answer is no, can there be more than one proper class? There are theories with a set of all sets, that set is its own powerset, there the identity and tautology is between the set and itself, its own powerset. That's for intuitionism to resolve, the singular excluded excluded middle and the double entendre. Double entendre: a double entendre. The proper class is the ur-element, it's at once all and nothing, and the void and null. Its dualistic nature is its singular nature, and vice versa. Ross F. === Subject: Re: induction vs Cantor <8JqdnQxnYLcA6zrcRVn-3w@giganews.com> <41a7dbeb$0$20860$afc38c87@news.optusnet.com.au> <1fRrd.107121$T02.77638@twister.rdc-kc.rr.com> <41B40665.2DB637D2@tiki-lounge.com> <41B4DED8.947CA9BB@tiki-lounge.com> RF> The reals are great. How the would YOU know THAT? You don't even know a basic definition of what they are. RF> Their simple assumption as a continuum There is nothing simple about it TO YOU, idiot. To the rest of us, there is a simple first-order axiomatization, but it is hardly a simple assumption as a continuum; it is an assumption that they are an ordered field that is complete in the sense of containing all of the things you can produce from it by the operation of taking limits of COUNTABLY infinite sequences of things already in it. RF> offers a necessary tool for the pursuit of many and much mathematics. RF> Most of the fundamental results about real numbers are in place RF> without set theory, Yes, there are axioms defining the reals without mentioning sets. RF> eg via Euclid. To prove that Euclid is an example of that, you would have to give an example FROM Euclid of some results about reals. RF> The (set of) real numbers is the indefinite contiguous sequence, In modern set theories, you CAN'T just gloss over that in parentheses: you have to PICK A SIDE: is the class of all reals proper, or is it a set???? RF> the point set. You use zero, and iota, and integral multiples of iota, RF> and those are all the non-negative real numbers. It is a consequence of the fact that first-order languages are countable that you can model any consistent first-order theory countably, so if there are denumerably many iota-terms, yes, there is a way you can make them serve as a model. BUt that does NOT mean that they really are the reals. RF> Mapping the integers to the reals in this way escapes the RF> consequences of Cantor's first proof, which some apply RF> to the rational numbers, and I to neither. NOthing escapes the proof. No matter how many iota's you use to represent the reals, they will STILL all have denumerably-long-bit-string representations as well, and all you have to do to reproduce the proof in the iota- context is define the function that returns the nth decimal place (or bit) of a real as output, give the right number of iotas as input. RF> Zorn's Lemma, the well-ordering principle, RF> or the Axiom of Choice allow RF> methods to guarantee enumerability of the sets. They DO NOT, dumbass. All these things are NON- constructive. You do get an enumeration or well-ordering from them but you NEVER get a METHOD! RF> Well-order the reals and inductively select elements to RF> inject into the integers, Obviously, this is not possible; you run out of natnums. RF> for the integers, the existence of integer n guarantees RF> the existence of integer n+1. And the existence of a term with n iota's guarantees the existence of a term with n+1 iota's. But nothing guarantees that you can inject the reals into either of those. RF> Do you use ZF with classes? NObody uses ZF with classes. ZF DOES NOT HAVE classes. In ZF, EVERYTHING IN THE ENTIRE UNIVERSE is a set. Proper classes DO NOT EXIST. RF> What's the class of all classes? A contradiction in terms, THAT'S what it is. RF> There are theories with a set of all sets, that set is its own RF> powerset, there the identity and tautology is between RF> the set and itself, its own powerset. You haven't actually studied any set theories with a universal set and you DON'T know WTF you are talking about. All these theories have some VERY counter-intuitive consequences at very simple levels. === Subject: Re: induction vs Cantor <8JqdnQxnYLcA6zrcRVn-3w@giganews.com> <41a7dbeb$0$20860$afc38c87@news.optusnet.com.au> <1fRrd.107121$T02.77638@twister.rdc-kc.rr.com> <41B40665.2DB637D2@tiki-lounge.com> <41B4DED8.947CA9BB@tiki-lounge.com> > RF> The reals are great. > How the would YOU know THAT? > You don't even know a basic definition of what they are. > RF> Their simple assumption as a continuum > There is nothing simple about it TO YOU, idiot. > To the rest of us, there is a simple first-order > axiomatization, but it is hardly a simple assumption > as a continuum; it is an assumption that they are > an ordered field that is complete in the sense of containing > all of the things you can produce from it by the operation of > taking limits of COUNTABLY infinite sequences of things > already in it. Erm, for every cauchy sequence... or for every dedikind cut... is second order. The first order theory of the real field is just the theory of real closed fields axiomatised via ordered field + square roots of positive elements + roots of odd degree polynomials. === Subject: Anyone can check if these homework questions are correct I have a couple of homework questions, appreciate if someone can take the time to check my answers. Q1 Given that voltages v1 = 12 sin 100 pi t and v2 = 20 sin ( 100 pi t + (pi / 3) ). State the minimum value and the phase angle (relative to v1) of the resultant voltage v1 + v2 by writing the sum as a single sinusoid. My answer 28 sin ( 100 pi t + 0.667 ) In an ac circuit i = 100 sin 20 pi t amperes and v = 50 sin (20 pi t - ( pi / 6 )) volts Instantaneous power p, is given by p = vi watts Use products-to-sums formulae to express p in a form involving only one sinusoid and hence state the maximum power. My answer 2165 - 2500 cos ( 20 pi t - (pi / 6) ) TIA === Subject: Multiplication of negative numbers I'm going to be hated for bringing this up... but I'm not content with many answers that I have read in the past. Why does the multiplication of two negative numbers return a positive number? I know everyone here can give me an algebraic proof, but how is it that one can give a definition of multiplication using multiplication? This is akin to saying that the definition : An apple is an apple. is correct. Let's put down the axioms of arithmetic aside and say that we only have addition. We know that 5+5=10, and because of the need to show (say) an individual's debt, we introduce negative numbers. It is quite easy to show from this that 5+(-5)=0. But what of multiplication? What exactly *is* multiplication? Multiplication is a shorthand for addition. So instead of having to write: (5+5+5+5+5) = 25, we can simply use the shorthand: 5*5 = 25. That is all it is... shorthand. It is generally understood that within multiplication, one number is a number to be multiplied, and the other is the multiplier. In 3*5, we can call 3 the multiplied, and 5 the multiplier, and represent this as (3+3+3+3+3). On the other hand, we can call 5 the multiplied, and 3 the multiplier, and can be represented as (5+5+5). Both give us the same answer. So what about multiplying a negative number? -5*3 = (-5) + (-5) + (-5) = -15. In this situation, it is easy to always call the positive number the multiplier, so that you can add any number (-/+) to itself a positive number of times. Alright... we are still within our existing axioms. But now a problem has come up... What happens when multiplying two negative numbers? The first number could be called the multiplied, and the second the multiplier. So: in -5*-3, -5 is the multiplied, and -3 is the multiplier. It doesn't sound right to add something to itself negative 3 times. Actually, this phenomenon doesn't occur anywhere in the real world. Nevertheless, it seems that it must be defined. So... even though it doesn't occur in the real world, we feel a need to define it. Maybe we could define it as, instead of with a negative times a positive (e.g., -5*3 = (-5 + -5 + -5)), we could define multiplication of two negatives to use subtraction rather than addition, so: (-5)*(-3)= (-5) - (-5) - (-5) = 5. We'll call this intuitive multiplication, or IM. But, of course, IM is not commutative. (-3)*(-5) is not equal to (-5)*(-3). So, we would could come up with a seperate definition for multiplying negatives so that multiplication will be commutative. If we defined multiplication so that, when two negatives are multiplied (e.g., -5*-3), we can say that the answer will be the same as (-(-5*3)). This way, multiplication can be commutative, same as addition. We can call this kind of multiplication: consistent multiplication, or CM. But why does multiplication need to be commutative? Simply because it is based off of addition? Well, subtraction is based off of addition, but no one cares that it isn't commutative. Same for division. Really, we can go so far to say that addition is NOT commutative, because (-3) + 5 is NOT equal to (-5) + 3. Addition is only commutative when dealing solely with positive numbers. The reason is that when negative numbers become involved, commutative laws do not always apply. So why did we decide to go with CM, rather than IM? After all, if we adopted IM rather than CM, there would exist a number 'i' such that i^2 = -1, therefore eliminating the need for a whole new set of numbers (complex numbers). If we had adopted IM rather than CM, would the world of mathematics be a better place? Can anyone explain to me if chosing between IM and CM was arbitrary, or if the decision to use CM was really necessary after all? I am not a math wiz. Actually, I hate mathematics. I can't stand the fact that everyone likes to think Mathematics is an absolute truth, when really it is all just as subjective as, say, religion. But beyond this, maybe there is something here that I am overlooking completely. Can anyone help out? Maybe I'm wrong about Math really being subjective, but I really need someone to shed some light on these kinds of things. Again, I'm not looking for a proof of multiplication using multiplication. That is circular logic. If you can't provide some deeper understanding, than please don't reply. === Subject: Re: Multiplication of negative numbers How's this for a physical interpretation of what's going on? Imagine that you have a rectangular piece of cardboard having dimensions m and n, divided into unit squares. One side is colored blue and the other is colored red. Place the retangle blue face up showing positive area. Now change the sign of n by flipping the cardboard theough a side of lenght m. Now the red face is up, which may be interpreted as flipping the area from positiv to negative when a side is so inverted. Next flip the cardboard again through a side of length n, changing the sign of the other factor m. The blue side has come back up; the area with two negative dimensions is positive. --OL === Subject: Re: Multiplication of negative numbers at 02:09 PM, afazio@gmx.net (Alfred Fazio) said: >Why does the multiplication of two negative numbers return a positive >number? Because it's defined that way. Why is it defined that way? Because that preserves some nice properties of addition and multiplication. >But what of multiplication? What exactly *is* >multiplication? Multiplication is a shorthand for addition. No. If you want to know the area of a rectangular field you multiply the length by the width; if you want to know the are of a circle you multiply Pi by the square of the radius. In neither case is the multiplication a shorthand for addition. >Maybe we could define it as, instead of with a negative times a >positive (e.g., -5*3 = (-5 + -5 + -5)), we could define >multiplication of two negatives to use subtraction rather than >addition, so: (-5)*(-3)=(-5) - (-5) - (-5) = 5. We'll call this >intuitive multiplication, or IM. That would be bizarre, because there is nothing intuitive about such a definition. Further, it fails to address the question of how to multiple negative numbers that are not integers. >So why did we decide to go with CM, rather than IM? Because CM is elegant and usefull; IM is neither. >After all, if we adopted IM rather than CM, there would exist a >number 'i' such that i^2 = -1, therefore eliminating the need for a >whole new set of numbers (complex numbers). No. You don't understand the reasons why complex numbers are important. >Can anyone explain to me if chosing between IM and CM was arbitrary, Probably not, since you'd need to first learn background information that you have no interest in learning. >Actually, I hate mathematics. That makes it unlikely that you will ever invest the time and effort into developing any insights into why things are the way that they are and whether they could be improved. You're in the position of someone who says I hate music and writing about how much better the Dvorak Cello Concerto would have been had it been scored for bagpipes. >I can't stand the fact that everyone likes to think Mathematics is an >absolute truth, when really it is all just as subjective as, say, >religion. Given your dislike of Mathematics, your beliefs are irrelevant; you obviously have no insight either into what everyone thinks or into what is true. >But beyond this, maybe there is something here that I am overlooking >completely. Much, but your lack of objectivity will forever prevent you from seeing it. >Can anyone help out? No. You can't help a person who doesn't want to be helped. >Maybe I'm wrong about Math really being subjective, but I really need >someone to shed some light on these kinds of things. Why? Wouldn't it be better to worry about a subject that you don't hate? >If you can't provide some deeper understanding, than please don't >reply. If you aren't willing to read the replies that people post, then more than a little presumptuous. Usenet is a public forum; others have as much right to post as you do, even if you don't like what they have to say. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail 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 not === Subject: Re: Multiplication of negative numbers > Really, we can go so far to say that addition is NOT commutative, > because (-3) + 5 is NOT equal to (-5) + 3. ??? (-3) + (+5) = (+5) + (-3) = 2. Just curious, what's your math background. Doesn't look like anything above kindergarten. === Subject: Re: Multiplication of negative numbers > I'm going to be hated for bringing this up... but I'm not content with > many answers that I have read in the past. > Why does the multiplication of two negative numbers return a positive > number? At a very intuitive level, the - in a negative number encodes the property of Going Backwards. Going Backwards from Going Backwards sends you Forward. This is intended to be simply intuitive and in no way represents anything recognizable as a mathematical property. Norm === Subject: Re: Multiplication of negative numbers >> I'm going to be hated for bringing this up... but I'm not content with >> many answers that I have read in the past. >> Why does the multiplication of two negative numbers return a positive >> number? A few basic facts from algebra: -1 x a = -a -(-a) = a from here +1 === Subject: Re: Multiplication of negative numbers > Why does the multiplication of two negative numbers return a positive > number? I know everyone here can give me an algebraic proof, but how > is it that one can give a definition of multiplication using > multiplication? I like to think of the integers as constructed out of the natural numbers. The natural numbers are the set {0, 1, 2, ...}. The integers are defined as follows: The integer +0 is the identity function that maps the natural number 0 to the natural number 0, the natural number 1 to the natural number 1, and so on. The integer +1 is the function that maps 0 to 1, 1 to 2, and so on. The integer -1 is the function that maps 1 to 0, 2 to 1, and so on. With this way of thinking, it is quite all right to define addition and multiplication of integers in terms of addition and multiplication of natural numbers. There is nothing circular about it. Actually, addition of integers is defined in terms of function composition, which may seem like an advanced idea to you. +1 + +2 would be the application of the function +2 to the result of the function +1. I forget how to define multiplication in this scheme. But I'm sure it something equally constructivist. And the properties of equality, greater than, addition, and multiplication of integers are proven, not taken as axioms. -- Mostly economics: r c v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Multiplication of negative numbers > Again, I'm not looking for a proof of multiplication using > multiplication. That is circular logic. If you can't provide some > deeper understanding, than please don't reply. I don't really know whether I should answer, but at least I'm gonna make myself short (is it ok to say that in english?) :-) You could define -1 * -1 to be whatever You want, also -1, but it would be inconvinient, that is integers with such multiplication wouldn't have nice algebraic structure. As You described it, the Right way of multiplying positive integers somehow seems to come from God. He doesn't however privide us with a way to multiply -1 and -1 so, having no choice, we invent our own way (which, as a matter of fact, has so nice properties that we do think it comes from God, but that is reflection a posteriori :-) sirix. === Subject: Re: Multiplication of negative numbers > But why does multiplication need to be commutative? Simply because it > is based off of addition? Yes. That is why numerical multiplication commutes. There are other kinds of multiplication (for example matrix multiplication) that does not commute. The rationals inherit their commutativity from the integers and the reals from the rationals. Well, subtraction is based off of addition, > but no one cares that it isn't commutative. Same for division. > Really, we can go so far to say that addition is NOT commutative, > because (-3) + 5 is NOT equal to (-5) + 3. > Addition is only > commutative when dealing solely with positive numbers. Nonsense. See above. > Again, I'm not looking for a proof of multiplication using > multiplication. That is circular logic. If you can't provide some > deeper understanding, than please don't reply. a*b + a*(-b) = a*(b - b) = 0 (distributive law) a*b = -(a*(-b)) (transposition) (-a)*(-b) + a*(-b) = (-a + a)*(-b) = 0*b = 0 (distributive law) (-a)*(-b) = -(a*(-b)) (transposition) a*b = -(a*(-b)) = (-a)*(-b) (substitution). The distrubitive law is as much a property of addition as it is of multiplication. Bob Kolker === Subject: Re: Multiplication of negative numbers - Fast first comment In 1843 the Irish mathematician and physicist R. W. Hamilton invented an algebra in which the four main operations of addition, subtraction, multiplication and division by a non-zero number are present and have the well-known properties, =except that= multiplication is not commutative. It is the so-called quaternion algebra, bearing this name because it is a 4-dimensional number system. For more information just search the WWW for quaternion and =enjoy=, despite your hatred against mathematics. It may pass over! This was quite a revolution in mathematics at that time. >I'm going to be hated for bringing this up... but I'm not content with >many answers that I have read in the past. >Why does the multiplication of two negative numbers return a positive >number? I know everyone here can give me an algebraic proof, but how >is it that one can give a definition of multiplication using >multiplication? This is akin to saying that the definition : An >apple is an apple. is correct. Let's put down the axioms of >arithmetic aside and say that we only have addition. We know that >5+5=10, and because of the need to show (say) an individual's debt, we >introduce negative numbers. It is quite easy to show from this that >5+(-5)=0. But what of multiplication? What exactly *is* >multiplication? Multiplication is a shorthand for addition. So >instead of having to write: (5+5+5+5+5) = 25, we can simply use the >shorthand: 5*5 = 25. That is all it is... shorthand. It is generally >understood that within multiplication, one number is a number to be >multiplied, and the other is the multiplier. In 3*5, we can call >3 the multiplied, and 5 the multiplier, and represent this as >(3+3+3+3+3). On the other hand, we can call 5 the multiplied, and 3 >the multiplier, and can be represented as (5+5+5). Both give us the >same answer. So what about multiplying a negative number? -5*3 = >(-5) + (-5) + (-5) = -15. In this situation, it is easy to always call >the positive number the multiplier, so that you can add any number >(-/+) to itself a positive number of times. Alright... we are still >within our existing axioms. But now a problem has come up... What >happens when multiplying two negative numbers? The first number could >be called the multiplied, and the second the multiplier. So: in >-5*-3, -5 is the multiplied, and -3 is the multiplier. It doesn't >sound right to add something to itself negative 3 times. Actually, >this phenomenon doesn't occur anywhere in the real world. >Nevertheless, it seems that it must be defined. So... even though it >doesn't occur in the real world, we feel a need to define it. Maybe >we could define it as, instead of with a negative times a positive >(e.g., -5*3 = (-5 + -5 + -5)), we could define multiplication of two >negatives to use subtraction rather than addition, so: (-5)*(-3)= (-5) >- (-5) - (-5) = 5. We'll call this intuitive multiplication, or IM. > But, of course, IM is not commutative. (-3)*(-5) is not equal to >(-5)*(-3). So, we would could come up with a seperate definition for >multiplying negatives so that multiplication will be commutative. If >we defined multiplication so that, when two negatives are multiplied >(e.g., -5*-3), we can say that the answer will be the same as >(-(-5*3)). This way, multiplication can be commutative, same as >addition. We can call this kind of multiplication: consistent >multiplication, or CM. >But why does multiplication need to be commutative? Simply because it >is based off of addition? Well, subtraction is based off of addition, >but no one cares that it isn't commutative. Same for division. >Really, we can go so far to say that addition is NOT commutative, >because (-3) + 5 is NOT equal to (-5) + 3. Addition is only >commutative when dealing solely with positive numbers. The reason is >that when negative numbers become involved, commutative laws do not >always apply. So why did we decide to go with CM, rather than IM? >After all, if we adopted IM rather than CM, there would exist a number >'i' such that i^2 = -1, therefore eliminating the need for a whole new >set of numbers (complex numbers). If we had adopted IM rather than >CM, would the world of mathematics be a better place? Can anyone >explain to me if chosing between IM and CM was arbitrary, or if the >decision to use CM was really necessary after all? >I am not a math wiz. Actually, I hate mathematics. I can't stand the >fact that everyone likes to think Mathematics is an absolute truth, >when really it is all just as subjective as, say, religion. But >beyond this, maybe there is something here that I am overlooking >completely. Can anyone help out? Maybe I'm wrong about Math really >being subjective, but I really need someone to shed some light on >these kinds of things. >Again, I'm not looking for a proof of multiplication using >multiplication. That is circular logic. If you can't provide some >deeper understanding, than please don't reply. === Subject: How to find the X/Y of two Cauchy Random Variables? Hi all, Supppose X and Y are two Cauchy Random variables, how to prove that Z=X/Y is Normal(0, 1) distribution? === Subject: Re: How to find the X/Y of two Cauchy Random Variables? > Hi all, > Supppose X and Y are two Cauchy Random variables, how to prove that Z=X/Y is > Normal(0, 1) distribution? What makes you think that it's true? Are you sure you don't mean the other way around (ratio of two standard normals is cauchy)? Glen === Subject: Re: How to find the X/Y of two Cauchy Random Variables? >Supppose X and Y are two Cauchy Random variables, how to prove that Z=X/Y is >Normal(0, 1) distribution? It isn't. I think what you should be proving is that if X and Y are independent Normal(0,1), then X/Y is Cauchy. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: how to understand that Cauchy distribution does not have mean and variance? The text says that Cauchy distribution's mean does not exist, and the variance exists but = +inf? What do this mean: does not exist, exist but =+inf? Looks to me they have no difference... But from the graphical plotting Cauchy distribution's pdf, the Cauchy distribution obviously centered around 0, it should have a mean = 0, right? === Subject: Re: how to understand that Cauchy distribution does not have mean and variance? > The text says that Cauchy distribution's mean does not exist, and the > variance exists but = +inf? What do this mean: does not exist, exist but > =+inf? Looks to me they have no difference... > But from the graphical plotting Cauchy distribution's pdf, the Cauchy > distribution obviously centered around 0, it should have a mean = 0, right? === Subject: Re: how to understand that Cauchy distribution does not have mean and variance? > The text says that Cauchy distribution's mean does not exist, and the > variance exists but = +inf? What do this mean: does not exist, exist but > =+inf? Looks to me they have no difference... > But from the graphical plotting Cauchy distribution's pdf, the Cauchy > distribution obviously centered around 0, it should have a mean = 0, right? So you want to measure a mean. You make measurements and find the mean of that. As you make more measurements the mean should converge to some number, ie your series of sample means with more and more measurements might look something like this. 102 148 136 130 128 127 132 130 131 127 130 131 130 131 130 So it seems to be converging to 130. With the Cauchy distribution this doesn't happen. The means just go all over the place forever. Same with the variance. It's also possible to have a distribution with a mean but no variance. I would expect the sample means to converge very slowly, but given enough meansurments it would converge. === Subject: Re: how to understand that Cauchy distribution does not have mean and variance? > The text says that Cauchy distribution's mean does not exist, and the > variance exists but = +inf? What do this mean: does not exist, exist but > =+inf? Looks to me they have no difference... That's a rather odd distinction. If they define the variance in terms of the average squared distance form the mean, how do they get around the apparent problem that they're saying the average squared distance from something that doesn't exist itself exists? > But from the graphical plotting Cauchy distribution's pdf, the Cauchy > distribution obviously centered around 0, it should have a mean = 0, right? No. Read up about improper integrals. The limits involved in evaluating the integral for the mean don't converge. However, it does indeed have a median of 0. The integral for the mean does have a Cauchy principal value of 0, though. Glen === Subject: Re: how to understand that Cauchy distribution does not have mean and variance? >The text says that Cauchy distribution's mean does not exist, and the >variance exists but = +inf? What do this mean: does not exist, exist but >=+inf? Looks to me they have no difference... If the mean does not exist, then the variance is difficult to define. However, if you try to calculate the second moment of a Cauchy distribution, you will find it is infinite. >But from the graphical plotting Cauchy distribution's pdf, the Cauchy >distribution obviously centered around 0, it should have a mean = 0, right? The median (and mode) is 0, but if you try to calculate the mean (the first moment) you will get infinity minus infinity, which is not well defined. === Subject: Re: how to understand that Cauchy distribution does not have mean and variance? Think about the integral that defines the expected value or mean of a distribution. Even if the distribution is symmetric about the origin, does the integral necessarily exist? === Subject: Re: how to understand that Cauchy distribution does not have mean and variance? > Think about the integral that defines the expected value or mean of a > distribution. > Even if the distribution is symmetric about the origin, does the > integral necessarily exist? Intuitively, one [at least this one] expects that something called a distribution to have certain properties, among them the property that it describes how things are distributed and therefore that there's some sense in which it describes all things which I understand to mean that the integral of the distribution is unity so that the distribution's value at a given value of its independent variables expresses the proportion of things with the properties expressed by the values of the independent variables. Usually this is done by normalizing the function if necessary by dividing it by the value of its integral over the entire range of independent variables. Clearly the Cauchy distribution doesn't have this property so it doesn't represent a probability distribution in the normal sense. But AIUI the ratio of its values at two points represents the ratio of randomly selected things with the two x-values. Is this correct? Norm === Subject: Please please help with log(z) I would sincerely appreciate any help, and please read until the end if you would : I am ashamed to say that even though I am almost finished my graduate complex variables course, I still don't think I fully understand the function log(z). If someone could please tell me where I am right or wrong in the each of the following, I would be very grateful. In particular there are some questions I ask later that I would really like to know the answers to as well. So I will start from scratch, to see if I understand it even from the beginning : log is defined as the function from C {0} ---> C by log(z) = ln|z| + i*arg(z) where we must specify the range as some horizontal strip of height 2pi, since log is multivalued. That is, if we try to solve the equation e^z = w for z, we find that { ln|w| + i*arg(w) + 2pi*i*k k is an integer} are all solutions. So, if we take the log of some complex number, we must specify an interval of 2pi for the argument. In my text, it says that For any branch of log(z), we have e^(log(z)) = z. I don't quite understand what this means. Isn't it true that e^(log(z)) = z always? That is, log(z) is multivalued, that is true, but for whatever value you pick for log(z), e to that value will still be z because e^(2pi*i*k) = 1 for k an integer, and 2pi*i*k is only what the different values of log(z) differ by. So can't I say that e^(log(z)) = z, period? However, log(e^z) does not equal z necessarily, since log is multivalued. log(e^z) = z once we have picked a branch. Occasionally I will see different branches so I get confused. Is -pi < arg(z) < pi a branch or is -pi <= arg(z) < pi a branch? Well clearly the first one is, but is the second one a branch? Here are two examples where I see two totally different branches that don't make sense to me : In my Complex analysis book, when we define log, arg(z) takes values in the interval [y_0, y_0 + 2pi [, so the half open half closed interval. However, later on, when we do real valued integrals using residue theorem, (in particular, the real valued integral 0 to infinity of [x^(-c)]/[1+x] where 0 < c < 1), and I quote, Let G = {z : z is not 0, and 0 < arg(z) < 2pi}; define a branch of the logarithm on G by putting l(re^(it)) = log(r) + it where 0 < t < 2pi. So why in the first case we have a half closed half open interval, and in the second case we have an open interval, defining a branch? On a slightly different topic, why is it that in log(z), when you approach pi from the top you get a different value for log(z) than when you approach from the bottom? Symbolically why is this true? (That is, proving it by writing limits) much for your help, Isaac === Subject: Re: Please please help with log(z) >I would sincerely appreciate any help, and please read until the end if you >would : I am ashamed to say that even though I am almost finished my >graduate complex variables course, I still don't think I fully understand >the function log(z). If someone could please tell me where I am right or >wrong in the each of the following, I would be very grateful. In particular >there are some questions I ask later that I would really like to know the >answers to as well. So I will start from scratch, to see if I understand it >even from the beginning : >log is defined as the function from C {0} ---> C Wrong already, although this one it seems you were aware of. >log(z) = ln|z| + i*arg(z) >where we must specify the range as some horizontal strip of height 2pi, >since log is multivalued. No, the range does not have to be a horizontal strip! For example, suppose that c is a simple curve connecting the origin to the point at infinity (more precise definition available on request) and let O be the complement of c. Then there is a branch of the logarithm defined in O; the range will be a horizontal strip if and only if c is a _ray_. > That is, if we try to solve the equation e^z = w >for z, >we find that { ln|w| + i*arg(w) + 2pi*i*k k is an integer} are all >solutions. So, if we take the log of some complex number, we must specify >an interval of 2pi for the argument. >In my text, it says that For any branch of log(z), we have e^(log(z)) = z. >I don't quite understand what this means. Isn't it true that e^(log(z)) = z >always? That is, log(z) is multivalued, that is true, but for whatever >value you pick for log(z), e to that value will still be z because >e^(2pi*i*k) = 1 for k an integer, and 2pi*i*k is only what the different >values of log(z) differ by. So can't I say that e^(log(z)) = z, period? Saying e^(log(z)) = z, period is just a little sloppy, because it might look as though you think that the expression log(z) has a well-defined meaning. >However, log(e^z) does not equal z necessarily, since log is multivalued. >log(e^z) = z once we have picked a branch. >Occasionally I will see different branches so I get confused. Is -pi < >arg(z) < pi a branch or is -pi <= arg(z) < pi a branch? Well clearly the >first one is, but is the second one a branch? What is the _definition_ of branch of the logarithm that you're using? The definition I use is this: Suppose that O is an open set, g is holomorphic in O, and e^g(z) = z for all z in O. Then g is a branch of the logarithm. With that definition it should be easy to answer your question. If that's not the definition you're using, what _is_ the definition? >Here are two examples where I >see two totally different branches that don't make sense to me : In my >Complex analysis book, when we define log, arg(z) takes values in the >interval [y_0, y_0 + 2pi [, so the half open half closed interval. However, >later on, when we do real valued integrals using residue theorem, (in >particular, the real valued integral 0 to infinity of [x^(-c)]/[1+x] >where 0 < c < 1), and I quote, Let G = {z : z is not 0, and 0 < arg(z) < >2pi}; define a branch of the logarithm on G by putting l(re^(it)) = log(r) + >it where 0 < t < 2pi. So why in the first case we have a half closed half >open interval, and in the second case we have an open interval, defining a >branch? Not sure what the question is. But the answer may well be contained in the answer to my question: what is the _definition_ of branch of the logarithm? Definitions are good things to know. >On a slightly different topic, why is it that in log(z), when you approach >pi from the top you get a different value for log(z) than when you approach >from the bottom? Symbolically why is this true? (That is, proving it by >writing limits) I doubt that pi is really the point you're referring to here. In any case, what you say is true or not true, depending on which branch of log you're talking about - you need to specify that before any question about the value of log(z) for some particular z makes sense. >much for your help, >Isaac ************************ David C. Ullrich === Subject: Re: Please please help with log(z) Note: Tony = Isaac. Apologies for that. === Subject: Re: Please please help with log(z) >>I would sincerely appreciate any help, and please read until the end if >>you >>would : I am ashamed to say that even though I am almost finished my >>graduate complex variables course, I still don't think I fully understand >>the function log(z). If someone could please tell me where I am right or >>wrong in the each of the following, I would be very grateful. In >>particular >>there are some questions I ask later that I would really like to know the >>answers to as well. So I will start from scratch, to see if I understand >>it >>even from the beginning : >>log is defined as the function from C {0} ---> C > Wrong already, although this one it seems you were aware of. Oy. If you would, please tell me as precise a definition of log as you can. I got that definition straight from my textbook and I quote The function log : C {0} ---> C, with range y_0 <= Im (log(z)) < y_0 + 2pi, is defined by log(z) = log|z| + i*arg(z), where arg(z) takes values in [y_0,y_0 + 2pi[ and log|z| is the usual log. I guess I see that what you say is that out some curve from 0 to infinity so you won't get a horizontal strip in that case. >>by >>log(z) = ln|z| + i*arg(z) >>where we must specify the range as some horizontal strip of height 2pi, >>since log is multivalued. > No, the range does not have to be a horizontal strip! > For example, suppose that c is a simple curve connecting > the origin to the point at infinity (more precise definition > available on request) and let O be the complement of c. Then > there is a branch of the logarithm defined in O; the range > will be a horizontal strip if and only if c is a _ray_. >> That is, if we try to solve the equation e^z = w >>for z, >>we find that { ln|w| + i*arg(w) + 2pi*i*k k is an integer} are all >>solutions. So, if we take the log of some complex number, we must specify >>an interval of 2pi for the argument. >>In my text, it says that For any branch of log(z), we have e^(log(z)) = >>z. >>I don't quite understand what this means. Isn't it true that e^(log(z)) = >>always? That is, log(z) is multivalued, that is true, but for whatever >>value you pick for log(z), e to that value will still be z because >>e^(2pi*i*k) = 1 for k an integer, and 2pi*i*k is only what the different >>values of log(z) differ by. So can't I say that e^(log(z)) = z, period? > Saying e^(log(z)) = z, period is just a little sloppy, because it > might look as though you think that the expression log(z) has > a well-defined meaning. Ok, but is it true that no matter what value for log(z) we pick, e^(log(z)) = z? On the other hand, it is not true that log(e^z) = z, unless we choose a branch first. A branch to me meant we specify an interval of 2pi for the argument. But now I guess it means that we can just take some curve from 0 to infinity and then this can also give us all of our complex numbers that we can take the log of. >>However, log(e^z) does not equal z necessarily, since log is multivalued. >>log(e^z) = z once we have picked a branch. >>Occasionally I will see different branches so I get confused. Is -pi < >>arg(z) < pi a branch or is -pi <= arg(z) < pi a branch? Well clearly the >>first one is, but is the second one a branch? > What is the _definition_ of branch of the logarithm that you're > using? > The definition I use is this: Suppose that O is an open set, g is > holomorphic in O, and e^g(z) = z for all z in O. Then g is a branch > of the logarithm. With that definition it should be easy to answer > your question. If that's not the definition you're using, what > _is_ the definition? But like I said above, doesn't e^(log(z)) = z for any choice of log(z)? I guess I'm getting confused because am I even making any sense when I say any choice of log(z)? In the definition of log(z), is log(z) single-valued? If it were, then saying for any choice of log(z) doesn't make any sense since there is one choice of log(z). >>Here are two examples where I >>see two totally different branches that don't make sense to me : In my >>Complex analysis book, when we define log, arg(z) takes values in the >>interval [y_0, y_0 + 2pi [, so the half open half closed interval. >>However, >>later on, when we do real valued integrals using residue theorem, (in >>particular, the real valued integral 0 to infinity of [x^(-c)]/[1+x] >>where 0 < c < 1), and I quote, Let G = {z : z is not 0, and 0 < arg(z) < >>2pi}; define a branch of the logarithm on G by putting l(re^(it)) = log(r) >>it where 0 < t < 2pi. So why in the first case we have a half closed >>half >>open interval, and in the second case we have an open interval, defining a >>branch? > Not sure what the question is. But the answer may well be contained > in the answer to my question: what is the _definition_ of branch > of the logarithm? > Definitions are good things to know. >>On a slightly different topic, why is it that in log(z), when you >>approach >>pi from the top you get a different value for log(z) than when you >>approach >>from the bottom? Symbolically why is this true? (That is, proving it by >>writing limits) > I doubt that pi is really the point you're referring to here. In any > case, what you say is true or not true, depending on which branch of > log you're talking about - you need to specify that before any > question about the value of log(z) for some particular z makes sense. >>much for your help, >>Isaac > ************************ > David C. Ullrich === Subject: Re: Please please help with log(z) >I would sincerely appreciate any help, and please read until the end if >you >would : I am ashamed to say that even though I am almost finished my >graduate complex variables course, I still don't think I fully understand >the function log(z). If someone could please tell me where I am right or >wrong in the each of the following, I would be very grateful. In >particular >there are some questions I ask later that I would really like to know the >answers to as well. So I will start from scratch, to see if I understand >it >even from the beginning : >log is defined as the function from C {0} ---> C >> Wrong already, although this one it seems you were aware of. >Oy. If you would, please tell me as precise a definition of log as you can. >I got that definition straight from my textbook and I quote The function >log : C {0} ---> C, with range y_0 <= Im (log(z)) < y_0 + 2pi, is defined >by log(z) = log|z| + i*arg(z), where arg(z) takes values in [y_0,y_0 + >2pi[ and log|z| is the usual log. What book is this? If they said this was _a_ function log that would not be so bad - it's a perfectly valid definition of something which is (almost) a branch of the logarithm. But if they actually say this is _the_ function log, without making it clear that they're defining one of many possible log functions, that's not good. See, there is a thing that deserves to be called _the_ complex logarithm. Saying exactly what it is involves talking about analytic continuation and/or Riemann surfaces - whatever it is, it's certainly not a _function_ with domain C{0}. But what _the_ logarithm function is doesn't really matter to the answer to your questions below about why this is a branch of log and that isn't. What matters to that is exactly what the phrase branch of the logarithm means. I asked three times and you didn't answer - I'm going to ask once again, and if you don't feel like replying there's really no point in continuing this discussion, since you're using terms but refusing to tell me what you mean by them: _What_ is the definition of branch of the logarithm that you're using? >I guess I see that what you say is that >out some curve from 0 to infinity so you won't get a horizontal strip in >that case. >by >log(z) = ln|z| + i*arg(z) >where we must specify the range as some horizontal strip of height 2pi, >since log is multivalued. >> No, the range does not have to be a horizontal strip! >> For example, suppose that c is a simple curve connecting >> the origin to the point at infinity (more precise definition >> available on request) and let O be the complement of c. Then >> there is a branch of the logarithm defined in O; the range >> will be a horizontal strip if and only if c is a _ray_. > That is, if we try to solve the equation e^z = w >for z, >we find that { ln|w| + i*arg(w) + 2pi*i*k k is an integer} are all >solutions. So, if we take the log of some complex number, we must specify >an interval of 2pi for the argument. >In my text, it says that For any branch of log(z), we have e^(log(z)) = >z. >I don't quite understand what this means. Isn't it true that e^(log(z)) = >z >always? That is, log(z) is multivalued, that is true, but for whatever >value you pick for log(z), e to that value will still be z because >e^(2pi*i*k) = 1 for k an integer, and 2pi*i*k is only what the different >values of log(z) differ by. So can't I say that e^(log(z)) = z, period? >> Saying e^(log(z)) = z, period is just a little sloppy, because it >> might look as though you think that the expression log(z) has >> a well-defined meaning. >Ok, but is it true that no matter what value for log(z) we pick, e^(log(z)) >= z? >On the other hand, it is not true that log(e^z) = z, unless we choose a >branch first. A branch to me meant we specify an interval of 2pi for the >argument. But now I guess it means that we can just take some curve from 0 >to infinity and then this can also give us all of our complex numbers that >we can take the log of. >However, log(e^z) does not equal z necessarily, since log is multivalued. >log(e^z) = z once we have picked a branch. >Occasionally I will see different branches so I get confused. Is -pi < >arg(z) < pi a branch or is -pi <= arg(z) < pi a branch? Well clearly the >first one is, but is the second one a branch? >> What is the _definition_ of branch of the logarithm that you're >> using? >> The definition I use is this: Suppose that O is an open set, g is >> holomorphic in O, and e^g(z) = z for all z in O. Then g is a branch >> of the logarithm. With that definition it should be easy to answer >> your question. If that's not the definition you're using, what >> _is_ the definition? >But like I said above, doesn't e^(log(z)) = z for any choice of log(z)? I >guess I'm getting confused because am I even making any sense when I say >any choice of log(z)? In the definition of log(z), is log(z) >single-valued? If it were, then saying for any choice of log(z) doesn't >make any sense since there is one choice of log(z). >Here are two examples where I >see two totally different branches that don't make sense to me : In my >Complex analysis book, when we define log, arg(z) takes values in the >interval [y_0, y_0 + 2pi [, so the half open half closed interval. >However, >later on, when we do real valued integrals using residue theorem, (in >particular, the real valued integral 0 to infinity of [x^(-c)]/[1+x] >where 0 < c < 1), and I quote, Let G = {z : z is not 0, and 0 < arg(z) < >2pi}; define a branch of the logarithm on G by putting l(re^(it)) = log(r) >+ >it where 0 < t < 2pi. So why in the first case we have a half closed >half >open interval, and in the second case we have an open interval, defining a >branch? >> Not sure what the question is. But the answer may well be contained >> in the answer to my question: what is the _definition_ of branch >> of the logarithm? >> Definitions are good things to know. >On a slightly different topic, why is it that in log(z), when you >approach >pi from the top you get a different value for log(z) than when you >approach >from the bottom? Symbolically why is this true? (That is, proving it by >writing limits) >> I doubt that pi is really the point you're referring to here. In any >> case, what you say is true or not true, depending on which branch of >> log you're talking about - you need to specify that before any >> question about the value of log(z) for some particular z makes sense. >much for your help, >Isaac >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Please please help with log(z) >>I would sincerely appreciate any help, and please read until the end if >>you >>would : I am ashamed to say that even though I am almost finished my >>graduate complex variables course, I still don't think I fully >>understand >>the function log(z). If someone could please tell me where I am right >>or >>wrong in the each of the following, I would be very grateful. In >>particular >>there are some questions I ask later that I would really like to know >>the >>answers to as well. So I will start from scratch, to see if I >>understand >>it >>even from the beginning : >log is defined as the function from C {0} ---> C > Wrong already, although this one it seems you were aware of. >>Oy. If you would, please tell me as precise a definition of log as you >>can. >>I got that definition straight from my textbook and I quote The function >>log : C {0} ---> C, with range y_0 <= Im (log(z)) < y_0 + 2pi, is >>defined >>by log(z) = log|z| + i*arg(z), where arg(z) takes values in [y_0,y_0 + >>2pi[ and log|z| is the usual log. > What book is this? If they said this was _a_ function log that would > not be so bad - it's a perfectly valid definition of something which > is (almost) a branch of the logarithm. But if they actually say this > is _the_ function log, without making it clear that they're defining > one of many possible log functions, that's not good. Marsden and Hoffman Basic Complex Analysis. This is exactly below and give my definition) > See, there is a thing that deserves to be called _the_ complex > logarithm. Saying exactly what it is involves talking about > analytic continuation and/or Riemann surfaces - whatever it > is, it's certainly not a _function_ with domain C{0}. > But what _the_ logarithm function is doesn't really matter to > the answer to your questions below about why this is a branch > of log and that isn't. What matters to that is exactly what > the phrase branch of the logarithm means. I asked three times > and you didn't answer - I'm going to ask once again, and if > you don't feel like replying there's really no point in > continuing this discussion, since you're using terms but > refusing to tell me what you mean by them: > _What_ is the definition of branch of the logarithm that > you're using? I guess I don't really have a precise definition in my mind. I would say that a branch of the logarithm is a domain in the complex plane where log is an analytic function. You said earlier that Suppose that O is an open set, g is holomorphic in O, and e^g(z) = z for all z in O. Then g is a branch of the logarithm. How does this definition deal with e ^( ln|z| + i*arg(z) + 2pi*i*k) = z for all integers k? >>I guess I see that what you say is that >>out some curve from 0 to infinity so you won't get a horizontal strip in >>that case. >>by >log(z) = ln|z| + i*arg(z) >where we must specify the range as some horizontal strip of height 2pi, >>since log is multivalued. > No, the range does not have to be a horizontal strip! > For example, suppose that c is a simple curve connecting > the origin to the point at infinity (more precise definition > available on request) and let O be the complement of c. Then > there is a branch of the logarithm defined in O; the range > will be a horizontal strip if and only if c is a _ray_. >> That is, if we try to solve the equation e^z = w >>for z, >>we find that { ln|w| + i*arg(w) + 2pi*i*k k is an integer} are all >>solutions. So, if we take the log of some complex number, we must >>specify >>an interval of 2pi for the argument. >In my text, it says that For any branch of log(z), we have e^(log(z)) = >>z. >>I don't quite understand what this means. Isn't it true that e^(log(z)) >>= >>z >>always? That is, log(z) is multivalued, that is true, but for whatever >>value you pick for log(z), e to that value will still be z because >>e^(2pi*i*k) = 1 for k an integer, and 2pi*i*k is only what the different >>values of log(z) differ by. So can't I say that e^(log(z)) = z, period? > Saying e^(log(z)) = z, period is just a little sloppy, because it > might look as though you think that the expression log(z) has > a well-defined meaning. >>Ok, but is it true that no matter what value for log(z) we pick, >>e^(log(z)) >>= z? >>On the other hand, it is not true that log(e^z) = z, unless we choose a >>branch first. A branch to me meant we specify an interval of 2pi for the >>argument. But now I guess it means that we can just take some curve from >>to infinity and then this can also give us all of our complex numbers that >>we can take the log of. >>However, log(e^z) does not equal z necessarily, since log is >>multivalued. >>log(e^z) = z once we have picked a branch. >Occasionally I will see different branches so I get confused. Is -pi >>< >>arg(z) < pi a branch or is -pi <= arg(z) < pi a branch? Well clearly >>the >>first one is, but is the second one a branch? > What is the _definition_ of branch of the logarithm that you're > using? > The definition I use is this: Suppose that O is an open set, g is > holomorphic in O, and e^g(z) = z for all z in O. Then g is a branch > of the logarithm. With that definition it should be easy to answer > your question. If that's not the definition you're using, what > _is_ the definition? >>But like I said above, doesn't e^(log(z)) = z for any choice of log(z)? I >>guess I'm getting confused because am I even making any sense when I say >>any choice of log(z)? In the definition of log(z), is log(z) >>single-valued? If it were, then saying for any choice of log(z) doesn't >>make any sense since there is one choice of log(z). >>Here are two examples where I >>see two totally different branches that don't make sense to me : In my >>Complex analysis book, when we define log, arg(z) takes values in the >>interval [y_0, y_0 + 2pi [, so the half open half closed interval. >>However, >>later on, when we do real valued integrals using residue theorem, (in >>particular, the real valued integral 0 to infinity of [x^(-c)]/[1+x] >>where 0 < c < 1), and I quote, Let G = {z : z is not 0, and 0 < arg(z) >>< >>2pi}; define a branch of the logarithm on G by putting l(re^(it)) = >>log(r) >>+ >>it where 0 < t < 2pi. So why in the first case we have a half closed >>half >>open interval, and in the second case we have an open interval, defining >>a >>branch? > Not sure what the question is. But the answer may well be contained > in the answer to my question: what is the _definition_ of branch > of the logarithm? > Definitions are good things to know. >>On a slightly different topic, why is it that in log(z), when you >>approach >>pi from the top you get a different value for log(z) than when you >>approach >>from the bottom? Symbolically why is this true? (That is, proving it by >>writing limits) > I doubt that pi is really the point you're referring to here. In any > case, what you say is true or not true, depending on which branch of > log you're talking about - you need to specify that before any > question about the value of log(z) for some particular z makes sense. >>much for your help, >Isaac >> ************************ > David C. Ullrich > ************************ > David C. Ullrich === Subject: Re: Please please help with log(z) >>I would sincerely appreciate any help, and please read until the end if >you >would : I am ashamed to say that even though I am almost finished my >graduate complex variables course, I still don't think I fully >understand >the function log(z). If someone could please tell me where I am right >or >wrong in the each of the following, I would be very grateful. In >particular >there are some questions I ask later that I would really like to know >the >answers to as well. So I will start from scratch, to see if I >understand >it >even from the beginning : >log is defined as the function from C {0} ---> C > Wrong already, although this one it seems you were aware of. >>Oy. If you would, please tell me as precise a definition of log as you >can. >I got that definition straight from my textbook and I quote The function >log : C {0} ---> C, with range y_0 <= Im (log(z)) < y_0 + 2pi, is >defined >by log(z) = log|z| + i*arg(z), where arg(z) takes values in [y_0,y_0 + >2pi[ and log|z| is the usual log. >> What book is this? If they said this was _a_ function log that would >> not be so bad - it's a perfectly valid definition of something which >> is (almost) a branch of the logarithm. But if they actually say this >> is _the_ function log, without making it clear that they're defining >> one of many possible log functions, that's not good. >Marsden and Hoffman Basic Complex Analysis. This is exactly Sure enough. I think this is a little hideous, for reasons I've explained. >What would you say is the defintion? The definition of _what_? The definition of log(z) for complex z? That's not something that has a simple definition. The definition of log? Again, for our purposes I wouldn't _give_ a definition of log, I'd give a definition of branch of log, which btw I've already done. >(I talk about branch >below and give my definition) >> See, there is a thing that deserves to be called _the_ complex >> logarithm. Saying exactly what it is involves talking about >> analytic continuation and/or Riemann surfaces - whatever it >> is, it's certainly not a _function_ with domain C{0}. >> But what _the_ logarithm function is doesn't really matter to >> the answer to your questions below about why this is a branch >> of log and that isn't. What matters to that is exactly what >> the phrase branch of the logarithm means. I asked three times >> and you didn't answer - I'm going to ask once again, and if >> you don't feel like replying there's really no point in >> continuing this discussion, since you're using terms but >> refusing to tell me what you mean by them: >> _What_ is the definition of branch of the logarithm that >> you're using? >I guess I don't really have a precise definition in my mind. I would say >that a branch of the logarithm is a domain in the complex plane >where log is an analytic function. Close, but not acceptable, because it assumes that there _is_ a single thing called log, and we're just choosing a domain for it. That's not how it works. >You said earlier that Suppose that O is >an open set, g is >holomorphic in O, and e^g(z) = z for all z in O. Then g is a branch >of the logarithm. Yes. Look up logarithm, branch of in the index of your book. You find essentially this definition on a later page (except they waffle and instead of saying that this _is_ the definition of branch of log they say something about how this might be regarded as a branch of log... if they're going to be talking about what is and what is not a branch of log they need to give a _definition_, darn it.) >How does this definition deal with e ^( ln|z| + >i*arg(z) + 2pi*i*k) = z for all integers k? I don't understand the question - what does it mean for a definition to deal with an equation? >[...] >Here are two examples where I >see two totally different branches that don't make sense to me : In my >Complex analysis book, when we define log, arg(z) takes values in the >interval [y_0, y_0 + 2pi [, so the half open half closed interval. >However, >later on, when we do real valued integrals using residue theorem, (in >particular, the real valued integral 0 to infinity of [x^(-c)]/[1+x] >where 0 < c < 1), and I quote, Let G = {z : z is not 0, and 0 < arg(z) >< >2pi}; define a branch of the logarithm on G by putting l(re^(it)) = >log(r) >+ >it where 0 < t < 2pi. So why in the first case we have a half closed >half >open interval, and in the second case we have an open interval, defining >a >branch? Why they define those functions the way they do: who knows. Why the first is not a branch of the logarithm and the second is: to answer this question we do need a _definition_ of branch of the logarithm. If we use the definition I give above it's clear why the first is not a branch of the logarithm, right? ************************ David C. Ullrich === Subject: Re: Please please help with log(z) >>I would sincerely appreciate any help, and please read until the end >>if >>you >>would : I am ashamed to say that even though I am almost finished my >>graduate complex variables course, I still don't think I fully >>understand >>the function log(z). If someone could please tell me where I am right >>or >>wrong in the each of the following, I would be very grateful. In >>particular >>there are some questions I ask later that I would really like to know >>the >>answers to as well. So I will start from scratch, to see if I >>understand >>it >>even from the beginning : >log is defined as the function from C {0} ---> C > Wrong already, although this one it seems you were aware of. >Oy. If you would, please tell me as precise a definition of log as you >>can. >>I got that definition straight from my textbook and I quote The >>function >>log : C {0} ---> C, with range y_0 <= Im (log(z)) < y_0 + 2pi, is >>defined >>by log(z) = log|z| + i*arg(z), where arg(z) takes values in [y_0,y_0 + >>2pi[ and log|z| is the usual log. > What book is this? If they said this was _a_ function log that would > not be so bad - it's a perfectly valid definition of something which > is (almost) a branch of the logarithm. But if they actually say this > is _the_ function log, without making it clear that they're defining > one of many possible log functions, that's not good. >>Marsden and Hoffman Basic Complex Analysis. This is exactly > Sure enough. I think this is a little hideous, for reasons I've > explained. >>What would you say is the defintion? > The definition of _what_? The definition of log(z) for complex z? > That's not something that has a simple definition. The definition > of log? Again, for our purposes I wouldn't _give_ a definition > of log, I'd give a definition of branch of log, which btw > I've already done. >>(I talk about branch >>below and give my definition) > See, there is a thing that deserves to be called _the_ complex > logarithm. Saying exactly what it is involves talking about > analytic continuation and/or Riemann surfaces - whatever it > is, it's certainly not a _function_ with domain C{0}. > But what _the_ logarithm function is doesn't really matter to > the answer to your questions below about why this is a branch > of log and that isn't. What matters to that is exactly what > the phrase branch of the logarithm means. I asked three times > and you didn't answer - I'm going to ask once again, and if > you don't feel like replying there's really no point in > continuing this discussion, since you're using terms but > refusing to tell me what you mean by them: > _What_ is the definition of branch of the logarithm that > you're using? >>I guess I don't really have a precise definition in my mind. I would say >>that a branch of the logarithm is a domain in the complex plane >>where log is an analytic function. > Close, but not acceptable, because it assumes that there _is_ a > single thing called log, and we're just choosing a domain for > it. That's not how it works. >>You said earlier that Suppose that O is >>an open set, g is >>holomorphic in O, and e^g(z) = z for all z in O. Then g is a branch >>of the logarithm. > Yes. Look up logarithm, branch of in the index of your book. > You find essentially this definition on a later page (except > they waffle and instead of saying that this _is_ the definition > of branch of log they say something about how this might be > regarded as a branch of log... if they're going to be talking > about what is and what is not a branch of log they need to give > a _definition_, darn it.) >>How does this definition deal with e ^( ln|z| + >>i*arg(z) + 2pi*i*k) = z for all integers k? > I don't understand the question - what does it mean for a > definition to deal with an equation? >>[...] >Here are two examples where I >>see two totally different branches that don't make sense to me : In my >>Complex analysis book, when we define log, arg(z) takes values in the >>interval [y_0, y_0 + 2pi [, so the half open half closed interval. >>However, >>later on, when we do real valued integrals using residue theorem, (in >>particular, the real valued integral 0 to infinity of [x^(-c)]/[1+x] >>where 0 < c < 1), and I quote, Let G = {z : z is not 0, and 0 < >>arg(z) >>< >>2pi}; define a branch of the logarithm on G by putting l(re^(it)) = >>log(r) >>+ >>it where 0 < t < 2pi. So why in the first case we have a half closed >>half >>open interval, and in the second case we have an open interval, >>defining >>a >>branch? > Why they define those functions the way they do: who knows. > Why the first is not a branch of the logarithm and the second > is: to answer this question we do need a _definition_ of > branch of the logarithm. If we use the definition I give > above it's clear why the first is not a branch of the logarithm, > right? This is not clear to me. Let G = {z : z is not 0, and 0 < arg(z) < 2pi} and let g(re^(it)) = log(r) + it where 0 < t < 2pi. I do not understand why g is a branch of the logarithm (by your definition) but if 0 < t <= 2pi, then g is not a branch of the logarithm. I am grateful for your explanation. > ************************ > David C. Ullrich === Subject: Re: Please please help with log(z) |Let G = {z : z is not 0, and 0 < arg(z) < 2pi} and let g(re^(it)) = |log(r) + it where 0 < t < 2pi. I do not understand why g is a branch of the |logarithm (by your definition) but if 0 < t <= 2pi, then g is not a branch |of the logarithm. I am grateful for your explanation. You appear to be describing a second function g (let's call it g_2) that we could define thus: g_2(z) is the w satisfying 00 from above of g_2(1+ui) is 0. The kind of confusion you're describing is normally well dispelled by paying more careful attention to definitions, and that often seems to be about the only way to treat it. The fact that the definition of branch is chosen so as to exclude g_2 is somewhat of the nature of a technical detail, but don't underestimate the value of getting such technical details straight. You will find that there are many situations in mathematics where there are multiple possible choices of definition that differ only in some detail, and it may be that none seems obviously better than any other, but where the theory develops much more elegantly later on if you make certain choices of definition. You will also find that when you're reading a well-written exposition, the author will have taken the time to make good choices of definition for you, and that it will typically be worth taking seriously that they mean by their definitions *exactly* what they say. Caratheodory defines the principal value of the log function at z to be the w such that e^w=z and -pi0}. Note, however, that I don't see him refer to anything but ANALYTIC branches of log f from then on. This is consistent with other textbooks that use phrases like analytic branch of log. The most general concept of inverse function just isn't all that nice unless one assumes some additional property like continuity or analyticity. However the definitions are set up, there are key lemmas that come up later, telling you that there exist analytic functions under suitable conditions, provided that the domain is a simply connected region. For instance, if f is analytic on R and z_0 is some point in R, then there exists a function g that is also analytic on R satisfying g(z_0)=0 and g'(z)=f(z) on R. Applying this to f(z)=1/z gives us a branch of the log on R, provided 0 is not in R. There's no direct way of applying this lemma to get functions like the one you alluded to (that I called g_2). If you want, later, to consider how a function g analytic on a region behaves in limits approaching the boundary of R, fine, but get the easy part first. Now that I think of it, it occurs to me that I have myself a topological question related to this. Suppose R is a maximal simply- connected region (i.e. open set) in the complex plane, not containing 0, where by maximal I mean simply that there is no region R' satisfying the same conditions such that R is a proper subset of R'. Is R necessarily the complement of a curve running that he would provide on request.) ;-) Keith Ramsay === Subject: Re: Please please help with log(z) >[...] >Now that I think of it, it occurs to me that I have myself a >topological question related to this. Suppose R is a maximal simply- >connected region (i.e. open set) in the complex plane, not >containing 0, where by maximal I mean simply that there is no >region R' satisfying the same conditions such that R is a proper >subset of R'. Is R necessarily the complement of a curve running >from 0 to infinity? I don't think so. R is simply connected if and only if the complement in the extended plane (ie including infinity) is connected, so the complement must be a minimal connected compact subset of the extended plane containing 0 and infinity. A curve is the obvious example of such a thing but surely there are others. Oh: for example the complement could be a topologist's sine curve thingie: That's connected, not a curve, and it's a minimal connected compact set, because it consists of the closure of a simple curve - omit a point of the curve and it's not connected, omit one of the other points and it's not closed. >that he would provide on request.) ;-) >Keith Ramsay ************************ David C. Ullrich === Subject: Re: Please please help with log(z) >[...] >Let G = {z : z is not 0, and 0 < arg(z) < 2pi} and let g(re^(it)) = >log(r) + it where 0 < t < 2pi. I do not understand why g is a branch of the >logarithm (by your definition) but if 0 < t <= 2pi, then g is not a branch >of the logarithm. I am grateful for your explanation. The definition is this: A branch of the logarithm is a function L, holomorphic in an open set O, such that e^L(z) = z. (This is equivalent to just requiring that L be continuous, it turns out.) What is the domain of the second example? _Is_ the second example holomorphic in its domain? (Is it even continuous?) >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Please please help with log(z) I have to now go to my complex variables class (seriously), I will respond afterwards. === Subject: Re: Please please help with log(z) > I would sincerely appreciate any help, and please read until the end if you > would : I am ashamed to say that even though I am almost finished my > graduate complex variables course, I still don't think I fully understand > the function log(z). If someone could please tell me where I am right or > wrong in the each of the following, I would be very grateful. In particular > there are some questions I ask later that I would really like to know the > answers to as well. So I will start from scratch, to see if I understand it > even from the beginning : > log is defined as the function from C {0} ---> C by > log(z) = ln|z| + i*arg(z) > where we must specify the range as some horizontal strip of height 2pi, > since log is multivalued. OK, so that's the definition of log(z) that you use. Please note there are some problems here: first you say that the domain of log is C{0}, but right after that you say that we must specify the range as some horizontal strip of height 2pi; therefore, the domain is *not* C{0}. Finally, you say that log is multivalued. Now, I understand the meaning of the assertion every non-zero complex number has infinitely many logarithms, but what's a multivalued function? > In my text, it says that For any branch of log(z), we have e^(log(z)) = z. > I don't quite understand what this means. Isn't it true that e^(log(z)) = z > always? That is, log(z) is multivalued, that is true, but for whatever > value you pick for log(z), e to that value will still be z because > e^(2pi*i*k) = 1 for k an integer, and 2pi*i*k is only what the different > values of log(z) differ by. So can't I say that e^(log(z)) = z, period? No. If the definition was log(z) is the only w in the given horizontal strip such that exp(w) = z then, yes, there would be nothing to prove here. But since you define it as log(z) = ln|z| + i*arg(z) then some proof must be provided. > However, log(e^z) does not equal z necessarily, since log is multivalued. > log(e^z) = z once we have picked a branch. I would say log(e^z) = z whenever z belongs to the chosen horizontal strip. > Occasionally I will see different branches so I get confused. Is -pi < > arg(z) < pi a branch or is -pi <= arg(z) < pi a branch? Well clearly the > first one is, but is the second one a branch? They both are. If you want to work with an analytic function, then the second one will not do, since the domain of an analytic function must always be an open set. > Here are two examples where I > see two totally different branches that don't make sense to me : In my > Complex analysis book, when we define log, arg(z) takes values in the > interval [y_0, y_0 + 2pi [, so the half open half closed interval. However, > later on, when we do real valued integrals using residue theorem, (in > particular, the real valued integral 0 to infinity of [x^(-c)]/[1+x] > where 0 < c < 1), and I quote, Let G = {z : z is not 0, and 0 < arg(z) < > 2pi}; define a branch of the logarithm on G by putting l(re^(it)) = log(r) + > it where 0 < t < 2pi. So why in the first case we have a half closed half > open interval, and in the second case we have an open interval, defining a > branch? Because both are branches, but in order to apply the residue theorem you must work with analytic functions and therefore the domain of your function has to be an open set. > On a slightly different topic, why is it that in log(z), when you approach > pi from the top you get a different value for log(z) than when you approach > from the bottom? Symbolically why is this true? (That is, proving it by > writing limits) As stated, the question has no meaning; you must say first of which branch are you talking about (and that's not the only problem). Suppose that you have picked up the horizontal strip -pi < Im z < pi (since we're talking about limits here, you can replace one of the <'s by a <=, but that will make no difference). Consider those z of the form exp(ti), with t in ]-pi,pi[. Then lim_{t -> pi} exp(ti) = lim_{t -> pi} exp(-ti) = -1. But log(exp(ti)) = t, and so lim_{t -> pi} log(exp(ti)) = pi i and lim_{t -> -pi} log(exp(ti)) = -pi i. Note that both limits are logarithms of -1. I hope that this helps. Jose Carlos Santos === Subject: Re: Please please help with log(z) Note: Tony = Isaac. Sorry for that. === Subject: Re: Please please help with log(z) >> I would sincerely appreciate any help, and please read until the end if >> you would : I am ashamed to say that even though I am almost finished my >> graduate complex variables course, I still don't think I fully understand >> the function log(z). If someone could please tell me where I am right or >> wrong in the each of the following, I would be very grateful. In >> particular there are some questions I ask later that I would really like >> to know the answers to as well. So I will start from scratch, to see if >> I understand it even from the beginning : >> log is defined as the function from C {0} ---> C by >> log(z) = ln|z| + i*arg(z) >> where we must specify the range as some horizontal strip of height 2pi, >> since log is multivalued. > OK, so that's the definition of log(z) that you use. Please note there > are some problems here: first you say that the domain of log is C{0}, > but right after that you say that we must specify the range as some > horizontal strip of height 2pi; therefore, the domain is *not* C{0}. > Finally, you say that log is multivalued. Now, I understand the > meaning of the assertion every non-zero complex number has infinitely > many logarithms, but what's a multivalued function? Why is the domain not C{0}? By multivalued I guess I mean that there are many values w such that e^w = z. I know that's true, but that would require me to say first that log is the inverse of e. I'm just getting confused again I think. >> In my text, it says that For any branch of log(z), we have e^(log(z)) = >> z. I don't quite understand what this means. Isn't it true that >> e^(log(z)) = z always? That is, log(z) is multivalued, that is true, but >> for whatever value you pick for log(z), e to that value will still be z >> because e^(2pi*i*k) = 1 for k an integer, and 2pi*i*k is only what the >> different values of log(z) differ by. So can't I say that e^(log(z)) = >> z, period? > No. If the definition was log(z) is the only w in the given horizontal > strip such that exp(w) = z then, yes, there would be nothing to prove > here. But since you define it as log(z) = ln|z| + i*arg(z) then some > proof must be provided. >> However, log(e^z) does not equal z necessarily, since log is multivalued. >> log(e^z) = z once we have picked a branch. I said above that e^(log(z)) = z, period, not log(e^z) = z, period. I understand that log(e^z) doesn't equal z unless we specify a branch. But doesn't e^(log(z)) = z no matter what value of log(z) we pick? > I would say log(e^z) = z whenever z belongs to the chosen horizontal > strip. >> Occasionally I will see different branches so I get confused. Is -pi < >> arg(z) < pi a branch or is -pi <= arg(z) < pi a branch? Well clearly the >> first one is, but is the second one a branch? > They both are. If you want to work with an analytic function, then the > second one will not do, since the domain of an analytic function must > always be an open set. >> Here are two examples where I see two totally different branches that >> don't make sense to me : In my Complex analysis book, when we define log, >> arg(z) takes values in the interval [y_0, y_0 + 2pi [, so the half open >> half closed interval. However, later on, when we do real valued >> integrals using residue theorem, (in particular, the real valued integral >> 0 to infinity of [x^(-c)]/[1+x] where 0 < c < 1), and I quote, Let G = >> {z : z is not 0, and 0 < arg(z) < 2pi}; define a branch of the logarithm >> on G by putting l(re^(it)) = log(r) + it where 0 < t < 2pi. So why in >> the first case we have a half closed half open interval, and in the >> second case we have an open interval, defining a branch? > Because both are branches, but in order to apply the residue theorem you > must work with analytic functions and therefore the domain of your > function has to be an open set. >> On a slightly different topic, why is it that in log(z), when you >> approach pi from the top you get a different value for log(z) than when >> you approach from the bottom? Symbolically why is this true? (That is, >> proving it by writing limits) > As stated, the question has no meaning; you must say first of which > branch are you talking about (and that's not the only problem). Suppose > that you have picked up the horizontal strip -pi < Im z < pi (since we're > talking about limits here, you can replace one of the <'s by a > <=, but that will make no difference). Consider those z of the form > exp(ti), with t in ]-pi,pi[. Then > lim_{t -> pi} exp(ti) = lim_{t -> pi} exp(-ti) = -1. > But log(exp(ti)) = t, and so > lim_{t -> pi} log(exp(ti)) = pi i > and > lim_{t -> -pi} log(exp(ti)) = -pi i. > Note that both limits are logarithms of -1. > I hope that this helps. > Jose Carlos Santos === Subject: Re: Please please help with log(z) >I would sincerely appreciate any help, and please read until the end if >you would : I am ashamed to say that even though I am almost finished my >graduate complex variables course, I still don't think I fully understand >the function log(z). If someone could please tell me where I am right or >wrong in the each of the following, I would be very grateful. In >particular there are some questions I ask later that I would really like >to know the answers to as well. So I will start from scratch, to see if >I understand it even from the beginning : >log is defined as the function from C {0} ---> C by >log(z) = ln|z| + i*arg(z) >where we must specify the range as some horizontal strip of height 2pi, >since log is multivalued. >>OK, so that's the definition of log(z) that you use. Please note there >>are some problems here: first you say that the domain of log is C{0}, >>but right after that you say that we must specify the range as some >>horizontal strip of height 2pi; therefore, the domain is *not* C{0}. >>Finally, you say that log is multivalued. Now, I understand the >>meaning of the assertion every non-zero complex number has infinitely >>many logarithms, but what's a multivalued function? > Why is the domain not C{0}? Well, *you* should know what's the domain of whatever function you're colling log. What I can assure you is that there's no continuous function log:C{0} --> C such that exp(log z) = z for each z in C{0}. > By multivalued I guess I mean that there are > many values w such that e^w = z. I know that's true, but that would require > me to say first that log is the inverse of e. I'm just getting confused > again I think. Yes, you are. If you state that you're talking about a function log:C{0} --> C, then to each complex number z corresponds a specific complex number log(z). Just one, not lots of them. I suggest that you stop thinking in terms of multivalued functions. >In my text, it says that For any branch of log(z), we have e^(log(z)) = >z. I don't quite understand what this means. Isn't it true that >e^(log(z)) = z always? That is, log(z) is multivalued, that is true, but >for whatever value you pick for log(z), e to that value will still be z >because e^(2pi*i*k) = 1 for k an integer, and 2pi*i*k is only what the >different values of log(z) differ by. So can't I say that e^(log(z)) = >z, period? >>No. If the definition was log(z) is the only w in the given horizontal >>strip such that exp(w) = z then, yes, there would be nothing to prove >>here. But since you define it as log(z) = ln|z| + i*arg(z) then some >>proof must be provided. >However, log(e^z) does not equal z necessarily, since log is multivalued. >log(e^z) = z once we have picked a branch. > I said above that e^(log(z)) = z, period, not log(e^z) = z, period. I > understand that log(e^z) doesn't equal z unless we specify a branch. But > doesn't e^(log(z)) = z no matter what value of log(z) we pick? No. You are *not* picking a value of log(z). You are *defining* log(z) as ln|z| + i*arg(z) for a certain choice of arg(z). Therefore, the statement exp(log(z)) = z needs a proof. Jose Carlos Santos === Subject: Re: Please please help with log(z) >> I would sincerely appreciate any help, and please read until the end if you >> would : I am ashamed to say that even though I am almost finished my >> graduate complex variables course, I still don't think I fully understand >> the function log(z). If someone could please tell me where I am right or >> wrong in the each of the following, I would be very grateful. In particular >> there are some questions I ask later that I would really like to know the >> answers to as well. So I will start from scratch, to see if I understand it >> even from the beginning : >> log is defined as the function from C {0} ---> C by >> log(z) = ln|z| + i*arg(z) >> where we must specify the range as some horizontal strip of height 2pi, >> since log is multivalued. >OK, so that's the definition of log(z) that you use. Please note there >are some problems here: first you say that the domain of log is C{0}, >but right after that you say that we must specify the range as some >horizontal strip of height 2pi; therefore, the domain is *not* C{0}. >Finally, you say that log is multivalued. Now, I understand the >meaning of the assertion every non-zero complex number has infinitely >many logarithms, but what's a multivalued function? >> In my text, it says that For any branch of log(z), we have e^(log(z)) = z. >> I don't quite understand what this means. Isn't it true that e^(log(z)) = z >> always? That is, log(z) is multivalued, that is true, but for whatever >> value you pick for log(z), e to that value will still be z because >> e^(2pi*i*k) = 1 for k an integer, and 2pi*i*k is only what the different >> values of log(z) differ by. So can't I say that e^(log(z)) = z, period? >No. If the definition was log(z) is the only w in the given horizontal >strip such that exp(w) = z then, yes, there would be nothing to prove >here. But since you define it as log(z) = ln|z| + i*arg(z) then some >proof must be provided. >> However, log(e^z) does not equal z necessarily, since log is multivalued. >> log(e^z) = z once we have picked a branch. >I would say log(e^z) = z whenever z belongs to the chosen horizontal >strip. >> Occasionally I will see different branches so I get confused. Is -pi < >> arg(z) < pi a branch or is -pi <= arg(z) < pi a branch? Well clearly the >> first one is, but is the second one a branch? >They both are. If you want to work with an analytic function, then the >second one will not do, since the domain of an analytic function must >always be an open set. What's _your_ definition of branch of log? >> Here are two examples where I >> see two totally different branches that don't make sense to me : In my >> Complex analysis book, when we define log, arg(z) takes values in the >> interval [y_0, y_0 + 2pi [, so the half open half closed interval. However, >> later on, when we do real valued integrals using residue theorem, (in >> particular, the real valued integral 0 to infinity of [x^(-c)]/[1+x] >> where 0 < c < 1), and I quote, Let G = {z : z is not 0, and 0 < arg(z) < >> 2pi}; define a branch of the logarithm on G by putting l(re^(it)) = log(r) + >> it where 0 < t < 2pi. So why in the first case we have a half closed half >> open interval, and in the second case we have an open interval, defining a >> branch? >Because both are branches, but in order to apply the residue theorem you >must work with analytic functions and therefore the domain of your >function has to be an open set. >> On a slightly different topic, why is it that in log(z), when you approach >> pi from the top you get a different value for log(z) than when you approach >> from the bottom? Symbolically why is this true? (That is, proving it by >> writing limits) >As stated, the question has no meaning; you must say first of which >branch are you talking about (and that's not the only problem). Suppose >that you have picked up the horizontal strip -pi < Im z < pi (since >we're talking about limits here, you can replace one of the <'s by a ><=, but that will make no difference). Consider those z of the form >exp(ti), with t in ]-pi,pi[. Then > lim_{t -> pi} exp(ti) = lim_{t -> pi} exp(-ti) = -1. >But log(exp(ti)) = t, and so > lim_{t -> pi} log(exp(ti)) = pi i >and > lim_{t -> -pi} log(exp(ti)) = -pi i. >Note that both limits are logarithms of -1. >I hope that this helps. >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: Please please help with log(z) >Occasionally I will see different branches so I get confused. Is -pi < >arg(z) < pi a branch or is -pi <= arg(z) < pi a branch? Well clearly the >first one is, but is the second one a branch? >>They both are. If you want to work with an analytic function, then the >>second one will not do, since the domain of an analytic function must >>always be an open set. > What's _your_ definition of branch of log? I do not have one; I was working with what seems to be Isaac's definition of branch of log. The concept that I use which is closest to this one is the concept of logarithmic function. What I mean by this is a function l from A into C, where A is a subset of C{0} and such that, for each z in A, exp(l(z)) = z. Jose Carlos Santos === Subject: Re: Please please help with log(z) >>Occasionally I will see different branches so I get confused. Is -pi < >>arg(z) < pi a branch or is -pi <= arg(z) < pi a branch? Well clearly the >>first one is, but is the second one a branch? >They both are. If you want to work with an analytic function, then the >second one will not do, since the domain of an analytic function must >always be an open set. >> What's _your_ definition of branch of log? >I do not have one; Well then you shouldn't say anything about branches of log. (It's a perfectly standard term, with a standard definition, by the way.) >I was working with what seems to be Isaac's >definition of branch of log. What seems most likely to me otoh is that he's simply ignoring the definition that's given in the book. >The concept that I use which is closest >to this one is the concept of logarithmic function. What I mean by >this is a function l from A into C, where A is a subset of C{0} and >such that, for each z in A, exp(l(z)) = z. >Jose Carlos Santos ************************ David C. Ullrich === Subject: Finite fields, rational functions question I'm trying to show that (F_p)(x,y) is a finite extension of (F_p)(x^p,y^p), but not a simple extension (these are rational function fields). I think I've got the first part. Let q(x^p,y^p) be a rational function in x^p,y^p. Well, let's even for now say polynomial, rather than rational function since the step to rational function is trivial I think. And let's start out with a polynomial in just x^p. Well, if q = a_m x^(pm) + a_(m-1) x^(p(m-1)) + ... + a_0, then I know that in a finite field of characterstic p, every element has a p-th root. So, if (b_i)^p = a_i, then we have q = (b_m x^m + b_(m-1) x^(m-1) + ... + b_0)^p, I skipped a few steps, but this is just Froebenius homomorphism. So I have shown that every element in (F_p)[x^p] is an element of (F_p)[x]. It is not hard (I don't think it is) to see that we can just use the same exact argument to show that every element of (F_p)(x^p,y^p) is in (F_p)(x,y). However, of course the inclusion does not work the other way. Now, I have no idea how to show it's not a simple extension. I guess the only thing I would do is try to show a contradiction. So, suppose that it was a simple extension. i.e. (F_p)(x^p,y^p)(A) = (F_p)(x,y). Then what? I am guessing that it has to do with the fact that we are working with 2 variables (right?). Is it true that (F_p)(x^p) is a simple extension of (F_p)(x) ? Any suggestions are appreciated, Tony === Subject: Re: Finite fields, rational functions question > I'm trying to show that (F_p)(x,y) is a finite extension of > (F_p)(x^p,y^p), > but not a simple extension (these are rational function fields). I think > I've got the first part. Let q(x^p,y^p) be a rational function in > x^p,y^p. Well, let's even for now say polynomial, rather than rational > function since > the step to rational function is trivial I think. And let's start out > with > a polynomial in just x^p. Well, if q = a_m x^(pm) + a_(m-1) x^(p(m-1)) + > ... + a_0, then I know that in a finite field of characterstic p, every > element has a p-th root. So, if (b_i)^p = a_i, then we have q = (b_m x^m > + b_(m-1) x^(m-1) + ... + b_0)^p, I skipped a few steps, but this is just > Froebenius homomorphism. So I have shown that every element in (F_p)[x^p] > is an element of (F_p)[x]. It is not hard (I don't think it is) to see > that we can just use the same exact argument to show that every element of > (F_p)(x^p,y^p) is in (F_p)(x,y). However, of course the inclusion does > not work the other way. I don't understand why you need anything like this: Writing k for F_p, k(x^p,y^p) < k(x^p,y) < k(x,y) and each extension is simple of degree p. Hence the whole extension is of degree p^2. > Now, I have no idea how to show it's not a simple extension. I guess the > only thing I would do is try to show a contradiction. So, suppose that it > was a simple extension. i.e. (F_p)(x^p,y^p)(A) = (F_p)(x,y). Then what? > I am guessing that it has to do with the fact that we are working with 2 > variables (right?). Is it true that (F_p)(x^p) is a simple extension of > (F_p)(x) ? > Any suggestions are appreciated, Any element z in k(x,y) has z^p in k(x^p,y^p); so the extension generated by z has degree p, and cannot be the whole of k(x,y). -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Change in meaning of inertial motion in Newton to Einstein Inertial compensation in REST LIF? NO! Inertial compensation in REST LNIF? YES! The Question is: What is The Question? John A. Wheeler We want to make Einstein's GR look as much like Maxwell's EM as possible. Actually like non-Abelian Yang-Mills since gravity carries its own charge like W bosons and gluons but unlike photons. Lots of papers on this of course including tetrads and also Ashtekar's (more on that down the line). Maxwell connection for parallel transport in the internal fiber space is the vector potential Au with the U(1) gauge covariant derivative Du = ,u - ieAu Au is not a U(1) tensor it transforms as Au -> Au' = Au + Chi,u this is the NON-TENSOR connection transform relative to the U(1) internal symmetry group. The Maxwell field tensor Fuv is the curl of Au AVOID RIGID CONNECTIONS! They are not BACKGROUND INDEPENDENT. Similarly, Levi-Civita connection for parallel transport of vector fields in curved spacetime base space (no torsion no Poltorak RIGID affine connection etc, just plain vanilla 1916 GR) has a GCT SPACETIME SYMMETRY GROUP gauge covariant derivative, symbolically like Du* = ,u - {LC} ... example, for any GCT first rank tensor we get the second rank tensor DuB^v = B^v,u + {LC}^vuwB^w The NON GCT TENSOR gauge transform analogous to Au -> Au' = Au + Chi,u in Maxwell's EM is precisely {LC} = N -> {LC}' = N' = XXXN + XY where X is the GCT Jacobian matrix and Y is a partial derivative of X. I can provide the indices but they are unwieldy. Just as Au is not a U(1) tensor so also is {LC} not a GCT tensor. Z's attempt to use Newtonian inertial compensation here is wrong. The shift from Newton to Einstein is a profound change in the meaning of inertial motion in which the idea of an objective Newtonian force of gravity in an inertial frame is completely eliminated. Hence there is nothing to compensate! The inertial g-force, which is {LC}^i00, i = 1,2,3 in the REST LNIF of a is always caused by the latter. The inertial g-force indeed compensates the non-gravity force in the REST LNIF. But that is not what Z proposes. Z wants to compensate a gravity force in a REST LIF - a profoundly wrong idea! That is, Z proposes that in the REST LIF where {LC} = 0 that {LC} = T + N = 0 T =/= 0 where T is a GCT tensor of rank 3. Do not confuse this with torsion. Z is only talking 1916 GR. This is profoundly wrong. What is correct is that in the REST LNIF {LC)^i00 + (External Non-Gravity Force)^i = 0 That is, the inertial g-force in the non-inertial frame exactly timelike geodesic. For example, here on surface of the Earth we are in a local non-inertial frame (LNIF) from the electrical reaction forces and quantum Fermi-Dirac pressure of the rock on which we stand. That's why we feel weight. Z's deep error is to apply inertial compensation not to the REST LNIF of the test object where it does apply, to the REST LIF where it does not apply. Back to the connections and Stoke's theorem & Bohm-Aharonov Effect The local vector potential U(1) EM connection Au is not a local classical observable, but it is a nonlocal quantum observable because from Stoke's theorem the closed loop line integral of Au is the magnetic this causes a fringe shift in a double slit experiment with electrons passing through a region free from magnetic field but with nonzero Au connection field. Similarly in general relativity, where the macro-quantum vacuum coherence, which makes Einstein's cosmological constant near zero, is a giant quantum wave, there will be an analogous fringe shift! tidal stretch-squeeze geodesic deviation curvature 4th rank GCT tensor Ruvwl is the analog to the Fuv Maxwell field tensor. Ruvwl is the GCT covariant curl of the Levi-Civita {LC}uvw connection just like Fuv is the curl of Au. Note that Ruvwl has physical dimensions of 1/Area, and the {LC} has physical dimensions 1/Length. The line integral and the surface integral in the generalized Stoke's theorem of manifold topology (independent of metric) i.e. DeRham-Hodge integral of p-Cartan form about a closed p-boundary of a p+1 co-form = integral of the exterior derivative p + 1 form of the p-form over the bounded p + 1 co-form (manifold) Gives the curvature flux through any bounded area in curved space-time. This is analogous to the magnetic flux. What is the analog of the fringe shift? Is it the change in the orientation of a vector parallel transported around the closed loop boundary of that arbitrary 2-surface in curved spacetime? If you do a macro-quantum interference experiment, you will get a Berry phase shift as well. How can you do that? That's where metric engineering the fabric of spacetime for warp, wormhole and weapon W^3 comes in. Another story coming soon to a computer screen near you. === Subject: Re: Philosopher/marketer is stumped by this Probability Problem. Plz help. <41AB7A43.3070508@netscape.net> > [bcc'd to OP, who e-mailed me] > You asked what is the *probability* that one method is better than the > other, but I suspect that that is not what you really want. In the > frequentist paradigm, one poses hypotheses in terms of unknown > parameters which are not random variables, hence it makes no sense to > speak of the probability of a hypothesis. Rather, one decides whether > an observed event is excessively improbable in the case that a > hypothesis is true; if so, one rejects the hypothesis and accepts the > alternative. > Suppose you want to ask the question, Is there evidence that two > methods differ in their efficacy? The frequentist method is as > follows. Let p1 and p2 be the yields of the two methods. You pose > the null hypothesis and its alternative, > H0: p1 = p2 > H1: p1 != p2 > Under the null hypothesis, the test statistic Z = (Q1-Q2) / sqrt(Qc > (1-Qc) / n) has approximately a standard normal distribution, where the > Q's are defined as in my post and n is the combined sample size (your > Y1+Y2). ... Oops! The test statistic is Z = (Q1-Q2) / sqrt(Qc (1-Qc) (1/Y1 + 1/Y2)) === Subject: Re: Philosopher/marketer is stumped by this Probability Problem. Plz help. <41AB7A43.3070508@netscape.net> >I'm looking for a general solution to the following problem: >Suppose you have two urns, each containing an equivalent VERY LARGE >(maybe not infinite, but very large) number of balls. Each urn has >only red and green balls in it. >Suppose you draw from urn A 11 times, with replacement, and you get 7 >green balls. So the ratio of green balls to total draws in the sample >is 7/11. >Suppose you draw from urn B 15 times, with replacement, and you get 8 >green balls. So the ratio, for urn B, of green balls to total draws >in the sample is 8/15. >Assuming random sampling, and given these samples, what is the >probability that urn A contains more green balls than urn B? >I am actually looking for the general solution. Given a sample from >urn A of X1/Y1, and a sample from urn B of X2/Y2 (all Xs and Ys are >integers), what is the probability that there are more greens in urn A >than in urn B. > If you actually want a *probability*, you need to specify a prior joint > distribution on the fractions of green balls in the two urns. Use > Bayesian updating. > It is also possible to do a frequentist hypothesis test: Y1 and Y2 > should be slightly larger (but not necessarily too much so), and use the > test statistic (Q1-Q2) / sqrt(Qc (1-Qc) / (Y1+Y2)), which has > approximately a standard normal distribution. Here, Qi = Xi / Yi, > and Qc = (X1+X2) / (Y1+Y2) Oops! The test statistic is (Q1 -Q2) / sqrt(Qc (1-Qc) (1/Y1 + 1/Y2)). === Subject: Re: random 1 to 3 to random 1 to 5 > One of my friend asked me this question which he was asked during a > job interview. Couldn't figure it out and very curious about the > solution (or this just mpossible? ) > The question is, given a function that returns evenly distributed > random number from 1 to 5, design a new function based on this > function, and return evenly distributed random number from 1 to 3. I would invoke the thing three times in a row and associate the first selection with the numbers 1-5, the second selection with 6-10 an the third with 11-15 and divide the result by three. OK, I figured this out in a minute but would NEVER have been able to do so during a job interview with mr/mrs personnel officer watching me, waiting for my defeat. There is no end to the abuse employers will go these days: I have had to sign a PISS form prior to even being interviewed. Well maybe I am an alcoholic! do you really care if I solve your problem? I view all of this as abuse: and it doesn't matter for the outcome anyway since the personnel person will hire people whom he(she) likes anyway but has to rationalize his(her) decisions with some pseudo-scientific rationalization. In the end he/she will hire: people ;ike him/her self! These would not be programmers of course but they would make great personnel officers. Now to the core of the matter: Is the person who can solve puzzles like the above in a minute the better candidate? I don't think so. Maybe this person has trained in goat wolf and cabbage puzzles all his life! Or maybe he is a member of Mensa which means that he can deduce the general term of any series with complete confidence from a mere three terms! I will never forget the scathing sarcasm of my algebra professor when he ridiculed certain puzzles in the newspaper asking the reader to complete a certain given series! I have heard that MICROSOFT uses puzzles like the above to evaluate potential job candidates so I assume they can cross any river without problems, maybe that's why their software is so popular. Now, there is such a thing as deep thinking and it requires time and an environment free of disturbances, much different from the environment of the job interview. There are also KNOWLEDGE and EXPERIENCE which one can actually measure in a scientific manner - provided one has some oneself. Finally, how a person will behave or fit in within a group cannot be deduced from superficial interpersonal behaviour in any way. Jentje Goslinga === Subject: Re: random 1 to 3 to random 1 to 5 > Now to the core of the matter: Is the person who can solve > puzzles like the above in a minute the better candidate? I believe this trend got started because Microsoft interviews programmers with logic puzzle, why are manhole covers round, and the like. But this is bad logic. Microsoft makes terrible software! What people should do is find out how Microsoft interviews their MARKETING people, and emulate that! === Subject: Re: random 1 to 3 to random 1 to 5 > One of my friend asked me this question which he was asked during a > job interview. Couldn't figure it out and very curious about the > solution (or this just mpossible? ) > The question is, given a function that returns evenly distributed > random number from 1 to 5, design a new function based on this > function, and return evenly distributed random number from 1 to 3. > rand1to3 = (rand1to5 - 1)/2 + 1 I'm guessing the OP meant to say integer rather than number (otherwise it's a kind of silly question). === Subject: Re: random 1 to 3 to random 1 to 5 > uh, f(x) =(x+1)/2? That will upset the ditribution. In the simplest example, if the 1-5 outputs all occur at equal frequency over a sufficiently long time interval, then the frequencys of the 1-3 outputs will be 1: 40% 2: 40% 3: 20% === Subject: Re: random 1 to 3 to random 1 to 5 ETAtAhUAxBJIWxM51G2i0uLGKaCpuJshQ2UCFHBkr2Qp/3fEVg4RMWkdEskPHtQo rphenry, I could not reach you via email. See what I subsequently posted for the discrete case. --OL === Subject: Re: semi-infinite number > Anyway, it seems that semi-infinite or semiinfinite is a short way of > saying infinite in one direction but not the other (e.g. along a > ray or a halfplane). In such contexts it is a perfectly reasonable > construction. That might be one meaning, but I think it's more usual as an engineering term. The engineering meaning, as I recall it explained to me by my father is something like this: A magnitude is semi-infinite if it is so large that increasing it would make no practical difference. === Subject: Re: semi-infinite number I'll take it to mean so many that no one can be bothered to count them. > There is a semi-infinite number of .... > Anyone know how many that would be? === Subject: I dont quite understand this as an equivalence relation Equivalence Relation === Subject: re:I dont quite understand this as an equivalence relation You seemed to have left something out. Where is the thing that you are questioning? *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Graph Coloring In any maximal planar graph it is theoretically possible for as many as three vertices to simultaneously have degree = (n-1). But is seems that the practical limit is only two vertices. Is it possible to prove that for all n, there can be only two vertices with degree = (n-1) in a maximal planar graph? thank you, === Subject: Re: Graph Coloring 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > In any maximal planar graph it is theoretically possible for as many > as three vertices to simultaneously have degree = (n-1). But is seems > that the practical limit is only two vertices. Is it possible to > prove that for all n, there can be only two vertices with degree = > (n-1) in a maximal planar graph? Three vertices with degree n-1, and any three other vertices, would form a K_3,3 subgraph. Therefore the only maximal planar graphs with three degree-(n-1) vertices are those with fewer than three other vertices: that is, K_3, K_4, and the unique 5-vertex maximal planar graph. -- David Eppstein Computer Science Dept., Univ. of California, Irvine http://www.ics.uci.edu/~eppstein/ === Subject: ? explicit formula of some special matrices Hi: Many numerical analysis for some algorithm solving DEs turn out to be study the eigen system of the discritized system matrix. These matrices have some psecial structures such as banded. Question is, though I searched many numerical books and linear algebra books, I just cannot find one that explicitly shows how to derive the eigenvalues and the eigenvectors of those special matrices. Can anyone suggect a book or show how to derive those? by Cheng Cosine Dec/05/2k4 UT === Subject: Compare randomness of shuffles - possible? A shuffle of cards is a probability distribution on the permutation group of the cards. What are the methods to compare shuffles in terms of how well they do? For example: randomness? === Subject: Re: Compare randomness of shuffles - possible? Sometime back I was also trying to solve this, and now u have reopened the question in my mind. > A shuffle of cards is a probability distribution on the permutation > group of the cards. What are the methods to compare shuffles in terms > of how well they do? For example: randomness? === Subject: Re: Compare randomness of shuffles - possible? >A shuffle of cards is a probability distribution on the permutation >group of the cards. What are the methods to compare shuffles in terms >of how well they do? For example: randomness? Take the standard deviation of the probabilities of all different permutations. It will be 0 when the probabilities are all equal -- a fair shuffle. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Compare randomness of shuffles - possible? A shuffle of cards is a probability distribution on the permutation group of the cards. What are the methods to compare shuffles in terms of how well they do? For example: randomness? === Subject: Planning in card shuffling To my understanding I suppose a shuffle of cards is a probability distribution on the permutation group of the cards. Given a deck of cards and a number of shuffles is there a method to find a combination of those shuffles that can achieve a certain requirements, for example: A full house on top of the deck with probability greater than a half. === Subject: Re: Planning in card shuffling There's a result called seven shuffles suffice which some mathematician calculated about 10 years ago or so. Based on one model of shuffling, I think they found that seven shuffles would completely randomize the deck by some measure. Kind of the opposite of what you're asking, to unrandomize a deck. Skilled magicians use perfect shuffles for some card tricks: divide the deck exactly in half, and shuffle so that cards from left and right hands alternate exactly. This is NOT a randomizing procedure, and I think it cycles back to the original permutation in four shuffles. - Randy === Subject: Re: Planning in card shuffling > There's a result called seven shuffles suffice which some > mathematician calculated about 10 years ago or so. Based on one model > of shuffling, I think they found that seven shuffles would completely > randomize the deck by some measure. Kind of the opposite of what you're > asking, to unrandomize a deck. > Skilled magicians use perfect shuffles for some card tricks: divide the > deck exactly in half, and shuffle so that cards from left and right > hands alternate exactly. This is NOT a randomizing procedure, and I > think it cycles back to the original permutation in four shuffles. > - Randy You will get back to the original arrangement after executing some number of perfect shuffles, but note that there are two different shuffles here, depending on whether the top card is on top or in second place in the shuffle, as below with 6 cards Original deck: ABCDEF Technique 1 Cut the deck in two: ABC DEF Shuffle the stacks together: ADBECF Technique 2 Cut the deck: ABC DEF Shuffle the stacks together: DAEBFC In this case, technique 1 requires 4 shuffles to get back to the original order and technique 2 requires 3 shuffles. For n = 52, the orders of the two operations are 8 and 52. It's an interesting exercise to find an expression for the orders for general n. Rick === Subject: Re: Planning in card shuffling > To my understanding I suppose a shuffle of cards is a probability > distribution on the permutation group of the cards. Given a deck of > cards and a number of shuffles is there a method to find a combination > of those shuffles that can achieve a certain requirements, for > example: A full house on top of the deck with probability greater than > a half. If you don't have any information about the initial state of the deck, applying a permutation to it's elements (cards) gives nothing better--or as they sometimes say in computer science, garbage in-garbago out. On the other hand, if you have some knowledge about the initial state of the deck, you could use it to obtain rather impressive looking results from your shuffles100% of the time. === Subject: Re: Planning in card shuffling <3LqdnThLjoVeXyncRVn-sA@comcast.com> >about the initial state of the deck Assume the distribution of input is random, output should be like I have to shuffle X times with permutation Y, Z, probability >0.5? === Subject: Computational problem in symmetric group using GAP By using GAP - http://www.gap-system.org/ How can i express a given element in a symmetric group as a product of certain fixed elements in the same symmetric group, if possible by using GAP? Examples of real implementations (by GAP, or Maple?) are mostly welcomed! === Subject: How much lossless compression is possible in images? Assuming ideal compression that removes every trace of redundancy from an image, how much lossless compression might really be possible? I know this is a difficult question because so much depends on image content, but I'm just trying to get an idea of how close current lossless compression methods are to the ideal. Visually images seem to be very highly redundant, although current lossless algorithms only manage about 50% compression. But assuming time and space were not constraints, how far could one theoretically go? -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? > Assuming ideal compression that removes every trace of redundancy from > an image, how much lossless compression might really be possible? I > know this is a difficult question because so much depends on image > content, but I'm just trying to get an idea of how close current > lossless compression methods are to the ideal. Visually images seem to > be very highly redundant, although current lossless algorithms only > manage about 50% compression. But assuming time and space were not > constraints, how far could one theoretically go? Time and space are not constraints? Okay then . . . Fix some binary representation x of your image, and some universal Turing machine U. Begin computations of U(0), U(1), U(00), U(01), U(10), and so on up to U(smax) where smax encodes Print x and stop. After BB(n) steps (where n is the number of computations you're running), all the computations that eventually halt will have halted. Now look at the output of all the computations, and pick the (lexigraphically) first string s such that U(s) halted leaving x on the tape. The function BB() is known as the Busy Beaver function. For a typical, say 4 megabyte, image, you'll have to run your computations for about BB(10^10^7) steps. This number is kind of big, and just to compute it is kind of hard. [And Jmes Hrris is kind of unlikely to prove Fermt's Lst Theorem.] The length of the chosen, minimal s is known as the Kolmogorov complexity of x. It is the ultimate measure of how much information is contained in a string, though it is kind of hard to compute in practice. === Subject: Re: How much lossless compression is possible in images? > Time and space are not constraints? Okay then . . . > Fix some binary representation x of your image, and some universal > Turing > machine U. Begin computations of U(0), U(1), U(00), U(01), U(10), and > so > on up to U(smax) where smax encodes Print x and stop. After BB(n) > steps > (where n is the number of computations you're running), all the > computations > that eventually halt will have halted. Now look at the output of all > the > computations, and pick the (lexigraphically) first string s such that > U(s) > halted leaving x on the tape. > The function BB() is known as the Busy Beaver function. For a typical, > say > 4 megabyte, image, you'll have to run your computations for about > BB(10^10^7) > steps. This number is kind of big, and just to compute it is kind of > hard. > [And Jmes Hrris is kind of unlikely to prove Fermt's Lst Theorem.] > The length of the chosen, minimal s is known as the Kolmogorov > complexity of > x. It is the ultimate measure of how much information is contained in > string, though it is kind of hard to compute in practice. So how much compression could you get? -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? >So how much compression could you get? I don't think you have to wait for BB(10^10^7) if you accept a good compression that halts early, given that the ideal algorithm for that image is consise, its a heuristic to expect it to run mostly useful cycles, even on the polynomial side of exponential complexity. Besides you can't tell what BB(x) x>20 is, you have make some heuristic decision when to halt. (of course in your ideal no complexity limits you can tell the value of BB). what compression? on half of the images (99.99999999999% of 4GB images are random pixels) you will get none. the algorithms number to match the data will just be a UTM copy algorithm with the 4GB input string hard wired. Program(487948744894984984 .. 4mb long) = 4 mb image on the other half of images (99.999999999999% of 4GB images are random pixels), you will save a few bytes. of the 0.00000000000001% of images with detail, probably quite good, maybe 10 to 100 times better than jpeg, 10% of original image size for photos. (guess). pictures of cities with intrinsic detail should compress well under a 'selectable algorithm'. Herc === Subject: Re: How much lossless compression is possible in images? what would be interesting about these images if you had the computing technology to do it, is that lossy distortion would be once only. if you decompress a lossy image then compress it again you get the exact image (given that image is a good local minimum of the program sizes to represent it). compare this to jpg, they are terrible to work with, every time you open the image, do a change and save it as jpg the distortion builds up and up. an analogy to digital copies vs analog, once you accept the distortion / approximation of the digitizing process the image will never distort again. Herc === Subject: Re: How much lossless compression is possible in images? > Assuming ideal compression that removes every trace of redundancy from > an image, how much lossless compression might really be possible? I > know this is a difficult question because so much depends on image > content, but I'm just trying to get an idea of how close current > lossless compression methods are to the ideal. Visually images seem to > be very highly redundant, although current lossless algorithms only > manage about 50% compression. But assuming time and space were not > constraints, how far could one theoretically go? > -- > Transpose hotmail and mxsmanic in my e-mail address to reach me directly. It depends totally on image content. An image consisting of random pixels will not compress at all (0% compression) using a generic algorithm; an image consisting of entirely one colour will compress almost to zero (almost 100% compression). A photos maximum compression will vary between 0% and 100% depending upon the amount of random detail in the photo. === Subject: Re: How much lossless compression is possible in images? ... > An image consisting of random pixels will not compress at all (0% > compression) using a generic algorithm; an image consisting of entirely one > colour will compress almost to zero (almost 100% compression). If you use a generic algorithm it is pretty likely that an image consisting of random pixels will expand and not compress. -- 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: How much lossless compression is possible in images? > If you use a generic algorithm it is pretty likely that an image consisting > of random pixels will expand and not compress. Actually, the chance that a random image will become smaller after compression with any given algorithm is always 0.5. The change that it will become larger is thus also 0.5. Half of all random images will be smaller; the other half will be larger. This is true no matter what algorithm is used, as long as it is lossless (reversible). -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? > If you use a generic algorithm it is pretty likely that an image consisting > of random pixels will expand and not compress. > Actually, the chance that a random image will become smaller after > compression with any given algorithm is always 0.5. The change that it > will become larger is thus also 0.5. Half of all random images will be > smaller; the other half will be larger. This is true no matter what > algorithm is used, as long as it is lossless (reversible). Wrong, see my other reply. -- 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: How much lossless compression is possible in images? Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >Actually, the chance that a random image will become smaller after >compression with any given algorithm is always 0.5. Why? Consider the algorithm that precedes all input strings with a 1-bit. It makes all strings longer. Perhaps you don't consider that a compression algorithm. So now consider the algorithm that replaces all strings starting with a sequence of ten or more zeros with a byte count of the zeros (up to 255) followed by the rest of the string, and the other strings with a 0 byte followed by the original string. That only make 1 in 1024 strings smaller. So maybe you mean an algorithm that, on average, doesn't change the length (which is the best kind of algorithm). Maybe it makes one very long string small, and lots of others just lightly bigger. This will obviously not make 50% of images smaller. -- Richard === Subject: Re: How much lossless compression is possible in images? > Why? Because compression only works if some messages of a given length are more probable than others of that same length. If messages are completely random, they are all equally probable. And if random messages must be preserved bit-by-bit, it is not possible to represent them with fewer bits than they originally contained. The entropy of the messages must be preserved. > Consider the algorithm that precedes all input strings with a 1-bit. > It makes all strings longer. This isn't really a compression algorithm in the sense being discussed here. A compression algorithm is simply a substitution algorithm. It performs a one-for-one substitution of variable-length messages from one set A for fixed-length messages from another set B. Set A is designed such that the messages it contains corresponding to the most probable messages in B are shorter than the fixed-length of the messages in B; all other messages in A are longer. Since the most probable messages in B are translated to shorter messages in A, the algorithm effectively compresses messages. But this only works if the messages in B are not all equally probable. If they are equally probable (as they must be if they are random), there is no way to compress them, because a longer message from A is just as likely to be substituted as a shorter message. Note that the total length of all messages in set A must be at least equal to the total length of all messages in set B. So if random substitution of messages from A is forced by equal probability for all messages from B (random messages in B), then the net result is zero compression. > Perhaps you don't consider that a compression algorithm. I don't. A more rigorous definition can be found above. > So maybe you mean an algorithm that, on average, doesn't change the > length (which is the best kind of algorithm). All algorithms will produce output of equal or greater length than input if the input is random. > Maybe it makes one > very long string small, and lots of others just lightly bigger. This > will obviously not make 50% of images smaller. It will, however, guarantee that the average length change for all messages is zero, which is nearly the same thing from a practical standpoint. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? > It depends totally on image content. > An image consisting of random pixels will not compress at all (0% > compression) using a generic algorithm; an image consisting of entirely one > colour will compress almost to zero (almost 100% compression). > A photos maximum compression will vary between 0% and 100% depending upon > the amount of random detail in the photo. My understanding has been that a Fourier transformation can greatly compress data, but that the best compression is achieved by analyzing and compressing the entire data set as a single unit, which is impractical in real life because it's computationally very difficult to analyze, say, an entire 400 MB of image data and produce a single transformation encompassing that entire file. If it were done, though, I presume it would be very compact indeed. Can anyone confirm this? I know that some existing systems also allow for a bit of image-dependent customization, but that it often isn't carried out because of the overhead. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <41b3ed05$0$9113$afc38c87@news.optusnet.com.au> gif uses run length encode r r r r -> 4r so its only suitable for a low color pallete jpg fourier transforms the image, then uses run length encode on that. often the final digits of each row are 0's, these are the high frequencies (the detail) and compression just ignores some of them. if you only chop 0's you get lossless compression, if you chop 00002000030001000 it gets lossy. quadtree is a recursive subdividing algorithm that could be useful for maps but never took off. i devised a composite quadtree gif system. the quadtree seperates the image into discernible simple sections, and the run length encode is tested at 360 different angles to find the best compression. basically it searches for line segments, suitable for vector type graphics, stickmen. what gets me is why they can't add a pollish function to jpeg decompression, obviously a big single color area with a border edge around it is not meant to get 'pixelated' at the border. apparently there is a fractal equation for every photo in existence. some amazing pictures of womens faces were constructed with equations (very high compression) if you can tailor the image somewhat. and the detail just keeps up as you zoom in.. computer games cheat, they store a library of tile bitmaps and surface everything, why download when you can put in a DVD of a million pictures and the computer just selects. Herc === Subject: Re: How much lossless compression is possible in images? > gif uses run length encode > r r r r -> 4r Wrong. Gif uses LZW. -- 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: How much lossless compression is possible in images? >> gif uses run length encode >> r r r r -> 4r >Wrong. Gif uses LZW. right, actually its selectable. Herc === Subject: Re: How much lossless compression is possible in images? >> gif uses run length encode >> >> r r r r -> 4r > >Wrong. Gif uses LZW. > right, actually its selectable. Wrong, it is not selectable, at least not in Gif89a, and as far as I know that is the last version ever published. If it were selectable there would have been a plethora of free Gif builders that used only RLE. But because Gif builders use LZW they fall under a patent and are not free to use (decoders for LZW *are* free to use). That is also the reason the W3 consortium has defined a new format: PNG. -- 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: How much lossless compression is possible in images? > But because Gif builders use LZW they fall under a patent and are not free > to use (decoders for LZW *are* free to use). The patent has expired. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? > But because Gif builders use LZW they fall under a patent and are not free > to use (decoders for LZW *are* free to use). > The patent has expired. I think it is too late. -- 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: How much lossless compression is possible in images? > I think it is too late. Too late for what? It's too late to file any patent-infringement claims, yes--since the patent has expired. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? > I think it is too late. > Too late for what? It's too late to file any patent-infringement > claims, yes--since the patent has expired. Too late to see a plethora of free gif encoders emerging. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: `Uncompressed' GIF files (was: Re: How much lossless compression is possible in images?) Originator: dmoews@ccrwest.org (David Moews) |[...] |Wrong, it is not selectable, at least not in Gif89a, and as far as I know |that is the last version ever published. If it were selectable there |would have been a plethora of free Gif builders that used only RLE. But |because Gif builders use LZW they fall under a patent and are not free |to use (decoders for LZW *are* free to use). That is also the reason the |W3 consortium has defined a new format: PNG. Although the GIF format specifies the use of an LZW decompressor, it is possible to write GIF files which, although readable by an LZW decompressor, are not actually compressed. Presumably, this does not infringe on LZW patents. This is the approach used by the developers of libungif . -- David Moews dmoews@xraysgi.ims.uconn.edu === Subject: Re: `Uncompressed' GIF files (was: Re: How much lossless compression is possible in images?) > |[...] > |Wrong, it is not selectable, at least not in Gif89a, and as far as I know > |that is the last version ever published. > Although the GIF format specifies the use of an LZW decompressor, it is > possible to write GIF files which, although readable by an LZW decompressor, > are not actually compressed. Oh, yes, that is very true. When you know enough about the LZW format and the actual format used, it is possible to present files that are not LZW compresses, but are still readable by a GIF/LZW decompressor. LZ compressions are pretty similar, there is only a slight variation. Standard LZ uses pointers back into previous material, and it is only a variation on how it is presented and how long the previous material is. LZW is only a slight change in that it uses a dictionary of previous material. The way how the dictionary is build, and the way how it is decided that the current dictionary is full, are part of the patent. If you have other ways to build a dictionary (and a way to encode it that can be understood by LZW decompressors), there is no patent infringement. I can imagine a few ways. And Eric Raymond is also not entirely stupid... But, RLE encoding is not decompressible by LZW decoders as far as I know. -- 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: How much lossless compression is possible in images? www.justsaying.com/rle.gif Gods don't make mistakes! Herc === Subject: Re: How much lossless compression is possible in images? > www.justsaying.com/rle.gif > Gods don't make mistakes! a. Description. The image data for a table based image consists of a sequence of sub-blocks, of size at most 255 bytes each, containing an index into the active color table, for each pixel in the image. Pixel indices are in order of left to right and from top to bottom. Each index must be within the range of the size of the active color table, starting at 0. The sequence of indices is encoded using the LZW Algorithm with variable-length code, as described in Appendix F. And see also Appendix F of that document. But I will look what your gif file amounts to. I suspect it will be a form of LZW. -- 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: How much lossless compression is possible in images? > > www.justsaying.com/rle.gif > > > Gods don't make mistakes! > a. Description. The image data for a table based image consists of a > sequence of sub-blocks, of size at most 255 bytes each, containing an > index into the active color table, for each pixel in the image. Pixel > indices are in order of left to right and from top to bottom. Each index > must be within the range of the size of the active color table, starting > at 0. The sequence of indices is encoded using the LZW Algorithm with > variable-length code, as described in Appendix F. > And see also Appendix F of that document. But I will look what your gif > file amounts to. I suspect it will be a form of LZW. The LZW patent has expired. === Subject: Re: How much lossless compression is possible in images? >But I will look what your gif >file amounts to. I suspect it will be a form of >LZW. >dik t. winter http://www.justsaying.com/rle.gif http://www.justsaying.com/lzw.gif here's 2 gifs with the same original image, simple checkers. Herc === Subject: Re: How much lossless compression is possible in images? > Assuming ideal compression that removes every trace of redundancy from > an image, how much lossless compression might really be possible? I > know this is a difficult question because so much depends on image > content, but I'm just trying to get an idea of how close current > lossless compression methods are to the ideal. Visually images seem to > be very highly redundant, although current lossless algorithms only > manage about 50% compression. But assuming time and space were not > constraints, how far could one theoretically go? > -- > Transpose hotmail and mxsmanic in my e-mail address to reach me directly. > It depends totally on image content. > An image consisting of random pixels will not compress at all (0% > compression) using a generic algorithm; an image consisting of entirely one > colour will compress almost to zero (almost 100% compression). I don't have the math at hand, but I believe that a truly random image would have at least some compressible areas. In a strange sense, a completely incompressible image would be as ordered as a single color image. === Subject: Re: How much lossless compression is possible in images? >> Assuming ideal compression that removes every trace of redundancy from >> an image, how much lossless compression might really be possible? I >> know this is a difficult question because so much depends on image >> content, but I'm just trying to get an idea of how close current >> lossless compression methods are to the ideal. Visually images seem to >> be very highly redundant, although current lossless algorithms only >> manage about 50% compression. But assuming time and space were not >> constraints, how far could one theoretically go? > -- >> Transpose hotmail and mxsmanic in my e-mail address to reach me >directly. >> It depends totally on image content. >> An image consisting of random pixels will not compress at all (0% >> compression) using a generic algorithm; an image consisting of entirely >one >> colour will compress almost to zero (almost 100% compression). >I don't have the math at hand, but I believe that a truly random image >would have at least some compressible areas. In a strange sense, a >completely incompressible image would be as ordered as a single color image. There's no such thing as an incompressible image, _until_ you specify what algorithm you're using. Given _any_ image I_0 there _is_ a lossless compression algorithm that compresses I_0 to a one-byte file: def compress(I): if I == I_0: return a single 0 byte else: return a 1, followed by I. I leave the decompressor as an exercise. ************************ David C. Ullrich === Subject: Re: How much lossless compression is possible in images? > There's no such thing as an incompressible image, _until_ you specify > what algorithm you're using. Given _any_ image I_0 there _is_ a > lossless compression algorithm that compresses I_0 to a one-byte file: > def compress(I): > if I == I_0: > return a single 0 byte > else: > return a 1, followed by I. > I leave the decompressor as an exercise. Your example also highlights a point about how the effectiveness of compression is to be measured. Since the *decompressor* is essential to recovering the original image from the output of the compressor, istm that an appropriate measure ought to include the size of the decompressor as well as the size of the compressor output. E.g., if we use the simple ratio r = (|image_compressed| + |decompressor|) / |image_original| , then of course your example has r > 1, rather than the desired r < 1. With that in mind, it might make sense to speak of an image as being absolutely incompressible if all compressors & decompressors have r >= 1 for that image. --r.e.s. === Subject: Re: How much lossless compression is possible in images? >> There's no such thing as an incompressible image, _until_ you specify >> what algorithm you're using. Given _any_ image I_0 there _is_ a >> lossless compression algorithm that compresses I_0 to a one-byte file: >> def compress(I): >> if I == I_0: >> return a single 0 byte >> else: >> return a 1, followed by I. >> I leave the decompressor as an exercise. >Your example also highlights a point about how the effectiveness of >compression is to be measured. Since the *decompressor* is essential >to recovering the original image from the output of the compressor, >istm that an appropriate measure ought to include the size of the >decompressor as well as the size of the compressor output. >E.g., if we use the simple ratio >r = (|image_compressed| + |decompressor|) / |image_original| , >then of course your example has r > 1, rather than the desired r < 1. >With that in mind, it might make sense to speak of an image as being >absolutely incompressible if all compressors & decompressors have >r >= 1 for that image. Nope. The length of the code for a decompressor depends on what language we're using. See my reply to guenther's reply to Tobin for a more detailed explanation. >--r.e.s. ************************ David C. Ullrich === Subject: Re: How much lossless compression is possible in images? >>Your example also highlights a point about how the effectiveness of >>compression is to be measured. Since the *decompressor* is essential >>to recovering the original image from the output of the compressor, >>istm that an appropriate measure ought to include the size of the >>decompressor as well as the size of the compressor output. >>E.g., if we use the simple ratio >>r = (|image_compressed| + |decompressor|) / |image_original| , >>then of course your example has r > 1, rather than the desired r < 1. >>With that in mind, it might make sense to speak of an image as being >>absolutely incompressible if all compressors & decompressors have >>r >= 1 for that image. > Nope. The length of the code for a decompressor depends on what > language we're using. See my reply to guenther's reply to Tobin > for a more detailed explanation. Interpret all compressors & decompressors more inclusively, please. --r.e.s. === Subject: Re: How much lossless compression is possible in images? >Your example also highlights a point about how the effectiveness of >compression is to be measured. Since the *decompressor* is essential >to recovering the original image from the output of the compressor, >istm that an appropriate measure ought to include the size of the >decompressor as well as the size of the compressor output. >E.g., if we use the simple ratio >r = (|image_compressed| + |decompressor|) / |image_original| , >then of course your example has r > 1, rather than the desired r < 1. >With that in mind, it might make sense to speak of an image as being >absolutely incompressible if all compressors & decompressors have >r >= 1 for that image. >> Nope. The length of the code for a decompressor depends on what >> language we're using. See my reply to guenther's reply to Tobin >> for a more detailed explanation. >Interpret all compressors & decompressors more inclusively, please. Huh? That's what I thought I was doing. _If_ we interpret it _less_ inclusively, restricting to one particular language, then what you said is right. But if we interpret it more inclusively, allowing any programming language, then there are no absolutely incompressible images by the definition above. >--r.e.s. ************************ David C. Ullrich === Subject: Re: How much lossless compression is possible in images? >>Your example also highlights a point about how the effectiveness of >>compression is to be measured. Since the *decompressor* is essential >>to recovering the original image from the output of the compressor, >>istm that an appropriate measure ought to include the size of the >>decompressor as well as the size of the compressor output. >E.g., if we use the simple ratio >>r = (|image_compressed| + |decompressor|) / |image_original| , >>then of course your example has r > 1, rather than the desired r < 1. >With that in mind, it might make sense to speak of an image as being >>absolutely incompressible if all compressors & decompressors have >>r >= 1 for that image. > > Nope. The length of the code for a decompressor depends on what > language we're using. See my reply to guenther's reply to Tobin > for a more detailed explanation. >>Interpret all compressors & decompressors more inclusively, please. > Huh? That's what I thought I was doing. > _If_ we interpret it _less_ inclusively, restricting to one > particular language, then what you said is right. But if > we interpret it more inclusively, allowing any programming > language, then there are no absolutely incompressible > images by the definition above. I don't care for the expression, but ... huh? The statement above, which I said might be true, is of the form S(x) = P, if Q(x) holds for all x in set B. In one case, B is the set A_L of all compressors & decompressors in language L, and in the other case B is the union of all the A_L; and I call the union more inclusive than any one of its subsets A_L. If S(x) is true when B is A, but not true when B is one of the A_L, then I would say it's true for the more inclusive set, wouldn't you? (The irony here is that I agree with what you say, otherwise; it would have better if I'd made the language issue explicit.) --r.e.s. === Subject: Re: How much lossless compression is possible in images? > There's no such thing as an incompressible image, _until_ you specify > what algorithm you're using. Given _any_ image I_0 there _is_ a > lossless compression algorithm that compresses I_0 to a one-byte file: > def compress(I): > if I == I_0: > return a single 0 byte > else: > return a 1, followed by I. > I leave the decompressor as an exercise. You can even do that with 255 images I_1 through I_255: def compress(I): if I = I_1: return a single 1 byte if I = I_2: return a single 2 byte ... if I = I_255: return a single 255 byte else: return a 0, followed by I -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: How much lossless compression is possible in images? <41b3ed05$0$9113$afc38c87@news.optusnet.com.au> <2of8r05du90qm548kmv6924elsqifm6pjr@4ax.com> message > Assuming ideal compression that removes every trace of redundancy from >> an image, how much lossless compression might really be possible? I >> know this is a difficult question because so much depends on image >> content, but I'm just trying to get an idea of how close current >> lossless compression methods are to the ideal. Visually images seem to >> be very highly redundant, although current lossless algorithms only >> manage about 50% compression. But assuming time and space were not >> constraints, how far could one theoretically go? > -- >> Transpose hotmail and mxsmanic in my e-mail address to reach me >directly. > It depends totally on image content. > An image consisting of random pixels will not compress at all (0% >> compression) using a generic algorithm; an image consisting of entirely >one >> colour will compress almost to zero (almost 100% compression). >I don't have the math at hand, but I believe that a truly random image >would have at least some compressible areas. In a strange sense, a >completely incompressible image would be as ordered as a single color image. > There's no such thing as an incompressible image, _until_ you specify You are joking Ullrich, are you not? It is amazing that a professional mathematician should utter such nonsense. There are binary strings which are not compressible under any circumstances. > what algorithm you're using. Given _any_ image I_0 there _is_ a > lossless compression algorithm that compresses I_0 to a one-byte file: > def compress(I): > if I == I_0: > return a single 0 byte > else: > return a 1, followed by I. > I leave the decompressor as an exercise. No Ullrich, you provide the decompressor. In trying seriously you will realise the humbling dimension of your blunder. > ************************ > David C. Ullrich === Subject: Re: How much lossless compression is possible in images? > You are joking Ullrich, are you not? It is amazing that a professional > mathematician should utter such nonsense. There are binary strings > which are not compressible under any circumstances. There are no such strings. It is always possible to devise an algorithm that compresses a given string, no matter what the pattern of bits. It's easy to do. If c is the specific string you wish to compress, then you define your algorithm such that 1 represents c, and 0 followed by m represents any other string m. This can be done for any arbitrary c. QED. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >> You are joking Ullrich, are you not? It is amazing that a professional >> mathematician should utter such nonsense. There are binary strings >> which are not compressible under any circumstances. >There are no such strings. It is always possible to devise an algorithm >that compresses a given string, no matter what the pattern of bits. Probably he's thinking of single-bit strings, which can't be compressed. But that's just pedantry. -- Richard === Subject: Re: How much lossless compression is possible in images? Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >You are joking Ullrich, are you not? It is amazing that a professional >mathematician should utter such nonsense. There are binary strings >which are not compressible under any circumstances. Suppose B is such a string, with length > 1 bit. Consider the compression algorithm which encodes B as a single 1 bit, and all other strings as 0 followed by the string. This algorithm compresses the allegedly incompressible string. -- Richard === Subject: Re: How much lossless compression is possible in images? <2of8r05du90qm548kmv6924elsqifm6pjr@4ax.com> Unfortunately Tobin, to talk about compression is meaningless if you do not specify how the length of the decompressor is included in the measurement of compression. Simply stating that pumpernikel is the compressed representation of Moby Dyck is a joke, not an algorithm. Things are only interesting if you can transmit pumpernikel and the decoding instructions at a lesser cost than transmitting the whole text of Moby Dyck. === Subject: Re: How much lossless compression is possible in images? > Unfortunately Tobin, to talk about compression is meaningless if you do > not specify how the length of the decompressor is included in the > measurement of compression. Simply stating that pumpernikel is the > compressed representation of Moby Dyck is a joke, not an algorithm. > Things are only interesting if you can transmit pumpernikel and the > decoding instructions at a lesser cost than transmitting the whole text > of Moby Dyck. The length of a compression algorithm designed to compress only one specific message to a single bit (maximal compression) will always be at least as long as the message being compressed. Since it provides compression for no other message, it can only be used once, which means that the length of the compressed message is effectively one bit plus the length of the algorithm. However, in more practical real-world scenarios, this situation does not obtain. An algorithm need only be communicated once. It may be able to significantly compress a vast number of messages. Dividing the length of the algorithm by the total volume of message data that it can compress may well cause it to shrink into insignificance. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <2of8r05du90qm548kmv6924elsqifm6pjr@4ax.com> <9v6cr09pcllb26tn5n6q8g8tsfitdrq8i7@4ax.com> Discussion, linux) > The length of a compression algorithm designed to compress only one > specific message to a single bit (maximal compression) will always be at > least as long as the message being compressed. That depends on your programming language. Nothing prevents me from using a formalism for Turing machines that assigns to the program outputting Moby Dick the index 0. You have to fix a formalism before really discussing compression. Even then, there's no reason your above claim is true. In fact, it can be false through means other than the cheap trick I use above. Consider the algorithm that maps the string 000...0 (5000 zeros) the code 0 and every other string prepends 1. The decoder can be very short indeed. It doesn't have to be 5000 bits long. -- Jesse F. Hughes I may have invented it, but Bill [Gates] made it famous. -- David Bradley, inventor of the Ctrl-Alt-Del reboot sequence === Subject: Re: How much lossless compression is possible in images? > Consider the algorithm that maps the string 000...0 (5000 zeros) > the code 0 and every other string prepends 1. The decoder can be > very short indeed. It doesn't have to be 5000 bits long. Such an algorithm wouldn't work for _any_ given string, but only for a string of zeros. I'm thinking of an algorithm optimized to compress a single, specific string maximally, without any regard for the actual bit pattern of the string. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <2of8r05du90qm548kmv6924elsqifm6pjr@4ax.com> <9v6cr09pcllb26tn5n6q8g8tsfitdrq8i7@4ax.com> <87vfbdx9ee.fsf@phiwumbda.org> Discussion, linux) >> Consider the algorithm that maps the string 000...0 (5000 zeros) >> the code 0 and every other string prepends 1. The decoder can be >> very short indeed. It doesn't have to be 5000 bits long. > Such an algorithm wouldn't work for _any_ given string, but only for a > string of zeros. I'm thinking of an algorithm optimized to compress a > single, specific string maximally, without any regard for the actual bit > pattern of the string. To which my other comments apply. You have to fix an encoding of Turing machines before any of this makes much sense. There's no reason I can't choose an encoding so that the machine which outputs Moby Dick is given index 0. -- By initially making it virtually impossible to maintain a heterogenous environment of Word 95 and Word 97 systems, Microsoft offered its customers that most eloquent of arguments for upgrading: the delicate sound of a revolver being cocked somewhere just out of sight. --Dan Martinez === Subject: Re: How much lossless compression is possible in images? > The length of a compression algorithm designed to compress only one > specific message to a single bit (maximal compression) will always be at > least as long as the message being compressed. Please enlighten me. Why would maximal compression be to a single bit? Will that bit turn out to be a 0, or a 1? If it matters, why not add it to the compression algorithm? If it doesn't matter, why have it at all? -- Alec McKenzie === Subject: Re: How much lossless compression is possible in images? > The length of a compression algorithm designed to compress only one > specific message to a single bit (maximal compression) will always be at > least as long as the message being compressed. > Please enlighten me. Why would maximal compression be to a single bit? That's the shortest possible output string. > Will that bit turn out to be a 0, or a 1? It doesn't matter. > If it matters, why not add it to the compression algorithm? > If it doesn't matter, why have it at all? I don't understand these questions. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? > Unfortunately Tobin, to talk about compression is meaningless if you do > not specify how the length of the decompressor is included in the > measurement of compression. Simply stating that pumpernikel is the > compressed representation of Moby Dyck is a joke, not an algorithm. > Things are only interesting if you can transmit pumpernikel and the > decoding instructions at a lesser cost than transmitting the whole text > of Moby Dyck. Ever heard of commercial telegraph codes? They weren't ever used for anything quite that dramatic, but there was some savings based on having sent the decoding instructions physically at a slower pace (perhaps in the luggage of the employee being relocated). -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: How much lossless compression is possible in images? >Unfortunately Tobin, to talk about compression is meaningless if you do >not specify how the length of the decompressor is included in the >measurement of compression. Simply stating that pumpernikel is the >compressed representation of Moby Dyck is a joke, not an algorithm. >Things are only interesting if you can transmit pumpernikel and the >decoding instructions at a lesser cost than transmitting the whole text >of Moby Dyck. Well, there's no point in trying to explain why this is specious to guenther, because he already Knows the Truth. For anyone else who's curious why this is not a valid objection: It's _impossible_ to include a _complete_ specification of the decompressor in _any_ compressed file. Say we include C code for the compressor. That code is meaningless unless we know in advance that it's C code, and know in advance how C code is to be interpreted. So we have to include the C specification in the file. But the C spec is written in English - it will be meaningless to someone who doesn't know English (and worse, it _could_ happen that the English specifying how to write a C compiler just _happens_ to be meaningful but mean something totally different in Martian.) So we need to include a specification of what all the English words mean. In what language shall that definition of the English language be written? Etc. It's impossible to define everything in terms of previously defined terms - _any_ communication relies on an _assumption_ that there is some common language, which assumption is impossible to actually verify. In practice we're not required to include the C specification, we're allowed to _assume_ that the user will know that the code for the decompressor is C and that he has a C compiler already. But allowing this assumption, while not allowing the assumption that the user knows the text of Moby Dick, is purely arbitrary - there's no objective way to draw the line mathematically. And if we assume that the user is using a language exactly like C, except that Moby_Dick expands to the text of Moby Dick, then transmitting pumpernikel along with the code for the decompressor is much more efficient than transmitting the complete text. ************************ David C. Ullrich === Subject: Re: How much lossless compression is possible in images? >>message > Assuming ideal compression that removes every trace of >redundancy from > an image, how much lossless compression might really be >possible? I > know this is a difficult question because so much depends on >image > content, but I'm just trying to get an idea of how close current > lossless compression methods are to the ideal. Visually images >seem to > be very highly redundant, although current lossless algorithms >only > manage about 50% compression. But assuming time and space were >not > constraints, how far could one theoretically go? > -- > Transpose hotmail and mxsmanic in my e-mail address to reach me >>directly. > It depends totally on image content. > An image consisting of random pixels will not compress at all (0% > compression) using a generic algorithm; an image consisting of >entirely >>one > colour will compress almost to zero (almost 100% compression). >I don't have the math at hand, but I believe that a truly random >image >>would have at least some compressible areas. In a strange sense, a >>completely incompressible image would be as ordered as a single >color image. >> There's no such thing as an incompressible image, _until_ you specify >You are joking Ullrich, are you not? It is amazing that a professional >mathematician should utter such nonsense. There are binary strings >which are not compressible under any circumstances. >> what algorithm you're using. Given _any_ image I_0 there _is_ a >> lossless compression algorithm that compresses I_0 to a one-byte >file: >> def compress(I): >> if I == I_0: >> return a single 0 byte >> else: >> return a 1, followed by I. >> I leave the decompressor as an exercise. >No Ullrich, you provide the decompressor. In trying seriously you will >realise the humbling dimension of your blunder. Um, in case _you're_ not joking: def decompress(data): if data == '0': return I_0 else: return data with first byte omitted >> ************************ >> David C. Ullrich David C. Ullrich === Subject: Re: How much lossless compression is possible in images? <41b3ed05$0$9113$afc38c87@news.optusnet.com.au> <2of8r05du90qm548kmv6924elsqifm6pjr@4ax.com> > There's no such thing as an incompressible image, _until_ you specify >You are joking Ullrich, are you not? It is amazing that a professional >mathematician should utter such nonsense. There are binary strings >which are not compressible under any circumstances. >> what algorithm you're using. Given _any_ image I_0 there _is_ a >> lossless compression algorithm that compresses I_0 to a one-byte >file: > def compress(I): >> if I == I_0: >> return a single 0 byte >> else: >> return a 1, followed by I. > I leave the decompressor as an exercise. >No Ullrich, you provide the decompressor. In trying seriously you will >realise the humbling dimension of your blunder. > Um, in case _you're_ not joking: > def decompress(data): > if data == '0': > return I_0 > else: > return data with first byte omitted >> ************************ > David C. Ullrich > David C. Ullrich You are not trying seriously Ullrich. You have not even started to define what you mean by compressing. If you care to check the literature, you will find that compression is said to have taken place when the representation of both the decompressing algorithm and the output are of lesser total length than the input. Apart from that, your construction fails for the input '0' since it does not compress it under any definition of compression you might wish to choose, therefore your statement is wrong in any case. === Subject: Re: How much lossless compression is possible in images? <41b3ed05$0$9113$afc38c87@news.optusnet.com.au> <2of8r05du90qm548kmv6924elsqifm6pjr@4ax.com> > There's no such thing as an incompressible image, _until_ you specify >You are joking Ullrich, are you not? It is amazing that a professional >mathematician should utter such nonsense. There are binary strings >which are not compressible under any circumstances. >> what algorithm you're using. Given _any_ image I_0 there _is_ a >> lossless compression algorithm that compresses I_0 to a one-byte >file: > def compress(I): >> if I == I_0: >> return a single 0 byte >> else: >> return a 1, followed by I. > I leave the decompressor as an exercise. >No Ullrich, you provide the decompressor. In trying seriously you will >realise the humbling dimension of your blunder. > Um, in case _you're_ not joking: > def decompress(data): > if data == '0': > return I_0 > else: > return data with first byte omitted >> ************************ > David C. Ullrich > David C. Ullrich You are not trying seriously Ullrich. You have not even started to define what you mean by compressing. If you care to check the literature, you will find that compression is said to have taken place when the representation of both the decompressing algorithm and the output are of lesser total length than the input. Apart from that, your construction fails for the input '0' since it does not compress it under any definition of compression you might wish to choose, therefore your statement is wrong in any case. === Subject: Re: How much lossless compression is possible in images? > You are not trying seriously Ullrich. You have not even started to > define what you mean by compressing. If you care to check the > literature, you will find that compression is said to have taken place > when the representation of both the decompressing algorithm and the > output are of lesser total length than the input. Said by whom? You're not compressing algorithms, you're compressing messages. If you want to include the length of the algorithm, then you must include it with the cumulative lengths of all possible messages, not just one message. And in that case, there obviously will always be an increase in size ... because random data cannot be compressed. A compression algorithm in this sense always increases redundancy. But in practice you don't write an algorithm for all possible messages, but only for certain, highly probable messages, and in this specific case the total length is greatly reduced, even with the length of the algorithm. So compression occurs. Since compression is meaningless if all messages are equally probable (and thus random), one must assume that some subset of all possible messages are more probable than others when discussing any type of compression. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <41b3ed05$0$9113$afc38c87@news.optusnet.com.au> <2of8r05du90qm548kmv6924elsqifm6pjr@4ax.com> > Said by whom? You're not compressing algorithms, you're compressing > messages. Here's a challenge for you. You have to stay inside your home for a week. Throw away anything edible. Now write down an order for the food you will eat during that week. Now call at the grocer's. According to your bull there is an algorithm whicht codifies that order into the word Rumpelstilskin. You may only pronounce the word Rumpelstilskin once into the phone and must then hang up. Now sit and wait for your delivery. ËAlready hungry? Hint: the decoding instructions, or the definition for the encoding, are part of the message. Exceptions to this rule in real life practical use are of no interest to theory. === Subject: Re: How much lossless compression is possible in images? > Hint: the decoding instructions, or the definition for the encoding, > are part of the message. No. If the definiton of the encoding is part of the message, then the only way to know how the message is to be decoded is to decode the message, which means that no communication via messages is possible at all. In fact, the encoding method is know by both ends of a communication channel before any messages are exchanged. Otherwise no communication of information can take place--all signals exchanged over the channel are noise. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? >Here's a challenge for you. You have to stay inside your home for a >week. Throw away anything edible. Now write down an order for the food >you will eat during that week. Now call at the grocer's. According to >your bull there is an algorithm whicht codifies that order into the >word Rumpelstilskin. You may only pronounce the word >Rumpelstilskin once into the phone and must then hang up. Now sit and >wait for your delivery. =BFAlready hungry? >Hint: the decoding instructions, or the definition for the encoding, >are part of the message. Exceptions to this rule in real life practical >use are of no interest to theory. So perfectly compressed messages are perfectly encrypted? Rich === Subject: Re: How much lossless compression is possible in images? >Here's a challenge for you. You have to stay inside your home for a >week. Throw away anything edible. Now write down an order for the food >you will eat during that week. Now call at the grocer's. According to >your bull there is an algorithm whicht codifies that order into the >word Rumpelstilskin. You may only pronounce the word >Rumpelstilskin once into the phone and must then hang up. Now sit and >wait for your delivery. =BFAlready hungry? >Hint: the decoding instructions, or the definition for the encoding, >are part of the message. Exceptions to this rule in real life practical >use are of no interest to theory. > So perfectly compressed messages are perfectly encrypted? > Rich Not as far as I understand. Good encryption requires that it be unbreakable even if the algorithm used is known to an attacker. On the other hand, well encrypted strings should be incompressible because they should be devoid of any statistical features, which are considered weaknesess in cryptology. === Subject: Re: How much lossless compression is possible in images? >Here's a challenge for you. You have to stay inside your home for a >>week. Throw away anything edible. Now write down an order for the >food >>you will eat during that week. Now call at the grocer's. According >>your bull there is an algorithm whicht codifies that order into >the >>word Rumpelstilskin. You may only pronounce the word >>Rumpelstilskin once into the phone and must then hang up. Now sit >and >>wait for your delivery. =BFAlready hungry? >Hint: the decoding instructions, or the definition for the encoding, >>are part of the message. Exceptions to this rule in real life >practical >>use are of no interest to theory. > So perfectly compressed messages are perfectly encrypted? >Not as far as I understand. Good encryption requires that it be >unbreakable even if the algorithm used is known to an attacker. On the >other hand, well encrypted strings should be incompressible because >they should be devoid of any statistical features, which are considered >weaknesess in cryptology. Yes. This is my understanding as well. What confuses me is that I understood your post about ordering food to imply that some *imperfectly* compressed messages are perfectly encrypted! How can this be? Rich === Subject: Re: How much lossless compression is possible in images? > Yes. This is my understanding as well. What confuses me is that I understood > your post about ordering food to imply that some *imperfectly* compressed > messages are perfectly encrypted! How can this be? You can perfectly encrypt a completely uncompressed and highly redundant message using a simple one-time pad. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? >Here's a challenge for you. You have to stay inside your home for a >>week. Throw away anything edible. Now write down an order for the >food >>you will eat during that week. Now call at the grocer's. According >to >>your bull there is an algorithm whicht codifies that order into >the >>word Rumpelstilskin. You may only pronounce the word >>Rumpelstilskin once into the phone and must then hang up. Now sit >and >>wait for your delivery. =BFAlready hungry? >Hint: the decoding instructions, or the definition for the encoding, >>are part of the message. Exceptions to this rule in real life >practical >>use are of no interest to theory. >> So perfectly compressed messages are perfectly encrypted? >>Not as far as I understand. Good encryption requires that it be >unbreakable even if the algorithm used is known to an attacker. On the >other hand, well encrypted strings should be incompressible because >they should be devoid of any statistical features, which are considered >weaknesess in cryptology. > Yes. This is my understanding as well. What confuses me is that I understood > your post about ordering food to imply that some *imperfectly* compressed > messages are perfectly encrypted! How can this be? > Rich Well, then you got it wrong. My challenge had nothing to do with either perfection of compression nor with encryption in any form. Mr. OP here keeps babbling that for any given string an algorithm can be deviced which will compress it. That statement can only be upheld under the assumption that the decompressor has not to be accounted for in any way in the process of meassuring the compression rate. Unfortunately any discussion based on such an assumption is nothing but contemptible prattle. The story with the grocery order is just an attempt at and creates an algorithm that will compress it. Then, he compresses the order and transmits the output to the grocer. The grocer, let's call him Mr. Li is in the very unfortunate situation that he cannot even start guessing at the meaning of the message. Even if he knows beforehand that Mr. M is a crank reputed for his pecularity and always compresses his messages with freshly and specially devised algorithms he has absolutely no clue as to what algorithm was applied (remember that the algorithm has most recently been invented by Mr. M for this ocassion). Therefore he has zero possibilities of decoding the message. As a result of this situation Mr. M will be one ravenous crackpot during the following week. There are only two possibilities for the resolution of the problem a) Mr Li always knows in advance what Mr. M will order, b) Mr. M transmits along with the output of his compressor, the essential information required by Mr. Li in the process of decoding. In case a) the processes of compressing and transmitting the information are superfluous a circumstance which makes a discussion of compressibility nothing but contemptible prattling. Case b) the interesting one, denies the OP's ridiculous musings. I would point out to you that there are no further cases, and that the whole thing has indeed nothing to do with either perfection nor encryption. === Subject: Re: How much lossless compression is possible in images? > So perfectly compressed messages are perfectly encrypted? It is possible to devise an algorithm that will compress any specific message to a single bit. If the entropy of the portion of the algorithm held confidential is at least equal to the length of the original message, this constitutes unbreakable encryption. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? ... >> def compress(I): >> if I == I_0: >> return a single 0 byte >> else: >> return a 1, followed by I. >> >> I leave the decompressor as an exercise. ... > def decompress(data): > if data == '0': > return I_0 > else: > return data with first byte omitted ... > You are not trying seriously Ullrich. You have not even started to > define what you mean by compressing. If you care to check the > literature, you will find that compression is said to have taken place > when the representation of both the decompressing algorithm and the > output are of lesser total length than the input. And if *you* care to check the literature you will find that a compression algorithm is an algorithm that compresses a particular kind of input. David's algorithm satisfies that definition. > Apart from that, your construction fails for the input '0' since it > does not compress it under any definition of compression you might wish > to choose, therefore your statement is wrong in any case. So? That input is not in the class of inputs for which the compressor is intended. By your definition compressors and compression algorithms do not exist. -- 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: How much lossless compression is possible in images? > ... > >> def compress(I): > >> if I == I_0: > >> return a single 0 byte > >> else: > >> return a 1, followed by I. > >> >> I leave the decompressor as an exercise. > ... > > def decompress(data): > > if data == '0': > > return I_0 > > else: > > return data with first byte omitted > ... > > You are not trying seriously Ullrich. You have not even started to > > define what you mean by compressing. If you care to check the > > literature, you will find that compression is said to have taken place > > when the representation of both the decompressing algorithm and the > > output are of lesser total length than the input. > And if *you* care to check the literature you will find that a compression > algorithm is an algorithm that compresses a particular kind of input. > David's algorithm satisfies that definition. Well said Winter however absurdly tautologic it is. An algorithm which does not compress any kind of input wouldn't be a compressor would it? And an algorithm which compresses any kind of input does not exist, believing in such would rather put you on the wacky side of crankdom. So there should be no doubt that we agree here, however, pray tell me how does your statement affect my objection to Ullrich's construction? > > Apart from that, your construction fails for the input '0' since it > > does not compress it under any definition of compression you might wish > > to choose, therefore your statement is wrong in any case. > So? That input is not in the class of inputs for which the compressor > is intended. By your definition compressors and compression > algorithms do not exist. Sorry Winter you seem to have lost the thread of our little discussion here. Ullrich stated that There's no such thing as an incompressible image, _until_ you specify what algorithm you're using. Given _any_ Image I_0 there _is_ a lossless compression algorithm that compresses I_0 to a one-byte file and then procedes to show how to construct such an algorithm. Note how he uses the word any and makes it clear that he is not excluding any inputs at all, as opposed to what you are now saying. His statement is wrong, because for the Input '0' his construction of a compressing algorithm fails. Well it's really trivial, but this is sci.math after all, isn't it? > 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: How much lossless compression is possible in images? >> ... >> >> def compress(I): >> >> if I == I_0: >> >> return a single 0 byte >> >> else: >> >> return a 1, followed by I. >> > >> I leave the decompressor as an exercise. >> ... >> > def decompress(data): >> > if data == '0': >> > return I_0 >> > else: >> > return data with first byte omitted >> ... >> > You are not trying seriously Ullrich. You have not even started to >> > define what you mean by compressing. If you care to check the >> > literature, you will find that compression is said to have taken >place >> > when the representation of both the decompressing algorithm and >the >> > output are of lesser total length than the input. >> And if *you* care to check the literature you will find that a >compression >> algorithm is an algorithm that compresses a particular kind of input. >> David's algorithm satisfies that definition. >Well said Winter however absurdly tautologic it is. An algorithm which >does not compress any kind of input wouldn't be a compressor would it? >And an algorithm which compresses any kind of input does not exist, >believing in such would rather put you on the wacky side of crankdom. >So there should be no doubt that we agree here, however, pray tell me >how does your statement affect my objection to Ullrich's construction? >> > Apart from that, your construction fails for the input '0' since >> > does not compress it under any definition of compression you might >wish >> > to choose, therefore your statement is wrong in any case. >> So? That input is not in the class of inputs for which the >compressor >> is intended. By your definition compressors and compression >> algorithms do not exist. >Sorry Winter you seem to have lost the thread of our little discussion >here. Ullrich stated that There's no such thing as an incompressible >image, _until_ you specify what algorithm you're using. Given _any_ >Image I_0 there _is_ a lossless compression algorithm that compresses >I_0 to a one-byte file and then procedes to show how to construct such >an algorithm. Note how he uses the word any and makes it clear that >he is not excluding any inputs at all, as opposed to what you are now >saying. His statement is wrong, because for the Input '0' his >construction of a compressing algorithm fails. Well it's really >trivial, but this is sci.math after all, isn't it? Uh, yes, it's sci.math. Hence we should be able to tell the difference between There exists an algorithm which will compress any input (um, the English is actually ambiguous here, what we mean is There exists an algorithm A such that for every input I, A compresses I.) and Given an input, there exists an algorithm which will compress it. The first is easily seen to be false by a counting argument, while the second is true, as I've shown. The _point_ to the second is exactly as I said: People shouldn't talk about incompressible input as though this had some absolute meaning - it doesn't. It doesn't mean anything _until_ we have specified what algorithm we're using. (heh-heh: On a related topic, for years I saw people around here discuss Kolomogorov complexity without reference to a programming language, as though length of the smallest algorithm implementing a function had some absolute meaning. Was a great relief one day when I looked at something Chaitin had written and saw that he _began_ by saying that of course we need to first specify a programming language before any of these concepts make sense...) >> dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, >+31205924131 >> home: bovenover 215, 1025 jn amsterdam, nederland; >http://www.cwi.nl/~dik/ ************************ David C. Ullrich === Subject: Re: How much lossless compression is possible in images? >> ... >> >> def compress(I): >> >> if I == I_0: >> >> return a single 0 byte >> >> else: >> >> return a 1, followed by I. >> > >> I leave the decompressor as an exercise. >> ... >> > def decompress(data): >> > if data == '0': >> > return I_0 >> > else: >> > return data with first byte omitted >> ... >> > You are not trying seriously Ullrich. You have not even started to >> > define what you mean by compressing. If you care to check the >> > literature, you will find that compression is said to have taken >place >> > when the representation of both the decompressing algorithm and >the >> > output are of lesser total length than the input. > And if *you* care to check the literature you will find that a >compression >> algorithm is an algorithm that compresses a particular kind of input. >> David's algorithm satisfies that definition. >Well said Winter however absurdly tautologic it is. An algorithm which >does not compress any kind of input wouldn't be a compressor would it? >And an algorithm which compresses any kind of input does not exist, >believing in such would rather put you on the wacky side of crankdom. >So there should be no doubt that we agree here, however, pray tell me >how does your statement affect my objection to Ullrich's construction? >> > Apart from that, your construction fails for the input '0' since >it >> > does not compress it under any definition of compression you might >wish >> > to choose, therefore your statement is wrong in any case. > So? That input is not in the class of inputs for which the >compressor >> is intended. By your definition compressors and compression >> algorithms do not exist. >Sorry Winter you seem to have lost the thread of our little discussion >here. Ullrich stated that There's no such thing as an incompressible >image, _until_ you specify what algorithm you're using. Given _any_ >Image I_0 there _is_ a lossless compression algorithm that compresses >I_0 to a one-byte file and then procedes to show how to construct such >an algorithm. Note how he uses the word any and makes it clear that >he is not excluding any inputs at all, as opposed to what you are now >saying. His statement is wrong, because for the Input '0' his >construction of a compressing algorithm fails. Well it's really >trivial, but this is sci.math after all, isn't it? > Uh, yes, it's sci.math. Hence we should be able to tell the > difference between > There exists an algorithm which will compress any input > (um, the English is actually ambiguous here, what we mean > is There exists an algorithm A such that for every input I, > A compresses I.) > and > Given an input, there exists an algorithm which will > compress it. > The first is easily seen to be false by a counting argument, > while the second is true, as I've shown. You have shown nothing until you tell us what you mean by will compress it > The _point_ to the second is exactly as I said: People > shouldn't talk about incompressible input as though this > had some absolute meaning - it doesn't. It doesn't mean > anything _until_ we have specified what algorithm we're > using. Wrong Ullrich, people shouldn't talk about anything at all, unless they have anything truly interesting to say. Want to know something Ullrich? Although I've been acting the role of a prick as far as you might be concerned, I'm quite interested in what you may have to say regarding compression (providing you bother to take a bit of a deeper look at it). > (heh-heh: On a related topic, for years I saw people > around here discuss Kolomogorov complexity without > reference to a programming language, as though length > of the smallest algorithm implementing a function had > some absolute meaning. Was a great relief one day when > I looked at something Chaitin had written and saw > that he _began_ by saying that of course we need to > first specify a programming language before any of > these concepts make sense...) Exactly Ullrich, just as your just said. >> dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, >+31205924131 >> home: bovenover 215, 1025 jn amsterdam, nederland; >http://www.cwi.nl/~dik/ > ************************ > David C. Ullrich === Subject: Re: How much lossless compression is possible in images? >> ... >> >> def compress(I): >> >> if I == I_0: >> >> return a single 0 byte >> >> else: >> >> return a 1, followed by I. >> > >> I leave the decompressor as an exercise. >> ... >> > def decompress(data): >> > if data == '0': >> > return I_0 >> > else: >> > return data with first byte omitted >> ... >> > You are not trying seriously Ullrich. You have not even started to >> > define what you mean by compressing. If you care to check the >> > literature, you will find that compression is said to have taken >place >> > when the representation of both the decompressing algorithm and >the >> > output are of lesser total length than the input. > And if *you* care to check the literature you will find that a >compression >> algorithm is an algorithm that compresses a particular kind of input. >> David's algorithm satisfies that definition. >Well said Winter however absurdly tautologic it is. An algorithm which >does not compress any kind of input wouldn't be a compressor would it? >And an algorithm which compresses any kind of input does not exist, >believing in such would rather put you on the wacky side of crankdom. >So there should be no doubt that we agree here, however, pray tell me >how does your statement affect my objection to Ullrich's construction? >> > Apart from that, your construction fails for the input '0' since >it >> > does not compress it under any definition of compression you might >wish >> > to choose, therefore your statement is wrong in any case. > So? That input is not in the class of inputs for which the >compressor >> is intended. By your definition compressors and compression >> algorithms do not exist. >Sorry Winter you seem to have lost the thread of our little discussion >here. Ullrich stated that There's no such thing as an incompressible >image, _until_ you specify what algorithm you're using. Given _any_ >Image I_0 there _is_ a lossless compression algorithm that compresses >I_0 to a one-byte file and then procedes to show how to construct such >an algorithm. Note how he uses the word any and makes it clear that >he is not excluding any inputs at all, as opposed to what you are now >saying. His statement is wrong, because for the Input '0' his >construction of a compressing algorithm fails. Well it's really >trivial, but this is sci.math after all, isn't it? > Uh, yes, it's sci.math. Hence we should be able to tell the > difference between > There exists an algorithm which will compress any input > (um, the English is actually ambiguous here, what we mean > is There exists an algorithm A such that for every input I, > A compresses I.) > and > Given an input, there exists an algorithm which will > compress it. > The first is easily seen to be false by a counting argument, > while the second is true, as I've shown. You have shown nothing until you tell us what you mean by will compress it > The _point_ to the second is exactly as I said: People > shouldn't talk about incompressible input as though this > had some absolute meaning - it doesn't. It doesn't mean > anything _until_ we have specified what algorithm we're > using. Wrong Ullrich, people shouldn't talk about anything at all, unless they have anything truly interesting to say. Want to know something Ullrich? Although I've been acting the role of a prick as far as you might be concerned, I'm quite interested in what you may have to say regarding compression (providing you bother to take a bit of a deeper look at it). > (heh-heh: On a related topic, for years I saw people > around here discuss Kolomogorov complexity without > reference to a programming language, as though length > of the smallest algorithm implementing a function had > some absolute meaning. Was a great relief one day when > I looked at something Chaitin had written and saw > that he _began_ by saying that of course we need to > first specify a programming language before any of > these concepts make sense...) Exactly Ullrich, just as your just said. >> dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, >+31205924131 >> home: bovenover 215, 1025 jn amsterdam, nederland; >http://www.cwi.nl/~dik/ > ************************ > David C. Ullrich === Subject: Re: How much lossless compression is possible in images? >> Assuming ideal compression that removes every trace of redundancy from >> an image, how much lossless compression might really be possible? I >> know this is a difficult question because so much depends on image >> content, but I'm just trying to get an idea of how close current >> lossless compression methods are to the ideal. Visually images seem to >> be very highly redundant, although current lossless algorithms only >> manage about 50% compression. But assuming time and space were not >> constraints, how far could one theoretically go? > -- >> Transpose hotmail and mxsmanic in my e-mail address to reach me > directly. >> It depends totally on image content. >> An image consisting of random pixels will not compress at all (0% >> compression) using a generic algorithm; an image consisting of entirely > one >> colour will compress almost to zero (almost 100% compression). > I don't have the math at hand, but I believe that a truly random image > would have at least some compressible areas. In a strange sense, a > completely incompressible image would be as ordered as a single color > image. Well, you would be wrong. At least one definition of random is that an image can't be compressed. I know of no definition of random which would allow every or even most random images to have at least some compressible areas. And what is completely incompressible image supposed to mean? Here's a little mathematical titbit, which follows immediately from the pigeon hole principle: The average image size (averaged across all possible images) of images compressed using lossless algorithms must be equal to or greater than the uncompressed image size. Indeed, for all practical lossless compression algorithms, the huge majority of all all images are larger when compressed than they are before compression. Its just the tiny minority of images which get smaller are those that have characteristics often found in photos - such as large areas with only moderate colour variability. === Subject: Re: How much lossless compression is possible in images? > Well, you would be wrong. At least one definition of random is that an > image can't be compressed. More precisely perhaps, the average compression possible for random images approaches zero as the number of images approaches infinity. However, since a given random image may (unpredictably) contain any degree of redundancy, virtually all random images are compressible to some extent using some algorithms. Put still another way, no algorithm exists (or can be devised) that will compress a random string of bits more than 50% of the time. In fact, no matter what the algorithm is, it will produce a shorter output string 50% of the time, and a longer output string the other 50% of the time. > I know of no definition of random which would > allow every or even most random images to have at least some compressible > areas. See above. > The average image size (averaged across all possible images) of images > compressed using lossless algorithms must be equal to or greater than the > uncompressed image size. Yes. Compression algorithms work only because all possible images are not equally probable. For every possible image that can be compressed, there exists some other image that will increase in size by the same amount after compression, if the compression is lossless (reversible). In real life, image data corresponding to pictorial representations is far more probable than random image data (even though the number of non-pictoral images possible vastly exceeds the number of pictoral images possible). So compression works. > Indeed, for all practical lossless compression algorithms, the huge majority > of all all images are larger when compressed than they are before > compression. Its just the tiny minority of images which get smaller are > those that have characteristics often found in photos - such as large areas > with only moderate colour variability. Yes. A different way of stating what I've said above. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? ... > Put still another way, no algorithm exists (or can be devised) that will > compress a random string of bits more than 50% of the time. Even 50% cannot be reached. > In fact, no > matter what the algorithm is, it will produce a shorter output string > 50% of the time, and a longer output string the other 50% of the time. Wrong. It will produce a shorter string less than 50% of the time, and it will produce an equal sized or longer string in the remaining cases. See David Ullrich's excellent algorithm that reduces exactly one bitstring in size and increases the size of all others. As another example, take the algorithm that leaves strings starting with (binary) 1 alone, that encodes a particular string that starts with 00 as plain 00, and prepends 01 to all other strings starting with 0. One string is shortened, 50 % stays equal in size and the remainder is lengthened. -- 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: How much lossless compression is possible in images? > Even 50% cannot be reached. True, but it can be very closely approached. The longer the message, the closer the approach, but as long as the message is of finite length, it will never be 50% exactly. > Wrong. It will produce a shorter string less than 50% of the time, and it > will produce an equal sized or longer string in the remaining cases. I left out the details, but I suspected someone would grab the opportunity to show how much more he knew, thus saving me some bandwidth. > See > David Ullrich's excellent algorithm that reduces exactly one bitstring > in size and increases the size of all others. Average the changes in size, and you'll find that the total reduction in size for the one string equals the total increase for all other strings. This is at least the theoretical case. It's easy to cheat with a real algorithm, though. > As another example, take > the algorithm that leaves strings starting with (binary) 1 alone, that > encodes a particular string that starts with 00 as plain 00, and prepends > 01 to all other strings starting with 0. One string is shortened, 50 % > stays equal in size and the remainder is lengthened. See above. The total change in length will converge on zero. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? ... > See > David Ullrich's excellent algorithm that reduces exactly one bitstring > in size and increases the size of all others. > Average the changes in size, and you'll find that the total reduction in > size for the one string equals the total increase for all other strings. Do you think so? One bitstring is reduced in size to 1 bit, and all other bitstrings increase by 1 bit. > This is at least the theoretical case. It's easy to cheat with a real > algorithm, though. What do you mean by theoretical and real here? And what do you mean with cheating? > As another example, take > the algorithm that leaves strings starting with (binary) 1 alone, that > encodes a particular string that starts with 00 as plain 00, and prepends > 01 to all other strings starting with 0. One string is shortened, 50 % > stays equal in size and the remainder is lengthened. > See above. The total change in length will converge on zero. In what sense? Suppose the particular string is 'k' bits. It will be reduced by 'k-1' bits. All other strings that start with 0 are lengthened by 2 bits. There are 2^(m-1) such strings of size m. So when we consider the total change in length it is '2^m - k + 1'. I do not see any sense in which this converges to zero. Even if we divide by the number of strings considered ('2^m') and take the average, it will converge to 1, not 0. So in the limit this algorithm will increase the size on average by 1 bit. -- 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: How much lossless compression is possible in images? > Do you think so? Yes. > One bitstring is reduced in size to 1 bit, and all other > bitstrings increase by 1 bit. Yes. So one string is reduced by 1,000,000 bits, and a million other strings increase by 1 bit. The net change is zero. > What do you mean by theoretical and real here? The perfect algorithms of theory, versus real algorithms that need not follow all aspects of theory. > In what sense? Suppose the particular string is 'k' bits. It will be > reduced by 'k-1' bits. All other strings that start with 0 are lengthened > by 2 bits. There are 2^(m-1) such strings of size m. So when we consider the > total change in length it is '2^m - k + 1'. I do not see any sense > in which this converges to zero. Even if we divide by the number of > strings considered ('2^m') and take the average, it will converge to 1, > not 0. So in the limit this algorithm will increase the size on average > by 1 bit. In the real world, I'll concede that it would not converge on exactly zero. A compression algorithm is an algorithm that converts fixed-length bit strings of unequal probabilities to variable-length bit strings such that the most probable input strings have the shortest output strings. The total number of information bits for all output messages cannot be less than the total number of information bits for all input messages, but offhand I don't see why it must be greater, at least in theory. In the real-world, algorithms always add a few bits to the total on output. output strings (is it part of the output information or not?). -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? > Do you think so? > Yes. > One bitstring is reduced in size to 1 bit, and all other > bitstrings increase by 1 bit. > Yes. So one string is reduced by 1,000,000 bits, and a million other > strings increase by 1 bit. The net change is zero. > What do you mean by theoretical and real here? > The perfect algorithms of theory, versus real algorithms that need not > follow all aspects of theory. > In what sense? Suppose the particular string is 'k' bits. It will be > reduced by 'k-1' bits. All other strings that start with 0 are lengthened > by 2 bits. There are 2^(m-1) such strings of size m. So when we consider the > total change in length it is '2^m - k + 1'. I do not see any sense > in which this converges to zero. Even if we divide by the number of > strings considered ('2^m') and take the average, it will converge to 1, > not 0. So in the limit this algorithm will increase the size on average > by 1 bit. > In the real world, I'll concede that it would not converge on exactly > zero. > A compression algorithm is an algorithm that converts fixed-length bit > strings of unequal probabilities to variable-length bit strings such > that the most probable input strings have the shortest output strings. > The total number of information bits for all output messages cannot be > less than the total number of information bits for all input messages, > but offhand I don't see why it must be greater, at least in theory. In > the real-world, algorithms always add a few bits to the total on output. > output strings (is it part of the output information or not?). > -- > Transpose hotmail and mxsmanic in my e-mail address to reach me directly. Here is some advice for you lout. Read about Kraft's inequality. Read the comp compression faq. Ask pertinent questions in that forum. After you're done, it will still be your choice to make an ass of yourself === Subject: Re: How much lossless compression is possible in images? > Do you think so? > Yes. > One bitstring is reduced in size to 1 bit, and all other > bitstrings increase by 1 bit. > Yes. So one string is reduced by 1,000,000 bits, and a million other > strings increase by 1 bit. The net change is zero. > What do you mean by theoretical and real here? > The perfect algorithms of theory, versus real algorithms that need not > follow all aspects of theory. > In what sense? Suppose the particular string is 'k' bits. It will be > reduced by 'k-1' bits. All other strings that start with 0 are lengthened > by 2 bits. There are 2^(m-1) such strings of size m. So when we consider the > total change in length it is '2^m - k + 1'. I do not see any sense > in which this converges to zero. Even if we divide by the number of > strings considered ('2^m') and take the average, it will converge to 1, > not 0. So in the limit this algorithm will increase the size on average > by 1 bit. > In the real world, I'll concede that it would not converge on exactly > zero. > A compression algorithm is an algorithm that converts fixed-length bit > strings of unequal probabilities to variable-length bit strings such > that the most probable input strings have the shortest output strings. > The total number of information bits for all output messages cannot be > less than the total number of information bits for all input messages, > but offhand I don't see why it must be greater, at least in theory. In > the real-world, algorithms always add a few bits to the total on output. > output strings (is it part of the output information or not?). > -- > Transpose hotmail and mxsmanic in my e-mail address to reach me directly. Here is some advice for you lout. Read about Kraft's inequality. Read the comp compression faq. Ask pertinent questions in that forum. After you're done, it will still be your choice to make an ass of yourself === Subject: Re: How much lossless compression is possible in images? > Here is some advice for you lout. Read about Kraft's inequality. Read > the comp compression faq. Ask pertinent questions in that forum. > After you're done, it will still be your choice to make an ass of > yourself Why do you seem to limit your posts to others to personal attacks? -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <5skfr0pb3p24nn2lthl7amb9eai1bf9rtr@4ax.com> > Here is some advice for you lout. Read about Kraft's inequality. Read > the comp compression faq. Ask pertinent questions in that forum. > After you're done, it will still be your choice to make an ass of > yourself > Why do you seem to limit your posts to others to personal attacks? Do not be unfair even if you do not like the way I talk to you. It seems to me that I am the only one to have actually attempted to answer your original question. Appart from that, I have tried to stop even more noise from entering this discussion. Noise which among others you keep introducing by constantly stating totally baseless bull. You might notice that others have tried to tell you the same thing in a much more civil manner than I, but you simply keep ignoring them. > -- > Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? > Do not be unfair even if you do not like the way I talk to you. I'm not being unfair, I'm making a simple observation. > It seems to me that I am the only one to have actually attempted to > answer your original question. I don't remember. All I remember is your ceaseless vituperation, which effectively eclipsed anything else you had to say. > Noise which among others > you keep introducing by constantly stating totally baseless bull. > You might notice that others have tried to tell you the same thing in a > much more civil manner than I, but you simply keep ignoring them. Your post was only a single paragraph, and yet you still managed to devote half of it to personal attacks. Perhaps a change in technique is indicated? -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <5skfr0pb3p24nn2lthl7amb9eai1bf9rtr@4ax.com> <6m4mr0p6iml9reaphldhmufd1pksbi4t6q@4ax.com> > It seems to me that I am the only one to have actually attempted to > answer your original question. > I don't remember. All I remember is your ceaseless vituperation, which > effectively eclipsed anything else you had to say. Guess what Larry? That's my main contention with you. You are here mainly for the prattling. The escence seems to be quite beyond your abilities... eh ur ... interest. === Subject: Re: How much lossless compression is possible in images? > Guess what Larry? My name isn't Larry. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <5skfr0pb3p24nn2lthl7amb9eai1bf9rtr@4ax.com> <6m4mr0p6iml9reaphldhmufd1pksbi4t6q@4ax.com> <8qumr09f34mehjaj9v2n7hhao9srbck532@4ax.com> See what I mean? === Subject: Re: How much lossless compression is possible in images? > Here is some advice for you lout. Here some advice for you. Do not post things multiple times, how easy help. -- 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: How much lossless compression is possible in images? > > Here is some advice for you lout. > Here some advice for you. Do not post things multiple times, how easy also > help. it seems to be happening quite a lot since they introduced the beta of the new version. Some posts seem to get lost too. It is not my intention to flood. As for the nasty style, well you can't please everybody all of the time, Winter. I'm sorry for any distress it may cause to you personally, since you are not among those I dislike or intend to be uncivil to. I find empty minded prattling of the kind the OP can't seem to quit posting, as annoying as others find my style. === Subject: Re: How much lossless compression is possible in images? > it seems to be happening quite a lot since they introduced the beta of > the new version. Use a dedicated newsgroup service. Many are available for only a few dollars a month. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? > it seems to be happening quite a lot since they introduced the beta of > the new version. > Use a dedicated newsgroup service. Many are available for only a few > dollars a month. And there are even available for free. -- 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: How much lossless compression is possible in images? ... > One bitstring is reduced in size to 1 bit, and all other > bitstrings increase by 1 bit. > Yes. So one string is reduced by 1,000,000 bits, and a million other > strings increase by 1 bit. The net change is zero. Or one string is reduced by 1,000 bits and a million other strings increasy by 1 bit. The net change is non-zero. You are assuming, without reason, that all strings are equally long. > What do you mean by theoretical and real here? > The perfect algorithms of theory, versus real algorithms that need not > follow all aspects of theory. What is a perfect algorithm of theory? I have honestly no idea. > In what sense? Suppose the particular string is 'k' bits. It will be > reduced by 'k-1' bits. All other strings that start with 0 are lengthened > by 2 bits. There are 2^(m-1) such strings of size m. So when we consider Sorry I meant here, there are 2^m - 1 such string of size <= m (m >= k). > the total change in length it is '2^m - k + 1'. I do not see any sense And this changes to '2^(m+1) - k + 1'. Other changes apply. > in which this converges to zero. Even if we divide by the number of > strings considered ('2^m') and take the average, it will converge to 1, > not 0. So in the limit this algorithm will increase the size on average > by 1 bit. > In the real world, I'll concede that it would not converge on exactly > zero. With the modification above it would not converge even to a number close to 1. It will diverge. > A compression algorithm is an algorithm that converts fixed-length bit > strings of unequal probabilities to variable-length bit strings such > that the most probable input strings have the shortest output strings. Why fixed-length bit strings? Almost all compression algorithms work on variable length input, not on fixed length input. And when you consider probabilities, the optimal algorithm for the input can be expected to *decrease* the size in general. > The total number of information bits for all output messages cannot be > less than the total number of information bits for all input messages, > but offhand I don't see why it must be greater, at least in theory. That entirely depends on the algorithm. With most compression algorithms the number of information bits for all input messages will *increase* because the compression algorithm has in most cases to store some additional information in the output stream to let the decompressor work. Consider the easiest scheme of all: huffman encoding. If the input is not totally random, the size of the message will become shorter, but the coding table has also to be included in the output, which may make some of the output strings longer. In addition there are the totally random strings that inherently grow longer by the addition of the coding table. I cannot see why it must be smaller or equal, because you are adding in additional information. Now if your compression uses a fixed coding table things might be different. Off-hand I do not know, but I expect that in that case indeed total sizes might be equal. > but offhand I don't see why it must be greater, at least in theory. In > the real-world, algorithms always add a few bits to the total on output. > output strings (is it part of the output information or not?). Yes, of course. It is necessary to make the decompressor work. Consider run-length encoding on bytes. You need a marker to show that the next byte is a count and not a plain byte. On the other hand, when the next plain byte is the marker you have to do other things (that increase the code) to let that show. So suppose you use 255 as marker, the next byte when it has the value 0..254 as a count of 4..258, and a plain byte of 255 just be doubled. Now you know that all strings that do not contain the 255 byte will decrease in size or stay equal. All strings that contain the 255 byte and do not contain a repetition of 4 or more equal bytes will increase in size. For the remainder it can go either way. I *think* that in the end the total size of all possible messages will be larger. On the other hand, the additional information put in the stream by the compressor is needed for decompression. -- 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: How much lossless compression is possible in images? > What is a perfect algorithm of theory? I have honestly no idea. To me, it's an algorithm that accomplishes its compression without any overhead. > Why fixed-length bit strings? For simplicity. > Now if your compression uses a fixed coding table things > might be different. Off-hand I do not know, but I expect that in that > case indeed total sizes might be equal. In theory a compression algorithm just performs a simple substitution. It should not have to produce a total number of output bits for all possible strings that is greater than the total number of input bits for all possible strings. But I haven't considered this in depth and I may be missing something. This would have to be a perfect algorithm, as well, that is, one that adds no overhead. > Yes, of course. It is necessary to make the decompressor work. Consider > run-length encoding on bytes. You need a marker to show that the next > byte is a count and not a plain byte. On the other hand, when the next > plain byte is the marker you have to do other things (that increase the > code) to let that show. So suppose you use 255 as marker, the next > byte when it has the value 0..254 as a count of 4..258, and a plain byte > of 255 just be doubled. Now you know that all strings that do not > contain the 255 byte will decrease in size or stay equal. All strings > that contain the 255 byte and do not contain a repetition of 4 or more > equal bytes will increase in size. For the remainder it can go either way. > I *think* that in the end the total size of all possible messages will be > larger. The question is, does there exist an algorithm that does not add any overhead? I think a simple substitution table would qualify, but I'm not sure, nor am I sure how the length of strings is to be treated (is it counted as part of the encoded information in the string?). -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <41b3ed05$0$9113$afc38c87@news.optusnet.com.au> <41b4079b$0$20858$afc38c87@news.optusnet.com.au> bla bla > Put still another way, no algorithm exists (or can be devised) that will > compress a random string of bits more than 50% of the time. In fact, no > matter what the algorithm is, it will produce a shorter output string > 50% of the time, and a longer output string the other 50% of the time. You started asking a question, now you are making wild assertions. Where is the math to prove the above bull? > Yes. Compression algorithms work only because all possible images are > not equally probable. Oh so there are actually images that are more probable than others? Give an example. If an image can be represented by a binary string of length n then there are 2^n such possible images. Each one of them is equally probable. Most compression algorithms work by ASSUMING that some images are more probable than others and assign them shorter representations, thus compressing those images which are more probable under the assumed restrictions. [snip more absurd prattling] === Subject: Re: How much lossless compression is possible in images? > You started asking a question, now you are making wild assertions. There's nothing wild about the assertion. > Where is the math to prove the above bull? Shannon provided most of it. I'm not a mathematician. However, I do understand the concepts, which are practically self-evident. > Oh so there are actually images that are more probable than others? In real-world image processing, yes, very much so. The set of all possible images of n bits in size contains 2^n images, which is extraordinarily large for the types of images being manipulated today (the image on the monitor of my PC, for example, is only one of 10^13871462 possible images that can be displayed). Only a very, very small fraction of these are useful pictorial images--most are random pixels. (It is sometimes interesting to consider exactly which images are contained in these sets.) Anyway, all image compression algorithms correctly assume that certain subsets of all possible images are vastly more probable than all other images. These probable images share a lot of common characteristics, such as very high redundancy, that can be exploited by the algorithms to produce significant compression for the targeted subset of images. These same algorithms will produce marginally _larger_ output files for the much more random images that they are unlikely to ever be called upon to compress (the increase in size would be very small, though, since there are so many such images). All of this is basic information theory (basic because even I understand it). > Give an example. See above. > If an image can be represented by a binary string of > length n then there are 2^n such possible images. Each one of them is > equally probable. Not in the real world. Only certain images are probable in the real world of pictorial image processing. All image compression algorithms are designed to provide high compression for this subset of highly probable images, at the expense of negative compression for all other possible images (which vastly outnumber the pictorial images but are extremely improbable). > Most compression algorithms work by ASSUMING that > some images are more probable than others and assign them shorter > representations, thus compressing those images which are more probable > under the assumed restrictions. Yes. The assumption is correct, which is why compression algorithms work. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <41b3ed05$0$9113$afc38c87@news.optusnet.com.au> <41b4079b$0$20858$afc38c87@news.optusnet.com.au> > You started asking a question, now you are making wild assertions. > There's nothing wild about the assertion. > Where is the math to prove the above bull? > Shannon provided most of it. I'm not a mathematician. However, I do > understand the concepts, which are practically self-evident. Now now,you snipped the statement that is being argued. Are you being plain unpolite here or are you being falacious ? You stated (i quote) In fact,no matter what the algorithm is, it will produce a shorter output string 50% of the time, and a longer output string the other 50% of the time. This is wildly asserted bull, Shannons paper is available for download, please indicate where he proved that nonsense. > Oh so there are actually images that are more probable than others? > In real-world image processing, yes, very much so. The set of all Exactly. Under the assumpition of restrictions i.e. real-world image processing (I'm willing to suppose that you mean something like the kind of images we are interested in dealing with) certain images are more probable. Without restrictions no image is more probable than any other. The distinction is crucial in getting an answer to your original question. Which is that basically, the limit of compressibility of an image depends on the model (the set of conditions that are being assumed), so that your question cannot be satisfactorily be answered. However, once you establish a model you can measure the entropy of any image (under that model) which is an exact measure of the compressibility. > All of this is basic information theory (basic because even I > understand it). It seems that you got most of it right but you need to work on it a bit more. > Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >> I don't have the math at hand, but I believe that a truly random image >> would have at least some compressible areas. >Well, you would be wrong. At least one definition of random is that an >image can't be compressed. I know of no definition of random which would >allow every or even most random images to have at least some compressible >areas. Suppose you choose some compression algorithm - simple run-length encoding of bytes for example. Suppose it uses a byte for the length. Any sequence of 3 or more equal bytes will be compressible. Any reasonably large image will have such a sequence in it, so it will have a part that can be compressed. (A large image that didn't have such a sequence would be most unlikely, if it really was random.) So it's true that sufficiently large random images will have some compressible areas. This doesn't do you any good in practice, because you need to distinguish the compressed areas from non-compressed areas that just happen to have the same values. You need extra bits for this, so you won't end up with any compression overall. >The average image size (averaged across all possible images) of images >compressed using lossless algorithms must be equal to or greater than the >uncompressed image size. The unix compress program always produces a file which is the same size as the original or smaller, which it achieves by not compressing the file if compression would not make it smaller. However, it appends .Z to the filename if it compresses it, and you have to take those extra bytes into acccount when determining the real compression ratio. -- Richard === Subject: Re: How much lossless compression is possible in images? > Assuming ideal compression that removes every trace of redundancy from >> an image, how much lossless compression might really be possible? I >> know this is a difficult question because so much depends on image >> content, but I'm just trying to get an idea of how close current >> lossless compression methods are to the ideal. Visually images seem to >> be very highly redundant, although current lossless algorithms only >> manage about 50% compression. But assuming time and space were not >> constraints, how far could one theoretically go? > -- >> Transpose hotmail and mxsmanic in my e-mail address to reach me > directly. > It depends totally on image content. > An image consisting of random pixels will not compress at all (0% >> compression) using a generic algorithm; an image consisting of entirely > one >> colour will compress almost to zero (almost 100% compression). > I don't have the math at hand, but I believe that a truly random image > would have at least some compressible areas. In a strange sense, a > completely incompressible image would be as ordered as a single color > image. > Well, you would be wrong. At least one definition of random is that an > image can't be compressed. I know of no definition of random which would > allow every or even most random images to have at least some compressible > areas. > And what is completely incompressible image supposed to mean? How about an image consisting of random pixels [that] will not compress at all? === Subject: Re: How much lossless compression is possible in images? > Assuming ideal compression that removes every trace of redundancy > from >> an image, how much lossless compression might really be possible? I >> know this is a difficult question because so much depends on image >> content, but I'm just trying to get an idea of how close current >> lossless compression methods are to the ideal. Visually images seem > to >> be very highly redundant, although current lossless algorithms only >> manage about 50% compression. But assuming time and space were not >> constraints, how far could one theoretically go? > -- >> Transpose hotmail and mxsmanic in my e-mail address to reach me > directly. > It depends totally on image content. > An image consisting of random pixels will not compress at all (0% >> compression) using a generic algorithm; an image consisting of entirely > one >> colour will compress almost to zero (almost 100% compression). > I don't have the math at hand, but I believe that a truly random image > would have at least some compressible areas. In a strange sense, a > completely incompressible image would be as ordered as a single color > image. Well, you would be wrong. At least one definition of random is that an > image can't be compressed. I know of no definition of random which would > allow every or even most random images to have at least some compressible > areas. > And what is completely incompressible image supposed to mean? > How about an image consisting of random pixels [that] will not compress at > all? What about the old pigeonhole principle: Any lossless encoding scheme that compresses at least one image will increase the size of some other image. === Subject: Re: How much lossless compression is possible in images? > How about an image consisting of random pixels [that] will not compress at > all? It is always possible to devise an algorithm that will compress any one given image, no matter how random it may appear to be. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <41b3ed05$0$9113$afc38c87@news.optusnet.com.au> <41b4079b$0$20858$afc38c87@news.optusnet.com.au> >It is always possible to devise an algorithm that will compress any one >given image, no matter how random it may appear to be. that in itself sounds like an algorithm, so you may find the algorithm you find is bigger than the image. Herc === Subject: Re: How much lossless compression is possible in images? > that in itself sounds like an algorithm, so you may find the algorithm > you find is bigger than the image. Not if it is designed only to compress one specific image. A generalized algorithm to generate such an algorithm for any arbitrary image would indeed be unwieldy, although I'm too lazy to try to figure out a lower bound for its size. -- Transpose hotmail and mxsmanic in my e-mail address to reach me directly. === Subject: Re: How much lossless compression is possible in images? <41b3ed05$0$9113$afc38c87@news.optusnet.com.au> <41b4079b$0$20858$afc38c87@news.optusnet.com.au> The point here is any compression algorithm, or system of compression algorithms, operating on the total set of inputs, cannot get a total compression over the entire set. Consider 2MB 1000 pixels X 1000 pixels Raw Data Compressed File 1 66 2 44 3 2 4 1 5 .. .. 16^1,000,000 There are 16^1,000,000 possible picture files, so there must be 16^1,000,000 distinct compressed files. You can't have 16^1,000,000 distinct files unless the average file length is over 1MB. The average file length of the original 16^1,000,000 images is also 1MB. Therefore, not every picture gets compressed. Finding a particular compression algorithm for that image doesn't help in the global scheme of things, because you have to say what the algorithm is, and it wont be small. The length of the algorithm + the compressed file will tend to be larger than the original file. The exceptions are easily compressible data, lots of flat areas. Herc === Subject: Questions of Higher Order Derivatives; and their Inverses Can someone provide me with a better proof than this; somehow I missed this relation in undergraduate calculus: (d^2s/dt^2)(dt/ds)^3 = -d^2t/ds^2 I differentiate the unit tangent vector, T =dr/ds, with respect to t: dT/dt = d(dr/ds)/dt = d(dr/dt)(dt/ds)/dt = (d^2r/dt^2)(dt/ds) + (d^2t/ds^2)(ds/dt)(dr/dt) = (d^2r/ds^2)(ds/dt) multiply by (dt/ds): (dT/dt)(dt/ds) = d^2r/ds^2 = (d^2r/dt^2)(dt/ds)^2 + (d^2t/ds^2)(dr/dt) (A) In literature, in matrix form (stated without proof, easy enough to do), I find: | (ds/dt) (d^2s/dt^2)| d^2r/ds^2 = | | * (dt/ds)^3 | (dr/dt) (d^r/dt^2)| or, = (d^2r/dt^2)(dt/ds)^2 - (dr/dt)(d^2s/dt^2)(dt/ds)^3 (B) if Equation (A) = Equation (B) then, (d^2s/dt^2)(dt/ds)^3 = -(d^2t/ds^2) I extend it this way: (ds/dt)(dt/ds)^2 = dt/ds (d^2s/dt^2)(dt/ds)^3 = -d^2t/ds^2 (d^3s/dt^3)(dt/ds)^4 = d^3t/ds^3 (d^4s/dt^4)(dt/ds)^5 = -d^4t/ds^4 I also have a similar proof for: (d^2r/dt^2)(dt/ds)^2 = (d^2r/ds^2) But have not known of these simple relations; nor have been able to make a simple proof of it, using the definition of differentiation or at least using: Du^n/dx = [(1/n)u ^(n-1)]dx I really don't have the necessary understanding of differentials, and differentiation, to prove the relations directly, or to even ask the proper questions, I guess; can someone help? Charles === Subject: Re: Skolem's Paradox and why is math the way it is? |I have a question about the universe of discourse, could you please expand |what you mean when you say that there are larger realms of discourse than set |theory? Well, partly I had in mind that there isn't just the one system of sets (the cumulative hierarchy). There are also the non-well-founded sets and the like. Restricting to just well-founded sets is a way of ensuring a certain kind of good behavior, but surely not absolutely necessary. Also although there are often ways of representing mathematical objects as sets, it's somewhat artificial to do so in many cases. Meanwhile, there are constructions in category theory that appear to make sense, but are too big to be sets as they stand. Keith Ramsay === Subject: Re: Skolem's Paradox and why is math the way it is? > |I have a question about the universe of discourse, could you please expand > |what you mean when you say that there are larger realms of discourse than set > |theory? > Well, partly I had in mind that there isn't just the one system of > sets (the cumulative hierarchy). There are also the non-well-founded > sets and the like. Restricting to just well-founded sets is a way of > ensuring a certain kind of good behavior, but surely not absolutely > necessary. > Also although there are often ways of representing mathematical objects > as sets, it's somewhat artificial to do so in many cases. > Meanwhile, there are constructions in category theory that appear to > make sense, but are too big to be sets as they stand. > Keith Ramsay That's an interesting point about the cumulative hierarchy and the non-well-founded sets. Where I equated the cumulative hierarchy with the projectively completed ring of integers in the theory of ubiquitous ordinals, then when I consider ubiquitous ordinals with no foundation their ordinal value would not in the naive construction differ from their order type in the simple case where A={A}. Thus they would all coincide with U, the universal set, as ordinals, or there would be a lot of sets outside of the cumulative hierarchy, the concept of which can be used to represent the integers as mechanistic sets. Category theory, with its objects and arrows, or functions, does not seem to be outside of the realm of discourse of set theory. Please provide an example or to ZF of any abstraction of primary objects if applicable. What's the point of a fundamental set theory if everything is not a set? People often use ZF with classes, if they don't then they have to stab out their figurative eyes every time they see the words collection of all sets, which regenerate painfully. Then, when they're using that, ZF with classes, say class of all classes and their eardrums explode. Skolem. As has been discussed in this and other recent threads, and rather continuously for a hundred eternally, it can be easily seen that room for improvement in foundations exists. There are a wide variety of logical theories, most using damningly self-reflexive non-logical axioms, that plainly admit the universal set and as well other non-well-founded sets. cumulative hierarchy as ubiquitous ordinals is that all of the infinite ordinals are non-well-founded. That seems a decent naive resolution. Then, you can easily assume the sets are well-founded or not, and if so then there are ubiquitous naturals, else ordinals. Infinity: too big. Ross F. === Subject: Re: Skolem's Paradox and why is math the way it is? |> I didn't say anything about standards of proof, but since you ask, we |> do of course have standards. They are just informal, not set by a fixed |> axiom system. | |But different axioms systems are different. For instance you can have |a non-well founded set theory, then you usually have collection |instead of repalcement, but in a normal set theory class a proof of |collection is required. It's a mistake, though, to assume that when people are using two different axiom systems, they are using them intending for the sentences of the language to have two different interpretations, one for each axiom system. Sometimes they are only adding more or less of the things that they believe to be true under one given interpretation. It's ordinarily impossible to capture one's knowledge of mathematics in a formal system. If one believes that the axioms are true, and that the deductive rules are sound, then one normally believes that they are also consistent, but a system strong enough to capture a person's knowledge of mathematics presumably is strong enough for Goedel's incompleteness theorem to apply. So they believe the formal system to be consistent, but it's impossible to prove in the system. Hence if I start working in ZFC+con(ZFC), it's fairly unlikely that I'm doing so intending for the sentences I'm working with to have a different meaning from what I intended when I was just working with ZFC. I merely added one more thing that I believe to be true to the axioms. Keith Ramsay === Subject: Re: GCH vs. Axiom of Choice. |Ah, I should have been more clear. L is a model (right?), It's a proper class of sets, the sets satisfying a certain condition. (Being constructible.) Another way to describe it is that it's the minimal model of ZF containing all the ordinals. | ZF is an |axiom system (right?). GCH is true for L (you said so in your post |right?). You also said that we prove that GCH is true for L? But |what do we need to prove it? ZF is strong enough by itself to prove that GCH holds in L. ZF is strong enough to prove that V=L holds in L, and that V=L->GCH. | For instance if we had to assume ZFC+GCH |to prove it, then it wouldn't mean much IMO. Heh, it's sort of amusing when to prove that some fact holds true when relativized to a model one needs to use the same fact not relativized to the model. I can see how that might seem underwhelming, but it's not necessarily trivial or uninteresting. For instance, one had to prove that the axioms of ZF hold in L, too, and obviously we need to assume some of them in order to prove they hold in L. | I assume that ZF can |prove only limited things about L, since ZF is incomplete, but if I |add GCH to ZF then can ZF+GCH prove something new about L? I'm guessing it probably can, but I don't know of a good example. My guess is that some assumption intermediate between ZF and ZF+V=L is sufficient to prove (say) that aleph_1 is the same as aleph_1 relative to L. Certainly V=L implies that! On the other hand, there aren't any first order properties of L that can be proven in ZF+GCH but not in ZF. In fact, for an arbitrary sentence X in the language of ZF, the following are equivalent: (a) ZF |- (V=L->X) (b) ZF + V=L |- X (c) ZF |- L |= X. (d) ZF + V=L |- L |= X (a)->(b) is a case of the deduction theorem. (b)->(c) holds because it's a theorem of ZF that V=L relativized to L is true Given a model M of ZF, we can consider constructibility as relativized to M. The elements of M satisfying this relativized constructibility form a submodel L_M, which is another model of ZF. If we repeat this process, to get L_{L_M}, it's possible to prove that we don't shrink the model any; L_{L_M} = L_M. So L_M is a model of V=L, the axiom stating that every set is in L. So anything that's a consequence of V=L holds relative to L. If X is a theorem of ZF+V=L, then that fact can be proven in ZF, and then applied to L. (c)->(d) is obvious. To prove (d)->(a), assume first that (a) is false. By Goedel's completeness theorem, there would exist a model of ZF+V=L+~X. Suppose M is a model of ZF+V=L+~X. Since the L relativized to M is the same as M, we get that M |= ~(L|=X). Hence M is also a model of ZF+V=L in which L|=X fails. Since GCH is a consequence of V=L, (c') ZF + GCH |- L|= X is also equivalent. | How about |ZF+(~GCH), can that also prove things about L? Since ~GCH is so weak, I don't think there are many theorems about L in ZF+(~GCH) that aren't also theorems in ZF. We can deduce from ~GCH the existence of sets not in L, of course. | Do the truths of L |vary depending on what other axioms L is defined with? I'm trying to |be more clear, but I apologize in advance if I failed again. How would whether something is true of L depend on axioms? Of course, what can be proven to be true of L from a set of axioms does depend on the axioms! I'm not familiar with the details, but I gather that some large cardinal axioms (I think the existence of a measurable cardinal is strong enough) imply a relatively detailed account of the structure of L. Note that the existence of a measurable cardinal is inconsistent with V=L. This real 0# that is believed to be outside of L encodes some of the structure of L (if it exists). Keith Ramsay === Subject: Re: GCH vs. Axiom of Choice. > If you have two axioms A and B such that > Con(ZF)=>[Con(ZF+A)&Con(ZF+B)], then are things that ZF+A proves about > L consistent with the things that ZF+B proves about L? Even if > Con(ZF) does not imply Con(ZF+A+B)? > > No, not necessarily. > So what is true in L depends on whether L is made in ZF or ZFC or some > other axioms system that is relatively consistent with ZF, right? Not what is *true* in L, but what is *provably true* in L, depends on which axiom system you're working with, yes. === Subject: Re: GCH vs. Axiom of Choice. > > > If you have two axioms A and B such that > Con(ZF)=>[Con(ZF+A)&Con(ZF+B)], then are things that ZF+A proves about > L consistent with the things that ZF+B proves about L? Even if > Con(ZF) does not imply Con(ZF+A+B)? > > No, not necessarily. > > So what is true in L depends on whether L is made in ZF or ZFC or some > other axioms system that is relatively consistent with ZF, right? > Not what is *true* in L, but what is *provably true* in L, depends on > which axiom system you're working with, yes. I asked if the things PROVABLY true in L by ZF+A are CONSISTENT with the things PROVABLY ture in L by ZF+B, and you said No, not necessarily, but without an example. Goedel's completeness says that you can add axioms to any system (like ZF) to prove any statement such that both it and it's negation are unprovable without causing a contradiction as long as the axiom you add is a statement of the unprovable statement itself, right? Well, that's a bit silly for proving things about the system itself because you'll clearly get different results depending on whether you add the statement or it's negation. But I'm, thinking that no matter what you add to ZF to maintain it's relative consistency, that you either prove X about L or you don't, where X is true of L, but that you can never prove ~X about L as long as X is true of L. But since Goedel mentioned that EVERY consistent statement can be proved with the right axioms, that seems like we'd be able to prove every truth of L actually proved by using every relatively consistent axiom system. A related example is that someone said that you can make a model of Union, Pairing, Empty Set, Foundation, Infinity, Separation, and Powers given Union, Pairing, Empty Set, Foundation, Infinity, Separation, and Powers + Replacement, clearly that's devoid of interest if you had to assume consistentcy of the second system, so I assumed it meant you just assumed the truth of the second system, but if the first system is large enough to make a model of itself (which since it contains omega seems reasonable given Skolem-Lowenheim) then maybe the truth of the model existing from the second system shows that the model exists and is consistent without begging the question. === Subject: Re: GCH vs. Axiom of Choice. > > > If you have two axioms A and B such that > Con(ZF)=>[Con(ZF+A)&Con(ZF+B)], then are things that ZF+A proves about > L consistent with the things that ZF+B proves about L? Even if > Con(ZF) does not imply Con(ZF+A+B)? > > No, not necessarily. > > So what is true in L depends on whether L is made in ZF or ZFC or some > other axioms system that is relatively consistent with ZF, right? > > Not what is *true* in L, but what is *provably true* in L, depends on > which axiom system you're working with, yes. > I asked if the things PROVABLY true in L by ZF+A are CONSISTENT with > the things PROVABLY ture in L by ZF+B, and you said No, not > necessarily, but without an example. Goedel's completeness says that > you can add axioms to any system (like ZF) to prove any statement such > that both it and it's negation are unprovable without causing a > contradiction as long as the axiom you add is a statement of the > unprovable statement itself, right? Well, that's a bit silly for > proving things about the system itself because you'll clearly get > different results depending on whether you add the statement or it's > negation. But I'm, thinking that no matter what you add to ZF to > maintain it's relative consistency, that you either prove X about L or > you don't, where X is true of L, but that you can never prove ~X about > L as long as X is true of L. But since Goedel mentioned that EVERY > consistent statement can be proved with the right axioms, that seems > like we'd be able to prove every truth of L actually proved by using > every relatively consistent axiom system. False axioms can be consistent with ZF. You could add a false axiom to ZF and the results that you could prove to be true in L with this axiom system might not in fact be true in L. > A related example is that someone said that you can make a model of > Union, Pairing, Empty Set, Foundation, Infinity, Separation, and > Powers given Union, Pairing, Empty Set, Foundation, Infinity, > Separation, and Powers + Replacement, clearly that's devoid of > interest if you had to assume consistentcy of the second system, so I > assumed it meant you just assumed the truth of the second system, but > if the first system is large enough to make a model of itself (which > since it contains omega seems reasonable given Skolem-Lowenheim) then > maybe the truth of the model existing from the second system shows > that the model exists and is consistent without begging the question. You can prove in ZF that Con(Z), without assuming Con(ZF), yes. === Subject: Re: GCH vs. Axiom of Choice. > False axioms can be consistent with ZF. You could add a false axiom to > ZF and the results that you could prove to be true in L with this > axiom system might not in fact be true in L. Could you please be so kind as to explain what a false axiom is? I'm familiar that somone has proven (in ZF?) that Con(ZF) => Con(ZFC) as well as Con(ZF) => Con(ZF+~C), so I didn't think C or ~C was considered false, just relatively independant axioms. What is an example of a false axiom that is (relativily?) consistent with ZF? > A related example is that someone said that you can make a model of > Union, Pairing, Empty Set, Foundation, Infinity, Separation, and > Powers given Union, Pairing, Empty Set, Foundation, Infinity, > Separation, and Powers + Replacement, clearly that's devoid of > interest if you had to assume consistentcy of the second system, so I > assumed it meant you just assumed the truth of the second system, but > if the first system is large enough to make a model of itself (which > since it contains omega seems reasonable given Skolem-Lowenheim) then > maybe the truth of the model existing from the second system shows > that the model exists and is consistent without begging the question. > You can prove in ZF that Con(Z), without assuming Con(ZF), yes. === Subject: Re: GCH vs. Axiom of Choice. |I'm |familiar that somone has proven (in ZF?) Only finitistic reasoning is actually required. | that Con(ZF) => Con(ZFC) That someone was Goedel. |as |well as Con(ZF) => Con(ZF+~C), That someone was Cohen. | so I didn't think C or ~C was |considered false, just relatively independant axioms. Are you assuming that every statement X for which we can prove Con(ZF)=>Con(ZF+X) and Con(ZF)=>Con(ZF+~X) is neither true nor false? I don't mean it as a rhetorical question. I don't assume anything like that, and to me this seems like a nonsequitur. If you believe such a thing, and you don't know why you believe it, it's liable to be a source of some confusion. Keith Ramsay === Subject: Re: GCH vs. Axiom of Choice. > False axioms can be consistent with ZF. You could add a false axiom > to > ZF and the results that you could prove to be true in L with this > axiom system might not in fact be true in L. > Could you please be so kind as to explain what a false axiom is? I'm > familiar that somone has proven (in ZF?) that Con(ZF) => Con(ZFC) as > well as Con(ZF) => Con(ZF+~C), so I didn't think C or ~C was > considered false, just relatively independant axioms. What is an > example of a false axiom that is (relativily?) consistent with ZF? Well, I believe that every sentence in the first-order language of set theory is either true or false. ~AC would be an example of a false axiom consistent with ZF. If you don't believe this, you shouldn't really use the notion of truth at all, just provability in this or that formal system. === Subject: Re: GCH vs. Axiom of Choice. > False axioms can be consistent with ZF. You could add a false axiom > to > ZF and the results that you could prove to be true in L with this > axiom system might not in fact be true in L. > Could you please be so kind as to explain what a false axiom is? I'm > familiar that somone has proven (in ZF?) that Con(ZF) => Con(ZFC) > as > well as Con(ZF) => Con(ZF+~C), so I didn't think C or ~C was > considered false, just relatively independant axioms. What is an > example of a false axiom that is (relativily?) consistent with ZF? > Well, I believe that every sentence in the first-order language of set > theory is either true or false. ~AC would be an example of a false > axiom consistent with ZF. If xey is fixed as either true or false in a domain of discourse for every x and y in the domain of discourse then yes every FOL sentance is either true or false of that same domain of discourse, it's just a fact of combinatorics. If you start with an unknown domain of discourse that satisifies ZF and then consider a sub domain that is L then you can assume AC or ~AC in the original domain and I don't see how it can effect the subdomain's truth. But you might be able to prove more. > If you don't believe this, you shouldn't really use the notion of > truth at all, just provability in this or that formal system. I'm not sure what you are saying that there is to believe. Why is AC true and ~AC false? You didn't explain that. But assuming one only talked about provability, are the theorems of ZF+AC that are about L inconsistent with the therorems of ZF+~AC that are about L? Clearly the theorems of ZF+AC about V are different than the theorems of ZF+~AC about V, but I want to know about L. === Subject: Re: GCH vs. Axiom of Choice. But assuming one only talked about provability, are the theorems of > ZF+AC that are about L inconsistent with the therorems of ZF+~AC that > are about L? Clearly the theorems of ZF+AC about V are different than > the theorems of ZF+~AC about V, but I want to know about L. I have just realized if ZF+~AC could prove a certain proposition is true in L, then it would be consistent with ZFC that that proposition was true in L. However, you might like to try the other example. Suppose ZF+Con(ZF) is consistent. Then ZF+Con(ZF) is consistent and proves Con(ZF) is true in L, whereas ZF+~Con(ZF) is consistent and proves ~Con(ZF) is true in L. === Subject: Re: GCH vs. Axiom of Choice. False axioms can be consistent with ZF. You could add a false > axiom > to > ZF and the results that you could prove to be true in L with this > axiom system might not in fact be true in L. > Could you please be so kind as to explain what a false axiom is? > I'm > familiar that somone has proven (in ZF?) that Con(ZF) => Con(ZFC) > as > well as Con(ZF) => Con(ZF+~C), so I didn't think C or ~C was > considered false, just relatively independant axioms. What is an > example of a false axiom that is (relativily?) consistent with ZF? > Well, I believe that every sentence in the first-order language of > set > theory is either true or false. ~AC would be an example of a false > axiom consistent with ZF. > If xey is fixed as either true or false in a domain of discourse > for every x and y in the domain of discourse then yes every FOL > sentance is either true or false of that same domain of discourse, it's > just a fact of combinatorics. If you start with an unknown domain of > discourse that satisifies ZF and then consider a sub domain that is L > then you can assume AC or ~AC in the original domain and I don't see > how it can effect the subdomain's truth. But you might be able to > prove more. > If you don't believe this, you shouldn't really use the notion of > truth at all, just provability in this or that formal system. > I'm not sure what you are saying that there is to believe. Why is AC > true and ~AC false? You didn't explain that. I just happen to believe it. Another example you might like better: Con(ZF) is true and ~Con(ZF) is false. However ~Con(ZF) is consistent with ZF. > But assuming one only talked about provability, are the theorems of > ZF+AC that are about L inconsistent with the therorems of ZF+~AC that > are about L? Clearly the theorems of ZF+AC about V are different than > the theorems of ZF+~AC about V, but I want to know about L. I'm not sure. === Subject: Re: GCH vs. Axiom of Choice. |But assuming one only talked about provability, are the theorems of |ZF+AC that are about L inconsistent with the therorems of ZF+~AC that |are about L? What do you mean by about L? It's a theorem of ZF+~AC that there exist sets not in L. Is that about L? In another posting I outlined a proof that for any sentence X in the language of ZF, we have ZF|-(L|=X) if and only if ZF+(V=L)|-(L|=X), and since AC is a consequence of V=L, these two are also equivalent to ZFC|-(L|=X). I believe it's also the case that for every such X, ZF+~AC|-(L|=X) if and only if ZF|-(L|=X). I don't think that would be too hard to prove, but might require something like redoing Cohen's proof of the relative consistency of ~AC. If I'm correct, then ZFC|-(L|=X) if and only if ZF+~AC|-(L|=X). Is that the kind of result you want? What's the motivation for the line of inquiry? Keith Ramsay === Subject: Re: GCH vs. Axiom of Choice. > But assuming one only talked about provability, are the theorems of > ZF+AC that are about L inconsistent with the therorems of ZF+~AC that > are about L? This question means nothing at all. You are in need of rest and recuperation. Have a beer or two, think about nice things unconnected with set theory. === Subject: Re: GCH vs. Axiom of Choice. I don't think the question is meaningless. He means, is there proposition relativized to L that you can prove in ZFC but disprove in ZF+~AC. I don't know whether this is the case or not. === Subject: Re: GCH vs. Axiom of Choice. I'm sure you mean well, but that's hardly constructive advice. I haven't yet understood what Rupert said, and you haven't helped me with that. I'm allergic to beer, and you didn't tell me why what I said was meaningless. I don't even understand what L even is, but some people seem to think it's well defined enough to have every oFOL sentenance with one binary relation be either true or false of it, which is pretty well-defined in my book if that's the case. My naive understanding is that it's the smallest intended universe that satisfies the ZF axioms. === Subject: Re: GCH vs. Axiom of Choice. |I don't even understand what L even is, I gave you one definition, that it's the smallest class of sets *that contains all ordinals* and satisfies ZF. For a more concrete definition and treatment you can read about it in Cohen's book about the axiom of choice. L is a hierarchy indexed by ordinals but with only one set being added at each step. If alpha is a limit ordinal, L_{alpha+n} for integers n are the subsets of {L_beta : beta < alpha} that are definable by formulas in the language of ZF with all the quantifiers restricted to {L_beta : beta < alpha}. L is the proper class consisting of all the sets that occur as L_alpha for some ordinal alpha. L satisfies AC because everything in it is well-ordered by the order in which the elements appear in the L_alpha. |but some people |seem to think it's well defined enough to have every oFOL sentenance |with one binary relation be either true or false of it, which is pretty |well-defined in my book if that's the case. I doubt you'll find anyone who believes that about L who doesn't believe it because they also believe that about V, the cumulative hierarchy. At first glance, the hierarchical definition of L is less problematic because the sets at each step are being defined by formulas quantifying over previous steps. But we still have to start with a belief in the well-definedness of the sequence of all ordinals, and once one believes in the ordinals, belief in arbitrary sets is not usually seen as a problem. | My naive understanding is |that it's the smallest intended universe that satisfies the ZF axioms. No. I don't know what you mean by intended, but if you mean what I would mean by intended, there's just one intended universe, V. V_0 is {{}}. For each ordinal alpha, V_alpha+1 is the power set of V_alpha. If alpha is a limit ordinal, V_alpha is the union of the V_beta for beta < alpha. Many sets are added at each successor ordinal. V is the class of sets that appear at some stage V_alpha for an ordinal alpha. Keith Ramsay === Subject: Re: GCH vs. Axiom of Choice. > |but some people > |seem to think it's well defined enough to have every oFOL sentenance > |with one binary relation be either true or false of it, which is pretty > |well-defined in my book if that's the case. > I doubt you'll find anyone who believes that about L who doesn't > believe it because they also believe that about V, the cumulative > hierarchy. At first glance, the hierarchical definition of L is less > problematic because the sets at each step are being defined by > formulas quantifying over previous steps. But we still have to start > with a belief in the well-definedness of the sequence of all ordinals, > and once one believes in the ordinals, belief in arbitrary sets is > not usually seen as a problem. I think there are probably people (by which I mean people who know what they're talking about) who might take the position that any sentence of LST relativized to L is either true or false, but not any sentence period. One reason this might be attractive is that you can't, by forcing, change the truth value of a sentence relativized to L. It's certainly not *my* position. I think the honest way of describing the position would be to say you think the powerset axiom is really false--that there isn't any completed whole that contains, say, all the reals. I guess that's not directly inconsistent with thinking that you *can* collect (say) all the countable ordinals into a completed whole. === Subject: Re: Set Theory |> |> As a result, people usually think one of two things, either that |> |> the continuum problem is hard, or that it's not a problem |> |> to solve. Platonists think of the problem as hard. It's a theorem |> |> of ZFC that c=aleph_x for some ordinal x. A Platonist would |> |> (ordinarily at least) believe that, and think that we just don't know |> |> yet which ordinal x is, whether it's 1, 2, or something else. On |> |> the other hand, there are people who think that the question is |> |> somehow undefined, and that it's meaningless to ask which |> |> ordinal x really is. |> | |> |If the universe of discourse is the heirarchy of sets you mentioned on |> |another thread, then is x a specific ordinal, or is it still unkown? |> It's a specific ordinal, although we may not know whether it's 1, |> 2, 3, or something else. |> The only way for it to make sense to say that the universe of discourse |> is indeed the cumulative hierarchy is to agree to at least part of the |> Platonists' position on it. | |Can you elaborate, I'm don't even understand what you are saying. |Things like agree to at least PART seems vague to me. Indeed, I was being a bit vague. The point is that lots of the standard philosophical positions opposed to Platonism would not agree that it makes sense to say that (we have decided) the cumulative hierarchy is our domain of discourse. The cumulative hierarchy is the class of pure, well-founded sets. I think that's fine. But some people consider it just fictional. The reason for the vagueness is that there are positions differing both from the ones treating it as a complete fiction and Platonism. It would be possible for a constructivist, for example, to accept the cumulative hierarchy as a valid construction without buying into Platonism. | What do you |have to do to consider that universe of discourse? I find this kind of how question hard to answer, because I can't tell what sort of answer you're looking for. If we want to take the integers as our universe of discourse (for doing number theory, say), what do we have to do? In very mundane terms, we do what we did in elementary school-- learn to count, learn to do operations on them, perhaps learn mathematical induction if we're lucky, and so on. At that point, we know what the integers are, to the extent that is needed to take them to be our universe of discourse in a more formal way. For the cumulative hierarchy it's similar. We perhaps acquaint ourselves with hereditarily finite sets {{},{{{}}}} something like the way we got acquainted with integers. Then maybe we choose to suppose that the hereditarily finite sets can be regarded as a whole as a larger kind of set. We can define subsets of it, families of subsets, and so on. Eventually someone describes the notion of rank, and we ask how high in rank it makes sense to let sets go. The way you ask this kind of question makes me uneasy, because I keep getting the impression you are essentially expecting a kind of answer that's actually impossible, without being able to articulate exactly what kind of expectation you have, so that we can hold it up to the light and show just why it's impossible to satisfy. It relates to the boostrapping problem I mentioned months back. It's difficult to explain how it is possible to build a foundation for a subject without the foundation itself being arbitrary, or it's being built on top of a still more fundamental foundation (which then still needs to be explained). In the case of the integers, I feel confident that the foundation is quite okay, and that we know what the integers are even by just a very informal introduction to them. | To me it's very |confusing, like if you assume that fewer sets exist by having GCH, |then you get choice sets that don't exist in other versions, is that |because the sets that can lack a choice function are between the |cardinals omege, 2^ omega, 2^(2^omega), etc? I'm not surprised you find it confusing, since I find the statement of the question confusing! It seems to me that you're imagining a mathematician changing the assumptions he's using, and this somehow going hand-in-hand with assuming a different class of sets as the domain of discourse. These two ideas *might* go through the mathematician's mind together, but there's no obvious relationship. To me, GCH is just another one of those conjectures that's probably untrue, but whose consequences we could explore anyway. Is what you're asking about a scenario where someone is considering a model M of ZF not satisfying GCH, and then shifts to considering a submodel N that is a model of ZF+GCH? Keith Ramsay === Subject: A problem in convexity I'm stuck in this problem: Let S be an infinite subset of the d-dimensional Euclidian space and let u be a point in the interior of convex hull of S. Prove that one can choose 2d points v1,...,v2d in S such that u lies in the i