mm-50 === Do negative numebrs exist in the real world or are they just a purely> symbolical and mathematical construct?I don't know. Do positive numbers exist in the real world?-- === > Do negative numebrs exist in the real world or are they just a purely> symbolical and mathematical construct?I don't know. Do positive numbers exist in the real world?-- Excellent answer. Some numbers have direction. This even can be true in thereal world. If something is countable or measured, then the opposite count ormeasure may also occur. One of them is the opposite of the other (obviously). G C === > intersection of the surface with the given plane. Find equations for the> line tangent to C at the point P. Plane x = 2; P (2, 1, 5) A solution I looked at first found f(2) = 5, then the partial derivative> of> f[sub-y](2,1) = 2. Then it turned it into the following two equations> x = 2> The tangent line is in the plane x=2, which explains this equation.> equation> z - 5 = 2(y - 1)> is the line (in the yz plane) through (1,5) satisfying dz/dy=2.> LH> === Part of the problem involves this:....C is the curve of intersection of the cylinder y - 2z^2 = 1 and theplane z = x + 1 from (0,3,1) to (1,9,2).Their solution to parametrizing it is: r(u) = (u - 1)i + (1 + 2u^2)j + uk u E [1,2]My question is what steps did they go to reach that. How did they find theintersection? === Part of the problem involves this:....C is the curve of intersection of the cylinder y - 2z^2 = 1 and the>plane z = x + 1 from (0,3,1) to (1,9,2).Their solution to parametrizing it is: r(u) = (u - 1)i + (1 + 2u^2)j + uk u E [1,2]My question is what steps did they go to reach that. How did they find the>intersection?By renaming z to u.John Mitchell === ...> Should circles be drawn by simple rules that check which neighbor to> turn on next, or by using cosine and sine functions? ...I don't know. Of course, in computer graphics, one can useBresenham's circle-drawing algorithm, which uses fast, simple, precise techniques much like those of his line-drawing algorithm.Here is another simple method that draws a round-looking thing but not a circle: Choose m so that m(m+1) is approximately equal to desired diameter. For the first quadrant, draw a line up n and left 1, then (each time from previous line's endpoint) a line up n-1 and left 2, then a line up n-2 and left 3, etc., and finally up 1 and left n. Reflect for other quadrants.-jiw === I am not sure about this, but I was just wondering if anyone knows anything>about this: if a function is bounded above, is the derivative bounded>above? > I guess yes, if the function is differentiable continuously in every domain point.No - consider -abs(sqrt(abs(x))) at x=0. === > So far it seems that all solutions to B6 were based on a discrete version.The reduction to the discrete situation is trivial, opens up many extraproof techniques, and doesn't seem to rule out any proof techniques. Soit's natural to express solutions in the discrete situation.Here, for example, is a discrete version of your proof. Fix real numbers r_1, r_2, etc. For t>0 define A(t) = sum_i[t=-r_i]. Then B(t)-A(t) = sum_i[t>=r_i>=-t], which isbetween 0 and n, so (B(t)-A(t))^2 <= n(B(t)-A(t)). Furthermore, |r_i| =int_{t>0} ([t0} (sum_i[t0} (A(t)+n-B(t)).Similarly, |r_i+r_j| = int_{t>0} ([t>=r_i][t<-r_j] + [t>=-r_i][t=r_j] + [t=-r_j]), so sum_{i,j} |r_i+r_j| = int_{t>0} (2(n-A(t))(n-B(t)) + 2A(t)B(t)) = int_{t>0} (n^2 + (B(t)+A(t)-n)^2 - (B(t)-A(t))^2) >= int_{t>0} (n^2 + 0 - n(B(t)-A(t))) = n sum_i |r_i|.---D. J. Bernstein, Associate Professor, Department of Mathematics,Statistics, and Computer Science, University of Illinois at Chicago === > So far it seems that all solutions to B6 were based on a discrete version.The reduction to the discrete situation is trivial, opens up many extra> proof techniques, and doesn't seem to rule out any proof techniques. So> it's natural to express solutions in the discrete situation.On the other hand, Pop's solution works quite generally to show that if X and Y are independent identically distributed random variables, then E|X+Y= E|X|. The discrete inequality is a special case of this inequality.-- A. === >So far it seems that all solutions to B6 were based on a discrete version.> The reduction to the discrete situation is trivial, opens up many extra> proof techniques, and doesn't seem to rule out any proof techniques. So> it's natural to express solutions in the discrete situation.> I never said the contrary.Ciprianhi, I need help with a question from an old exam at my school.Find all k-transposable numbers (and the corresponding k, where k is aninteger, of course)where a k transposable number is a number where when you move the leftmostdigit and move it all the way to the right, the number is k times as large. also part two of the question asked about the reverse question -- moving therightmost digit all the way too the left. find all, say, kk - transposablenumbers.Any help would be appreciated.Jason === > hi, I need help with a question from an old exam at my school.Find all k-transposable numbers (and the corresponding k, > where k is an integer, of course)> where a k transposable number is a number where when you move > the leftmost digit and move it all the way to the right, the > number is k times as large.also part two of the question asked about the reverse question -- > moving the rightmost digit all the way too the left. find all, > say, kk - transposable numbers....Have you already noted that k*(d*10^j+e)=10*e+d (where d is a leading digit and e is a j-digit number in base 10), from which e = d*(k*10^j -1)/(10-k) ? Also k*d < 10, e must be an integer with the same factors as d*(k*10^j -1)/(10-k), and when k=2, d must be even. You will need a few more ideas besides these, but should be able to readily characterize solutions at k=1, show there are no solutions at k=2, find solutions at k=3 with j=5, 11, 17, 23..., etc.-jiw === > Try talk.origins if you want to see the sparks fly. By comparison,> sci.math is tame. And there are groups far worse that> talk.origins.>Which ones?talk.abortion ?> Oooh... Good answer. === > |Has been tried, it will not work. If he does not get attention he starts > |and so he must be correct, but does not receive the proper attention. > when was this occasion that everyone ignored jsh for a period of a > couple of months? > | > |Long, long ago. > when was it? was it after jsh started posting? was it during a time > period that's recorded in the google archive well enough to determine > whether you're telling the truth?ignored James. There are always people that will reply, and it would nowbe worse than at that time (a larger public you know). But if I rememberright, almost all of the regular responders refrained from responding forJSH where he gleefully tells that he has silenced all objectors to hismath (ignoring those he had not answered, and also ignoring that theanswers to those he did answer were not really answers). In it, of course,was the complete exposition of his rambling theory.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === |Has been tried, it will not work. If he does not get |claims that nobody disagrees, and so he must be correct, but |does not receive the proper attention. when was this occasion that everyone ignored jsh for a period of a couple of months? | |Long, long ago. when was it? was it after jsh started posting? was it during a time period that's recorded in the google archive well enough to determine whether you're telling the truth?||everyone ignored James. There are always people that will reply, and|it would now be worse than at that time (a larger public you know).|But if I remember right, almost all of the regular responders|refrained from responding for some time. It did not stop the flow of|that he has silenced all objectors to his math (ignoring those he had|not answered, and also ignoring that the answers to those he did|answer were not really answers). In it, of course, was the complete|exposition of his rambling theory.so almost all of the regular responders refrained from responding forsome time, and then found themselves unable to refrain any longerbecause they did not have the self-control to ignorekindergarten-level taunting. it would be hilarious to compare theactual length of time that they heroically managed to refrain fromresponding to the period of a couple of months that you seem to beclaiming.-- === > so almost all of the regular responders refrained from responding for> some time, and then found themselves unable to refrain any longer> because they did not have the self-control to ignore> kindergarten-level taunting. it would be hilarious to compare the> actual length of time that they heroically managed to refrain from> responding to the period of a couple of months that you seem to be> claiming.He seems to be claiming a couple of months only because in one ofyour posts you said a couple of months. Dik didn't ever explicitlymonths and it was to this that Dik first responded.In any case, I won't defend or criticize those who respond to JamesHarris and in recent months, I've responded directly to him more oftenthan previously. Certainly, I've poked fun at him passively foryears, by using JSH quotations in my .sigs. But I know that theendless JSH activity is a great annoyance to many people that comehere for loftier reading (Penthouse Letters would be an example ofloftier reading).In any case, I don't find your ability to put words in Dik's mouth andto beat up strawmen particularly edifying either.-- Jesse F. HughesThat's what's annoying about Usenet as some loser will state a case,get their ass kicked, but STILL keep coming back as if nothinghappened. -- James Harris explains his strategy. === >[...] But I know that the>endless JSH activity is a great annoyance to many people that come>here for loftier reading (Penthouse Letters would be an example of>loftier reading).He-heh. Do they still have those? I never thought I'd write apost to sci.math...************************David C. Ullrich <87u1477nex.fsf@phiwumbda.org> === >[...] But I know that the>endless JSH activity is a great annoyance to many people that come>here for loftier reading (Penthouse Letters would be an example of>loftier reading). He-heh. Do they still have those? I never thought I'd write a> post to sci.math...How would I know?Since moving to Holland (Pornography Capital of the Free World,according to the promotional material I received), I've unfortunatelymust learn Dutch.-- Sale or rental of this disc is ILLEGAL. If you have rented orpurchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China === ... > |everyone ignored James. There are always people that will reply, and > |it would now be worse than at that time (a larger public you know). > |But if I remember right, almost all of the regular responders > |refrained from responding for some time. It did not stop the flow of > |that he has silenced all objectors to his math (ignoring those he had > |not answered, and also ignoring that the answers to those he did > |answer were not really answers). In it, of course, was the complete > |exposition of his rambling theory. > so almost all of the regular responders refrained from responding for > some time, and then found themselves unable to refrain any longer > because they did not have the self-control to ignore > kindergarten-level taunting. it would be hilarious to compare the > actual length of time that they heroically managed to refrain from > responding to the period of a couple of months that you seem to be > claiming.This is a gross misrepresentation. I have refrained from posting inhis first entry to the newsgroup. I think I never did attack James onin a direct response to an answer by him (but sometimes I piggy-backedon other responders). (And I may have sinned of late.)What is happening is that James posts something new. There are a fewpeople that question his findings. James answers, but actually doesnot answer. Similar questions are posted. James goes into insult mode,never even answering the questions, only providing examples. Someregulars step in clarifying the first questions were not the correctquestions, and showing what the actual questions should be. And Jamescomes in with more insult. Actually finding the gaps in James'reasoning is quite a challenge...-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > [.snip.]> Try talk.origins if you want to see the sparks fly. By comparison,> sci.math is tame. And there are groups far worse that> talk.origins.>Which ones?talk.abortion ?alt.revisionismAlan-- Defendit numerus === >[...]> Also, you seem to know a lot more maths than most people, but judging by the> threads you are no expert. Why don't you get some help, get yourself> thinking straighter, then you might be able to learn a lot more about maths,> and even do some interesting work for people to appreciate?I haven't claimed to be an expert!!!Guffaw. How many times have you explained that the only explanationis that the professional mathematicians' brains are just not wiredright, somehow they're just unable to follow these proofs whichare so simple to you?************************David C. Ullrich <3c65f87.0312091243.1feb11da@posting.google.com> <3c65f87.0312100650.6135dec8@posting.google.com> <55cftv4ff1aqnc5k8k15alnf4eeoihu84c@4ax.com> === >[...]> Also, you seem to know a lot more maths than most people, but judging by the> threads you are no expert. Why don't you get some help, get yourself> thinking straighter, then you might be able to learn a lot more about maths,> and even do some interesting work for people to appreciate?>I haven't claimed to be an expert!!! Guffaw. How many times have you explained that the only explanation> is that the professional mathematicians' brains are just not wired> right, somehow they're just unable to follow these proofs which> are so simple to you?Ignoring right vs. wrong wiring, I think I agree with thatassessment. Somehow, I'm just unable to follow proofs which areself-evident to James (not that I'm a professional mathematician).-- Yup, as far as I'm concerned, if you live out your lives smiling theentire time full of pride in your *believed* accomplishments, when younever had any, well that's ok with me. --James Harris, a man of remarkable accomplishments. === > Ignoring right vs. wrong wiring, I think I agree with that> assessment. Somehow, I'm just unable to follow proofs which are> self-evident to James (not that I'm a professional mathematician).The correct way to handle a proof from a crank: Start at the top, read up to the first sentence that is either wrong or where you don't understand what he means. Takes very little time usually. A technique I like but that doesn't work with cranks: Without bothering whether the proof is correct or incorrect or makes any sense at all, make slight changes to the proof that don't change its correctness (or incorrectness) but make it prove something that is well known to be wrong. Many proofs that a^n + b^n = c^n has no solutions in positive integers a, b, c and n >= 3 can easily be modified to the case n=2 as well, which is a strong indication to anyone but the crank that there is something wrong with the proof. <3c65f87.0312091243.1feb11da@posting.google.com> <3c65f87.0312100650.6135dec8@posting.google.com> <55cftv4ff1aqnc5k8k15alnf4eeoihu84c@4ax.com> <87r7zb7n5w.fsf@phiwumbda.org> === > Ignoring right vs. wrong wiring, I think I agree with that> assessment. Somehow, I'm just unable to follow proofs which are> self-evident to James (not that I'm a professional mathematician). The correct way to handle a proof from a crank: Start at the top, read > up to the first sentence that is either wrong or where you don't > understand what he means. Takes very little time usually. But what effect does this have? It doesn't convince the crank thathis argument is wrong (not that I know of any technique that worksin this regard). The only advantage that this technique has over theone below is: it takes less time to fail to achieve your goal than theother one.> A technique I like but that doesn't work with cranks: Without bothering > whether the proof is correct or incorrect or makes any sense at all, > make slight changes to the proof that don't change its correctness (or > incorrectness) but make it prove something that is well known to be > wrong. Many proofs that a^n + b^n = c^n has no solutions in positive > integers a, b, c and n >= 3 can easily be modified to the case n=2 as > well, which is a strong indication to anyone but the crank that there is > something wrong with the proof.-- Jesse HughesCertainly he who can digest a second or third fluxion neednot, methinks, be squeamish about any point in divinity. George Berkeley, 1734 <87ekvcc1g4.fsf@phiwumbda.org> === |> If he does not get attention he starts |posting a multitude of> be correct, but does not receive the proper attention. when> was this occasion that everyone ignored jsh for a period of a> couple of months?> |> |Long, long ago.> |> when was it? was it after jsh started posting? was it during a> time period that's recorded in the google archive well enough to> determine whether you're telling the truth?> |> |Like Dik, I remember something like the experiment you advocate who the hell said i was advocating something? this entire message of> yours seems to be based on a bizarre hallucination.Fair enough. I assumed you were the person who said earlier in thethread, If everyone would ignore him, he *might* go away after acouple weeks or months. Insulting him only inspires him. I assumedthat, in part, because of the odd tone in your post (how do I knowyou're telling the truth?) but the assumption was wrong.You didn't advocate ignoring JSH as far as I can tell, so my post is aresponse to a claim you've never made.But if you do make it, I'm ready.-- Jesse HughesBut nothing's being Dr. Ullrich is a particular case of something'sbeing such that nothing is it: (Ex)~(Ey)(y = x) -- John Correy on the failings of first order logic === |> when was this occasion that everyone ignored jsh for a|> period of a couple of months?|> ||> |Long, long ago.|>|> when was it? was it after jsh started posting? was it during a|> time period that's recorded in the google archive well enough to|> determine whether you're telling the truth?|> ||> |Like Dik, I remember something like the experiment you advocate|>|> who the hell said i was advocating something? this entire message|> of yours seems to be based on a bizarre hallucination.||Fair enough. I assumed you were the person who said earlier in the|thread, If everyone would ignore him, he *might* go away after a|couple weeks or months. Insulting him only inspires him. I assumed|that, in part, because of the odd tone in your post (how do I know|you're telling the truth?) but the assumption was wrong.ok, so you think it's odd to be skeptical of laughably implausiblebull.-- <87fzfsa1oo.fsf@phiwumbda.org> === > when was this occasion that everyone ignored jsh for a> period of a couple of months?> |> |Long, long ago.> |> when was it? was it after jsh started posting? was it during a> time period that's recorded in the google archive well enough to> determine whether you're telling the truth?> |> |Like Dik, I remember something like the experiment you advocate> |> who the hell said i was advocating something? this entire message> of yours seems to be based on a bizarre hallucination.> |> |Fair enough. I assumed you were the person who said earlier in the> |thread, If everyone would ignore him, he *might* go away after a> |couple weeks or months. Insulting him only inspires him. I assumed> |that, in part, because of the odd tone in your post (how do I know> |you're telling the truth?) but the assumption was wrong. ok, so you think it's odd to be skeptical of laughably implausible> bull.There's a difference between your belligerent tone and its skepticalmotivations. How do I know you're telling the truth? [paraphrased]sounds like more than disinterested skepticism.Anyway, laughably implausible bull or not, my laughablyimplausible (but foggy) memory agrees with Dik's laughably implausiblememory.-- Yup, you guessed it. If worse comes to worse, I *will* turn to theArmy to help me with mathematicians. And then mathematicians don'tthink the NSA or CIA can save your asses, as generals LIKE me. -- James Harris's latest foray into mathematical logic. === snip> James, as one sufferer to another possible sufferer. I think you shouldget> some help for the mental state you seem to be in. You assume too much. I *have* consulted mental health professionals. Why are you accepting online diagnosis from posters or even my own> comments about my fears of mental illness? Shouldn't you be *more* sensitive on the subject? Worse, by accepting their diagnoses, you're simply giving posters who> like to use accusations of mental illness more power. Don't just assume that because you guess that another person is> mentally ill that you have the expertise to make that medical> diagnosis, and especially don't just assume that you know what their> medical history is--like whether or not they've consulted mental> health professionals.> Also, you seem to know a lot more maths than most people, but judging bythe> threads you are no expert. Why don't you get some help, get yourself> thinking straighter, then you might be able to learn a lot more aboutmaths,> and even do some interesting work for people to appreciate? I haven't claimed to be an expert!!! I've found some math that interests me, which I believe *should*> interest mathematicians if their own claims about pure math and> appreciation of beauty in mathematics are to be believed. However, they've responded with insults and personal attacks like> accusations of mental illness. But actual mental health professionals have never diagnosed me with> any mental illness other than a brief bit of depression back when I> was in college. It pains me that I need to share that information because of people> like you Pat, who open the door to spurious attacks by taking them> seriously. Just because I like to post about my math research on several> newsgroups, can get belligerent at times, and at times question my own> mental health, it doesn't make me mentally ill. Mental health professionals are tasked with making such diagnoses and> people who decide to act as psychiatrists or psychologists online to> give their own medical diagnoses, usually as a means to attack another> person, are a danger to others because unfortunately, people *do* take> them seriously.> James HarrisJames, I'm not necessarily saying you are diagnosed mentally ill or wouldreceive such a diagnoses from a psychiatrist or psychiatrists if you were tomake contact with mental health services now or in the future.That diagnosis is based on psychiatric concepts generally based onobservations made in a clinical environment by someone with a license topractice psychiatry. And such concepts, it has been argued, are flawed on anumber of levels. But, that does not discount the observation that somepeople do behave in a way that is seen as 'strange' or sometimes evenunacceptable to other people, and some of the people who behave like thatindeed do have diagnostic labels. All I said was I think you should get some help. You could get help from afriend. All, I've said is I'm diagnosed as mentally ill, and I've mixed witha lot of people closely who are also diagnosed as mentally ill, and I see insome of them to varying degrees behaviour that is similar to your behaviourfrom my impressions of reading your posts. Also, my impression is you aregetting obsessional, and I recognize a bit of that in myself from time totime, and I try to guard against it. So, I do feel some empathy towards youbecause you do seem to have some of the problems that I have and otherpeople I know have.Yes, of course I'm sensitive to these issues, there are a lot ofstereotypes concerning the 'mad.' But, you are not doing me any favours,James, are you? You seemed to be patronizing me with your comments aboutpeople stepping on ants, referring to other posters having no concern forpeople 'like me,' which I feel, could have been based on some empathytowards people diagnosed like I am, but mainly I think you were trying toget me on your side. Do you think all people 'like me' are just morons thatyou can persuade so easily? Is evidence of this patronizing attitude alsolinked to your *assumption* that I can't come to an independent judgementabout your behaviour? and I am so easily persuaded by others here?James, it is common for people to call each other nutter, psycho, etc. whensomeone is acting like your acting, and to a certain extent it fostersstereotypes about the 'mad' in the publics' mind. But, James, if you do haveconcern for people like me which your statement about stepping on ants wasan attempt to convey to me, then you should realize that you are encouragingwords like psycho to be banded about and directed at you by yourunacceptable behaviour. So, if you do have some empathy towards people likeme you should at least make some sort of effort to modify your behaviour andlessen the insults. Do you not agree? === > Also, my impression is you are> getting obsessional, Getting? GETTING?!?!V.-- homepage: cs utk edu tilde lastname === > No, that is wrong. The Indian philosophical thought - Sanatana> dharma, or the way of life beyond the scope of time - is completely> different from the modern and dominant Jewish [...]The Judeo-Christian ideology defines thought in the West today, but is India> the West? Clearly to an Eastern thinker, the ego/moral emphasis of> Christianity, Judaism and Islam is not a viable doctrine.contempt black skin. === contempt black skin.And thats why most important menifestation of god in hinduism likevishnu,shiva,ram,krishna have black skin.Stupid propoganda machine learn thefacts right before spewing venom against hinduism. === contempt black skin.> And thats why most important menifestation of god in hinduism like> vishnu,> shiva,ram,krishna have black skin.Stupid propoganda machine learn the> facts right before spewing venom against hinduism.That may be so. But, one is disadvantage if one's skin color is kala inIndia. Why?Dan === > No, that is wrong. The Indian philosophical thought - Sanatana> dharma, or the way of life beyond the scope of time - is completely> different from the modern and dominant Jewish thinking [...]This frame of mind, of course, serves to lend additional credenceto the otherwise unbelievable notion that the Swastika actuallyoriginated in India. === > Knowledge is wasted upon low-minded and deliberate fools.> Well, 5000 years without flight in the presence of all that> knowledge: clearly it must have been wasted on the whole> subcontinent all those millennia.> Mortals neither designed nor made vimans, though some aspects of their> design were known [...]Whatever. Clearly the knowledge must have been wasted on the wholesubcontinent all those millennia.It's a good thing the Wright Brothers were not mortal, or we'd stillbe taking the chariot to work.> Meanwhile, the first airplanes used by Deccan Airways were made> and purchased from whom?> We are talking about the past, with reference to the future. The present is> just a bad idea.Wow. I've seen people with a touch of nostalgia before. But don't youthink calling the 1940's the future or present is going a little bitoverboard?Meanwhile, I wonder how many temples have been erected to the Godsknown as McDonald and Douglas; the savior of India's 5000 yearflight-impairment whose insights were communicated to the massesby their emissary: Deccan Airways? Oops, did I get that spellingright? We don't want to get the golden arches in the mix here.What about temples devoted to the Gods known as Apollo -- Apollo 17,Apollo 16, Apollo 15, etc., that is? === > What about temples devoted to the Gods known as Apollo -- Apollo 17,> Apollo 16, Apollo 15, etc., that is?Rockets are for fireworks. === theorem. Here is the problem:integral over C (counter-clockwise) of [ (3xy + y^2)dx + (2xy +5x^2)dy ] C: (x - 1)^2 + (y + 2)^2 = 1I got as far as: (double integral) 7x dxdyMy book found the solution by taking the centroid of the circle andthen multiplying by the area to get 7[pi]I was wondering if there was another way to compute the doubleintegral without using centroids and how would I do that. === Its messy, butYou solve the curve to get y1=f(x),y2 =f(x) and the limits of xThen bung them into the double integral integrating over y from y1 to y2then over x.The think about maths problems is finding the neatest solution, thats whatthe author has doone.-- Bruce Harveybruce@bearsoft.co.ukThe Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft> theorem. Here is the problem: integral over C (counter-clockwise) of [ (3xy + y^2)dx + (2xy +> 5x^2)dy ]> C: (x - 1)^2 + (y + 2)^2 = 1 I got as far as: (double integral) 7x dxdy My book found the solution by taking the centroid of the circle and> then multiplying by the area to get 7[pi] I was wondering if there was another way to compute the double> integral without using centroids and how would I do that.> === Its messy, but You solve the curve to get y1=f(x),y2 =f(x) and the limits of x Then bung them into the double integral integrating over y from y1 to y2> then over x. The think about maths problems is finding the neatest solution, thats what> the author has doone. -- Bruce Harvey> bruce@bearsoft.co.uk> The Alternative Physics Site> http://users.powernet.co.uk/bearsoft> theorem. Here is the problem: integral over C (counter-clockwise) of [ (3xy + y^2)dx + (2xy +> 5x^2)dy ]> C: (x - 1)^2 + (y + 2)^2 = 1 I got as far as: (double integral) 7x dxdy My book found the solution by taking the centroid of the circle and> then multiplying by the area to get 7[pi] I was wondering if there was another way to compute the double> integral without using centroids and how would I do that.>X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hB9KwL510666; === > Now what if some person says they found a brilliant gem, should they> go through college courses, and learn a lot of techniques in the> analysis of gems?> Or can't they just holler out that they found something?> Bad analogy.> Someone says they've found a gem. Lots of people look at it. Everyone > who knows about gems says, That's just a lump of coal.> Would that mean that it really was just a lump of coal, or does it mean > that the discoverer can carry on shouting that he's being ignored?Yeah, but here I can show the find:dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,>sqrt(y-1))],S(x,1) = 0, p(x, y) = floor(x) - S(x, y) - 1, and S(x,y) is the sum of dS from dS(x,2) to dS(x,y).Reference: http:// mathforprofit.blogspot.com/25, which is the number of primes up to 100 or p(10,3) = 4, and those>primes are 2, 3, 5 and 7.So it's more like someone finding a gem, and having experts claim it's>glass, only to then have that person cut glass with it.I want critical thinkers to search on prime counting function and>see the methods that mathematicians have on record for counting prime>numbers to compare with what I just gave and see for yourselves how>obvious it is that what I have isn't just junk.Remember, mathematicians are fighting to totally dismiss my result as>unimportant to justify not putting it in math references.>James HarrisMy math discoveries, found for profit>http:// mathforprofit.blogspot.com/That's just a lump of coal.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBADQQO17430; === >prove if x is an irrational number then sqroot of x is irrational??Any rational number has a finite number of digits, so when multiplied by itself it would have at most twice a many while any irrational number requires infinite digits. === >prove if x is an irrational number then sqroot of x is irrational??>Any rational number has a finite number of digitsYou mean, like 1/3 = 0.3333333333333333...?Doug === >prove if x is an irrational number then sqroot of x is irrational??Any rational number has a finite number of digits, so when> multiplied by itself it would have at most twice a many while any> irrational number requires infinite digits.What do you mean any rational number has a finite number of digits?For example, in base 10, the number 1/3 has infinitely many digits.Or if you mean finitely many different digits (so that 1/3 has onedigit), then pi has only digits.ThomasX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBAEB6g20736; === Does anyone know a little about the history of this problem (that{n*x} is dense in [0,1) if x is irrational)? I've seen in some placesthat it was first considered by Chebyshev. Was he the first to proveit? And how did Kronecker get involved in approximation theory withhis beliefs about irrational numbers? Any information would beappreciated.MarcusX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBAHQtO03689; === Sorry to take this discussion further afield than it already is, butI just wanted to comment on this point. I disagree with Herman thattheory must always come before computation. Many times computationprovides an understanding of what motivates the theory. It's alot prettier to present the theory first as if it was discoveredfully formed. But that often obscures the computations and historicaldevelopments that motivated it. Pedagogically that is not helpful forthe student.As for the claim that all mathematics courses should be rigorous,well that depends on what you mean by rigorous. People usemathematics for very different purposes. A course should be taughtappropriate to the needs of the students. For some students,proficiency in computation is far more important than absolute rigorof every single step - assuming that is even possible.No doubt young students should be exposed to more of the facets ofmathematics than just computation. But a good early education inmathematics should introduce them to all of it's aspects; it'shistory, computational techniques, foundations, and even philosophy.Marcus>There are rigorous books, starting with high school Euclid,>Landau's _Foundations of Analysis_ (number systems), texts>on logic, set theory, abstract algebra, and real analysis,>all of which are readable. Those who can learn these>subjects should learn them before computation without any>idea of what it means, and should. I would put off real>analysis until proofs are understood; otherwise, both have>to be learned at the same time. But proofs HAVE BEEN taught>to fifth graders, and I believe it can be done earlier.One does not always have to be complete, but all mathematics>courses need to be rigorous, with no exceptions.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBANttD02210; === ...what is a mouse set?X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBB4DuM21249; === hi could you please help me with these problems please.the answers.>1. y = -x y = 2x2. 3x+y=6 y+2=x3. x-y=6 2x-4y+284. 4x=3y+44 x+y=-3THANKS..> === > hi could you please help me with these problems please.the answers.> 1. y = -x> y = 2x 2. 3x+y=6> y+2=x 3. x-y=6> 2x-4y+28 4. 4x=3y+44> x+y=-3> THANKS..> hi could you please help me with these problems please.the answers.>For all the problems do this.Solve for y in both equations. This has already been done in #1.Set the two solutions for y equal.Solve the resulting equation for x.Use that value for x and one of the solutions for y, to get the value of y.> 1. y = -x> y = 2x>For example, setting the two solutions = gives -x = 2xWhich I leave for you to solve.> 2. 3x+y=6Thus y = ??1> y+2=xand y = ??2setting the two equal, solve??1 = ??2 3. x-y=6> 2x-4y+28 4. 4x=3y+44> x+y=-3> THANKS.. === > What can you add in this context for Jove, Saturn, Uranus and Neptune? The perihelion advance is too small, and the orbits toopoorly determined, for any verification.> I would be more interested in cases where GR predicts a perihelionadvance, but there is none observed. Are there any such cases to thebest of your knowledge Dr. Van Flandern? Yes, a few binary star systems have observed pericentermotions different from GR predictions. These are mainly cases of twocontributing-but-unequal masses. However, other effects such as tidalforces or mass transfer have not yet been ruled out. See, e.g.,http://www.gsanctuary.com/general_relativity.html. -|Tom|-Tom Van Flandern - Washington, DC - see our web site on replacementastronomy research at http://metaresearch.org === What can you add in this context for Jove, Saturn, Uranus and Neptune? The perihelion advance is too small, and the orbits too> poorly determined, for any verification.Dear Dr. Tom Van FlandernWhy so strongly have varied the legitimated quantities of planetary masses between 1980 and 1990? Or it can be, that the modern legitimated quantities of planetary masses too poorly determined, for any modern verification... ?What can you tell about changes in the measuring techniques of quantities of planetary masses between 1980 and 1990? Sincerely yours, Aleksandr I would be more interested in cases where GR predicts a perihelion> advance, but there is none observed. Are there any such cases to the> best of your knowledge Dr. Van Flandern? Yes, a few binary star systems have observed pericenter> motions different from GR predictions. These are mainly cases of two> contributing-but-unequal masses. However, other effects such as tidal> forces or mass transfer have not yet been ruled out. See, e.g.,> http://www.gsanctuary.com/general_relativity.html. -|Tom|-> Tom Van Flandern - Washington, DC - see our web site on replacement> astronomy research at http://metaresearch.org === What can you add in this context for Jove, Saturn, Uranus and Neptune? The perihelion advance is too small, and the orbits too> poorly determined, for any verification.I would be more interested in cases where GR predicts a perihelion> advance, but there is none observed. Are there any such cases to the> best of your knowledge Dr. Van Flandern? Yes, a few binary star systems have observed pericenter> motions different from GR predictions. These are mainly cases of two> contributing-but-unequal masses. However, other effects such as tidal> forces or mass transfer have not yet been ruled out. See, e.g.,> http://www.gsanctuary.com/general_relativity.html. -|Tom|-> Dr. Van Flandern:To what extent you think the website you quote (http://www.gsanctuary.com/general_relativity.html) can be a reliablesource of information, when it is asserted there that the universe isonly 6,000 years old and the speed of light is dependent on theposition of its source and path?I may as well read the Bible then and forget about science...Tom Van Flandern - Washington, DC - see our web site on replacement> astronomy research at http://metaresearch.org === > Define:f(0)=0> f(x+1)=(f(x)+1)/2> f(1/x)=1-f(x)I think this function is defined for every integer n and for every> number of the form 1/n, n<>0, n integer. It's increasing for x>=0, but> I don't see why it's defined for every rational >=0.> Anyway, this is not a counter example, you didn't show a monotonic> function such that the set of points where it's not differentiable is> not null.> AmandaOnce you define it for every number of the form 1/n, then you candefine it for every number 1+1/n, 2+i/n, ... 1 / (1 + 1/n), ect. Factis, he has defined f for all rationals. === >It's well known that if a real valued function f is monotonic on a>compact interval I then f is integrable over I and the set of>discontinuities of f on I is countable. I'm not sure, but it seems to>me that if we combine this fact with the Fundamental Theorem of>Integral Calculus (in the form that deals with the derivative of an>integral)and consider Lebesgue Integrability Criterion, then we come>to the conclusion that the set of elements of I at which f is not>differentiable has measure zero.Is this conclusion true?The _conclusion_ is indeed true - a monotonic function is > differentiable except on a set of measure zero. But I don't> follow how you think you've proved this at all. (if you can> give a proof that's that simple you can publish it. But you're> going to need to explain the argument a little more carefully> first...)IIRC, there's a proof very early on in Riesz-Nagy. Since it IS veryearly on, it can't use much measure theory. I suppose it is possibleto prove a theorem a.e. without having measure theory--you justdirectly define a set of measure zero--but I don't have the book infront of me and I don't know how they handle this.They also have one of the standard examples of an everywherecontinuous, nowhere differentiable function.--Ron Bruck === >It's well known that if a real valued function f is monotonic on a>compact interval I then f is integrable over I and the set of>discontinuities of f on I is countable. I'm not sure, but it seems to>me that if we combine this fact with the Fundamental Theorem of>Integral Calculus (in the form that deals with the derivative of an>integral)and consider Lebesgue Integrability Criterion, then we come>to the conclusion that the set of elements of I at which f is not>differentiable has measure zero.Is this conclusion true?The _conclusion_ is indeed true - a monotonic function is > differentiable except on a set of measure zero. But I don't> follow how you think you've proved this at all. (if you can> give a proof that's that simple you can publish it. But you're> going to need to explain the argument a little more carefully> first...)Well, when I looked more carefully into what I thought might be aconclusions I cited, the conclusion does not follow....(at least, Ican't show it follows). Anyway, I had a good guess...Well, that happened to Fermat, too, right? He thought he had provedthat famous theorem and when he got carefully into the details, he sawthere was a mistake. nothing...And, still, FLT is named after him, whonever proved that theorem...Amanda === >It's well known that if a real valued function f is monotonic on a>compact interval I then f is integrable over I and the set of>discontinuities of f on I is countable. I'm not sure, but it seems to>me that if we combine this fact with the Fundamental Theorem of>Integral Calculus (in the form that deals with the derivative of an>integral)and consider Lebesgue Integrability Criterion, then we come>to the conclusion that the set of elements of I at which f is not>differentiable has measure zero.Is this conclusion true?> The _conclusion_ is indeed true - a monotonic function is > differentiable except on a set of measure zero. But I don't> follow how you think you've proved this at all. (if you can> give a proof that's that simple you can publish it. But you're> going to need to explain the argument a little more carefully> first...)Well, when I looked more carefully into what I thought might be a>conclusions I cited, the conclusion does not follow....(at least, I>can't show it follows). Anyway, I had a good guess...>Well, that happened to Fermat, too, right? He thought he had proved>that famous theorem and when he got carefully into the details, he saw>there was a mistake. nothing...And, still, FLT is named after him, who>never proved that theorem...Yup. Too bad this theorem is already named for someone else.(Just teasing...)>Amanda************************David C. Ullrich === im trying to find all groups of order eight. i already know what thesegroups are, but i am trying to derive it, mainly from the sylowtheorems. will this approach work? for example, I was able to workwith groups of order 21=3*7, because I was able to find a semidirectproduct representation of the group, but 8=2^3 doesnt really lead meanywhere. is there another method?thanks === > im trying to find all groups of order eight. i already know what these> groups are, but i am trying to derive it, mainly from the sylow> theorems. will this approach work? Trouble is, as you say later, is that 2 is a prime powerso Sylow tells you nowt.> for example, I was able to work> with groups of order 21=3*7, because I was able to find a semidirect> product representation of the group, but 8=2^3 doesnt really lead me> anywhere. is there another method?Two possible attacks(i) the exponent of the group: this is the smallest n such that a^n = 1for all a in the group. For a group of order 8 it is 2, 4 or 8.Divide into these three cases.(ii) Use the centre. A standard theorem states that the centre Zof a p-group G is nontrivial. Split into cases accordingto the structure of the centre.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === |>Lastly, is there any converse of the It can be approximated too well|>so it must be transcendental? Can all transcedentals be approximated|>too well or are there known examples of transcendentals which cannot|>be approximated any better than an algebraic?||No. Consider the numbers whose classical continued |fraction has only 1's and 2's. These cannot be|better approximated than sqrt(2), and yet there must|be transcendentals, as there are continuum many.For each irrational r, let a(r) be the greatest lower boundof the C>0 such that there are only finitely many rationalsp/q satisfying |r-p/q| aleph0.But each infinitely long sequence of 01 notations can be mapped withsome natural number, for example:...000 <--> 1 ...001 <--> 2 ...010 <--> 3 ...011 <--> 4 ...100 <--> 5 ...101 <--> 6 ...110 <--> 7 ...111 <--> 8 ... Therefore we can conclude that 2^aleph0 = aleph0, and we come tocontradiction.(2^aleph0 >= aleph0) = {}, and we have a proof saying that BooleanLogic cannot deal with infinitely many objects.Doron === > Because I don’t know how to write my idea in the common formal> way, I am going to do it in a non-formal way, but I will do my best to> write it in the clearest way (PLEASE READ ALL OF IT, OTHERWISE IT> CAN'T BE UNDERSTOOD).So here it is: Let us check these lists. P(2) = {{},{0},{1},{0,1}} = 2^2 = 4 and also can be represented as: 00 > 01 > 10 > 11 > P(3) = {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8 and also can be represented as: 000 > 001 > 010 > 011 > 100 > 101 > 110 > 111 Let us call any full 01 list, combinations list. Now, let us use Cantor's Diagonalization method on some finitely long> combinations list, for example, the combinations list of number 3:000 > 001 > 010 > 011 > 100 > 101 > 110 > 111 We can change the order of the rows, and then use Cantor's> Diagonalization method, for example:001 > 011 > 010 > 000 > 101 > 100 > 111 > 110 The input for Cantor's Diagonalization method in the first example is> 000 and the output is 111.of course that choice was completely arbitrary. you could have startedwith any string. and any ordering of the strings. The input for Cantor's Diagonalization method in the second example is> 010 and the output is 101.> fine, so you're ignoring the bottom 2^n - n rows then?> In both examples we find that the result is already in the> combinations list, and this combination, which is already in the list,> is one of the combinations that Cantor's Diagonal does not cover.The number of the combinations, which are out of the range of Cantor's> diagonal is:2^n - n > so that's where that number came from. now i understand what it means.incidentally, you'd (still be wrong but) be better off saying domain. > Every column, which belongs to some combinations list is a sequence of> 01 notations, based on some periodic frequency changes, for example:the right column of number 3 combinations list, is based on 2^0(=1). Therefore the periodic frequency changes are 1, and the result in this> case is:> 01010101. > but a completely arbitrary ordering of the rows wouldn't look so nicewould it? i mean 01110100 is possible under some ordering, so what'syour point?> The result of the middle column is based on 2^1(=2), therefore the> sequence is:> 00110011. The result of the left column is based on 2^2(=4), therefore the> sequence is:> 00001111. and we get the full combinations list of number 3: 000 > 001 > 010 > 011 > 100 > 101 > 110 > 111 We can get a combinations list of infinitely many places, by using the> ZF Axiom of infinity induction, on the left side of our combinations> list, by using the induction on the power_value of each column, for> example:2^0, 2^1, 2^2, 2^3, ... > still passing to infinite constructions without demonstrating it's valid.> In this stage we have proven, by induction, that Cantor's diagonal> cannot cover any full 01 combinations list, finite or infinite.> what do you mean cover? at best you'd need transfinite induction wouldn't you to pass from afinite cardinal to a countably infinite cardinal?otherwise:the sum of 1/r from 1 to n is finite. therefore, using your assumption,the infinte sum is finite. prove that's a problem if you don't already know > Therefore its result is not a new combination (that has to be added to> the list).Because Cantor's diagonal cannot cover the full 01 combinations list,> (of aleph0 places for each combination) we can conclude that 2^aleph0> aleph0.But each infinitely long sequence of 01 notations can be mapped with> some natural number, for example:...000 <--> 1 > ...001 <--> 2 > ...010 <--> 3 > ...011 <--> 4 > ...100 <--> 5 > ...101 <--> 6 > ...110 <--> 7 > ...111 <--> 8 > ... > what does the string entirely made of ones get sent to? i can see how thatworks for strings with a finite number of ones in them:the string (x_i) i in N gets sent tosum x_i.2^ibut where do the ones with infinitely many 1's get sent to? > Therefore we can conclude that 2^aleph0 = aleph0, and we come to> contradiction.(2^aleph0 >= aleph0) = {}, and we have a proof saying that Boolean> Logic cannot deal with infinitely many objects.> DoronI think that at best you've proven the set of finite subsets of acountable set is countable. But that's not surprising is it?You've not got the infinite subsets in your enumeration.if you insist on doing this. note that it is not required to talk aboutall those sets and strings. just think about all strings of 0's and 1's,forget that set stuff, it's just confusing and unnecessary. === A simple question about mutual information:mutual information is defined as I(X;Y) = H(X) - H(X|Y)If X and Y aren't simple discrete random variables, but are defined onset of random discrete variables (X in {X1,X2, ....,Xn}, Y in {Y1,Y2,....,Yn}, with Xi and Yi discrete random variables), how can icalculate I(X;Y)? === I am working on the so called twisted Hall algebra of a Dynkin quiver, andI have a problem proving a fundamental relation among the genereatingelements. I am able to solve the problem if the following more generalholds:Let R be a ring and let M be a finite (yes: finite) R-module. Assume thatM is a direct sum of two simple R-modules S_1 and S_2. Then the number ofsubmodules N of M isomorphic to S_1 with M/N isomorphic to S_2 is the sameas the number of submodules N' of M isomorphic to S_2 with M/N' isomorphicto S_1.Is this true? Perhaps it is trivial but I am not able to prove it.-- Michael Knudsen [knudsen@imf.au.dk] === I am working on the so called twisted Hall algebra of a Dynkin quiver, and> I have a problem proving a fundamental relation among the genereating> elements. I am able to solve the problem if the following more general> holds:Let R be a ring and let M be a finite (yes: finite) R-module.Finite = finitely many elements?> Assume that> M is a direct sum of two simple R-modules S_1 and S_2. Then the number of> submodules N of M isomorphic to S_1 with M/N isomorphic to S_2That's redundant by the Jordan-Holder theorem.> is the same> as the number of submodules N' of M isomorphic to S_2 with M/N' isomorphic> to S_1.If S_1 is not isomorphic to S_2, then S_1 is the onlysubmodule of M isomorphic to S_1 (consider the projection onto S_2).Both numbers are one.If S_1 and S_2 are isomorphic, then both numbers must be the same :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === > Finite = finitely many elements?Yes.> That's redundant by the Jordan-Holder theorem.I do not understand that. The Jordan-Holder theorem states that twocomposition series of M are equivalent, right?> If S_1 is not isomorphic to S_2, then S_1 is the only> submodule of M isomorphic to S_1 (consider the projection onto S_2).> Both numbers are one.Oh! Is the following correct?Let N be a submodule of M isomorphic to S_1. Assume that N doesn't equalS_1. Then the projection N --> S_2 is non-zero. Since S_2 is simple theprojection is surjective and since M is the direct sum of S_1 and S_2 theprojection is injective. Thus N is isomorphic to S_2. $qedsymbol$Do you know about Ext groups? I am having trouble finding out theirrelation to finite length modules. I am working over a K-algebra, K afinite field. It know that Ext(S_1,S_2)=K, so there must be anindecomposable module N of length 2 and an exact sequence0 --> S_2 --> N --> S_1 --> 0Taking into account the assumptions Ext(S_1,S_1)=Ext(S_2,S_1)=0, we seethat there are just two isomorphism classes of modules of length 3 with 2composition factors of the form S_1 and one of the form S_2, namelyX=N+S_1 and Y=2S_1+S_2.I have not seen any litterature explaining this stuff in detail but I amsure that I can work something more general out myself, if somebody canhelp me with the example above. I cannot see a clear connection betweenExt groups and modules of finite length.-- Michael Knudsen [knudsen@imf.au.dk]-- Michael Knudsen === > Finite = finitely many elements?Yes.> That's redundant by the Jordan-Holder theorem.I do not understand that. The Jordan-Holder theorem states that two> composition series of M are equivalent, right?Yes, M has a composition series 0 <= S_1 <= Mwith factor groups S_1 and M/S_1 = S_2. So any composition serieshas S_1 on the bottom and S_2 up top or vice versa.> If S_1 is not isomorphic to S_2, then S_1 is the only> submodule of M isomorphic to S_1 (consider the projection onto S_2).> Both numbers are one.Oh! Is the following correct?Let N be a submodule of M isomorphic to S_1. Assume that N doesn't equal> S_1. Then the projection N --> S_2 is non-zero. Since S_2 is simple the> projection is surjectiveAnd so S_1 has a quotient isomorphic to S_2; only possibleif S_1 and S_2 are isomorphic (S_1 is simple).Do you know about Ext groups? I am having trouble finding out their> relation to finite length modules. I am working over a K-algebra, K a> finite field. It know that Ext(S_1,S_2)=K, so there must be an> indecomposable module N of length 2 and an exact sequence0 --> S_2 --> N --> S_1 --> 0right?OK ... there are |K|-1 non-splitting isomorphism classes of suchshortexactsequences.> Taking into account the assumptions Ext(S_1,S_1)=Ext(S_2,S_1)=0, we see> that there are just two isomorphism classes of modules of length 3 with 2> composition factors of the form S_1 and one of the form S_2, namely> X=N+S_1 and Y=2S_1+S_2.If we have composition series 0 <= A <= B <= M exactlyone quotient is S_2. If A is isomorphic to S_1 then B is adirect sum of S_1 and a simple. So B is S_1 (+) S_1 orS_1 (+) S_2. In the former case ext(S_2,B)=0 soM is S_1 (+) S_1 (+) S_2. In the fomer case ext(S_1,B)= ext(S_1,S_2) so the S_1 in A splits from M and M is S_1 (+)an extension of S_2 by S_1.If A is S_2, then M/A is S_1 (+) S_1. Now ext(S_1(+)S_1,S_2)= K^2 so there are nontrivial extensions. Now S_1(+)S_1has to have lots of submodules isomorphic to S_1. If each elementof this ext comes from ext(C,S_2) for some C which is a directsummand of S_1(+)S_1 we're in business (a complementarydirect summand will split off).... I'm sure that's true,but I don't have the time/inclination to work that out in deatil.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === > In the fomer case ext(S_1,B)> = ext(S_1,S_2) so the S_1 in A splits from M and M is S_1 (+)> an extension of S_2 by S_1.Hmmmm...shouldn't it be: M is S_1 (+) an extension of S_1 by S_2? This iswhat we would like to prove. Anyway, how do you get that? We know, as youassumption. How do you use that to get the desired decomposition?-- Michael Knudsen [knudsen@imf.au.dk] === > In the fomer case ext(S_1,B)> = ext(S_1,S_2) so the S_1 in A splits from M and M is S_1 (+)> an extension of S_2 by S_1.Hmmmm...shouldn't it be: M is S_1 (+) an extension of S_1 by S_2? This is> what we would like to prove. Anyway, how do you get that? We know, as you> But the last group is K by> assumption. How do you use that to get the desired decomposition?The point is that the characteristic class of the ses in Ext(S_1,B) comesfrom one in Ext(S_1,S_2). This comes from an ses looking like0 -> S_1 (+) S_2 -> S_1 (+) N -> S_1 -> 0where0 -> S_2 -> N -> S_1 -> 0is an ses.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === > If we have composition series 0 <= A <= B <= M exactly> one quotient is S_2. If A is isomorphic to S_1 then B is a> direct sum of S_1 and a simple.Why is that true? Do you say that the canonical sequence0 --> A --> B --> B/A --> 0splits? Is that always true?> So B is S_1 (+) S_1 or> S_1 (+) S_2. In the former case ext(S_2,B)=0 so> M is S_1 (+) S_1 (+) S_2.To conclude M = S_1(+)S_1(+)S_2 from Ext(S_2,B)=0 you need an exactsequence 0 --> B --> M --> S_2 --> 0, right? Do you know that sucha sequence exists?-- Michael Knudsen [knudsen@imf.au.dk] === > If we have composition series 0 <= A <= B <= M exactly> one quotient is S_2. If A is isomorphic to S_1 then B is a> direct sum of S_1 and a simple.Why is that true?'cos Ext(S_1,S_1) = 0 and Ext(S_2, S_1) = 0. No matterwhat B/A is, then Ext(B/A,A) = 0.> splits? Is that always true?> So B is S_1 (+) S_1 or> S_1 (+) S_2. In the former case ext(S_2,B)=0 so> M is S_1 (+) S_1 (+) S_2.To conclude M = S_1(+)S_1(+)S_2 from Ext(S_2,B)=0 you need an exact> sequence 0 --> B --> M --> S_2 --> 0, right? Do you know that such> a sequence exists?Yes. (By J-H if B = S_1(+)S_1, M/B is S_2).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === hi im a high school student and i was browsing some good ol' mathsites geared for prepping for the Putnam competition and there wasthis one question that said:evaluate the following integral:INT[1/(a^2+(x+1/x)^2),x,0,positive infinity]....(the integral is takenover the positve reals) for a>= 2I finally was able to evaluate the integral using non-complex analysismethods but out of plain mathematical curiosity i started to wonder ifthis integral could be evaluated by summing complex residues (residuetheorem). Im pretty knew to complex analysis since I have beenteaching myself it (pretty cool field of math) so i would be very muchobliged if someone could help me out by suggesting how to get started.what kind of contour should i choose of contour integration? etcthanks~sam r. === > hi im a high school student and i was browsing some good ol' math> sites geared for prepping for the Putnam competition and there was> this one question that said:> evaluate the following integral:> INT[1/(a^2+(x+1/x)^2),x,0,positive infinity]....(the integral is taken> over the positve reals) for a>= 2I finally was able to evaluate the integral using non-complex analysis> methods but out of plain mathematical curiosity i started to wonder if> this integral could be evaluated by summing complex residues (residue> theorem). Im pretty knew to complex analysis since I have been> teaching myself it (pretty cool field of math) so i would be very much> obliged if someone could help me out by suggesting how to get started.> what kind of contour should i choose of contour integration? etcYou haveINT[1/(a^2 + (x + 1/x)^2),x,0,+oo] = (1/2).INT[1/(a^2 + (x + 1/x)^2),x,-oo,+oo] = (1/2).INT[x^2/(a^2.x^2 + (x^2 + 1)^2),x,-oo,+oo] = (1/2).INT[x^2/(x^4 + (a^2 + 2)x + 1),x,-oo,+oo]Now, choose a big positive real number R and consider the path that goesfrom -R to R along a straight line and then goes along a semicircle, throughthe complex numbers with positive imaginary part, from R into -R.I hope that this helps,Jose Carlos Santos <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de> <1Ff9oJOLyx1$Ew6T@baesystems.com> === In message <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>, Bjoern[snip]> sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique.That's incomprehensible. Could you rephrase this, please?He means that any Turing machine can be represented by a tape fed to a> universal Turing machine, and the contents of that tape can be> represented by some number x.How can the content of the whole tape be represented with a singlenumber x?And where does he get the probably under a million from?[snip]> Non computable numbers is not a proof that irrationals exist, IMO,> that no halting function exists does not clearly define what the halting> number is, it states it is impossible. There is no gap on the number> line from non computable numbers.That's incomprehensible again.I think he's trying to argue that because there exist TMs for which we> can't determine to which of the sets halts or does not halt they> belong, the does not halt set must be empty.*scratches head* Does anyone understand this logic?Bye,Bjoern === > In message <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>, Bjoern[snip]> sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique.>That's incomprehensible. Could you rephrase this, please?> He means that any Turing machine can be represented by a tape fed to a> universal Turing machine, and the contents of that tape can be> represented by some number x.How can the content of the whole tape be represented with a single> number x?Because all but finitely many characters on the tape will be blank.That means that there are only countably many possible configurationsof the input tape. So we can associate each one with a unique integerx.That there are only finitely many non-blank characters on the inputtape is normally just a conventional assumption. So this argumentis a bit weak for the general case.In this case, the input tape encodes a Turing machine, and allTuring machines are finitely describeable. Accordingly the set ofTuring machines is countable and it follows that the set of canonicalencodings on tape is countable. Therefore each canonical encoding canbe associated with a unique integer x. John Briggs === > In message <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>, Bjoern [snip]> sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique.>That's incomprehensible. Could you rephrase this, please?> He means that any Turing machine can be represented by a tape fed to a> universal Turing machine, and the contents of that tape can be> represented by some number x. How can the content of the whole tape be represented with a single> number x?Because all but finitely many characters on the tape will be blank.> That means that there are only countably many possible configurations> of the input tape. So we can associate each one with a unique integer> x. That there are only finitely many non-blank characters on the input> tape is normally just a conventional assumption. So this argument> is a bit weak for the general case.In this case, the input tape encodes a Turing machine, and all> Turing machines are finitely describeable. Accordingly the set of> Turing machines is countable and it follows that the set of canonical> encodings on tape is countable. Therefore each canonical encoding can> be associated with a unique integer x.Bye,Bjoern === --www.StealthHostiing.com You rule Truman. http://tinyurl.com/iky4 Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! >Is the truman living in Townsville? I've been hearing stuff, yeah.Webmasters help the TRUEman by joining www.theBanner.net Current:1 Goal:1000----------------------------------------------------- ----------------------------- > In message <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>, Bjoern> [snip] sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique.>That's incomprehensible. Could you rephrase this, please?> He means that any Turing machine can be represented by a tape fed to a> universal Turing machine, and the contents of that tape can be> represented by some number x.> How can the content of the whole tape be represented with a single> number x? Because all but finitely many characters on the tape will be blank.> That means that there are only countably many possible configurations> of the input tape. So we can associate each one with a unique integer> x. That there are only finitely many non-blank characters on the input> tape is normally just a conventional assumption. So this argument> is a bit weak for the general case. In this case, the input tape encodes a Turing machine, and all> Turing machines are finitely describeable. Accordingly the set of> Turing machines is countable and it follows that the set of canonical> encodings on tape is countable. Therefore each canonical encoding can> be associated with a unique integer x.Any encoding technique that represents all TMs will suffice. Here youcan see a TM in action counting in binary, only 3 states :1 0 1 r 21 1 1 l 3.2 0 0 l 12 1 0 r 2.3 0 1 l 13 1 1 l 99.To represent this as an integer is not difficult, each transition link from state to statehas a limited number of options which can be encoded. The integer does nothave to be parsed directly on a standard TM (UTM), as long as there is *some* definedmapping from integers to the associated TM's output number.There are 2 * 2 * 4 possible links on a 3 state TM, that makes 6^16 possible TMs,just count through them, although for theory a UTM applied to a tape is elementary.Herc === What's the obsession with computers? A number being irrational has aperfectly clear definition and had one long before anyone had the ideathat machines could do any amount of logic.~ Chris === > What's the obsession with computers? A number being irrational has a> perfectly clear definition and had one long before anyone had the idea> that machines could do any amount of logic.>That definition is that they cannot be represented as integer over integer.The definition is *not* uncountable.Look at a typical irrational number, it CAN be counted to.Here you can see complicated numbers being output from simple Turing machines.I have calculated the index *number* of these machines, I would hazard a guessthat sqrt(2) would be encapsulated on a small TM, whose index number could befound. Aren't all the equations you write algorithmic?It seems odd Information Theory is wiped out of existence from a single 'proof'that is self referential, can anyone give me an actual number that isn't computable?Herc === Why pick sqrt(2)? By your logic, 1/3 is irrational because no Turing machinecan generate its decimal representation in finite time.And don't say decimal representation is the problem. There is nothingintrinsically natural about base 10, or even about base n (positionalnotation). Lots of other representations of numbers exist. If I am free topick the representation of the number, I can generate sqrt(2) in finitetime. I will use as my number representation one which uses the sequence ofnumbers in the polynomial which produces the number as its root. So 1/3 isthe root of 3x - 1 =0, its representation is (3,1). You can easily changethis representation to be a single number by mapping any arbitrary finitesequence of natural numbers to a single natural number (you might alreadyknow how to do this using prime multipliers). Similarly, sqrt(2) is the rootof x^2 - 2 = 0; its representation is (1,0-2). I can get a Turing machine togenerate the single number corresponding to this sequence. So in thisrepresentation, 1/3, sqrt(2), and all other roots of polynomials arecomputable by Turing machines. And this number representation is just asvalid as base 10 positional notation. With a little more work, I can make piand a whole lot of other irrational numbers (but not all) computable infinite time by a Turing machine, by picking an appropriate representationfor the number and asking the TM to generate it.0.3333... may be imposible to write out in full, but 1/3 is computable.1.414.... may be impossible to write out in full, but sqrt(2) is computable.You are talking about limitations of arbitrary representations of numbers,not the numbers themselves.Peter Webb What is an irrational number? Can you count to it?> Can you pin point it? There is no such thing. All numbers are the result of computable functions, sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique. That only leaves two types of numbers left that qualify for irrational,> non computable and random numbers. Non computable numbers is not a proof that irrationals exist, IMO,> that no halting function exists does not clearly define what the halting> number is, it states it is impossible. There is no gap on the number line> from non computable numbers. That leaves all numbers on the number line that cannot be encapsulatedwithin> a computing indexing system are just..... random. Irrational numbers are just random sequences, otherwise a computer willmap to them. And what exactly is an infinite sequence of random numbers? nothing, they> don't exist except for rationalisations about their finite limits. In alleventuality> with a small probability, on any representation system, each digit willflip to> a different digit, Shannon's noise demon at work. As an infinite sequenceany> random number is no longer defined. Cantor's proof is cyclic, given a finite space to represent allpossibilities of infinite sequences> there are more sequences than combinations, but they are only randomanyway. Herc> --> www.StealthHostiing.com You rule Truman.http://tinyurl.com/iky4> Hey Trueman...love the show. YOU ARE the Truman I heard him. Veryspooky!>Is the truman living in Townsville? I've been hearing stuff,yeah.> Webmasters help the TRUEman by joining www.theBanner.net Current:1Goal:1000> -------------------------------------------------------------- --------------------> === > Why pick sqrt(2)? By your logic, 1/3 is irrational because no Turing machine> can generate its decimal representation in finite time. And don't say decimal representation is the problem. There is nothing> intrinsically natural about base 10, or even about base n (positional> notation). Lots of other representations of numbers exist. If I am free to> pick the representation of the number, I can generate sqrt(2) in finite> time. I will use as my number representation one which uses the sequence of> numbers in the polynomial which produces the number as its root. So 1/3 is> the root of 3x - 1 =0, its representation is (3,1). You can easily change> this representation to be a single number by mapping any arbitrary finite> sequence of natural numbers to a single natural number (you might already> know how to do this using prime multipliers). Similarly, sqrt(2) is the root> of x^2 - 2 = 0; its representation is (1,0-2). I can get a Turing machine to> generate the single number corresponding to this sequence. So in this> representation, 1/3, sqrt(2), and all other roots of polynomials are> computable by Turing machines. And this number representation is just as> valid as base 10 positional notation. With a little more work, I can make pi> and a whole lot of other irrational numbers (but not all) computable in> finite time by a Turing machine, by picking an appropriate representation> for the number and asking the TM to generate it. 0.3333... may be imposible to write out in full, but 1/3 is computable.> 1.414.... may be impossible to write out in full, but sqrt(2) is computable.> You are talking about limitations of arbitrary representations of numbers,> not the numbers themselves. Peter WebbThis is the basis of the universal count to all numbers. 1/3 and sqrt(2) can appearon the same list of counted numbers, so what types of numbers are we missing?It doesn't matter if 0.3333. goes on indefinately or 1.414... When we countrationals we just index the *formula* to *make* that number. We encode thenumerator and denominator, not 0.3333333..Counting over the UTM parameter is no different to counting the numerator / denominator grid.Is it impossible to count all square roots, because most of them take up infinite space?No, sqrt(1), sqrt(2), sqrt(3), sqrt(4), sqrt(5)... will suffice. The square roots of integersare countable, only this list will *not* count out all rationals aswell. But UTM(Z) will countout *both* lists and more.There are no numbers themselves they are abstract concepts, we can only ever dealwith representations.Herc> What is an irrational number? Can you count to it?> Can you pin point it? There is no such thing. All numbers are the result of computable functions, sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique. That only leaves two types of numbers left that qualify for irrational,> non computable and random numbers. Non computable numbers is not a proof that irrationals exist, IMO,> that no halting function exists does not clearly define what the halting> number is, it states it is impossible. There is no gap on the number line> from non computable numbers. That leaves all numbers on the number line that cannot be encapsulated> within> a computing indexing system are just..... random. Irrational numbers are just random sequences, otherwise a computer will> map to them. And what exactly is an infinite sequence of random numbers? nothing, they> don't exist except for rationalisations about their finite limits. In all> eventuality> with a small probability, on any representation system, each digit will> flip to> a different digit, Shannon's noise demon at work. As an infinite sequence> any> random number is no longer defined. Cantor's proof is cyclic, given a finite space to represent all> possibilities of infinite sequences> there are more sequences than combinations, but they are only random> anyway. Herc === Why pick sqrt(2)? By your logic, 1/3 is irrational because no Turing machine> can generate its decimal representation in finite time.> And don't say decimal representation is the problem. There is nothing> intrinsically natural about base 10, or even about base n (positional> notation). Lots of other representations of numbers exist. If I am free to> pick the representation of the number, I can generate sqrt(2) in finite> time. I will use as my number representation one which uses the sequence of> numbers in the polynomial which produces the number as its root. So 1/3 is> the root of 3x - 1 =0, its representation is (3,1). You can easily change> this representation to be a single number by mapping any arbitrary finite> sequence of natural numbers to a single natural number (you might already> know how to do this using prime multipliers). Similarly, sqrt(2) is the root> of x^2 - 2 = 0; its representation is (1,0-2). I can get a Turing machine to> generate the single number corresponding to this sequence. So in this> representation, 1/3, sqrt(2), and all other roots of polynomials are> computable by Turing machines. And this number representation is just as> valid as base 10 positional notation. With a little more work, I can make pi> and a whole lot of other irrational numbers (but not all) computable in> finite time by a Turing machine, by picking an appropriate representation> for the number and asking the TM to generate it.> 0.3333... may be imposible to write out in full, but 1/3 is computable.> 1.414.... may be impossible to write out in full, but sqrt(2) is computable.> You are talking about limitations of arbitrary representations of numbers,> not the numbers themselves.> Peter WebbThis is the basis of the universal count to all numbers. 1/3 and sqrt(2) can appear> on the same list of counted numbers, so what types of numbers are we missing?It doesn't matter if 0.3333. goes on indefinately or 1.414... When we count> rationals we just index the *formula* to *make* that number. We encode the> numerator and denominator, not 0.3333333..Counting over the UTM parameter is no different to counting the numerator / denominator grid.It is different in a very important way. Not all UTM parameters encodenumbers. Some Turing machines both fail to halt and fail to produce thei'th digit of their output. And it is provable that there is nocomputable procedure to tell with certainty which do and which do not.That means that if you try to make a list of numbers in UTM parameterorder, that list will not be computable.By contrast, a list of all numerator/denominator pairs _is_ computable.Even if you decide to eliminate redundant pairs. John Briggs === > Why pick sqrt(2)? By your logic, 1/3 is irrational because no Turing machine> can generate its decimal representation in finite time.> And don't say decimal representation is the problem. There is nothing> intrinsically natural about base 10, or even about base n (positional> notation). Lots of other representations of numbers exist. If I am free to> pick the representation of the number, I can generate sqrt(2) in finite> time. I will use as my number representation one which uses the sequence of> numbers in the polynomial which produces the number as its root. So 1/3 is> the root of 3x - 1 =0, its representation is (3,1). You can easily change> this representation to be a single number by mapping any arbitrary finite> sequence of natural numbers to a single natural number (you might already> know how to do this using prime multipliers). Similarly, sqrt(2) is the root> of x^2 - 2 = 0; its representation is (1,0-2). I can get a Turing machine to> generate the single number corresponding to this sequence. So in this> representation, 1/3, sqrt(2), and all other roots of polynomials are> computable by Turing machines. And this number representation is just as> valid as base 10 positional notation. With a little more work, I can make pi> and a whole lot of other irrational numbers (but not all) computable in> finite time by a Turing machine, by picking an appropriate representation> for the number and asking the TM to generate it.> 0.3333... may be imposible to write out in full, but 1/3 is computable.> 1.414.... may be impossible to write out in full, but sqrt(2) is computable.> You are talking about limitations of arbitrary representations of numbers,> not the numbers themselves.> Peter Webb This is the basis of the universal count to all numbers. 1/3 and sqrt(2) can appear> on the same list of counted numbers, so what types of numbers are we missing? It doesn't matter if 0.3333. goes on indefinately or 1.414... When we count> rationals we just index the *formula* to *make* that number. We encode the> numerator and denominator, not 0.3333333.. Counting over the UTM parameter is no different to counting the numerator / denominator grid. It is different in a very important way. Not all UTM parameters encode> numbers. Some Turing machines both fail to halt and fail to produce the> i'th digit of their output. And it is provable that there is no> computable procedure to tell with certainty which do and which do not. That means that if you try to make a list of numbers in UTM parameter> order, that list will not be computable. By contrast, a list of all numerator/denominator pairs _is_ computable.> Even if you decide to eliminate redundant pairs.>True, there's no bijection, but as long as computable numbers cover reals thereare *not* infinitely more reals than computables.Herc === > All numbers are the result of computable functions, sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique.That doesn't mean that sqrt(2) is not irrational. Some Turingmachines can compute irrational numbers.> That only leaves two types of numbers left that qualify for irrational,> non computable and random numbers.But there are only countably many Turing machines, so there are onlycountably many computable numbers. But why does that mean there areonly countably many numbers? Who says that all numbers are the resultof computable functions?Indeed, easy proof that they are not: there are only countably manycomputable numbers, but uncountably many numbers. So some are notcomputable. Thomas === ---------------------------------------------------- ------------------------------ All numbers are the result of computable functions, sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique. That doesn't mean that sqrt(2) is not irrational. Some Turing> machines can compute irrational numbers.Right! rational is Z/Z That only leaves two types of numbers left that qualify for irrational,> non computable and random numbers. But there are only countably many Turing machines, so there are only> countably many computable numbers. But why does that mean there are> only countably many numbers? Who says that all numbers are the result> of computable functions?Process of elimination : if its not computable it is : random or non computable functionNeither of which are well defined numbers. Indeed, easy proof that they are not: there are only countably many> computable numbers, but uncountably many numbers. So some are not> computable.>The theory is to refute there are uncountable numbers, so the deduction isnot substantiated here.Herc === > But there are only countably many Turing machines, so there are only> countably many computable numbers. But why does that mean there are> only countably many numbers? Who says that all numbers are the result> of computable functions?Process of elimination : if its not computable it is : random or non computable functionNeither of which are well defined numbers.I don't know what a random number is. But the existence of noncomputable numbers is perfectly well defined.Unless you want to define well defined as computable, and thenwe're back where we began.> Indeed, easy proof that they are not: there are only countably many> computable numbers, but uncountably many numbers. So some are not> computable.The theory is to refute there are uncountable numbers, so the deduction is> not substantiated here.But then what you have done is not to prove that there are uncountablymany numbers. If you are correct, then you have uncovered acontradiction in ZFC, and I would urge you to put it instraightforward FOL form so that it is incontestable.Thomas === > But there are only countably many Turing machines, so there are only> countably many computable numbers. But why does that mean there are> only countably many numbers? Who says that all numbers are the result> of computable functions? Process of elimination : if its not computable it is : random or non computable function Neither of which are well defined numbers. I don't know what a random number is. But the existence of non> computable numbers is perfectly well defined.Computers can't output random sequences of digits. What you considerthe uncountable real numbers I call random superfluous useless non repeatable digitsequences that will succomb to Shannons noise theory and corrupt. Unless you want to define well defined as computable, and then> we're back where we began.not at all, if you can identify the position on the number line I will accept thatas a number, if you can also show that no computable function shares that positionthen you have established a proof that computable numbers C reals.(proper subset here) > Indeed, easy proof that they are not: there are only countably many> computable numbers, but uncountably many numbers. So some are not> computable. The theory is to refute there are uncountable numbers, so the deduction is> not substantiated here. But then what you have done is not to prove that there are uncountably> many numbers. If you are correct, then you have uncovered a> contradiction in ZFC, and I would urge you to put it in> straightforward FOL form so that it is incontestable.>perhaps, there are other avenues for developing applications in the search space offunctions.Herc === > I don't know what a random number is. But the existence of non> computable numbers is perfectly well defined.Computers can't output random sequences of digits. What you> consider the uncountable real numbers I call random superfluous> useless non repeatable digit sequences that will succomb to Shannons> noise theory and corrupt.I have no idea (still) what you mean by random. It seems to methough that you are saying useless because not computable. Well,maybe useless to you, but very useful to others.I'm sure that some people find quaternions useless. That doesn't makethem useless to others.> not at all, if you can identify the position on the number line I> will accept that as a number, if you can also show that no> computable function shares that position then you have established a> proof that computable numbers C reals. (proper subset here)Do you agree that the set of positions on the number line has theleast upper bound property?Thomas === ...> I don't know what a random number is. But the existence of non> computable numbers is perfectly well defined.May I introduce a new question.Suppose we have a computable system where we can define some computableirrational numbers. Call the set of numbers that can be defined with thiscomputable system, the set C.Question:- Is there a computable system as described above, for which all the axioms concering numbers, as layed down by the Ancient Greeks (I have to look this up), are valid for the set C?If the answer is NO, then Cantor's system is rather unavoidable, but ifthe answer is YES...Lucas === To extend,basically I'm reopening a 75 post thread from several months agowith Daryl McCollough (I recognise Jim Burns and BruceS my skepticcounterparts too).Richard was correct : He means that any Turing machine can be represented by a tape fed to a universal Turing machine, and the contents of that tape can be represented by some number x.Namely x is countable (an integer)Straight after I posted I realised a glaring flaw!Apply Cantors proof directly to the TM ordering, instead of his usual random list. DIGIT 1 2 3 4 5 6__________________________UTM(1) 4 3 6 4 2 4UTM(2) 7 4 3 4 3 2UTM(3) 0 1 0 1 1 1UTM(4) 1 2 2 2 2 2UTM(5) 7 7 7 7 7 7I am suggesting here all computable numbers covers all possible and all typesof numbers, but then we can take the diagonal :4 4 0 2 7and deduce a new 'irrational' not on the list5 5 1 3 8I don't think this is a contradiction though more an error of reasoning oninfinite lists.The diagonal proof is nothing more than you can always add one more, regardlessof what its applied to, completeness and infinite lists have to be more carefullyreasoned about.Essentially real numbers are divided into 2 types :Rational (integer over integer) and not RationalNot Rational can be divided into 2 types, computable (eg.sqrt(2)) and not computable.Computable can again be divided into 2 types : non computable functions and randomsequences of digits.My conclusion is that 'not computable' are not numbers, they are just random sequencesthat for any length can be computed and tallied in the computable number list. OtherwiseNon Computable Numbers are just theoretic assertions of non existence, ....Here is the beginning of a long thread about Non Computable Numbers,essentially 5 5 1 3 8 (the diagonal with a transformation applied) is non computablesince it doesn't fit on the TM list.Uncle Al, sqrt(2) isn't random, check your calculator.> More seriously, though, the reason it can't be computable is> because if it were, then we could compute the sequence H(n)> by computing its digits. We know that it's impossible to> compute whether the n-th Turing machine halts on input n> for all n, so this number r isn't computable.>please go through my above post,the sequence H(n) isn't computable.that has a different meaning to its result isn't computable.According to your logic, a non computable functionthat has a range onto the integers implies that integers are not computable.Ok, the busy beaver function values so far are 1, 2, 6, 2012,options for the 5th value are up to 555654434 (all figures made up)bb(n) is a non computable sequence but the values are all integers, theyhave to be integers because they are all finite and they are all a countof the number of 1s the TM outputs before it terminates. So why canyou say irrationals cannot be calculated by a TM because of non computablefunctions that map to irrationals, yet non computable functions also mapto integers, we cannot compute these values either by your logic. I havea 3 state TM that outputs all integers in binary if you want to check.Herchttp://tinyurl.com/yo9w === To extend,basically I'm reopening a 75 post thread from several months ago> with Daryl McCollough (I recognise Jim Burns and BruceS my skeptic> counterparts too).I'm surprised you remember me. I assumed at the time that youdid not read my post.> Richard was correct :> He means that any Turing machine can be represented by a tape > fed to a universal Turing machine, and the contents of that > tape can be represented by some number x.Namely x is countable (an integer)Straight after I posted I realised a glaring flaw!> Apply Cantors proof directly to the TM ordering, instead of > his usual random list.Cantor does not require a random list. Cantor does not require anything of the list of reals other than it be a list of reals.Since your TM ordering, below, is a list of reals, Cantor'sargument applies to it.The fact that it's (imagined to be) the output of eachTuring Machine makes the list more interesting than a randomlist, but it doesn't make the argument invalid. This is the DIGIT 1 2 3 4 5 6> __________________________> UTM(1) 4 3 6 4 2 4> UTM(2) 7 4 3 4 3 2> UTM(3) 0 1 0 1 1 1> UTM(4) 1 2 2 2 2 2> UTM(5) 7 7 7 7 7 7I am suggesting here all computable numbers covers all possible > and all types of numbers, but then we can take the diagonal :4 4 0 2 7and deduce a new 'irrational' not on the list5 5 1 3 8I don't think this is a contradiction though more an error of > reasoning on infinite lists.The diagonal proof is nothing more than you can always add one > more, regardless of what its applied to, completeness and > infinite lists have to be more carefully reasoned about.You make a good point that one must be careful about completeness,and infinite things. We are probably both thinking of the variousTales of the Infinite Hotel, all rooms full, yet still room foran infinite number of additional guests.However, I think that this argument has been carefully made.The argument isn't really you can always add one more. It'smore like All the guests cannot fit in the hotel. I'm going to make some parts of the argument more explicit.I claim there are no complete lists of the real numbers.Let's look at _all_ the lists of real numbers and see that_none_ of them are complete.Let Lists = the set of all lists of real numbers (that is,all one-to-one functions from the natural numbers to the reals).Define the function Cantor: Lists -> Reals as in Cantor'sproof. Note that for all L in Lists, Cantor(L) is a realnumber that is not listed in L (is not in the image of thefunction L). Remember that this is not defined to be true.Cantor first defined his function and then showed it hadthis property.We have shown that there are no complete lists of reals numbers.We have _not_ shown (so far) that there are more realsthan integers. That requires one last step, where I thinkCantor's brilliance really shines: define what we shallmean by sets (finite or infinite) being the same or differentsizes.(Here I depend on vague recollection. I may have details wrong,but I think my formulation should be functionally equivalent.)-- Two sets A and B are the same size, |A| = |B|, if and only ifthere exists a bijection between the two sets.-- If B is a subset of A, then A is at least the same sizeas B, that is, |B| =< |A|. (There is more that needs to be shown. For example, for all sets A and B, either |A| =< |B| or |B| =< |A|, andthat |A| =< |B| and |A| =< |B| together imply |A| = |B|. I thinkthose two points are all we have to assume, though.)We have shown _there does not exist_ a bijection betweenthe natural numbers and the reals. Remember we have lookedat _all_ the potential bijections and _all_ of them failedat at least one point. As we understand the size of infinite sets, that means N and R are different sizes.It's easy to show |N| =< |R|, since N is a subset of R.Thus |N| < |R|.> Essentially real numbers are divided into 2 types :> Rational (integer over integer) and not RationalNot Rational can be divided into 2 types, computable (eg.sqrt(2)) > and not computable.Computable can again be divided into 2 types : non computable > functions and random sequences of digits.Random is, I think, not well-defined here, unless you justmean not-computable, along the same lines as irrational.> My conclusion is that 'not computable' are not numbers, they are > just random sequences that for any length can be computed and > tallied in the computable number list. Otherwise Non Computable > Numbers are just theoretic assertions of non existence, ....Your conclusion just seems to be the assertion that non-computablenumbers are not real numbers. I think what you have done isre-define the reals as the set of all computable numbers.There is nothing wrong with defining a set of all computable numbers, and there is nothing logically wrong with calling it anything you want, the real numbers or Philip, if you want.But what the rest of us call the real numbers are defined completely separately from notions of computability. For example, I might not know whether the least upper boundof some complicated set is computable, but I do know that,if the set is bounded, the LUB exists. Under your scheme,I might not know whether the LUB exists, but I would knowthat, if it existed, it would be computable. You mightprefer to know the second and not the first, but generallyspeaking, knowing the first and not the second is moreuseful in real analysis.Jim Burns === For example, I might not know whether the least upper bound> of some complicated set is computable, but I do know that,> if the set is bounded, the LUB exists. Under your scheme,> I might not know whether the LUB exists, but I would know> that, if it existed, it would be computable. You might> prefer to know the second and not the first, but generally> speaking, knowing the first and not the second is more> useful in real analysis.>Its not more useful, its a superimposed mutually sustaining impression that mathematics liesoutside the realMs of computability.Your proof jumps to the fact of a contradiction, what is this contradiction?Is it self referencing? YESDoes the formulation of extra numbers occur purely depending on the original formulationof all numbers? YESDoes this contradiction rely on specifying the new number is *different* from the originalforumlation at finite set points? YESDoes the extra number reinsert into the original list to demontrate the imcompatibility? YESThe extra number is simply this :Given a number on a list, change it and put it back in the list without changing the list.Does the diagonal transform form a well defined number?NO -> computables = realsYES -> computables C realsThere is only 1 infinity type, the answer is NO, here's a similar exposition of diagonalisation.> Physicists hold the stage on the accepted view> because they are corrupt. hundred names attached to this discovery. We> don't need new science or maths its all there,> we already know everything. All that is> incomplete is the dispertion of facts. The> world expert on such topic KNOWS certain facts> but another expert on a different topic will> never union the information. Accepting peer> review is too trivial as a confirmation. Our society> is established enough now for public review.> our knowledge base? Only a few percent have> do we have to find to assume matter is just forces. What don't we know about the universe, everything> about the atom, our composition, our life, large> scale matter, forces. Hailing a unifying theory> to be when we already have all the mathematics for> the respective domains. What physical problem> can't we calculate? Physics is 50 years old,> the only way they retain their authoritative view> is by dispersing to the public a flawed assertion> on limitations of information systems. Within> a couple years your OS will download an upgrade> to Outlook with a third window under this text> body, being an Artificial Intelligence comment on> the post. That will be one way to stop the incessant> error of the majority of educated people perpetuating> the repurcussions of Godels Flawed Proof. Here is proof that it is the most elaborate dead end> in science. Most versions of the proof start with the assertion> that finding a contradiction in the negation of a> statement is a proof of the statement. Then the G> statement is formulated, and a mathematics is given> to formulate the self referencing statement. The G> statement is 'this theorom does not belong to the> universal model' or 'this statement has no proof'.> NOW assume 'this has no proof' is false, then it is> proven, so it must be true, uh oh a contradiction, that> means its true. Somehow it MEANS its true but everyone> accepts that as completely different to proving truth> and our minds are superior to logic. The statement> is taken as true however, and a systematic prover by> virtue of its expressive power is incapable of proving> it. STUPID. Set theory, natural language, number> theory all allow the expression of paradoxical statements.> But our thoerom prover is dissalowed this privalege,> the proof by contradiction rule is all encompassing> and theorom provers must give absolute truth or false> values to all theoroms. This is rubbish, the self> referencing statement 'this statement is false' is not> false. Dont run around yelling its a true statement. This is a second version of Godels limit popular with> physisits :> Enter Godel The man who showed once and for all that Russell's aim was impossible was,> Undecidable Propositions of Principia Mathematica and Related Systems. In> it he showed that a statement in a system could be made to refer to itself> was very complicated involving the mapping of prime numbers onto statements.> For example, Godelese for (x)(x=x) is the unique prime number code 28 X 311> X 58 X 78 X 1111 X 135 X 1711 X 199. A Godelian proof Here is a simpler proof that no number system can generate all the> statements which might be true within it. This proof is based on the> writings of A. W. Moore and Roger Penrose. #1. POINT TO PROVE: IT IS IMPOSSIBLE TO DERIVE ALL MATHEMATICAL TRUTH FROM> ANY SET OF SELF-EVIDENT AXIOMS. #2. IF ALL MATHEMATICAL TRUTHS CAN BE DERIVED FROM A CHOSEN SET OF AXIOMS,> THEN, IN PRINCIPLE, AN ALGORITHM A CAN BE CREATED TO TEST WHETHER OR NOT> ANY GIVEN THEOREM DERIVES FROM THE CHOSEN AXIOMS.84I.E.: WHETHER OR NOT IT IS> TRUE OR FALSE. #3. AT PRESENT WE DO NOT HAVE SUCH AN ALGORITHM. IF A CAN BE SHOWN TO BE> IMPOSSIBLE, THEN #1 IS ESTABLISHED. #4. LIST THE FACTUAL STATEMENTS WHICH CAN BE MADE ABOUT NUMBERS. EXAMPLES OF> SUCH STATEMENTS ARE X IS EVEN, X IS ODD, X IS PRIME,X IS LESS THAN> 100, ETC. #5. CREATE A TABLE OF SUCH STATEMENTS, BEGINNING WITH THE SIMPLEST AND> MOVING TO THE MORE COMPLEX. WE WILL CALL OUR STATEMENTS 1, 2, 3, 4... NOW WE> NOTE THAT OUR TABLE CAN REFER TO ITS OWN STATEMENTS. SUPPOSE STATEMENT 0> MEANS: X IS EVEN, STATEMENT 1 X IS ODD ETC... WE LET THE VERTICAL AXIS> REPRESENTS THE STATEMENT NUMBER. THE HORIZONTAL AXIS REPRESENTS ALL NUMBERS> FROM 0 TO INFINITY. WE THEN ASK OURSELVES FOR EACH NUMBER IN THE HORIZONTAL> AXIS, IS THE VERTICAL STATEMENT TRUE OF THIS NUMBER? WE WRITE Y BELOW IT> IF IT IS TRUE, AND N IF IT ISN'T: 0 1 2 3 .... 0 (EVEN) N N Y N... 1 (ODD) N N Y Y 2 (PRIME) N N Y Y... 3 (x<100 ) Y Y Y Y.... ... .................. #6. FOR ANY NATURAL NUMBER (HORIZONTAL LINE) WE NOW HAVE A METHOD OF> DECIDING IF THE VERTICAL STATEMENT IS TRUE. SINCE EVERY POSSIBLE STATEMENT> OF THE SYSTEM CAN APPARENTLY BE LISTED AND SINCE EVERY NATURAL NUMBER CAN> ALSO BE LISTED, IT APPEARS WE HAVE A COMPLETE SYSTEM OF NATURAL NUMBERS AND> AXIOMS. NOTICE THAT EACH STATEMENT ON THE VERTICAL AXIS PRODUCES ITS OWN> UNIQUE HORIZONTAL LINE OF Ys AND Ns. #7. CREATE A NEW WELL-DEFINED SEQUENCE OF Ys AND Ns BY FOLLOWING A DIAGONAL> ON THE CHART WE HAVE JUST CREATED. DO THIS BY TURNING EACH DIAGONAL ELEMENT> INTO ITS OPPOSITE. THE N AT 0/0 ON THE TABLE BECOMES A Y. THE Y AT 1/1> BECOMES AN N. THE Y AT 2/2 BECOMES AN N. THE Y AT 3/3 BECOMES AN N AND SO> FORTH. WE GET YNNN... DOES ANY STATEMENT WHICH HAS ALREADY BEEN GIVEN> PRODUCE THIS NEW SEQUENCE? #8. STATEMENT 0 DOESN'T BECAUSE IT HAS AN N WHERE THE NEW STATEMENT HAS Y.> 1, 2, AND 3 DON'T BECAUSE THEY HAVE Ys WHERE THE NEW STATEMENT HAS Ns. THIS> WOULD HOLD TRUE TO INFINITY IF WE COULD MAKE OUR TABLE THAT LONG, #9. WE KNOW WE LEGITIMATELY CREATED THIS NEW Y & N PATTERN, IE: IT IS TRUE.> YET NONE OF THE EXISTING AXIOM STATEMENTS PRODUCE THIS DIAGONAL STATEMENT. A> NEW AXIOM IS NEEDED TO EXPRESS THE DIAGONAL. 10. IF WE WRITE A NEW STATEMENT (CALL IT R) THAT INCLUDES A PROCEDURE FOR> MAKING THIS DIAGONAL , AT SPACE R/R A NEW DIAGONAL LETTER WILL APPEAR AND WE> WILL HAVE TO ADD STATEMENT S TO REPRESENT THIS NEW SEQUENCE. BUT AT S/S A> NEW DIAGONAL NUMBER WILL APPEAR, REQUIRING A STATEMENT T AND SO ON,> INFINITELY. 11. THEREFORE ALGORITHM A IS IMPOSSIBLE, WHICH IS THE PROOF REQUIRED BY #2.> IT IS IMPOSSIBLE TO AUTOMATICALLY DERIVE ALL POSSIBLE MATHEMATICAL TRUTH. Step 9 is erronous : #9. WE KNOW WE LEGITIMATELY CREATED THIS NEW Y & N PATTERN,> IE: IT IS TRUE. YET NONE OF THE EXISTING AXIOM STATEMENTS> PRODUCE THIS DIAGONAL STATEMENT. A NEW AXIOM IS NEEDED TO> EXPRESS THE DIAGONAL. The pattern is not legitimately created, it is obviously> self referencing and a has a paradoxical bit when it evaluates> its own number. Just because there's two steps in seeing the> plausibility in a theorom, one of the steps fails so the> theorem fails, not the whole encapsulation of theoroms. Herc> that's my 2 centsWonderful.. I picked up on the self referential paradox right away but Ifind it to be revelatory. Mathematics begins from the very self referenceof consciousness as unity self divided and thus self equal which is anassertion of a negation that has no actuality in fact. So all mathematicsis self referentially paradoxical. In fact consciousness itself is so.Only by negation can any divisive discernment take place even if thisnegation is entirely synthetic. Only with the introduction of a No can theYes be apprehended and in no other way. It is important that I point outthat this by no means therefore negates mathematics or logic or reason atall, but grounds them in our fundamental nature.--*._.') (_.' Raan === ...> -- Two sets A and B are the same size, |A| = |B|, if and only if> there exists a bijection between the two sets.If size is a natural number, then you can't take the size ofan infinite set (without additional axioms and definitions).At the moment you say that infinite sets are of the same sizewhen there exists a bijection, then you already introduce someof the choices Cantor made. But those are choices, not proofs.> -- If B is a subset of A, then A is at least the same size> as B, that is, |B| =< |A|.> (There is more that needs to be shown. For example, for> all sets A and B, either |A| =< |B| or |B| =< |A|, and> that |A| =< |B| and |A| =< |B| together imply |A| = |B|. I think> those two points are all we have to assume, though.) We have shown _there does not exist_ a bijection between> the natural numbers and the reals. Remember we have looked> at _all_ the potential bijections and _all_ of them failed> at at least one point.This is indeed proven.> As we understand the size of infinite sets, that means N> and R are different sizes. It's easy to show |N| =< |R|,> since N is a subset of R. Thus |N| < |R|.This is Cantor's choice, not a proof.You may also conclude that the logical system you are currentlyusing is not capable of definining all irrational numbers andleave it there. So, there are at least two choices you canmake:a. You may refer to R even when there is no logical system that can list all its elements (this is Cantor's choice).b. You must be aware that R is never be complete in your logical system (alternative to Cantor).I don't think choice b is very attractive, but to my opinionit is a way you can try to go.However, I always questioned if saying that |N| < |R|has any more meaning than saying that irrational numbersare green.Lucas === > I am suggesting here all computable numbers covers all possible and all types> of numbers, Surely that is false. There are more real numbers than there are computations. === To extend,basically I'm reopening a 75 post thread from several months ago>with Daryl McCollough (I recognise Jim Burns and BruceS my skeptic>counterparts too).Sorry, I don't recall what it was about.--Daryl McCullough === ------------------------------------------------ ----------------------------------To extend,basically I'm reopening a 75 post thread from several months ago>with Daryl McCollough (I recognise Jim Burns and BruceS my skeptic>counterparts too). Sorry, I don't recall what it was about.I think the Cantor diagonalisation across UTM(Z) is more a concern anyway,but the thread is here : http://tinyurl.com/yo9wI said reals are countable by UTM(Z)You said that it misses non computable numbers, like r = Halt(1) + 1/10 Halt(2) + 1/100 Halt(3)...since Halt(n) is unknown for all n.i.e. r is not computable and will not be listed in UTM(Z).I tried to argue r is just not known, you tried to show r is not computable at all (standard view),now I am now suggesting r it not existent! so Computables = Reals.Herc === > In message <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>, Bjoern <...> sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique.That's incomprehensible. Could you rephrase this, please?He means that any Turing machine can be represented by a tape fed to a > universal Turing machine, and the contents of that tape can be > represented by some number x.Well, let's forget about the Turing machine for a second. Let's justnote that sqrt(2) is computable in the sense that, armed with somefinite algorithm and an indefinite amount of storage, we can computean arbitrary number of digits of it.Ok ... that does sound a lot like a Turing machine after all. ;-)This stirs some vague recollection: there are some privilegedirrational numbers which can be specified with a finite amount ofinformation. Obviously _all_ irrational numbers cannot be sospecified, or they in fact would be countable. So most irrationalnumbers are poor lost souls which not only have non-repeating decimalrepresentations, but can't even be named in any meaningful way -- theyare unknowable. (This was the subject of some (IBM?) news releasewithin the past decade, possibly one of those over-hyped news releaseswhich appear regularly, repeating essentially known results as if werefresh revelation).Can somebody remind me what this concept is called? Do theseunknowable irrational numbers correspond to non-computable functions?> That only leaves two types of numbers left that qualify for irrational,> non computable and random numbers.Well, I would say that irrational numbers are not computable; why do>you think otherwise?Because he's confused about the halting problem.So am I, evidently -- would you refresh my memory of its significancehere?And, repeating the argument above, surely _some_ irrational numbersare in fact computable?> Non computable numbers is not a proof that irrationals exist, IMO,> that no halting function exists does not clearly define what the halting> number is, it states it is impossible. There is no gap on the number line> from non computable numbers.That's incomprehensible again.I think he's trying to argue that because there exist TMs for which we > can't determine to which of the sets halts or does not halt they > belong, the does not halt set must be empty.Ah yes ... irrational numbers correspond to the output of a machineswhich neither halts nor loops. As to what he is trying to say, OTOH,I would feel comfortable leaving it in the incomprehensible bin.Ah, knowledge, ah, intellect, ah, thought, I miss you ... you jade. === In message <3FD6F7A6.F52A40E@ix.urz.uni-heidelberg.de>, Bjoern<...> sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique.>That's incomprehensible. Could you rephrase this, please? He means that any Turing machine can be represented by a tape fed to a> universal Turing machine, and the contents of that tape can be> represented by some number x.Well, let's forget about the Turing machine for a second. Let's just> note that sqrt(2) is computable in the sense that, armed with some> finite algorithm and an indefinite amount of storage, we can compute> an arbitrary number of digits of it.Ok ... that does sound a lot like a Turing machine after all. ;-)Well, the lecture on computer science I once heard was several yearsago, so I don't know exactly what computable means...> This stirs some vague recollection: there are some privileged> irrational numbers which can be specified with a finite amount of> information. Obviously _all_ irrational numbers cannot be so> specified, or they in fact would be countable. So most irrational> numbers are poor lost souls which not only have non-repeating decimal> representations, but can't even be named in any meaningful way -- they> are unknowable. (This was the subject of some (IBM?) news release> within the past decade, possibly one of those over-hyped news releases> which appear regularly, repeating essentially known results as if were> fresh revelation).Can somebody remind me what this concept is called? Do these> unknowable irrational numbers correspond to non-computable functions?Does sound a bit like transcendental (sp?) numbers: irrational numberswhich aren't solutions of polynomial equation with rationalcoefficients.[snip rest]Bye,Bjoern === In sci.math, |-|erc What is an irrational number? Can you count to it?> Can you pin point it? There is no such thing.Dedekind cuts or Cauchy sequences are AFAIK adequate for the strictmathematician; most others are content with 53-bit approximations.(53? That's the number of bits in a standard 8-byte floating point'smantissa, including the hidden '1'. There's an IEEE spec forthis approximation but I'd have to dig for it.)These approximations are always multiples of a power of 2. It getsworse: the number of approximations is finite -- there are only 64bits to play with, after all, in total. (1 sign, 11 bits forthe excess-0x200 exponent, and 52 mantissa bits, the '1' beinghidden. By convention, a value of 0x0000000000000000 is taken tobe exactly 0.0 by most floating-point processors.) All numbers are the result of computable functions, sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique.An interesting viewpoint from an engineering standpoint; mathematicians,however, aren't constrained by Turing machines. :-)Most use a 64-bit integer with the IEEE mapping. (A 32-bit integercan also be mapped, with 1 sign bit, 8 bits for the exponent,23 bits for the mantissa.) That only leaves two types of numbers left that qualify for irrational,> non computable and random numbers.There are no irrational numbers, really, in the standard double.Of course this hasn't stopped engineering professionals fromusing M_PI as though it really were pi; if one constructs a bridgewith an arch of 100 m radius using M_PI, one gets an error of about100 * 2^-53 = 1.11 * 10^-14 m. Since this is about the size of 10atomic nuclei placed end-to-end this is more than adequate forthe vast majority of applications.It's also worth noting that 2^53/100 = 90 trillion (wedo want to be able to represent 1 cent :-) ), althoughthe guardbanding approximations might give headaches overabout 22 trillion. Since the GWP (gross world product,http://www.cia.gov/cia/publications/factbook/geos/ xx.html)is about 49 trillion this usually isn't much of a concern.In a pinch one can count pennies -- literally -- usinglong longs instead. Non computable numbers is not a proof that irrationals exist, IMO,Define exist. *No* numbers exist. However, given theusual rules of mathematics, irrational numbers can bedefined in a consistent fashion.> that no halting function exists does not clearly define what the halting> number is, it states it is impossible.Halting number?> There is no gap on the number line from non computable numbers.The standard double precision floating point approximation islittle more than a scattering of 2^64 points among the oceanof the numberline. That leaves all numbers on the number line that cannot be> encapsulated within a computing indexing system are just..... random.That sentence didn't parse too well. Could you clarify? Irrational numbers are just random sequences, otherwise a> computer will map to them.Far from random. Cauchy sequences, in particular, will convergeto a specific point. Which point, of course, may be a matterof debate, on occasion. And what exactly is an infinite sequence of random numbers? nothing,> they don't exist except for rationalisations about their finite> limits. In all eventuality with a small probability, on any> representation system, each digit will flip to a different digit,> Shannon's noise demon at work. As an infinite sequence any> random number is no longer defined. Cantor's proof is cyclic, given a finite space to represent all> possibilities of infinite sequences there are more sequences> than combinations, but they are only random anyway.Cantor's proof has apparently been thrown out, but replaced bya more rigorous demonstration of an uncountable infinity ofreal numbers within [0,1]. However, I'd have to look for thedetails.[.sigsnip]-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === What is an irrational number? Can you count to it?> Can you pin point it? There is no such thing.sqrt(2)http://mathforum.org/library/drmath/view/57117. htmlhttp://medialab.dyndns.org/bignum/> All numbers are the result of computable functions, sqrt(2) is the> result of turing machine(x), where x is some integer, probably> under a million with any crude mapping technique.The concepts irrational and transcendental are rigorousy definedand rigorously provable. Do some reading. If you think sqrt(2) isnot irrational, supply integers for the ratio. Note that sqrt(2) hasbeen taken to ridiculous lengths - though not to the 1.241 trillionplaces of pi by Yasumasa Kanada - to do statistical correlations.ftp://metalab.unc.edu/pub/docs/books/gutenberg/ etext94/2sqrt10a.zip sqrt(2) to 5 million decimal placeshttp://antwrp.gsfc.nasa.gov//htmltest/gifcity/ sqrt2.10mil sqrt(2) to 10 million decimal placeshttp://www.sciencenews.org/20021214/mathtrek.aspsqrt(A): 1) Choose a rough approximation G of sqrt(A). 2) Divide A by G and then average the quotient with G, G* = ((A/G)+G)/2 3) If G* is sufficiently accurate, stop. Otherwise, let G = G* andreturn to step 2.The number of correct decimal places roughly doubles with eachrepetition of step 2.> That only leaves two types of numbers left that qualify for irrational,> non computable and random numbers.Jesus you do whine. The digit occurances of sqrt(2) are demonstratedto be random (with one exception) with a huge battery of statisticalqualifiers including string poker hands (no flushes), but fractionallyless so than those of pi (also with the obvious excpetion).[snip]-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === Dear friends,How to express in elliptic functions sn, cn, or dn if the denominator of ourintegral is not easily being transform into a standard from ?For example, the denominator of the integral is sqrt of1 + c^3*x + c^2*x^2 - 1/4*x^4, where c is a constant.Natanael === > Dear friends,How to express in elliptic functions sn, cn, or dn if the denominator of our> integral is not easily being transform into a standard from ?> For example, the denominator of the integral is sqrt of> 1 + c^3*x + c^2*x^2 - 1/4*x^4, where c is a constant.NatanaelTransform the plane so that the zeros of the polynomial are thevertices of a rectangle. === I am looking for an example of a supermartingale (Xn)n=0,1,2,... such that Sup E[|Xn|]<+oo and Xn-> 0 a.s. , and yet Xn<0 with non-zero probability for some n.Any help is appreciated.Noel. === I am looking for an example of a supermartingale (Xn)n=0,1,2,... such that >Sup E[|Xn|]<+oo and Xn-> 0 a.s. , and yet Xn<0 with non-zero probability >for some n.Is that really all you want, or is there another condition you leftout? It seems to me very easy to give an example of a martingaledoing all this. For example take [0,1] as the probability space,and let X_n be -2^n times the characteristic function of [0,2^(-n)].>Any help is appreciated.Noel.************************David C. Ullrich === Is that really all you want, or is there another condition you left> out? It seems to me very easy to give an example of a martingale> doing all this. No, there is no mistake. This is exactly what I need.Noel. === Is that John Ennis again, (better known as soggy on sci.skeptic)?-jcr === Is that John Ennis again, (better known as soggy on sci.skeptic)?>Naaah, it's a kook called Herc. Ignore him, he pops up every now andthen when he forgets to take his meds. Really.-- Find out about Australia's most dangerous Doomsday Cult:http://users.bigpond.net.au/wanglese/pebble.htmYou can't fool me, it's turtles all the way down. === Can someone explain to me what Jackson's supposed Gricean defense ofassert the stronger means? It seems like assert the stronger wouldsometimes conflict with the maxim of quantity, e.g., if you only needto sayTHIS giraffe has a long neckyet assert the stronger holds that you sayALL giraffes have long necksif you know this second statement to be true. Is this what Jacksonmeans when he talks about asserting the stronger? This couldn't be it,because it would so often obviously conflict with the maxim ofquantity, right? Someone please clarify... === How do you produce the material conditional symbol that looks kind of likesupset but is longer and thinner? === > How do you produce the material conditional symbol that looks kind of like> supset but is longer and thinner?I'm not sure to have understood your question. If you want to know how to getit in a LaTeX document, the answer is Rightarrow or Longrightarrow. But,if I got it right, then the next time you want to know how to produce asymbol in a LaTeX document, read Scott Pakin's The Comprehensive LATEX SymbolList. You'll find it at CTAN at/info/symbols/comprehensive/Jose Carlos Santos === > How do you produce the material conditional symbol that looks kind of like> supset but is longer and thinner?>'P implies Q' in ascii? ->, -->, ==> === (also posted in sci.fractals but it seems to be dead in there)Please forgive my ignorance, but I've only just been introduced to the worldof fractals. I already have the box counting method understood, butI'mstuck with something else...If I were to randomly select a particular point in a given area (whether ornot that point contains the particular thing I'm looking at eg. say I waslooking at stars in an area of space, the point wouldn't necessarily have tohave a star there), how would I go about calculating the fractal dimensionfor that particular point? I've seen ways of calculating the fractaldimension for a range of other situations, but nothing that could help mewith this.Is it possible (or have I just confused you all)? If not, why not? If yes,I would appreciate it if you could let me know how or point me in thedirection of some relevant literature. I've searched google, but can't findanything that may help with this situation. === > (also posted in sci.fractals but it seems to be dead in there)> Please forgive my ignorance, but I've only just been introduced to the world> of fractals. I already have the box counting method understood, but> I'mstuck with something else...If I were to randomly select a particular point in a given area (whether or> not that point contains the particular thing I'm looking at eg. say I was> looking at stars in an area of space, the point wouldn't necessarily have to> have a star there), how would I go about calculating the fractal dimension> for that particular point? I've seen ways of calculating the fractal> dimension for a range of other situations, but nothing that could help me> with this.Is it possible (or have I just confused you all)? If not, why not? If yes,> I would appreciate it if you could let me know how or point me in the> direction of some relevant literature. I've searched google, but can't find> anything that may help with this situation.> I don't know whether this answers your question, but here goes:A. Dimension is a property of a set of points, not of individualpoints. By restricting the set under consideration to a neighborhoodof a given point, a local dimension can be calculated (or estimated,for sampled data). The dimension is determined by the number ofneighborhoods of a given diameter are needed to contain the set, asa function of the diameter of those neighborhoods, as that diametertends towards zero. Note that for sampled data, you don't reallyhave the requisite uncountable number of points, so the discretenature of your sample set has to be taken into account as youchoose the diameters of your neighborhoods that cover the set.B. If you choose a point not in the set under consideration, the dimension is that of the empty set. After all, at the chosen levelof detail, you can't see the set, so it may as well be empty. Myrecollection is that (-1) is the convention for the dimension ofthe empty set, but I may be mistaken.C. If you choose a point that is isolated in the set (that is, thereare no nearby points of that set other than the one you've chosen),the dimension is that of a single point: 0. Once again, at theresolution of interest, that point may as well be the only pointin the set.D. If you choose a point that is in a cluster of points, then takea neighborhood, and look at this URL: http://life.csu.edu.au/fractop/doc/notes/It contains a number of definitions that can probably be convertedto useful algorithms without too much pain, and there is also a linkto a software page, which I haven't done more with than to verifyits existence.Dale. === > ... stuff deleted ...> ... look at this URL: http://life.csu.edu.au/fractop/doc/notes/It contains a number of definitions that can probably be converted> to useful algorithms without too much pain, and there is also a link> to a software page, which I haven't done more with than to verify> its existence.Dale.> I just noted a bit of text located in the above-referenced page thatgives me pause, and suggests one look at the page with at least a milddegree of care: BEGIN QUOTED TEXT: The fractal dimension is a geometric description of an image. It has an integer value for topological sets and a non-integer value for fractal sets. Central to fractal geometry is the concept of self-similarity. END QUOTED TEXTOne of the measures, the Hausdorff dimension, is well-suited towhat the author calls topological sets, and does, in fact, yieldnon-integral dimension for the standard middle-thirds) Cantor set: dim_H(CantorSet) = log(2)/log(3).The reader will note that the various Cantor sets are indeedtopological sets, inasmuch as that term has any meaning at all.In addition, while self-similarity is a feature that is very usefulfor the dimension calculation to be tractable, it is by no means*central* to the concept of fractional dimension. It is true thatmany, if not most, of the sets of non-integral dimension that we'vemanaged to construct have been done via iteration of some basicconstruction (the middle-thirds Cantor set is an example). However,it is a vast overstatement (so vast that it is simply incorrect)to suggest that this self-similarity is central to the existenceof sets of non-integral dimension.That's my caveat wrt the above web page.Dale. === Assume b=a^t and study this function of t.-- Maxi === By chance I found his page, A Treatise on Class Theory this morning,and was quite impressed by it - and not least by the fact that it waswritten by a 15/16-year old! So now I'm really curious.. four yearshave passed since he used to hang out at sci.math, but some of youmight still know; what is he doing these days? Is he still into math,and if so, is he having a splendid career?Again, I'm just so curious.-Leander( Seraph-sama's page on Class Theory:http://members.tripod.com/~SeraphSama/class5.html ) === in our ongoing discussions and theoretical science writings:> http://groups.yahoo.com/messages/SarfattiScienceSeminar http://stardrive.org> http://www.1st-books.com - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -> - - - SarfattiScienceSeminar@YahooGroups.com> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/ === this rule:in pseudo-LaTeXd/dt(vec(A)dotvec(B))=d/dt(vec(A))dotvec(B)+d/dt( vec(B))dotvec(A)P.S. Wouldn't you like to have LaTex in this forum ?It looks so damn ugly unformatted !/Lasse === > P.S. Wouldn't you like to have LaTex in this forum ?> It looks so damn ugly unformatted !It looks ugly unformatted, which is why you shouldn't use it again. === > this rule:> in pseudo-LaTeX> d/dt(vec(A)dotvec(B))=d/dt(vec(A))dotvec(B)+d/dt(vec(B))dotvec (A)>It's the same thing you learned with the product rule for functionsd(f*g)/dt = f'*g + f*g'> P.S. Wouldn't you like to have LaTex in this forum ?No!> It looks so damn ugly unformatted !Exactly why, no! What's worse, it verges upon unreadable when much use asin above line. === I just wanted to check my solution for one more task ... ;-)###task: express the following term through an infinite sequence.solution:2/x/(x+3) = A/x + B/(x+3)<=> 2 = A(x+3) + Bx => A = 2/3; B = -2/3=> 2/x/(x+3) = 2/3 * ( 1/x - 1/(x+3) )= 2/3 * ( (1-(-x)-1)^(-1) - ( 1+(2+x) )^(-1) )= 2/3 * ( ( 1--(x-1) )^(-1) - ( 1--(2+x) )^(-1) )= 2/3 * ( ( 1-(1-x) )^(-1) - ( 1-(-x-2) )^(-1) )= 2/3 * ( sum( (1-x)^k, k, 0, infinity - sum( (-x-2)^k, k, 0,infinity) )= 2/3 * ( sum( (1-x)^k - (-x-2)^k, k, 0, infinity) )This equation is only valid for |1-x| < 1 and |-x-2| < 1 => |x+2| < 1###Is this correct?Karl === I just wanted to check my solution for one more task ... ;-)> ###> task: express the following term through an infinite sequence.> solution:> 2/x/(x+3) = A/x + B/(x+3)> <=> 2 = A(x+3) + Bx => A = 2/3; B = -2/3> => 2/x/(x+3) = 2/3 * ( 1/x - 1/(x+3) )> = 2/3 * ( (1-(-x)-1)^(-1) - ( 1+(2+x) )^(-1) )> = 2/3 * ( ( 1--(x-1) )^(-1) - ( 1--(2+x) )^(-1) )> = 2/3 * ( ( 1-(1-x) )^(-1) - ( 1-(-x-2) )^(-1) )> = 2/3 * ( sum( (1-x)^k, k, 0, infinity - sum( (-x-2)^k, k, 0,> infinity) )> = 2/3 * ( sum( (1-x)^k - (-x-2)^k, k, 0, infinity) )This equation is only valid for |1-x| < 1 and |-x-2| < 1 => |x+2| < 1> ###Is this correct?> There are no numbers x that satisfy both |1-x| < 1 and |x+2| < 1.So I would say it is not correct.How about expanding both terms in powers of the same thing?Both in powers of x, say, or powers of (x+1)? === F{} is a linear operator/transform if:F{A + B} = F{A} + F{B}But what about this?G{A*B} = G{A} * G{B}Are there operators for which this is true? are they well known? === F{} is a linear operator/transform if:>F{A + B} = F{A} + F{B}>But what about this?>G{A*B} = G{A} * G{B}Are there operators for which this is true? are they well known?Depends on which multiplication(s) you use. If o is the convolutionoperator (sort of multiplication between functions) and * is thenormal multiplication operator, thenL(f(x) o g(x)) = L(f(x)) * L(g(x))F(f(x) o g(x)) = F(f(x)) * F(g(x))where L(f(x)) is the Laplace transform and F(f(x)) is the Fouriertransform of f(x). === > F{} is a linear operator/transform if:>F{A + B} = F{A} + F{B}I would call that additive not linear.>But what about this?>G{A*B} = G{A} * G{B}I would call that multiplicative.Are there operators for which this is true? are they well known?Depends on which multiplication(s) you use. If o is the convolution> operator (sort of multiplication between functions) and * is the> normal multiplication operator, thenL(f(x) o g(x)) = L(f(x)) * L(g(x))> F(f(x) o g(x)) = F(f(x)) * F(g(x))where L(f(x)) is the Laplace transform and F(f(x)) is the Fourier> transform of f(x). === Is there a better group for computational geometry ? === > Is there a better group for computational geometry ?comp.graphics.algorithms or comp.theory are probably better than here, not so much because they're more topical but because they have better snr.-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, School of Information & Computer Science === I never know what to do to fill out the AMS Standard Cover Sheet. I thinkmy research interests don't fit neatly into any of the subject classifications.In the past, when I've tried to choose what seems like the nearest categoryand hope for the best, I've found that people who seriously attach themselvesto that category mean something entirely different by it and have expectationsfor others of that category that I never thought I was committing myself toby selecting it.As noted in the marginalia of the AMS Notices, the cover sheet is intendedto help departments process job applications. As such, it serves a purelybureaucratic purpose and could perhaps be forgiven on that basis if it werethen not also used to determine how one is to be treated.I don't know what they should do instead, but I am finding the codificationof primary and secondary interests to be a primary and secondary obstacleto putting together job applications.The one category that sounds sufficiently ambiguous for me to feel I'mnot misrepresenting myself is 00, i.e. General. Perhaps by its nature,after looking at publications with that classification in Math-Sci, I can'tfigure out what it means, but that doesn't matter. What does matter is whatassumptions a hiring committee will make about someone who describes his/herprimary interest as 00. And what would they assume?Maybe the real purpose of the AMS Standard Cover Sheet is to help weed outpeople with a low tolerance for things that don't make sense, things whichthey would certainly encounter in more severe forms after they get hired.If someone can look over my publications on Math-Sci and figure out whatthe appropriate codes are for my primary and secondary interests for theAMS Standard Cover Sheet, I would appreciate the advice. I don't seem tohave been programmed with this capability.Allan Adlerara@zurich.ai.mit.edu************************************ ***************************************** ** Intelligence Lab. My actions and comments do not reflect ** in any way on MIT. Moreover, I am nowhere near the Boston ** metropolitan area. ** ************************************************************** *************** === > I never know what to do to fill out the AMS Standard Cover Sheet. I think> my research interests don't fit neatly into any of the subject> classifications.> In the past, when I've tried to choose what seems like the nearest category> and hope for the best, I've found that people who seriously attach themselves> to that category mean something entirely different by it and have expectations> for others of that category that I never thought I was committing myself to> by selecting it.As noted in the marginalia of the AMS Notices, the cover sheet is intended> to help departments process job applications. As such, it serves a purely> bureaucratic purpose and could perhaps be forgiven on that basis if it were> then not also used to determine how one is to be treated.I don't know what they should do instead, but I am finding the codification> of primary and secondary interests to be a primary and secondary obstacle> to putting together job applications.The one category that sounds sufficiently ambiguous for me to feel I'm> not misrepresenting myself is 00, i.e. General. Perhaps by its nature,> after looking at publications with that classification in Math-Sci, I can't> figure out what it means, but that doesn't matter. What does matter is what> assumptions a hiring committee will make about someone who describes his/her> primary interest as 00. And what would they assume?Maybe the real purpose of the AMS Standard Cover Sheet is to help weed out> people with a low tolerance for things that don't make sense, things which> they would certainly encounter in more severe forms after they get hired.If someone can look over my publications on Math-Sci and figure out what> the appropriate codes are for my primary and secondary interests for the> AMS Standard Cover Sheet, I would appreciate the advice. I don't seem to> have been programmed with this capability.How about this: Take your papers published in the last 10 years, andsee what classification they have in Math Reviews. I did that, and itseems you are a 14, with 11 in second place. Now look up what thatis... 14=algebraic geometry, 11=number theory.Here we get hundreds (maybe even thousands?) of applications. Obviously no single person will read them all. A committee of tensubdivides the applications (how? perhaps by that number you arewriting there), one person reads your application...and unless thatperson thinks it is exceptional it may be the end for you. If he/sheDOES think it is exceptional, it will get additional readings. So...if an algebraic geometer is the one who reads your application, wouldthat make you happy or not?-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Problem :Use the implicit function theorem to show that as a subspace ofR^{n+1} an n-surface M is locally homeomorphic to an open subset of R^n.That is, for each p in M, there exists a neighborhood of p(call it O, which is a subset of M) and an open set U (subset of R^n)and a homeomorphismf : U --> OSteven Rossi === >Problem :Use the implicit function theorem to show that as a subspace of>R^{n+1} an n-surface M is locally homeomorphic to an open subset of R^n.>That is, for each p in M, there exists a neighborhood of p>(call it O, which is a subset of M) and an open set U (subset of R^n)>and a homeomorphismf : U --> O>Steven Rossi>The details depend on how your surface is defined. If it's defined (atleast locally) by an equation of the form F(x) = 0, with x in R^{n+1}and F: R^{n+1} -> R being non-singular on the zero-set M, the implicitfunction theorem says that F locally looks like a linear function L:R^{n+1} -> R. Clearly, the zero-set of a linear function is globally(and therefore locally) diffeomorphic to an open subset of R^{n+1}, sothe same is true for F. The phrase looks like means that there arelocally diffeomorphisms of the domain and range space that transform Fto L.John Mitchell === Problem :Use the implicit function theorem to show that as a subspace of>R^{n+1} an n-surface M is locally homeomorphic to an open subset of R^n.>That is, for each p in M, there exists a neighborhood of p>(call it O, which is a subset of M) and an open set U (subset of R^n)>and a homeomorphismf : U --> O>Steven Rossi> The details depend on how your surface is defined. If it's defined (at> least locally) by an equation of the form F(x) = 0, with x in R^{n+1}> and F: R^{n+1} -> R being non-singular on the zero-set M, the implicit> function theorem says that F locally looks like a linear function L:> R^{n+1} -> R. Clearly, the zero-set of a linear function is globally> (and therefore locally) diffeomorphic to an open subset of R^{n+1}, so> the same is true for F. The phrase looks like means that there are> locally diffeomorphisms of the domain and range space that transform F> to L.My surface is defined as follows :An n-surface M (which is a subset of R^{n+1}) is a non-empty subset suchthat M = f^{-1}(c) wheref : U ----> R (U is a subset of R^{n+1}) , f is C-infinity, and suchthat for all p in M, pis a regular point of f (i.e. it's not a critical point of f). That's all Iam given. John Mitchell === I found this on a web search, the result was stated without thedetails of derivation:gravitational field around a central mass m, the eccentricity (e) isgiven by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the totalenergy (kinetic plus potential), h = rv_t is the angular momentum, vis the total speed, v_t is the tangential component of the speed, andr is the radial distance from the center of the mass. (Note v_t is vsubscript t). === > I found this on a web search, the result was stated without the> details of derivation:gravitational field around a central mass m, the eccentricity (e) is> given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total> energy (kinetic plus potential), h = rv_t is the angular momentum, v> is the total speed, v_t is the tangential component of the speed, and> r is the radial distance from the center of the mass. (Note v_t is v> subscript t).> I'd suggest you reference Becker's book Introduction to TheoreticalMechanics, Section 10-5 in which he addresses Equation of the Orbitby the Energy Method. While the equations of the derivation are abit too much to post on Usenet, essentially Becker begins withexpressions for the total energy and angular-momentum : T + V = W mr^2*Theta(dot) = JAfter a bit of integration and routine substitutions, Becker imports apreviously derived relationship derived earlier in chapter 10: r = 1/((1/ep) - (1/p)Cos(Theta))and obtains: e = sqrt((2WJ^2)/mk^2) + 1)This is a remarkably similar to form as the equation that you posted,and by expanding to equivalent terms as that you posted, I suspectthat they will or should turn out to be equivalent. I wasn'tsufficiently ambitious to perform this final step. Harry C.p.s. My copy of Becker's text is dated 1954 (Lord, how time flies!),so you may have a difficult time locating a copy. If so, I'd be happy === > I found this on a web search, the result was stated without the> details of derivation: gravitational field around a central mass m, the eccentricity (e) is> given by e=sqrt(1 + 2Eh^2 / m^2)where E = v^2 / 2 - m/r is the total> energy (kinetic plus potential), h = rv_t is the angular momentum, v> is the total speed, v_t is the tangential component of the speed, and> r is the radial distance from the center of the mass. (Note v_t is v> subscript t).>Your m in the above is actually mu = G*M, the gravitationalparameter. It's possible that you've chosen a canonicalsystem of units where G = 1.You will find a derivation of this expression in Fundamentalsof Astrodynamics by Bate et al. The text is only about tenbucks on Amazon; it's a great buy. === Blindly stumbling around with a spreadsheet and the Mathworld website, I came across the following O(1) approximation to the prime counting function pi(x):If we let: L = ln(x)/ln(10)And then: a = L^[2/(3L)] b = 1 + 2/(17*[cosh(L-e^2)^(1/32)]) (where e is the Euler number 2.71828...)Then using a and b, let: k = 1 + ln(a/b)Then pi(x) ~= f(x) = ceiling[ k.x/ln(x) ]The actual values of pi(x) for the first 23 powers of ten are listed here: http://mathworld.wolfram.com/PrimeCountingFunction.htmlFor comparison, the table below lists values of this approximation for the same powers of ten. n: f(10^x) 1: 4 2: 25 3: 166 4: 1226 5: 9632 6: 78921 7: 666544 8: 5768728 9: 5089148010: 45533121311: 411916210112: 3760371344813: 34590521626914: 320253734849115: 2981601644542416: 27893717100364517: 262062535017659718: 2471326377429067519: 23383077255448046720: 221905062128788961021: 2111527576638762849222: 20140887349083212087623: 192537825911777012717624: 18442719150950880371139Carl === >Blindly stumbling around with a spreadsheet and the Mathworld website, I >came across the following O(1) approximation to the prime counting >function pi(x):If we let:> L = ln(x)/ln(10)And then:> a = L^[2/(3L)] b = 1 + 2/(17*[cosh(L-e^2)^(1/32)])> (where e is the Euler number 2.71828...)Then using a and b, let:> k = 1 + ln(a/b)Then> pi(x) ~= f(x) = ceiling[ k.x/ln(x) ]The actual values of pi(x) for the first 23 powers of ten are listed >here: http://mathworld.wolfram.com/PrimeCountingFunction.htmlFor comparison, the table below lists values of this approximation for >the same powers of ten.The error for 10^23 appears to be 57867510952120449. Why doyou call this an O(1) approximation? You have some reason tothink the error is never larger than 57867510952120450 or what?> n: f(10^x)> 1: 4> 2: 25> 3: 166> 4: 1226> 5: 9632> 6: 78921> 7: 666544> 8: 5768728> 9: 50891480>10: 455331213>11: 4119162101>12: 37603713448>13: 345905216269>14: 3202537348491>15: 29816016445424>16: 278937171003645>17: 2620625350176597>18: 24713263774290675>19: 233830772554480467>20: 2219050621287889610>21: 21115275766387628492>22: 201408873490832120876>23: 1925378259117770127176>24: 18442719150950880371139Carl************************David C. Ullrich === > The error for 10^23 appears to be 57867510952120449. Why do> you call this an O(1) approximation? You have some reason to> think the error is never larger than 57867510952120450 or what?My apologies.By training I am a computer scientist, and so had forgotten that the O(x) notation is more commonly interpreted as an indication of the size of difference from some required value.By O(1), I was defining the _computational_ complexity with regard to the variable 'x' and not defining the size of an error term.Simple formulae are not normally used for calculating, say, the Riemann Prime Counting function or the Logarithmic integral, as both are usually calculated by summing an infinite series.Summing an infinite series is computationally O(p) where p is some chosen limit of precision.Again, sorry for any confusion.Carl === >There are various versions of this. I give three possible solutions at the end. Please comment.>Infinitely many balls, each numbered (#1,#2,#3, etc.) are to be placed into a bucket, ten at a time, by the scheme given below. Immediately after each group of ten are placed in the bucket, one is removed and discarded. The process is as described below.>11am: Balls #1 - #10 placed into the bucket. Ball #1 is removed and discarded.>11:30am: Balls #11-#20 placed into the bucket. Ball #2 removed and>discarded.>11:45am: Balls #21-#30 placed into the bucket. Ball #3 removed and>discarded.11:52.5am: Balls #31-#40 placed into the bucket. Ball #4 is removed and>discarded.>Etc.>The process continues by halving the remaining time until 12 noon. Then ten are placed in and one is removed and discarded by the above scheme. The remaining time is halved again, etc. There is a flurry of activity just prior to 12 noon. The process does not continue at or beyond 12 noon.>Question: How many balls remain in the bucket at 12 noon?>There are three common, though not necessarily correct, replies.>1) In that the net gain is +9 balls per event, and there are infinitely>many events, there are infinitely remaining balls in the bucket.>2) None remain. Any given ball, say ball #k, is removed and discarded at a specific time prior to 12 noon.>3) The question is meaningless as the process can not be extended to or>beyond 12 noon.>Comments?>You are trying to deal with infinity, or limits to infinity, as if they were finite numbers. The ambiguity lies in what limit is requested.Let's say a transaction includes of both actions, putting in the ten new balls and removing the least numbered ball.Let x_n be the total cumulative number of balls placed in the basket through the nth transaction.Let y_n be the number of balls remaining in the basket just after the nth transaction.Let z_n be the smallest amongst the numbers on the balls remaining in the basket just after the nth transaction.(As others have noted, the time of the events is a red herring.)Thenz_n = n + 1, andy_n = x_n - z_n + 1.Since z_n approaches infinity as n does, in (2) you want to conclude that y_n approaches 0. Yet you are playing with the devil (as I think Cauchy said): you are subtracting one infinite limit from another. For it is clear thatx_n = 10n, whencey_n = 9n -> infinity as n -> infinity (your conclusion 1).In other words, it is simultaneously true that both the number of balls in the basket and the number of balls removed grow arbitrarily large as time approaches noon. It makes no sense to talk about what is in the basket at noon, as time is outside the realm of your thought experiment (your conclusion 3), which is not physically realizable.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === > In other words, it is simultaneously true that both the number of balls > in the basket and the number of balls removed grow arbitrarily large as > time approaches noon. It makes no sense to talk about what is in the > basket at noon, as time is outside the realm of your thought experiment ^^^^^^^ > (your conclusion 3), which is not physically realizable.I meant to type as *that* time, i.e., noon.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === I was wondering whether there be a name for a faithful functor F: C--> D such that given any isomorphism f': F(a) --> b' in D, there isunique object b of C such that for some isomorphism f: a --> b, F(f) =f'? In other words, when F is into the category of sets, given astructure on a and a bijection f from a into b, there is a uniquestructure on b such that f is an isomorphism from a into b. Thisseems to be the categorical notion with which you would want toreplace Bourbaki's notion of structure and morphism (as defined inTheories des Ensembles, Ch. IV) if you want all his results to workout yet (understandably) want to use a categorical approach. It iswhat I prefer to think of when he uses the concepts of morphism andstructure. === |I was wondering whether there be a name for a faithful functor F: C|--> D such that given any isomorphism f': F(a) --> b' in D, there is|unique object b of C such that for some isomorphism f: a --> b, F(f) =|f'? In other words, when F is into the category of sets, given a|structure on a and a bijection f from a into b, there is a unique|structure on b such that f is an isomorphism from a into b. This|seems to be the categorical notion with which you would want to|replace Bourbaki's notion of structure and morphism (as defined in|Theories des Ensembles, Ch. IV) if you want all his results to work|out yet (understandably) want to use a categorical approach. It is|what I prefer to think of when he uses the concepts of morphism and|structure.if i'm not misunderstanding what you said too badly, then it soundslike you really want to consider only the case where c and d aregroupoids rather than arbitrary categories. (in particular when youtake the objects of d to be the sets it seems that you want themorphisms to be the bijections rather than the arbitrary functions.)in that case i think the standard name for a functor of the typeyou're describing would be something like discrete fibration, andyes they're used for exactly the purpose that you're describing.-- === Consider the operator-d^2/dx^2 - F x + p(x) + q(x) on the L_2(0, +infty),whereF>0;p(x) -- real periodic function from L_{1, loc}, p(x+1) = p(x);q(x) -- decreasing function.Does the spectrum of such operator? What condition on the p and q forpreservation absolutely continuous cpectrum? === Please help me with my problem ,,,,Consider : X_1, X_2, ... is a sequence of independent random variables withfinite variances and a common distribution F such that F(0) = 0.What is the relationship between 2*E[(X_n)^2] and E[1 / ( (X_{n-1})^2 +(X_{n-2})^2 ) ]? Is one always at leastas large as the other? (E denotes expectation) If I replaced 2*E[(X_n)^2]by C*E[(X_n)^2], what value of C wouldstill make this true? (C is a real number)Henrique === (axiom-schema) is introduced...along with power set...pairing is> redundant, they advocate keeping it because it is elementary and> essential for the sequential development and even for a scheme that> shuns replacement, pairing makes sense. I think they are suggesting> that set theory even without replacement is strong enough for almost> all practical purposes. What do you think?Oh, I see, I'm used to the shunning replacement version, because I'm> used to GBN set theory. So I think of replacement+comprehension as> just GBN's comprehension, but of course, that isn't quite right, and> this is an example why.I believe that in GBN set theory, pairing *is* required, is it not?ThomasHa ha...I'm having trouble reading just ZF (Zermelo-Fraenkel, Ithink), haven't yet reached ZFC (with Choice)...let alone ZFC+ (withconstruction...the authors refer to Godel...and I've heard scarythings about that guy)...but I would very much like to know what GBNstands for. VNB starts with von Neuman, I forgot the other guy's name.Quine appears all the time in the footnotes. What did he do to logic?So much for the summer vacasion. === > Ha ha...I'm having trouble reading just ZF (Zermelo-Fraenkel, I> think), haven't yet reached ZFC (with Choice)...let alone ZFC+ (with> construction...the authors refer to Godel...and I've heard scary> things about that guy)...but I would very much like to know what GBN> stands for. VNB starts with von Neuman, I forgot the other guy's name.> Quine appears all the time in the footnotes. What did he do to logic?Goedel, Bernays, von Neuman. It's a different formulation of settheoretic axioms, now proved equivalent in the relevant ways, whichexplicitly has proper classes as a part of the system.Thomas === forgot what the little numbers at the top and botom of the integral sign arecalled.One of my problems has a triangle with verticies at (0,0), (1,1), and (0,1)and I had to do something to it to end up with this: [int from 0 to 1] {[integral from 0 to y] x dx} dyMy book showed it to me this way and I figured why couldn't I switch theorder of integration: [int from 0 to 1] {[int from x to 1] x dy} dxI solved both and the answers matched (I hope I integrated correctly). Thiswas easy because the shape I was integrating over was simple and had simplenumbers. My question is do you have any hints/tips/sites that show how todetermine the correct numbers/variables in the integrand. === <4ab51$3fd8af60$80dcc27d$31393@ msgid.meganewsservers.com>,> forgot what the little numbers at the top and botom of the integral sign are> called.One of my problems has a triangle with verticies at (0,0), (1,1), and (0,1)> and I had to do something to it to end up with this: [int from 0 to 1] {[integral from 0 to y] x dx} dyMy book showed it to me this way and I figured why couldn't I switch the> order of integration: [int from 0 to 1] {[int from x to 1] x dy} dxI solved both and the answers matched (I hope I integrated correctly). This> was easy because the shape I was integrating over was simple and had simple> numbers. My question is do you have any hints/tips/sites that show how to> determine the correct numbers/variables in the integrand.> It usually helps me a lot to first draw a sketch of the region of integration and label each boundary piece with its equation. === I have the following application: I have a force sensor probe that touches 3 points on a plate. At eachof the points, I can calculate an XYZ point representing the pointthat the probe detectes the plate.I can easily calculate the equation of the plane Ax+By+Cz+D = 0. WhatI really need is the rotation matrix/Euler angles as represented bythe plane so I can correctly translate positions along the slope ofthe plane.Internally, I am storring all my locations and frames of reference asXYZ points and Euler angles in terms of DZ, DX, DZ rotations.Any thoughts would be appreciated === | | [...]| | I hope this helps!Yes, very much. This got me interested. Do you know any good booksand/or webpages that explain these two different viewpoints ontensors and their relationship.Boris-- boris@uncommon-sense.net - He who hesitates is last. === I am trying to find a primitive function for y = (1 - x^2)^2 but havenow idea how to do. Now I am wondering if there is a formuladescribing how to find primitive functions for this and similarfunctions.--/Torbj.9arn Svensson DiazPlease visist this site. http://www.againsttcpa.com/ === I am trying to find a primitive function for y = (1 - x^2)^2 but have> now idea how to do. Now I am wondering if there is a formula> describing how to find primitive functions for this and similar> functions.Note (1 -x^2)^2 = 1 - 2*x^2 + x^4, and can be integrated term by term. === :I am trying to find a primitive function for y = (1 - x^2)^2 but haveTypo. It's supposed to be y = (1 - x^2)^(1/2).--/Torbj.9arn Svensson DiazPlease visist this site. http://www.againsttcpa.com/ === > :> I am trying to find a primitive function for y = (1 - x^2)^2 but haveTypo. It's supposed to be y = (1 - x^2)^(1/2).So this is an example where f and g both have algebraic primitives, buttheir composition does not.Integrals of square-roots of cubic or quartic polynomials are ellipticintegrals. === >I am trying to find a primitive function for y = (1 - x^2)^2 but haveTypo. It's supposed to be y = (1 - x^2)^(1/2).Trig substituion: let x = sin(t), etc. === >:>I am trying to find a primitive function for y = (1 - x^2)^2 but haveTypo. It's supposed to be y = (1 - x^2)^(1/2).Mathematica gives (x * sqrt(1-x^2) + arcsin(x)) / 2. Looks correct.Maybe a clever trigonometric substitution like x = sin t: Int[ (1 - x^2)^(1/2) dx ]= Int[ (cos t)^2 dt ]= Int[ 1/2 * ( 1 + cos 2x) ]= 1/2 * [ t + 1/2 * sin 2x ]= t/2 + sin(2x)/4= t/2 + (sin t cos t) / 2Substituting back gives:= arcsin(x) / 2 + (x cos(arcsin x)) / 2= arcsin(x) / 2 + x * Sqrt(1-x^2) / 2Presto! === Helping my son with his probability theory course, we have a problemwith the following exercise: A rat is in a chamber in a maze. There are 3 doors to the chamber. Door_1 will make the rat return to the chamber after 3 minutes. Door_2 will make the rat find its way out after 2 minutes. Door_3 will make the rat return to the chamber after 5 minutes. The probabilities that the rat takes the doors 1, 2 and 3 are resp. 1/2, 1/6 and 1/3. Find the expected value and the variance of the total time spent in the maze.This is an exercise on the theorem on conditional expected values: E[X] = Sum( all t, E[X|T=t] * P[T=t] )It happens very early in the course, just before the chapter onMarkov Chains.If we label the total time spent in the maze as X, and the firstdoor chosen as T, we have P[T=1] = 1/2 P[T=2] = 1/6 P[T=3] = 1/3We can calculate the expected value of X as follows: E[X|T=1] = 3+E[X] 3 minutes, then back to square one E[X|T=2] = 2 2 minutes and leaving the maze E[X|T=3] = 5+E[X] 5 minutes, then back to square oneUsing this, we get the equation E[X] = (3+E[X])*1/2 + 2*1/6 + (5+E[X])*1/3which produces E[X] = 21Then we *tried* the following to calculate E[X^2]: E[X^2|T=1] = (3+E[X])^2 E[X^2|T=2] = 4 E[X^2|T=3] = (5+E[X])^2Using the value E[X] = 21 ,we get E[X^2] = (3+21)^2*1/2 + 4*1/6 + (5+21)^2*1/3 = 514and thus VAR[X] = E[X^2] - E[X]^2 = 514 - 21^2 = 73I have ran a simulation (producing 10^8 values of X) of this and I got average[X] = 21.00034 average[X^2] = 878.86246 average[(X-21)^2] = 437.84824and indeed average[X^2] - (average[X])^2 = 437.84824This 437.84824 should be very close to the variance VAR[X] = E[X^2] - E[X]^2but it is not very close to 73 at all, so something is obviouslywrong and I suspect the error is in the second calculation.Anyone any idea how to correctly calculate E[X^2] and/orVAR[X]?Dirk (and Bert) Vdm === You have misstake when you calculate E[X^2].This formula wrongE[X^2] = (3+21)^2*1/2 + 4*1/6 + (5+21)^2*1/3.You have to write followingLet P -- measure on your probability space, s -- time of walkingE[X] = int s(w) P(w) = 21 Here you absolutely right.E[X^2] = int s^2(w) P(w) =int (s + 3)^2(w) P(w) * (1/2) +2^2 * (1/6) +int (s + 5)^2(w) P(w) * (1/3)=int s^2(w) P(w) * (1/2) + int 6 s(w) P(w) * (1/2) + int 3^2 P(w) * (1/2)+2/3 +int s^2(w) P(w) * (1/3) + int 10 s (w) P(w) * (1/3) + int 25 P(w) * (1/3)=E[X^2] * (1/2) + 3 * E[X] + 9/2+2/3+E[X^2] * (1/3) + (10/3) * E[X] + 25/3Thus,E[X^2] * (1/6) = 3*9 + 27*2 + (19/3) * E[X],E[X^2] = 879,E[X^2] - E[X]^2 = 438.> Helping my son with his probability theory course, we have a problem> with the following exercise: A rat is in a chamber in a maze. There are 3 doors to the chamber.> Door_1 will make the rat return to the chamber after 3 minutes.> Door_2 will make the rat find its way out after 2 minutes.> Door_3 will make the rat return to the chamber after 5 minutes.> The probabilities that the rat takes the doors 1, 2 and 3 are> resp. 1/2, 1/6 and 1/3.> Find the expected value and the variance of the total time spent> in the maze. This is an exercise on the theorem on conditional expected values:> E[X] = Sum( all t, E[X|T=t] * P[T=t] )> It happens very early in the course, just before the chapter on> Markov Chains. If we label the total time spent in the maze as X, and the first> door chosen as T, we have> P[T=1] = 1/2> P[T=2] = 1/6> P[T=3] = 1/3 We can calculate the expected value of X as follows:> E[X|T=1] = 3+E[X] 3 minutes, then back to square one> E[X|T=2] = 2 2 minutes and leaving the maze> E[X|T=3] = 5+E[X] 5 minutes, then back to square one> Using this, we get the equation> E[X] = (3+E[X])*1/2 + 2*1/6 + (5+E[X])*1/3> which produces> E[X] = 21 Then we *tried* the following to calculate E[X^2]:> E[X^2|T=1] = (3+E[X])^2> E[X^2|T=2] = 4> E[X^2|T=3] = (5+E[X])^2> Using the value> E[X] = 21 ,> we get> E[X^2] = (3+21)^2*1/2 + 4*1/6 + (5+21)^2*1/3 = 514> and thus> VAR[X] = E[X^2] - E[X]^2 = 514 - 21^2 = 73 I have ran a simulation (producing 10^8 values of X) of this and I got> average[X] = 21.00034> average[X^2] = 878.86246> average[(X-21)^2] = 437.84824> and indeed> average[X^2] - (average[X])^2 = 437.84824 This 437.84824 should be very close to the variance> VAR[X] = E[X^2] - E[X]^2> but it is not very close to 73 at all, so something is obviously> wrong and I suspect the error is in the second calculation. Anyone any idea how to correctly calculate E[X^2] and/or> VAR[X]? Dirk (and Bert) Vdm === Awhile back I posted this linier equation ---e^2 - (1/(n-2))/(1/n) = 0 I had no closed form for n at the time. The closed form for (n) in this case I just discovered ---n = (1/((1/((1/(e^2 - 7))*2))+3))+2 where n satisfies this equation--- e^2 - (1/(n-2))/(1/n) = 0 where n = 2.3130352854993313036361612469...Where the cf of n = [2:3,5,7,9,11,13,15,17,19,21,23,25,27,...]This cf contains all the odd integers ---->oo after even integer 2. As you will note, the last two terms on the right in the closed formfor (n) are the first two terms in its cf in reverse order.This cf. is a simple ordered progression of an irrational (n) evenmore significant than e's cf. because of its simplicity.Where the cf. of e = [2:1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1...] Strange how (e^2) has no order in its cf but the (n) in my equationhas.This constant (n) probably disserves a spot in OEIS because of itswell ordered continued fraction derived from e^2.Dan === Dan schrieb:Strange how (e^2) has no order in its cf but the (n) in my equation> has.> cf(e^2) = [ 7, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1, 1, 12, 54, 14, 1, 1, 15, 66, 17, 1, 1, 18, 78, 20, 1, 1, 21, 90, 23, 1, 1, 24, 102, 26, 1, 1, 27, 114, 29, 1, 1, 30, 126, 32, 1, ... ]computed by the gnu-program maximaGottfried Helms === Suppose I want to studdy a function f : G -> Hwhere H is the quaternion division algebra, hypercomplexnumbers of dimension 4 and G is a given (commutative) group,for example a finite one e.g. Z/nZ.Is there a kind of 'representation theory' with values on H ?What can be said for the 'caracter' of Z/nZ with walue inH ?I think they are of the formphi_k : l -> e^{ j*2*pi*k*l/n }for j a unit pure quaternion.Once you have fixed j, you get a complete setof caracter, and you can compute Fourier-like decomposition,with an inversion formula.Is it that simple ?I would like some pointer on classical studiesof this theory.G. === > As can be seen by the number of posts in this thread,> and the references to his web site in thousands of other posts,> a computer programmer, who took some data processing classes> at a third rate California college, has become a highly regarded expert> in math, physics, and other science disciplines, and> many people, who pretend to be rational, intelligent, open-minded> scientists (Or at least, pretend to have a scientific mind.),> frequently use this programmer as a major reference.So what-if it is true? What is your point?Suppose x is a real number and 0=0 compute:1: a[i]:=the largest integer k s.t. floor(k*x[i])=12: x[i+1]:=a[i]*x[i]-1It's not too difficult to see: (1) 2<=a[i]<=5 if i>0, and (2)x=1/a[0]+1/(a[0]*a[1])+... So a[i] is reasonably called the ith digit of x.In a very nice post from several years ago, Rob Johnson showed the expecteddistribution of the digits of a random x is [206,81,30,20]/337(tinyurl.com/ys8t). Thus p(a[i]=5) is 20/337 or about .059347. The exactdistribution of the first n digits of log(2) is as follows (according toPari/GP): n freq1000:[587,276,87,50]2000:[1179,537,178,106]3000:[ 1783,793,267,157]4000:[2404,1035,360,201]5000:[ 3008,1274,448,270]6000:[3618,1529,528,325]7000:[ 4240,1753,625,382]8000:[4886,1943,727,444]9000:[ 5490,2194,912,562]10000:[6113,2413,912,562]20000:[ 12236,4817,1780,1167]30000:[18337,7266,2678,1719]40000:[ 24449,9716,3545,2290]50000:[30564,12105,4443,2888]60000:[ 36692,14483,5368,3457]70000:[42847,16819,6291,4043]It is kind of curious to note the observed frequency of the digit 5 is *always*less than expected. Is there an explanation for this wierdness?rich === > Suppose x is a real number and 0=0 compute:1: a[i]:=the largest integer k s.t. floor(k*x[i])=1> 2: x[i+1]:=a[i]*x[i]-1It's not too difficult to see: (1) 2<=a[i]<=5 if i>0, and (2)> x=1/a[0]+1/(a[0]*a[1])+... So a[i] is reasonably called the ith digit of x.In a very nice post from several years ago, Rob Johnson showed the expected> distribution of the digits of a random x is [206,81,30,20]/337> (tinyurl.com/ys8t). Thus p(a[i]=5) is 20/337 or about .059347. The exact> distribution of the first n digits of log(2) is as follows (according to> Pari/GP): n freq1000:[587,276,87,50]> 2000:[1179,537,178,106]> 3000:[1783,793,267,157]> 4000:[2404,1035,360,201]> 5000:[3008,1274,448,270]> 6000:[3618,1529,528,325]> 7000:[4240,1753,625,382]> 8000:[4886,1943,727,444]> 9000:[5490,2194,912,562]> 10000:[6113,2413,912,562]> 20000:[12236,4817,1780,1167]> 30000:[18337,7266,2678,1719]> 40000:[24449,9716,3545,2290]> 50000:[30564,12105,4443,2888]> 60000:[36692,14483,5368,3457]> 70000:[42847,16819,6291,4043]It is kind of curious to note the observed frequency of the digit 5 is> *always*> less than expected. Is there an explanation for this wierdness?> Always? Or just when n ends in 000? === Is there an arithmetic compression formula or algorithm that can be used to storea numeric 64bit or numeric 32bit quantity in a compressed format into a numeric 16bit quantity?Then, when needed, decompress back into a 32bit or 64bit quantity?Is this a totally ridiculous request?? :-) :-)Guru === Is there an arithmetic compression formula or algorithm that can be used> to store a numeric 64bit or numeric 32bit quantity in a compressed format> into a numeric 16bit quantity?Then, when needed, decompress back into a 32bit or 64bit quantity?The number of different 16bit quantities is 65536. Applying yourdecompress to each of those will result in only one 32bit or 64bitquantity. In the case of decompressing 16bit to 32bit, this means thatat least 4294901760 32bit quantities can not result from yourdecompress.-- Daniel W. Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Is there an arithmetic compression formula or algorithm that can be used to store> a numeric 64bit or numeric 32bit quantity in a compressed format into a numeric > 16bit quantity?Then, when needed, decompress back into a 32bit or 64bit quantity?Is this a totally ridiculous request?? :-) :-)> The VAX instruction set allows you to a subset of the 32, 64 or 128 bitfloating point values (32 bit F_Floating, 64 bit D_floating or G_floatingand 128 bit H_floating) as six bit literals in the instruction stream.You get a 3 bit exponent and a 3 bit (plus the free bit fromnormalization) exponent. That allows you to encode 1/2 through 15/16in units of 1/16, 1 through 1 and 7/8 in units of 1/8 and so on up tointegers 64 through 120 in steps of 8.If you need to encode a number outside of that set of 64 values,you need to use a different encoding. John Briggs === > Then, when needed, decompress back into a 32bit or 64bit quantity?Lossless compression can't be achieved unlessyour number have some special properties.If your number is taken from a series with sometemporal coherence / scale coherence / spectral coherence(all these 3 are of course somewhat related !)you can perform a lossy compression with ratio 1/4and store the result in an array of 16 bit float.This will be much more efficient than makinga raw quantization of the data(e.g. cast your data from double to 16bit !).try google with 'dct' or 'wavelet transform'.Gabriel === Is there an arithmetic compression formula or algorithm that can be used> to store a numeric 64bit or numeric 32bit quantity in a compressed format> into a numeric 16bit quantity?Then, when needed, decompress back into a 32bit or 64bit quantity?Is this a totally ridiculous request?? :-) :-)Yes :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Goedel is to a discussion about the meaning of life and fate of the universeasNewton is to a discussion about how I got to the grocery store and what I amhaving for dinner.Painting with mathematics, logic, computer science, and cosmology, and thentrying to fit Life, the Universe, and Everything onto that canvasjust won't work. There is something else in the room - You. The perciever.That which sees. That which is asking the questions. That which is holdingthe brush.Turn towards the East, and seek the Vedas, the Buddhist texts, and Zenwritings.They have a lot to say about that-which-is-holding-the-brush.Artists and philosophers and even the laymen of the era use the currentknowledge of the day to fit their life, and draw conclusions about theworld around them. It happened with Aristotle (logic, rhetoric, taxonomy),with Copernicus (Man is Not the measure of all things), with Einstein (thereare no absolute truths), with quantum physics (there is no objectivereality). We read about these discoveries, we think about our lives and whatthey mean, and then we try to correlate. We try to live as if we know whatis really going on. We try to draw conclusions about life, the universe,and everything. === It was clear to me what Paul was talking about. Suppose you have> an algorithm that allows you to compute the state of the universe> at time t given the state of the universe at time 0. You still may> not be able to answer questions of the form If the universe is in state s0 at time 0, then will there ever> be a time in which the universe is in state s1? The quantification over all future times implicit in the question> can make the question undecidable even if the evolution is perfectly> algorithmic. Sure. Now how does an evolutionary process solve this? Paul makes an analogy with a Turing machine. Given a description of> a Turing machine and the tape at time 0, you can compute the state at> any future time. But you still may not be able to answer the question:> Does there exist a time at which the Turing machine halts? Paul seems to think that a nondeterministic Turing machine, or an> evolutionary process, are somehow able to solve this problem, which> they are not.There is some axiom that solves every halting problem. If you insist on asingle approach to developing mathematics you will run up against a Goedellimit and there will be some of these axioms that you never consider.Alternatively you can consider every possible axiom (at least in apotentially infinite universe). Many of these axioms are false but everyhalting problem will be solved by one of these.When I talk about a nondeterministic process I am talking about a processthat explores, not every possible axiom, but an ever expanding set ofmutually incompatible axioms. The crucial difference from a single pathapproach is that you never come to agreement about which of these everexpanding number of alternatives is correct. This is the price you pay foravoiding a Goedel limit.Such a process resembles biological evolution with its great and (given theresources) expanding number of species. I think the similarity of thesestructures is not coincidental. I think unlimited creativity requiresunlimited diversity and this is an implication of Goedel's theorem.--Paul BudnikMountain Math Softwarehttp://www.mtnmath.com408 353 3824 === > There is some axiom that solves every halting problem. If you insist on a> single approach to developing mathematics you will run up against a Goedel> limit and there will be some of these axioms that you never consider.> Alternatively you can consider every possible axiom (at least in a> potentially infinite universe). Many of these axioms are false but every> halting problem will be solved by one of these.But solved is a far cry from solved correctly, you know.> When I talk about a nondeterministic process I am talking about a process> that explores, not every possible axiom, but an ever expanding set of> mutually incompatible axioms. The crucial difference from a single path> approach is that you never come to agreement about which of these ever> expanding number of alternatives is correct. This is the price you pay for> avoiding a Goedel limit.But we can do that already, and model it right in ZFC. We don't haveto exceed any bounds to talk about that. Moreover, folks working inthe subject do.Thomas === Torkel Franzen says...> I take it you have in mind theorems of the form R is a>well-ordering, where R specifies a recursive (we may assume primitive>recursive) relation. How do you prove that it is not the case>that there is for every recursive ordinal alpha a primitive recursive>well-ordering R of order type alpha such that ZFC proves R is a>well-ordering?Is this question specifically for Paul, or can anyone play?Let R_n be the nth primitive recursive relation onnaturals such that ZFC proves that R_n is a well-ordering.Assume some primitive recursive pairing function mappingpairs of naturals onto naturals.Define a well-ordering R on naturals as follows: To compute R(n,m), decode n into a pair (i,j), and decode m into a pair (k,l). If i < k, then R(n,m) = true. If i > k, then R(n,m) = false. If i=k, then R(n,m) = R_i(j,l)It seems to me that R will be a primitive recursive well-ordering,but R will be larger than any well-ordering provable in ZFC. So(unless ZFC is inconsistent) ZFC does not prove that R is a well-ordering.--Daryl McCulloughIthaca, NY === > It seems to me that R will be a primitive recursive well-ordering,> but R will be larger than any well-ordering provable in ZFC. This looks good except that there is no reason to think that R isprimitive recursive. But this can be fixed by modifying theconstruction or by appealing to general results. === |anyone? anyone at all have a post thats against the grain in sci.math that|didn't get them verbally abused?I don't know exactly what you mean by against the grain, orwhat kind of reaction you count as verbal abuse.Bill Taylor and I have both explained why we don't entirely believethe axiom of choice without getting anything that I would call verbalabuse. The axiom of choice is standard enough a part of mainstreammathematics to make disbelief in it eccentric. In my case, it's thecredibility that constructivism as a philosophy of mathematics thatleads me not to wholly embrace nonconstructive assumptions likethe axiom of choice. I've posted about this many times without evergetting what I would call abuse. Only a tiny minority of mathematiciansor people who've examined the philosophy of mathematics areactually constructivists or even take it very seriously, so I would saythat this runs against the grain. Bill Taylor has different reasons; he'smore interested in things like predicativity.One could argue that if it doesn't get much of a negative response, itmust not be so much against the grain after all. This is one reason whyI wonder what exactly you mean by against the grain and abuse.|such a nice crowd according to you all must be 1000sI don't know what you mean by a 1000. Someone who bats 1000,as in basebasll?Keith Ramsay === [...]|> What third rate California college? Who rated it? What criteria?|>|Hey Wormley,|as you use this programmer's web site as your primary rederence,|it seems to me that you should know what college your resident expert|attended.So you want the credibility of sources to be *more* dependenton their credentials and occupation than they are now?Keith Ramsay === >Something is proven only>when there is concensus that the supplied evidence is>valid and the conclusions made are, by concensus, agreed too.Logic is not a popularity contest.-- Wolf Kirchmeir, Blind River ON CanadaNature does not deal in rewards or punishments, but only in consequences.(Robert Ingersoll) === Let a_n = n*Pi^2/6 - sum(1/(k*C(n+k-1,n)),k=1..infinity),where C(n+k-1,n) is binomial coefficient n+k-1 take n.Then {a_n} is a sequence of rational numbers which starts out {0,2,15/4,49/9,1025/144,5269/600,...}.Is there a formula, perhaps involving Bernoulli or Stirling numbers for a_n? Jim Buddenhagen-- To reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE === ...> of solutions for n=3...7; some solution times in seconds;> and some solution-sizes, ie, number of rectangles in solutions:> n #sols time Solution-sizes...> 8 >2330 >5400 14 16-27> (n=8 has run about 1.5 hours so far, on my 450MHz Athlon)n=8 finished this morning, after 48.5 cpu hours (1000 times longerthan at n=7), with 3434 solutions. I don't plan to devote the yearsof computer time that exhaustive searches for n>8 would take; however,I ran n=9 to 12 for a few seconds each with following results: n Smallest-solution-size-seen Run-time and sol-count in that time 9 9 (10s, 91)10 14 (10s, 34)11 19 (10s, 12)12 16 (70s, 46) I also tested 13 to 20 at about 15 minutes each but gotno solutions.Here's the reported 9-cover at n=9 --Cells Prime Corner Form 18 743 0 6 2x9 18 739 0 4 2x9 14 577 1 2 2x7 10 223 0 0 2x5 9 373 0 8 1x9 6 401 6 0 2x3 2 151 8 2 2x1 2 107 5 0 2x1 2 7 0 2 2x1(Corner = top left corner's row and column numbers)-jiw === > ...> the 3-by-3 case has at least 6 solutions (where I stopped counting);> but the 4-by-4 grid has (I believe I proved, perhaps) only 2 solutions.> (I am referring to the back-and-forth variation of the puzzle.)(You can try to find the 4-by-4 solutions yourself.> More fun, and not too hard, is trying to prove that there > are ONLY 2 solutions for the grid.> Extra credit if you find more than 2 solutions!)There are 4 solutions at n=4:1 2 3 4 aabb aabb aabb aabb> 8 7 6 5 aacc cdee aacd cdef> 9 10 11 12 aacc cdee aacd cdef> 16 15 14 13 ddcc ffee eecf ggehYeah, I missed that the combo of primes in the upper left could be addedinto a single prime.of solutions for n=3...7; some solution times in seconds;> and some solution-sizes, ie, number of rectangles in solutions:> n #sols time Solution-sizes> 3 9 0.001 3 5 6> 4 4 0.005 4 6 8> 5 154 0.15 3 5 7 9-14> 6 837 10 8 10 11-19> 7 24 175 11 13 15 17 19> 8 >2330 >5400 14 16-27> (n=8 has run about 1.5 hours so far, on my 450MHz Athlon)> -jiwOh, never mind then about n=8!...:)9, 4, 154, 837, 24,...Hmmm... not in the EIS yet. ... .thanks,Leroy Quet === > The Schoenfeld Theorem:> An infinite bounded sequence of random numbers contains all finite> bounded sequences of numbers.> [snip] Trivially disproven by example in Hofstadter's Godel, Escher, Bach.> It's only 777 pages long. If you look at each page for one second you> can find the table within 12 minutes. What are you babbling on about now? An infinite random sequence of integers bound by [n,m] contains all> finite sequences of integers within [n,m] (not necessarily bound by> n,m though). This is proven. It's called the SCHOENFELD THEOREM.Well, I am truly glad it is not called the Heymann theorem, because it hasby now been discredited thoroughly by more than one contributor.Why don't you get out whilr the boing is good?Franz Heymann === 6968== your libel andattributing it to someone else is just more libel. He has repeatedly citedmy interest in Callie to support his unfounded allegations that I am apedophile but Callie is not a child but is instead a woman. I have neverbeen hospitalized in mental institutions for accosting women (nor do Iaccost women as he purports), but only for daring to criticize Protestantand Catholic churches and daring to try to show people that God providesthem with their names (and for failing to abide by my parent's goodadvice). -Daryl S. Kabatoff HALOHA!!! === : at 01:28 PM, George Greene said: : >That is not only not interesting, it is the opposite of reasonable. : >If the object is itself something as trivial and structureLESS as a : >point, to begin with, asking ANYbody to think of it as something : >BIGGER (e.g., the collection of all arrows ending at it) is simply : >RIDICULOUS. : The duality principal in Projective Geometry was quite productive. It : is RIDICULOUS to pretend otherwise. : : -- : Shmuel (Seymour J.) Metz, SysProg and JOATI have NEVER pretended otherwise, and it is WORSE than ridiculous --it is just slanderous -- to suggest that I have.We were NOT TALKING about a projective geometry.The path category in question was over the AFFINE plane, NOTthe projective one. The examples from statistics that DE wastalking about represent the independent variable as the x-axisand the dependent variable as the y-axis of an AFFINE plane, NOTa projective one. Exploiting projective duality in THAT contextis NOT automatic; it requires some preliminary WORK. You haveto transform the affine plane INTO a projective one first.DE talked about vertical lines but the affine plane DOESN'T HAVEany of those.One way of making an affine plane projective is to add a lineat infinity, each of point of which becomes the new-intersection-point fora parallel-class of lines of the original affine plane.One of these classes will be parallel to the new line at infinity, which meansin some sense that the lines in this class won't fit into the newgeometry. THOSE become the vertical lines for purposes of the dualitythat DE was talking about. They essentially have to be defined out ofthe original affine plane EVEN though the points in them are still there.For a given parallel-class, the point-on-the-line-at-infinity where itslines now intersect can be located as the tangent of the angle thatthe class forms relative to the line at infinity (perpendicular=0,parallel = +/- pi/2, which has no tangent).Of course, adding a completely independent vertical line andforgetting about all the ones you had before doesn't seemto have much of an effect on all the other original pointsand lines in the affine plane, so you could (and working statisticians,apparently, do) speak of the point-line duality as applyingdirectly to them. But in the context of the originaldiscussion, which was about representing more complexcategories by simpler canonical ones, that is going toCAUSE MORE complexity than it cures. -- --- It's difficult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America === >Actually category theory started out as a way to define ``natural >transformation'' [There is a quote by Mac Lane to this effect. I don't >have the precise reference handy.]> ...>The fundamental importance of naturality in homology and >homotopy is impossible to overstate.> Well,if the path category in the Euclidean plane is> prototypical, what are the simplest natural transformations> that are most directly relevant to it?Natural transformations come only with functors.Anyway, I didn't mean to claim that path categories areprototypical categories (they are the test cases forweak infinity-categories though, in case, the suitable weak naturaltransformations will involve homotopies), but only as categoriesfound in nature whose objects are not generally thoughtof as sets with structures with morphisms preservingsuch structures. === The problem really only has one answer, despite the fact that itappears as a paradox. If we merely redescribe the same problem usingonly mathematical and set terms, we get the answer right away. Consider the following redescription:Let the set S0 = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 } {1}. Nowrecursively define the set Sn as follows:Sn = Sn+1 U { 10n+1, 10n+2, ..., 10n+10 } {10n+1}.Let S = lim n->inf (Sn).If you agree with the above as a mathematical description of theproblem, then Solution #2 is the only correct answer, for the reasonsgiven.Solution #1 is incorrect because it confuses Card(lim n->inf Sn) withlim n->inf Card(Sn).Solution #3 is avoided because no laws of physics are involved in thisredescription, and these set operations are defined on countableunions.Jonathan HoyleGene Codes Corporation === >There are various versions of this. I give three possible solutions at the>end. Please comment.>Infinitely many balls, each numbered (#1,#2,#3, etc.) are to be placed into>a bucket, ten at a time, by the scheme given below. Immediately after each>group of ten are placed in the bucket, one is removed and discarded. The>process is as described below.>11am: Balls #1 - #10 placed into the bucket. Ball #1 is removed>and discarded.>11:30am: Balls #11-#20 placed into the bucket. Ball #2 removed and>discarded.>11:45am: Balls #21-#30 placed into the bucket. Ball #3 removed and>discarded.>11:52.5am: Balls #31-#40 placed into the bucket. Ball #4 is removed and>discarded.>Etc.>The process continues by halving the remaining time until 12 noon. Then ten>are placed in and one is removed and discarded by the above scheme. The>remaining time is halved again, etc. There is a flurry of activity just>prior to 12 noon. The process does not continue at or beyond 12 noon.>Question: How many balls remain in the bucket at 12 noon?>There are three common, though not necessarily correct, replies.>1) In that the net gain is +9 balls per event, and there are infinitely>many events, there are infinitely remaining balls in the bucket.>2) None remain. Any given ball, say ball #k, is removed and discarded at a>specific time prior to 12 noon.>3) The question is meaningless as the process can not be extended to or>beyond 12 noon.>Comments?> Almost 4 years ago, there was a very long thread on this very topic.> See the thread starting at> http://groups.google.com/groups?threadm=38b0da2f.5767305@ news.globalcenter.net> As you mention, there are three ways to approach this problem; each> boils down to a question of convergence. If we have a sequence {a_n},> we might ask what happens to that sequence as n tends to infinity, and> thus, pass to the limit to define what happens at infinity. However,> to define a limit, we must first define a topology.But the question can be answered without appealing to any sort of>topology.> -------> There are standard topologies on both the reals and the integers, so> that if the a_n are reals or integers, we use those topologies.> For example, let a_n be the number of balls in the bucket after the n^th> placement/removal. That sequence converges to infinity as n tends to> infinity.But that has nothing to do with the question that was asked.The question asked was, how many balls? Counting how many balls arein the bucket at any given time would seem to have something to do withthe question. Just because we don't know which balls are in the bucketdoes not mean there are no balls in the bucket.Suppose the problem were asked in a slightly different way. Supposethat instead of ball #n being removed at step n, we remove ball #10n,that is, the last ball added. Here I don't think you will disagree thatthere are 9 balls added to the bucket at each step that stay there forall time. Thus, at noon, there are infinitely many balls.Suppose one ball is removed at each step, but we are not told which ballthat is. What happens at noon. All we know is that at step n, thereare 9n balls in the bucket. How many balls are in the bucket at noon?Does our knowledge of which balls are in the bucket alter how many ballsare in the bucket?> -------> Let us define a_n to be the function, after the n^th placement/removal,> which maps the set of balls to {0,1}, where 1 means that ball is in the> bucket and 0 means it is not. If we use the topology of pointwise> convergence on these functions, the sequence {a_n} converges to the> function which maps all the balls to 0, since at some point, each ball> is removed from the bucket, never to be put back.If we view a_n to be a function defined in the time domain rather than in>the domain of natural numbers, such that a_n(t) = 1 if ball n is in the>bucket at time t and a_n(t) = 0 otherwise, then we find that a_n(t_0) = 0>for each n, where t_0 = noon. Thus a(t_0) = sum_n=1^oo a_n(t_0) = 0.Although this reasoning leads to the same conclusion as your argument,>there is an essential difference. Your argument uses limits as t->t_0,>but mine does not. The only limit that appears in my argument is the one>that says the sum of an infinite collection of zeros is zero. Your>argument depends on justifying the step of introducing a pointwise>topology and using it to answer the original question, while my argument>depends on no such artifice.You are summing an infinite number of zeroes. Just because they arezeroes does not excuse you from having to take a limit, trivial as itmay be. Taking a limit involves a topology.> -------> let us use the discrete topology on this sequence. With this topology,> the functions converge to a function f if after some point, all the a_n> are equal to f. Thus, using this topology, the {a_n} do not converge.Again, I don't see that this argument has anything to do with the>question that was asked.It does in the sense that although we know that past step n there aremore that 9n balls in the bucket, the set of balls keeps changing.Thus, the set of balls in the bucket never settles to any limit. Itdoesn't settle to the empty set since there is an increasing number ofballs in the bucket, yet no particular ball is in the bucket forever.> Thus, each of the replies is correct depending on how you define the> convergence. Since the question asks, how many balls, it seems to me> I would say an infinite number. If the question were which balls,> balls.And this is strong evidence that there is something wrong in your>reasoning. If you know which balls, then you automatically know how>many balls. Every set has a cardinality. If your reasoning leads to>inconsistent answers, then your reasoning is wrong.If you know which balls, then you know how many balls; but if you knowhow many balls, you don't necessarily know which balls. Looking only atthe cardinality of the balls in the bucket, we get that there are aninfinite number of balls at noon. However, looking at which balls arein the bucket, we cannot say as that information keeps changing.Let us pose a different problem. Suppose we have two balls, #1 and #2.At each step above we swap which ball is in the bucket. At each stepthere is one ball in the bucket. How many balls are in the bucket atnoon? Which ball is in the bucket at noon? Although we can answer thefirst question pretty easily, we can't answer the second.Rob Johnson take out the trash before replying === > Almost 4 years ago, there was a very long thread on this very topic.> See the thread starting at> http://groups.google.com/groups?threadm=38b0da2f.5767305@ news.globalcenter.net> As you mention, there are three ways to approach this problem; each> boils down to a question of convergence. If we have a sequence {a_n},> we might ask what happens to that sequence as n tends to infinity, and> thus, pass to the limit to define what happens at infinity. However,> to define a limit, we must first define a topology.>But the question can be answered without appealing to any sort of>topology.> -------> There are standard topologies on both the reals and the integers, so> that if the a_n are reals or integers, we use those topologies.> For example, let a_n be the number of balls in the bucket after the n^th> placement/removal. That sequence converges to infinity as n tends to> infinity.>But that has nothing to do with the question that was asked.> The question asked was, how many balls? Counting how many balls are> in the bucket at any given time would seem to have something to do with> the question. Just because we don't know which balls are in the bucket> does not mean there are no balls in the bucket.Counting the balls at some time before noon has nothing to do with thebucket's contents at noon, if there are still balls to be moved in andout. Counting the balls at noon is what matters.Your approach reminds me of the old joke that when you have a hammer,every problem begins to resemble a nail. Here you have three hammers,slightly different in design, but they are all basically the same tool.Yet nowhere do you address the question of whether your tool is theappropriate one for the job. Not only that, your tools give conflictinganswers, which is evidence that at least some of them are beingincorrectly applied.> Suppose the problem were asked in a slightly different way. Suppose> that instead of ball #n being removed at step n, we remove ball #10n,> that is, the last ball added. Here I don't think you will disagree that> there are 9 balls added to the bucket at each step that stay there for> all time. Thus, at noon, there are infinitely many balls.Right answer, wrong reason. The reason there are infinitely many ballsat noon is that a_n(t_0) = 1 for each n that is not a multiple of 10, andthere are infinitely many such n. Therefore a(t_0) = sum_{n=0}^ooa_n(t_0) = +oo.> Suppose one ball is removed at each step, but we are not told which ball> that is. What happens at noon. All we know is that at step n, there> are 9n balls in the bucket. How many balls are in the bucket at noon?> Does our knowledge of which balls are in the bucket alter how many balls> are in the bucket?If the ball to be removed is selected at random at each step, then I canconclude that with probability 1, there are no balls left at noon.> -------> Let us define a_n to be the function, after the n^th placement/removal,> which maps the set of balls to {0,1}, where 1 means that ball is in the> bucket and 0 means it is not. If we use the topology of pointwise> convergence on these functions, the sequence {a_n} converges to the> function which maps all the balls to 0, since at some point, each ball> is removed from the bucket, never to be put back.>If we view a_n to be a function defined in the time domain rather than in>the domain of natural numbers, such that a_n(t) = 1 if ball n is in the>bucket at time t and a_n(t) = 0 otherwise, then we find that a_n(t_0) = 0>for each n, where t_0 = noon. Thus a(t_0) = sum_n=1^oo a_n(t_0) = 0.>Although this reasoning leads to the same conclusion as your argument,>there is an essential difference. Your argument uses limits as t->t_0,>but mine does not. The only limit that appears in my argument is the one>that says the sum of an infinite collection of zeros is zero. Your>argument depends on justifying the step of introducing a pointwise>topology and using it to answer the original question, while my argument>depends on no such artifice.> You are summing an infinite number of zeroes. Just because they are> zeroes does not excuse you from having to take a limit, trivial as it> may be. Taking a limit involves a topology.Didn't I already say that? Actually, I do see a way to formulate theargument that does not involve a limit at all, but that was not the pointI was trying to make in that paragraph, where I specifically said I wastaking a limit and I identified exactly where it was. The point is notwhether a topology is involved, but *which* topology is involved, and howit relates to the stated problem.> -------> let us use the discrete topology on this sequence. With this topology,> the functions converge to a function f if after some point, all the a_n> are equal to f. Thus, using this topology, the {a_n} do not converge.>Again, I don't see that this argument has anything to do with the>question that was asked.> It does in the sense that although we know that past step n there are> more that 9n balls in the bucket, the set of balls keeps changing.> Thus, the set of balls in the bucket never settles to any limit. It> doesn't settle to the empty set since there is an increasing number of> balls in the bucket, yet no particular ball is in the bucket forever.If no particular ball remains in the bucket until noon, and no ball isreturned to the bucket having once been removed, then no ball is in thebucket at noon.> Thus, each of the replies is correct depending on how you define the> convergence. Since the question asks, how many balls, it seems to me> I would say an infinite number. If the question were which balls,> balls.>And this is strong evidence that there is something wrong in your>reasoning. If you know which balls, then you automatically know how>many balls. Every set has a cardinality. If your reasoning leads to>inconsistent answers, then your reasoning is wrong.> If you know which balls, then you know how many balls; but if you know> how many balls, you don't necessarily know which balls. Oh, come, now. You know better than that. If I demonstrate if A thenB, it is useless to retort that I have not proved the converse, since Ihave not even stated the converse or relied on it in any way.If the answer to which balls is none, then the answer to how manyballs cannot possibly be infinitely many. This demonstrates thatsomething is wrong with the application of your chosen tools.>Looking only at> the cardinality of the balls in the bucket, we get that there are an> infinite number of balls at noon. No, we do not get that. Not unless you add an unwarranted assumptionconcerning continuity at noon. The function is obviously discontinuousat a great many points (each time balls are added or removed), so whyshould we expect continuity at noon? After all, noon (t=t_0) is a limitpoint of the set of discontinuities of the function a = sum a_n, andtherefore it should not be surprising that there is also a discontinuityat t_0.>However, looking at which balls are> in the bucket, we cannot say as that information keeps changing.Each of the changes takes place before noon, and the state of each ballis unchanging once that ball is removed.> Let us pose a different problem. Suppose we have two balls, #1 and #2.> At each step above we swap which ball is in the bucket. At each step> there is one ball in the bucket. How many balls are in the bucket at> noon? Which ball is in the bucket at noon? Although we can answer the> first question pretty easily, we can't answer the second.If we use b_1(t) and b_2(t) as the characteristic functions of the twoballs, then we find that in this case (unlike the original problem)b_1(t_0) and b_2(t_0) cannot be determined from the stated information,and therefore the approach that I described fails. But so what? Thisproblem obviously is not well-posed, but the original problem does nothave that defect. In that case we can determine a_n(t_0) for each n.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === >Depends on whether you pay by the ball or by the minute...Wait, both ways they get the same pay!The only guy I can find who is willing to work that hard wants to bepaid by the ball in real time. Unfortunately, he got snared in the WalMart roundup and is back in Mexico now.--Lynn