mm-1076 === >>Subject: The Size of Graham's Number >>Message-id:Trying to explain the sheer hugeness of Graham's Number in a popular >>way to someone, I kind of came up short. I have read on the net >>somewhere that even if all matter in the universe were converted to >>pen and paper, it wouldn't be enough to write the number down. But >>that's a pretty difficult concept to grasp as well. Is there an easier >>way to a) estimate the number of digits in Graham's Number and b) >>express this as something slighty less unfathomable (i.e. it >>wouldrequier a hard drive 40 times the size of our galaxy to store it >>or something) >> Not likely. Even a tiny number like 10**600 is unfathonable. >> See How big is a 2000-bit number? >> http://members.aol.com/mensanator666/fun/2000_bit.htm >>what is it? >a googol is 10^(10^100)) That's a googolplex. A googol is just 10^100. >think of a number with 100 zeroes, that's how many zeroes it has!! >a googol of monkeys tapping keys at a googol typewriters WILL come up >with the entire works of Shakespear. === Subject: Bessel Numbers, Borel Numbers. What are Bessel Numbers, and Borel Numbers? How do you generate them? -- ------------------------------- Patrick D. Rockwell === Subject: Re: Bessel Numbers, Borel Numbers. > What are Bessel Numbers, and Borel Numbers? How do you generate them? > -- > ------------------------------- > Patrick D. Rockwell What is Google, and how do you use it? === Subject: Re: Bessel Numbers, Borel Numbers. >> What are Bessel Numbers, and Borel Numbers? How do you generate them? >> -- >> ------------------------------- >> Patrick D. Rockwell > What is Google, and how do you use it? I already did! I couldn't find a straightforward answer that I could make sense of. I want to see if someone here can give me a better answer. -- ------------------------------- Patrick D. Rockwell === Subject: Re: Bessel Numbers, Borel Numbers. >> What are Bessel Numbers, and Borel Numbers? How do you generate them? >> What is Google, and how do you use it? > I already did! I couldn't find a straightforward answer that I could make > sense of. I want to see if someone here can give me a better > answer. My results from Google... Bessel numbers http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cg i?Anum=A006789 Borel's number http://www.cs.auckland.ac.nz/CDMTCS/chaitin/summer.html -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Bessel Numbers, Borel Numbers. >> What are Bessel Numbers, and Borel Numbers? How do you generate them? > -- > ------------------------------- > Patrick D. Rockwell > What is Google, and how do you use it? >> I already did! I couldn't find a straightforward answer that I could make >> sense of. I want to see if someone here can give me a better >> answer. > My results from Google... > Bessel numbers > http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cg > i?Anum=A006789 > Borel's number > http://www.cs.auckland.ac.nz/CDMTCS/chaitin/summer.html used Google to find that page. One of the ones that I couldn't make much sense of. The page gives these numbers 1,1,2,5,14,43,143,509,1922,7651,31965,139685,636712,3020203, 14878176,75982829,401654560,2194564531,12377765239, 71980880885,431114329728,2656559925883,16825918195484, 109439943234749,730365368850192 It says that Bessel numbers are the number of nonoverlapping partitions of an n-set into equivalence classes. That much I understand, but it then gives this generating function G.f. 1/(1-x-x^2/(1-2x-x^2/(1-3x-x^2/...))) (a continued fraction). Could this generate the above numbers? I downloaded a PDF file which defines Bessel numbers in the following way. Let the Bessel number of the second kind B(n, k) be the number of set partitions of [n] into k blocks of size one or two, and let the Bessel number of the first kind b(n, k) be the coefficient of xn?k in ?y(n)?1(?x) , where y(n)(x) is the nth Bessel polynomial. In this paper, we show that Bessel numbers satisfy two properties of Stirling numbers: The two kinds of Bessel numbers are related by inverse formulas, and both Bessel numbers of the first kind and the second kind form log-concave sequences. By constructing sign-reversing involutions, we prove the inverse formulas. We review Krattenthaler's injection for the log-concavity of Bessel numbers of the second kind, and give a new explicit injection for the log-concavity of signless Bessel numbers of the first kind. It then defines b(n,k) = (-1)^(n-k)*(2*n-k-1)!/((2^(n-k)(n-k)!(k-1)!) if 1<=k<=n and 0 if 1<=n<=k B(n,k)=n!/((2^(n-k)*(n-k)!(2k-n)!) if (n/2)<=k<=n, 0 otherwise. -- ------------------------------- Patrick D. Rockwell === Subject: Re: Bessel Numbers, Borel Numbers. >> What are Bessel Numbers, and Borel Numbers? How do you generate them? > -- >> ------------------------------- >> Patrick D. Rockwell > What is Google, and how do you use it? I already did! I couldn't find a straightforward answer that I could > make > sense of. I want to see if someone here can give me a better > answer. >> My results from Google... >> Bessel numbers >> http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cg >> i?Anum=A006789 >> Borel's number >> http://www.cs.auckland.ac.nz/CDMTCS/chaitin/summer.html > used Google to find that page. One of the ones that I couldn't make much > sense of. The page gives these numbers > 1,1,2,5,14,43,143,509,1922,7651,31965,139685,636712,3020203, > 14878176,75982829,401654560,2194564531,12377765239, > 71980880885,431114329728,2656559925883,16825918195484, > 109439943234749,730365368850192 > It says that Bessel numbers are the number of nonoverlapping > partitions of an n-set > into equivalence classes. > That much I understand, but it then gives this generating function > G.f. 1/(1-x-x^2/(1-2x-x^2/(1-3x-x^2/...))) (a continued fraction). Could > this generate the above numbers? > I downloaded a PDF file which defines Bessel numbers in the following way. > Let the Bessel number of the second kind B(n, k) be the number of set > partitions of [n] into k > blocks of size one or two, and let the Bessel number of the first kind > b(n, k) be the coefficient of xn?k > in ?y(n)?1(?x) , where y(n)(x) is the nth Bessel polynomial. In this > paper, we show that Bessel numbers > satisfy two properties of Stirling numbers: The two kinds of Bessel > numbers are related by inverse > formulas, and both Bessel numbers of the first kind and the second kind > form log-concave sequences. > By constructing sign-reversing involutions, we prove the inverse formulas. > We review Krattenthaler's > injection for the log-concavity of Bessel numbers of the second kind, and > give a new explicit injection for > the log-concavity of signless Bessel numbers of the first kind. > It then defines b(n,k) = (-1)^(n-k)*(2*n-k-1)!/((2^(n-k)(n-k)!(k-1)!) if > 1<=k<=n and 0 if 1<=n<=k > B(n,k)=n!/((2^(n-k)*(n-k)!(2k-n)!) if (n/2)<=k<=n, 0 otherwise. > -- > ------------------------------- > Patrick D. Rockwell I forgot to mention in my last message that I can't get either of the formulas that I quoted to generate any of the numgers that I listed in the sequence above. -- ------------------------------- Patrick D. Rockwell === Subject: Re: Bessel Numbers, Borel Numbers. >> What are Bessel Numbers, and Borel Numbers? How do you generate > them? > -- > ------------------------------- > Patrick D. Rockwell > What is Google, and how do you use it? > I already did! I couldn't find a straightforward answer that I could >> make >> sense of. I want to see if someone here can give me a better >> answer. > My results from Google... > Bessel numbers > http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cg > i?Anum=A006789 > Borel's number > http://www.cs.auckland.ac.nz/CDMTCS/chaitin/summer.html >> used Google to find that page. One of the ones that I couldn't make much >> sense of. The page gives these numbers >> 1,1,2,5,14,43,143,509,1922,7651,31965,139685,636712,3020203, >> 14878176,75982829,401654560,2194564531,12377765239, >> 71980880885,431114329728,2656559925883,16825918195484, >> 109439943234749,730365368850192 >> It says that Bessel numbers are the number of nonoverlapping >> partitions of an n-set >> into equivalence classes. >> That much I understand, but it then gives this generating function >> G.f. 1/(1-x-x^2/(1-2x-x^2/(1-3x-x^2/...))) (a continued fraction). Could >> this generate the above numbers? >> I downloaded a PDF file which defines Bessel numbers in the following >> way. >> Let the Bessel number of the second kind B(n, k) be the number of set >> partitions of [n] into k >> blocks of size one or two, and let the Bessel number of the first kind >> b(n, k) be the coefficient of xn?k >> in ?y(n)?1(?x) , where y(n)(x) is the nth Bessel polynomial. In this >> paper, we show that Bessel numbers >> satisfy two properties of Stirling numbers: The two kinds of Bessel >> numbers are related by inverse >> formulas, and both Bessel numbers of the first kind and the second kind >> form log-concave sequences. >> By constructing sign-reversing involutions, we prove the inverse >> formulas. We review Krattenthaler's >> injection for the log-concavity of Bessel numbers of the second kind, and >> give a new explicit injection for >> the log-concavity of signless Bessel numbers of the first kind. >> It then defines b(n,k) = (-1)^(n-k)*(2*n-k-1)!/((2^(n-k)(n-k)!(k-1)!) if >> 1<=k<=n and 0 if 1<=n<=k >> B(n,k)=n!/((2^(n-k)*(n-k)!(2k-n)!) if (n/2)<=k<=n, 0 otherwise. >> -- >> ------------------------------- >> Patrick D. Rockwell > I forgot to mention in my last message that I can't get either of the > formulas that I quoted > to generate any of the numgers that I listed in the sequence above. > -- > ------------------------------- > Patrick D. Rockwell I suspect you don't know what a generating function is. Its not a formula where you substitute some value for x and it provides the value of the fumction. Its a formula where the co-efficients on the polynomial expansion provide the terms you need ... sounds like you need to http://mathworld.wolfram.com/GeneratingFunction.html I don't think anybody is going to write out a lengthy explanation of Bessel or Borel numbers for you, just because you don't like (for reasons unspecified) the explanations that people have already published to the web. I think you are going to have to: * Tell us what you don't understand in the material already on the web (with pointers), and ask for assistance with those parts. OR * Otherwise rephrase your questions so they are more closed - ie, provide a question that can be meaningfully answered in a few lines OR * Give some context to your question, so the people in this newsgroup can see what is relevant and what you actually need to know about the numbers OR * Learn to use Google and think for yourself. === Subject: Re: .99999... still=/= 1 > It doesn't converge. That's not correct. It *does* converge. Converge as number of terms increases without bound means it has a limit as number of terms increases without bound, which in turn means that for every positive epsilon there's an N such that if you sum more than N terms it's closer than epsilon to the limit. Those are the standard mathematical definitions. Are you using some other definition of the same words that you haven't bothered to share with us? > .999... is similar to an asymptote. It never reaches 1. That's correct. For any finite number of terms, the partial sum is not exactly 1. But if you take as many terms as needed, it gets as close to 1 as you want, less than any positive epsilon whatsoever. If you draw a smooth curve through a plot of the partial sum as a function of the number of terms, you get an asymptote, the value of the partial sum getting as close to 1 as desired. (Asymptote properly refers only to continuous graphs, not to discrete functions such as partial sums. That's why I say you need to draw a smooth curve through the points in the graph before you can talk about an asymptote. You said similar to, which is correct here. A limit of a discrete sequence is similar to an asymptote of a continuous curve.) So if you understand asymptote as being similar to limit, why can't you understand that a series or sequence gets arbitrarily close to its limit value without ever exactly reaching it just as a smooth curve gets arbitrarily close to its asymptote without ever exactly reaching it? === Subject: Re: .99999... still=/= 1 >> It doesn't converge. >That's not correct. It *does* converge. Converge as number of terms >increases without bound means it has a limit as number of terms >increases without bound, which in turn means that for every positive >epsilon there's an N such that if you sum more than N terms it's closer >than epsilon to the limit. Those are the standard mathematical >definitions. Are you using some other definition of the same words that >you haven't bothered to share with us? >> .999... is similar to an asymptote. It never reaches 1. >That's correct. For any finite number of terms, the partial sum is not >exactly 1. But if you take as many terms as needed, it gets as close to >1 as you want, less than any positive epsilon whatsoever. If you draw a >smooth curve through a plot of the partial sum as a function of the >number of terms, you get an asymptote, the value of the partial sum >getting as close to 1 as desired. (Asymptote properly refers only to >continuous graphs, not to discrete functions such as partial sums. >That's why I say you need to draw a smooth curve through the points in >the graph before you can talk about an asymptote. You said similar >to, which is correct here. A limit of a discrete sequence is similar >to an asymptote of a continuous curve.) >So if you understand asymptote as being similar to limit, why can't you >understand that a series or sequence gets arbitrarily close to its >limit value without ever exactly reaching it just as a smooth curve >gets arbitrarily close to its asymptote without ever exactly reaching >it? Because the number 1 is the most mysterious number of all real numbers. You can multiply it by another number and that other number remains perfectly that same number. If you say an imperfect convergence number is the same as 1, it contaminates the entire real numbers with a new value of 1. The number 1 must remain as perfectly pure as possible which is, 1 = 1 . And, The Advanced Engineering Mathematics book on convergence clearly states that, If m-->oo lim Z_m = 0 then it converges otherwise it diverges. .999... = function ( 1 - 1/10^m ) = Z_m m-->oo lim (1 - 1/10^m ) = 1 So it diverges and this proves, .999... does not converge to 1 and it doesn't equal 1. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 > If > m-->oo > lim Z_m = 0 > then it converges otherwise it diverges. What is the meaning of it in that incomplete sentence? > .999... = function ( 1 - 1/10^m ) = Z_m That makes no sense whatsoever, because .999... is a constant, not something involving m which varies depending on the value of m. Perhaps you should write this instead: .99999..................99999 = 1 - 1/10^m |<- (m digits, each '9') ->| That would make sense because the sequence of 9's is finite, there are m of them, so the value of that finite sequence of 9's varies as m varies. > .999... does not converge to 1 and it doesn't equal 1. in the Subject field of this thread. We thought you meant what any mathematician would mean by that notation, namely: limit .99999..................99999 m->inf |<- (m digits, each '9') ->| but apparently that's not what you meant, so I've repeatedly asked you to say what you meant by that notation, and so-far you haven't answered. === Subject: Re: .99999... still=/= 1 >>If >>m-->oo >>lim Z_m = 0 >>then it converges otherwise it diverges. > What is the meaning of it in that incomplete sentence? The sequence {Z_m} of partial sums, you putz. You just don't listen and learn do you? Bob Kolker === Subject: Re: .99999... still=/= 1 Some one asked me this, .999... < X < 1 What is X? The way I approached this was to find the last higher order multi-dimensional infinitesimal digit before it converged to 1, using a multi-dimensional higher order differential of the 9/10^n term in the series. A differential can also be defined as a difference between comparable classes. What I got was something like this, n-->oo DIM (nth) d(^nth) (9/10^n) = where d(nth) is the higher order derivative as it approached infinity. 9 * 1^ (-->oo) * LOG (10) X = --------------------------------- 10^n Taking the Partial Sum view, 1^(-->oo) = 1^oo It looks like only the calculation of, (1^oo) * LOG (10) is the last thing that separates .999... from 1. So in other words, the last multi-dimensional differential calculation of the last digit, is the last digit in the series .999... before it converges to 1. This also shows you that, .999... =/= 1 because there is one last calcualtion separating .999... from 1. And right after that calculation it goes right back to, 9/10^n again never reaching 1. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 > Some one asked me this, >.999... < X < 1 >What is X? > The way I approached this was to find the last higher order >multi-dimensional infinitesimal digit before it converged to 1, using a >multi-dimensional higher order differential of the 9/10^n term in the series. > A differential can also be defined as a difference between comparable >classes. >What I got was something like this, >n-->oo >DIM (nth) >d(^nth) (9/10^n) = >where d(nth) is the higher order derivative as it approached infinity. Correction: 9 * 1^ (-->oo) * (LOG (10))^(-->oo) X = ---------------------------------------- 10^n >Taking the Partial Sum view, Correction: 1^(-->oo) = 1^oo LOG(10)^(-->oo) = 1^oo It looks like only the calculation of, (1^oo) * (1^oo) is the last thing that separates .999... from 1. So in other words, the last multi-dimensional >differential calculation of the last digit, is the last digit in the series >.999... before it converges to 1. > This also shows you that, >.999... =/= 1 because there is one last calcualtion separating .999... from >And right after that calculation it goes right back to, 9/10^n again never >reaching 1. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 In sci.math, S. Enterprize Company > Some one asked me this, >>.999... < X < 1 >>What is X? >> The way I approached this was to find the last higher order >>multi-dimensional infinitesimal digit before it converged to 1, using a >>multi-dimensional higher order differential of the 9/10^n term in the series. >> A differential can also be defined as a difference between comparable >>classes. >>What I got was something like this, >>n-->oo >>DIM (nth) >>d(^nth) (9/10^n) = >>where d(nth) is the higher order derivative as it approached infinity. > Correction: > > 9 * 1^ (-->oo) * (LOG (10))^(-->oo) > X = ---------------------------------------- > 10^n >>Taking the Partial Sum view, > Correction: > 1^(-->oo) = 1^oo > LOG(10)^(-->oo) = 1^oo > It looks like only the calculation of, (1^oo) * (1^oo) is the last thing > that separates .999... from 1. So in other words, the last multi-dimensional >>differential calculation of the last digit, is the last digit in the series >>.999... before it converges to 1. >> This also shows you that, >>.999... =/= 1 because there is one last calcualtion separating .999... from >>1. >>And right after that calculation it goes right back to, 9/10^n again never >>reaching 1. Fine. Are there Y and Z such that .999... < Y < X < Z < 1 ? If so, what are they? BTW, 1^oo can be one of the following: [1] 1. 1^N = 1 as N -> +oo. [2] oo. exp(1/x^2)^x -> +oo as x -> 0+. [3] e. (1 + 1/N)^N -> e as N -> +oo. [.sigsnip] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: .99999... still=/= 1 >> If >> m-->oo >> lim Z_m = 0 >> then it converges otherwise it diverges. >What is the meaning of it in that incomplete sentence? Look it up in the dictionary. >> .999... = function ( 1 - 1/10^m ) = Z_m >That makes no sense whatsoever, because .999... is a constant, not No it's not, it's a series. It's shown as a constant on calculators because they round off that value. >something involving m which varies depending on the value of m. >Perhaps you should write this instead: > .99999..................99999 = 1 - 1/10^m > |<- (m digits, each '9') ->| >That would make sense because the sequence of 9's is finite, there are >m of them, so the value of that finite No, there is oo repeating 9's. sequence of 9's varies as m >varies. >> .999... does not converge to 1 and it doesn't equal 1. Correction: .999... does converge to 1. It just doesn't equal 1. The lim SUM 9/10^m doesn't converge on this convergence test. The mth term has to be analyzed in this test. >in the Subject field of this thread. We thought you meant what any >mathematician would mean by that notation, namely: >limit .99999..................99999 >m->inf |<- (m digits, each '9') ->| >but apparently that's not what you meant, so I've repeatedly asked you >to say what you meant by that notation, and so-far you haven't >answered. .999... = .9999999999999999999999999999999999999999999999999999999999999999999999999 99999999999999999999999999999999999999999999999999999999999999999999999999 999999-->oo Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 sine qua non, if you're referring to the real number, 1.0000...; eh? you can't be really serious about your textbook author, wanting to be so circumspect about the physics of numbers; can you? > Because the number 1 is the most mysterious number of all real numbers. You > The Advanced Engineering Mathematics book on convergence clearly states that, --ils ducs d'Enron! http://tarpley.net/bush22.htm === Subject: Re: .99999... still=/= 1 >Summary: My method of proof, where I use a notation to represent >infinite-long decimal fractions which represent limits of series, and >make sure the series converges, and all my steps are valid >mathematically, is a valid form of proof. > Your method of proof is inconsistent with [1] floating points, > [2] Partial Sums, and [3] adding terms with infinity attached. (Bracketed numbers added by me so I can refer to three points below without breaking up the quoted text above.) [1] Floating point representations as given by calculators and other computing devices are nothing more than an approximation to a arithmetic result. Limits of series have nothing to do with such calculated approximations, so how can there be any inconsistency between the two? [2] The partial sum of the first n digits of the infinite series 0.9999999999999999999... is the exact value of the finite series (exact decimal fraction) 0.9999 ... 9999 |<- n 9's ->| which is equal to 1 - 10^(-n). Do you agree so-far? The exact value of the infinite series is the limit of the partial sum as N grows larger without bound. Thus 0.9999999999999999999... = limit as N->posInf of 1 - 10^(-N) = 1. Do you agree with that too? Where is the alleged inconsistency in my argument? [3] Please define what with infinity attached is supposed to mean as you've introduced that phrase here. It's not any standard mathematics phrasing that I've ever heard. > What is, [4]> .999... - ... = ? [5]> .999 > or [6]> .999 + oo = ? [7]> .999... [4] Some string of nonsense text you've typed for no apparent reason. [5] An exact decimal fraction, equal to the common fraction 999/1000. [6] Another string of nonsense text you've typed for no apparent reason. [7] Standard notation for the infinite series I mentionned in [2] above. whose value is equal to 1. The rest of your message is all complete nonsense text with no apparent meaning. If you would be so kind as to clearly define what any of your apparent nonsense text is supposed to mean, then I might be willing to read what you write and check if your claims are correct. But when you write nonsense text, and don't define any meaning for it, all I can reasonably do is just pass it by. I've done you the courtesy of defining what most people including mathematicians and including myself mean when we use the notation in line [7] above, even though you should have already known that meaning before you started this thread. Now will you please do us the courtesy of telling us what alternate meaning you apply to exactly the same [7] notation, and what you mean by the apparent nonsense of [4] and [6] above? === Subject: Re: .99999... still=/= 1 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2JbVW32373; > >> .999...* 10 - 9 = .999... >=> x * 10 - 9 = x where x = .999... >=> 10x - x = 9 = 9x <==> x = 1 >So why you claim that .999... != 1? > > That's right. > x = .999... > > and after his calculations are done he shows it to be equal to something > else. > > And another way to look at it let's set x = 1 to start with instead of > .999... > > > x = 1 > 10x = 10 > x = 1 > > Here we can see x = 1 before and after. > > And, > > x = .999... > > 10 *x = 9.999... > > x = .999... > > Here we can see x = .999... before and after. > > This is often given as a layman's justification (not proof) that .999... > =1. However mathematically, it's not quite right. >>It is sufficient for a proof. > You can't justify the > step 10x = 9.999... without appealing to a theorem that says you can > multiply each term of a convergent series by a constant, and the > resulting series will converge to the constant times the sum of the > original series. >>Such a theorem would be a tautology of little value, since the >>statement follows from the directly from distributive axiom a(b+c)= ab >>+ ac. >>If >> x = 0.99999... [repeating n times] >>Then, >> x = 9/10^1 + 9/10^2 + .... 9/10^n >> ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >>Let a = 10 >> 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n >> = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n >> = 9 + x > If you could prove that, then you'd already have enough understanding of > limits to know that .999... = 1. >>Limits are unnecessary. The proof is simply, >>x = 0.9_ >>10x = 9.9_ (follows directly from the distributive axiom) >> = 9 + x >> 9x = 9 >> x = 1 > Nope, this is incorrect. > The question was does .999... converge to 1, not 9.999... . >equation. Equations aren't used in convergence tests. > The basic convergence test is like this. >If, >m-->oo >Lim Z_m = 0 >then convergence otherwise divergence. >.999... is represented as (1 - 1/10^m) >Z_m = ( 1 - 1/10^m) >m-->oo >Lim Z_m = 1 > This means divergence. So not only does it not converge to 1 but it doesn't >equal 1. And even if it did converge it wouldn't EQUAL 1. There isn't even one >series in all of math that does converge that really equals that convergence >value. >.999... --> diverges and =/= 1 >Smart's Alt. Physics News Group >ht tp://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1S. Enterprize (Science Journal) >http://smart1234.s-enterprize.com / convergrence test according to your definition 1+ 1/2 + 1/3+1/4 +... would converge since 1/n--->0 as n--> infinity. It is generally accepted that the sum given is divergent. thus we have a counter example that disproves your convergence test. Please find an acceptable trust worthy test for convergence === Subject: Re: .99999... still=/= 1 >> > .999...* 10 - 9 = .999... >=> x * 10 - 9 = x where x = .999... >>=> 10x - x = 9 = 9x <==> x = 1 >So why you claim that .999... != 1? > >> That's right. >> x = .999... >> >> and after his calculations are done he shows it to be equal to >something >> else. >> >> And another way to look at it let's set x = 1 to start with instead of >> .999... >> >> >> x = 1 >> 10x = 10 >> x = 1 >> >> Here we can see x = 1 before and after. >> >> And, >> >> x = .999... >> >> 10 *x = 9.999... >> >> x = .999... >> >> Here we can see x = .999... before and after. >> >> This is often given as a layman's justification (not proof) that .999... >> =1. However mathematically, it's not quite right. >It is sufficient for a proof. >> You can't justify the >> step 10x = 9.999... without appealing to a theorem that says you can >> multiply each term of a convergent series by a constant, and the >> resulting series will converge to the constant times the sum of the >> original series. >Such a theorem would be a tautology of little value, since the >statement follows from the directly from distributive axiom a(b+c)= ab >+ ac. >If > x = 0.99999... [repeating n times] >Then, > x = 9/10^1 + 9/10^2 + .... 9/10^n > ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >Let a = 10 > 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n > = 9 + x >> If you could prove that, then you'd already have enough understanding of >> limits to know that .999... = 1. >Limits are unnecessary. The proof is simply, >x = 0.9_ >10x = 9.9_ (follows directly from the distributive axiom) > = 9 + x > 9x = 9 > x = 1 >> Nope, this is incorrect. >> The question was does .999... converge to 1, not 9.999... . >>equation. Equations aren't used in convergence tests. >> The basic convergence test is like this. >>If, >>m-->oo >>Lim Z_m = 0 >>then convergence otherwise divergence. >>.999... is represented as (1 - 1/10^m) >>Z_m = ( 1 - 1/10^m) >>m-->oo >>Lim Z_m = 1 >> This means divergence. So not only does it not converge to 1 but it >doesn't >>equal 1. And even if it did converge it wouldn't EQUAL 1. There isn't even >one >>series in all of math that does converge that really equals that convergence >>value. >>.999... --> diverges and =/= 1 >>Smart's Alt. Physics News Group >>http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >>S. Enterprize (Science Journal) >>http://smart1234.s-enterprize.com/ >convergrence test >according to your definition 1+ 1/2 + 1/3+1/4 +... would converge since >1/n--->0 as >n--> infinity. It is generally accepted that the sum given is divergent. >thus we have a counter example that disproves your convergence test. >Please find an acceptable trust worthy test for convergence The form of of a harmonic series is n= 1 to oo lim SUM 1/n Using The Root Convergence Test ____ lim n/ Z_n = L n-->oo L < 1 = converge L > 1 = diverge L = 1, test fails, no conclusion possible. At n-->0 it's divergent , L >1 n=0, L= 1 n>1, L < 1 = convergent. According to this convergence test, it is divergent, no conclusion possible , and convergent. It depends on the point of view you take, or you can just observe all cases. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 > A scientific calculator does approximate convergence with a certain > number significant digits. Just because a calculator emits a particular number of digits of output, doesn't say anything about how accurate the represented number is, i.e. how close to the correct answer it is. If you do a chain of calculations, the accuracy usually gets worse and worse, yet the digits output remains the same, looking like it's that accurate (all the way out to the last digit shown) when in fact many of the last digits shown are totally wrong. In fact with a long enough chain of calculations, you can produce a displayed result which may have a whole bunch of digits shown but not even one significant digit of accuracy. For example, start with the exact integer 2, then take square root, then take square root of that, etc., five square roots in a row. Now work your way backward by multiplying the value by itself five times in a row. The correct mathematical answer is eactly 2, but your calculator won't show that. Instead it'll show some exact value that starts out 2.00 but then somewhere has a bunch of non-zero digits, or starts out 1.99 but then somewhere has a bunch of non-nine digits. Do you honestly believe that the correct mathematical result is correctly shown to all the digits displayed on your calculator? The point I'm making is that you can't use the results given by such a calculator to establish truth of any mathematical fact. Now if you had a calculator that used interval arithmetic internally, and which displayed *only* as many digits as were accurate per the interval (lower&upper bounds on correct result), *then* you could simply look at the output and know that all digits except the last are accurate. But I'm not aware of any calculators that work that way. > I think my calculator about 32 significant digits. > 1/3 = .3333333333333333333333333333 > see that, > 1/3 = .333... No, from the output of your calculator you can't see anything like that except in your imagination, or if you happen to already know the correct result and are just ignoring what the calculator says. In that case, you got lucky, the output from the calculator happened to be accurate to the last digit shown. For a few classes of simple calculations, such as dividing small integer by small integer, that's generally true. But in general, even with all those 3's correct, that still doesn't give you any proof or assurance that the rest of the digits beyond that point will fulfill the obvious pattern you see within those limited digits. Try this on your calculator: 10000000000000000001/30000000000000000004 Now you know the correct mathematical result is not 1/3, so if you see 3333333333333333333333333333 and just assume the pattern of digits of 3 continues forever then you should know you are wrong. > Partial Sum Convergence does something like this. It uses a certain > number of finite significant digits, then it finds a converges value. And like the calculator, you can't really trust the value it gives you. It might be correct out to the number of digits it shows, or it might not be, and even if it's correct to that point you can't conclude any pattern of digits you see will be condinued forever in the mathematically correct result. > This convergence value is a number that the series APPROACHES NOT > EQUALS. Your language there is very sloppy. FIrst, do you mean the true mathematical limit or the approximate value given by the calculator? Second, by the series do you mean any finite segment of the series, i.e. any partial sum, or do you mean the whole series, which is defined as the limit of partial sum as number of terms increases without bound? If you mean exact mathematical value of limit, and value of whole infinite series, then you are wrong because those two ways of expression mean exactly the same thing hence are of course equal. If you mean exact value of limit but only an individual partial sum not the limit, then you are correct, no individual partial sum (in the series .3 + .03 + .003 + .0003 + ... we're discussing here) exactly equals the limit. If you mean the value produced by your calculator, then you are wrong because depending on which way roundoff error occurs, the result it gives you can be greater than the mathematical limit, or smaller, or exactly correct, and you have no way of knowing for any given problem which is the case, so any statement you make about it definitely one way or the other is a wrong statement. > .999... can NEVER = 1 When any competant mathematician uses the notation .999... he/she means the limit of the series, where the pattern of digits of 9 are continued forever without exception. That limit of the series is defined as the limit of partial sums. That limit is in fact exactly 1. You seem to mean something other than the limit of partial sums when you write that notation, and I've asked you several times to say what exactly you mean by your use of that notation, and still you have failed to tell me. So I ask you again, what do you mean by .999... as you use it just above, and as you used .99999... in the Subject of this thread. Do the two notations .999... and .99999... mean something different to you, or are they just two different amounts of effort put into denoting the same thing? In any case, what did *you* mean when you > Theorem 1 ( Divergence ) > If a series z_1 + z_2 + .... converges, then > lim z_m = 0 > m-->oo That is correct. Note that it's a one-way test. The converse is not true. (If you don't know what converse means, please look it up online.) > Hence if the series doesn't satisfy this condition, it diverges. Correct. That's the contrapositive, which is necessarily true if the original statement is true. (If you don't know what contrapositive means, look it up online.) > .999... = .9 + .09 + .... + ( 1 - 1/10^m) Your notation is unclear. What is that extra term at the very end? It appears to be either an error term or the individual mth term of the sequence that is being summed to make the series. If it's supposed to be the mth individual term, then you need to write: > .999... = .9 + .09 + .... + ( 1 - 1/10^m) + ... to indicate that the series doesn't stop at some point but continues on is *not* equal to the infinite series indicated by the left side, so If you intended the partial sum, you should have written: .9999999...99999999 = .9 + .09 + .... + ( 1 - 1/10^m) |<- (m digits) ->| If it's supposed to be the error term, that is the amount you have to fudge a partial sum to make up all the rest of terms you haven't included, then first of all that's not the correct error term, and the part before that isn't a partial sum, it's the notation for the entire series. You should have written: .999... = (.9 + .09 + .... + 0.000000...000000009) + 1/10^m |<- (m digits) ->| where the first parenthesized expression is the partial sum and the final expression is the error term. Note that if the right side (the mth partial sum plus the mth error term) is a constant, then the notation makes sense. In that case, the series converges if and only if the error term approaches zero, and if it converges then the constant right-side value is the limit. In this case, indeed the right side is a constant (independent of m), that constant value is 1, and the error term does indeed approach zero, so the limit of the series is exactly 1. The only difficult part of that proof is showing the right side has the same exact value for all values of m. That is you must show that .9 + 1/10 equals 1 (easy), and that (.9 + .09) + 1/100 also equals 1 (pretty easy), and that (.9 + .09 + .009) + 1/1000 also equals 1 (not too hard), etc. etc. etc. However this way of proving a series converges has virtually nothing to do with the divergence test you listed above, so I you must be confused to introduce it here. > Z_m = ( 1 - 1/10^m) If Z_m is supposed to be the mth term of the sequence you are summing, Z_m. > lim 1 - 1/10^m = 1 > m -->oo That's correct! The limit of partial sums (not individual terms) is exactly 1, proving the series converges to 1. > So, .999.... doesn't converge to 1. You must be awfully confused, given that you just finished proving that it *does* converge to 1. I suspect you are confusing the mth term with the mth partial sum. You didn't prove the mth term approaches 1, you proved the mth partial sum approaches 1. The mth term Z_m is 9/10^m which indeed approaches zero. === Subject: Re: .99999... still=/= 1 >> Z_m = ( 1 - 1/10^m) >If Z_m is supposed to be the mth term of the sequence you are summing, Nope, the convergence test shown in my Advanced Engineering Math book shows Z_M = z_1 + z_2 + z_3 + ... The total series is defined as, Z_m = 1 - 1/10^m The mth term isn't used in the test so 9/10^m isn't part of the convergence test. The test further states that if, m-->oo lim Z_m = 0 then it converges otherwise it diverges. m-->oo lim ( 1 - 1/10^m) = 1 .999... diverges so it doesn't equal 1 or converge to 1. >Z_m. >> lim 1 - 1/10^m = 1 >> m -->oo >That's correct! The limit of partial sums (not individual terms) is >exactly 1, proving the series converges to 1. >> So, .999.... doesn't converge to 1. >You must be awfully confused, given that you just finished proving that >it *does* converge to 1. I suspect you are confusing the mth term with >the mth partial sum. You didn't prove the mth term approaches 1, you >proved the mth partial sum approaches 1. The mth term Z_m is 9/10^m >which indeed approaches zero. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 > Z_M = > z_1 + z_2 + z_3 + ... That's utter nonsense because the first line (left side of equality) depends on M whereas the second line (right side of equality) doesn't depend on M. > Z_m = 1 - 1/10^m That's correct, assuming Z_m means the mth partial sum, i.e. 0.9 + 0.09 + ... + 0.00000000000...00000000009 |<- (m-1 '0' digits) ->| which partial sum can also be denoted as: 0.999999999...9999999999 |<- (m '9' digits) ->| > The test further states that if, > m-->oo > lim Z_m = 0 > then it converges otherwise it diverges. That's blatantly false!! You're saying if the partial sums converge to zero, then they converge, else they don't converge at all, in other words no series can ever converge to anything except zero. Here's a trivial counterexample: 1 + 0 + 0 + 0 + 0 + ... The first term is 1. Every term after the first is zero. Every partial sum is 1. Obviously this series converges to 1, and the limit of partial sums is 1: > m-->oo > lim Z_m = 1 Yet what you said above says it doesn't converge at all because the limit of partial sums isn't zero. I think you are totally confused, probably because you are using upper case Z_m to represent partial sum and lower case z_m to represent indiviual terms, yet you can't tell the difference between upper-case and lower-case when reading off your favorite calculator or WebSite etc. so you get the two all mixed up. I suggest you change notation: Use a_m to represent individual term, and S_m to represent partial sum, so you can tell the two letters apart and not mix them up in your mind. Also I suggest you tell us what you originally meant by .99999... in the Subject field, since you obviously don't mean what everyone else means by that notation, namely the limit of partial sums. === Subject: Re: .99999... still=/= 1 > You must be awfully confused, given that you just finished proving that > it *does* converge to 1. I suspect you are confusing the mth term with > the mth partial sum. You didn't prove the mth term approaches 1, you > proved the mth partial sum approaches 1. The mth term Z_m is 9/10^m > which indeed approaches zero. This fool (Enterprise) has no concept of limit. He does not know what the word means. Bob Kolker === Subject: Re: .99999... still=/= 1 >> You must be awfully confused, given that you just finished proving that >> it *does* converge to 1. I suspect you are confusing the mth term with >> the mth partial sum. You didn't prove the mth term approaches 1, you >> proved the mth partial sum approaches 1. The mth term Z_m is 9/10^m >> which indeed approaches zero. >This fool (Enterprise) has no concept of limit. He does not know what >the word means. >Bob Kolker Well at least take the word of the Advanced Engineering Math book. z_1 + z_2 + z_3 +... = Z_m The total series is used not the mth term. m-->oo lim (1 - 1/10^m) = 1 If, m-->oo lim Z_m = 0 then it converges otherwise diverges These are the words from the Advanced Engineering Math book. I hope you don't start calling this book foolish, too. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 > Well at least take the word of the Advanced Engineering Math book. > z_1 + z_2 + z_3 +... = Z_m > The total series is used not the mth term. You can't read either. A necessary condition for convergence is that the n-th term of the series (not the n-th partial sum, you putz!) go to zero. Bob Kolker === Subject: Re: .99999... still=/= 1 >> Well at least take the word of the Advanced Engineering Math book. >> z_1 + z_2 + z_3 +... = Z_m >> The total series is used not the mth term. >You can't read either. A necessary condition for convergence is that the >n-th term of the series (not the n-th partial sum, you putz!) go to zero. >Bob Kolker I looked at it again and I think you're right. z_m is the the mth term and it does equal 9/10^m. And it appears I was using the Partial Sums view of convergence. But even if .999... does converge to 1, .999... still=/= 1 .999...--->1 convergence The whole debate was whether or not .999... really EQUALS 1. It doesn't. Convergence doesn't mean equal to in any case. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 >> Well at least take the word of the Advanced Engineering Math book. > z_1 + z_2 + z_3 +... = Z_m > The total series is used not the mth term. >You can't read either. A necessary condition for convergence is that the >n-th term of the series (not the n-th partial sum, you putz!) go to zero. >Bob Kolker > I looked at it again and I think you're right. z_m is the the mth term and > it does equal 9/10^m. And it appears I was using the Partial Sums view of > convergence. But even if .999... does converge to 1, > .999... still=/= 1 > .999...--->1 convergence > The whole debate was whether or not > .999... really EQUALS 1. It doesn't. Convergence doesn't mean equal to in any > case. I see. You set up a test, fail the test, then declare the test to be meaningless. You're almost as good at this as James Harris. === Subject: Re: .99999... still=/= 1 > Well at least take the word of the Advanced Engineering Math book. > z_1 + z_2 + z_3 +... = Z_m > The total series is used not the mth term. >You can't read either. A necessary condition for convergence is that the >>n-th term of the series (not the n-th partial sum, you putz!) go to zero. >Bob Kolker >> I looked at it again and I think you're right. z_m is the the mth term >and >> it does equal 9/10^m. And it appears I was using the Partial Sums view of >> convergence. But even if .999... does converge to 1, >> .999... still=/= 1 >> .999...--->1 convergence >> The whole debate was whether or not >> .999... really EQUALS 1. It doesn't. Convergence doesn't mean equal to in >any >> case. >I see. You set up a test, fail the test, then declare the test to be >meaningless. You're almost as good at this as James Harris. First of all I didn't set up this test. Someone else did. They showed it as, m-->oo lim SUM 9/10^m = I took it to this, (1 - 10^m) = SUM 9/10^m m-->oo lim ( 1 - 1/10^m ) = 1 The test shows divergence relative to Partial SUMs. But I did agree, the mth term should be used in this test. And it shows convergence, but not perfectly equal to 1. I really didn't make any mistakes. You did, when you said, .999... = 1 .999... converges to 1 not equals 1. Why not admit your mistake? Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: .99999... still=/= 1 >> .999...* 10 - 9 = .999... >=> x * 10 - 9 = x where x = .999... >>=> 10x - x = 9 = 9x <==> x = 1 >So why you claim that .999... != 1? >> That's right. >> x = .999... > and after his calculations are done he shows it to be equal to >something >> else. > And another way to look at it let's set x = 1 to start with instead of >> .999... >> x = 1 >> 10x = 10 >> x = 1 > Here we can see x = 1 before and after. > And, > x = .999... > 10 *x = 9.999... > x = .999... > Here we can see x = .999... before and after. > This is often given as a layman's justification (not proof) that .999... >> =1. However mathematically, it's not quite right. >It is sufficient for a proof. >> You can't justify the >> step 10x = 9.999... without appealing to a theorem that says you can >> multiply each term of a convergent series by a constant, and the >> resulting series will converge to the constant times the sum of the >> original series. >Such a theorem would be a tautology of little value, since the >statement follows from the directly from distributive axiom a(b+c)= ab >+ ac. >If > x = 0.99999... [repeating n times] >Then, > x = 9/10^1 + 9/10^2 + .... 9/10^n > ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >Let a = 10 > 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n > = 9 + x >> If you could prove that, then you'd already have enough understanding of >> limits to know that .999... = 1. >Limits are unnecessary. The proof is simply, >x = 0.9_ >10x = 9.9_ (follows directly from the distributive axiom) > = 9 + x > 9x = 9 > x = 1 > Nope, this is incorrect. > The question was does .999... converge to 1, not 9.999... . >equation. Equations aren't used in convergence tests. > The basic convergence test is like this. >>If, >m-->oo >>Lim Z_m = 0 >then convergence otherwise divergence. >.999... is represented as (1 - 1/10^m) >Z_m = ( 1 - 1/10^m) >m-->oo >>Lim Z_m = 1 > This means divergence. So not only does it not converge to 1 but it >doesn't >>equal 1. And even if it did converge it wouldn't EQUAL 1. There isn't even >one >>series in all of math that does converge that really equals that convergence >>value. >>.999... --> diverges and =/= 1 >Smart's Alt. Physics News Group >>http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >>S. Enterprize (Science Journal) >>http://smart1234.s-enterprize.com/ >>convergrence test >according to your definition 1+ 1/2 + 1/3+1/4 +... would converge since >1/n--->0 as >n--> infinity. It is generally accepted that the sum given is divergent. >thus we have a counter example that disproves your convergence test. >Please find an acceptable trust worthy test for convergence > The form of of a harmonic series is > n= 1 to oo > lim SUM 1/n > Using The Root Convergence Test > ____ > lim n/ Z_n = L > n-->oo > L < 1 = converge > L > 1 = diverge > L = 1, test fails, no conclusion possible. > At n-->0 it's divergent , L >1 > n=0, L= 1 > n>1, L < 1 = convergent. > According to this convergence test, it is divergent, no conclusion possible , > and convergent. It depends on the point of view you take, or you can just > observe all cases. Are you ignoring your lsat post now? You're hopeless. === Subject: Re: .99999... still=/= 1 > >> .999...* 10 - 9 = .999... >=> x * 10 - 9 = x where x = .999... >=> 10x - x = 9 = 9x <==> x = 1 >So why you claim that .999... != 1? > > That's right. > x = .999... > > and after his calculations are done he shows it to be equal to >>something > else. > > And another way to look at it let's set x = 1 to start with instead > .999... > > > x = 1 > 10x = 10 > x = 1 > > Here we can see x = 1 before and after. > > And, > > x = .999... > > 10 *x = 9.999... > > x = .999... > > Here we can see x = .999... before and after. > > This is often given as a layman's justification (not proof) that .999... > =1. However mathematically, it's not quite right. >It is sufficient for a proof. >> You can't justify the > step 10x = 9.999... without appealing to a theorem that says you can > multiply each term of a convergent series by a constant, and the > resulting series will converge to the constant times the sum of the > original series. >Such a theorem would be a tautology of little value, since the >>statement follows from the directly from distributive axiom a(b+c)= ab >>+ ac. >If >> x = 0.99999... [repeating n times] >>Then, >> x = 9/10^1 + 9/10^2 + .... 9/10^n > ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >Let a = 10 > 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n >> = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n >> = 9 + x >> If you could prove that, then you'd already have enough understanding of > limits to know that .999... = 1. >Limits are unnecessary. The proof is simply, >>x = 0.9_ >>10x = 9.9_ (follows directly from the distributive axiom) >> = 9 + x >> 9x = 9 >> x = 1 > Nope, this is incorrect. > The question was does .999... converge to 1, not 9.999... . >equation. Equations aren't used in convergence tests. > The basic convergence test is like this. >If, >m-->oo >Lim Z_m = 0 >then convergence otherwise divergence. >.999... is represented as (1 - 1/10^m) >Z_m = ( 1 - 1/10^m) >m-->oo >Lim Z_m = 1 > This means divergence. So not only does it not converge to 1 but it >>doesn't >equal 1. And even if it did converge it wouldn't EQUAL 1. There isn't even >>one >series in all of math that does converge that really equals that >convergence >value. >.999... --> diverges and =/= 1 >Smart's Alt. Physics News Group >http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 >S. Enterprize (Science Journal) >http://smart1234.s-enterprize.com/ >>convergrence test >>according to your definition 1+ 1/2 + 1/3+1/4 +... would converge since >>1/n--->0 as >>n--> infinity. It is generally accepted that the sum given is divergent. >>thus we have a counter example that disproves your convergence test. >>Please find an acceptable trust worthy test for convergence >The form of of a harmonic series is >n= 1 to oo >lim SUM 1/n >Using The Root Convergence Test Correction : Use the Ratio Test The answer is still the same. Not this vvvvv > ____ >lim n/ Z_n = L >n-->oo >L < 1 = converge >L > 1 = diverge >L = 1, test fails, no conclusion possible. > At n-->0 it's divergent , L >1 > n=0, L= 1 > n>1, L < 1 = convergent. > According to this convergence test, it is divergent, no conclusion possible >and convergent. It depends on the point of view you take, or you can just >observe all cases. Smart's Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Vectors question 2 I am working through a question for a university assignment. I am not wanting to be given the answer, but am after some pointers. The question is: A train is travelling in a straight line at 24km/h. A movie stunt-man moves at 4km/h straight across its roof, at right angles to the direction of its motion. Draw a clear sketch of his resultant movement relative to the ground, and find the speed and direction of that movement. My take on this is as follows (way below is a very bad ASCII representation of the type of graph that is in my head...definetly not to any sort of scale) I first create a graph with x-axis distance travelled, and y-axis time. I can then plot u = 24i + j to show the speed that the train is travelling at. The line perpendicular to to u is the speed and direction of the stunt man, the length of that line will be 4. The line that forms the hyptonuse (v), represents the speed (length), and direction (x,y) that the stuntman has travelled. Am I on the right track (no pun intended)? If not, could you offer some suggestions. ^ | | 0 | 0 0 | 0 0 | v 0 0 (24,1) | 0 0 | 0 0 | 0 0 u | 0 0 |0____________________________> Please note, this is an assignment, so I am not wanting anyone to solve it for me. I do not want any reason to be accused of wrongdoings. Cassandra === Subject: THANKS!!! (calculus) has turned out to be much easier then the first half. I think tonight is the last night I can afford to give to my assignment (other essays beckon). All in all I had a thoroughly enjoyable weekend and am looking forward to the next assignment. double checking my work, and most importantly, patiently explaining concepts to me. Cassandra Thompson === Subject: Lines, points and planes Aah, too tired to explain too much. I have answered the following question but fear I am very incorrect. Q) Find the equation of the plane that passes through the origin and is perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) A) Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real number. v = 3i + j - k Q1 = (0, 0, 0) a(x [CapitalEth] x0) + b(y [CapitalEth] y0) + c(z [CapitalEth] z0) = 0 3(x [CapitalEth] 0) + 1(y [CapitalEth] 0) [CapitalEth] 1(z [CapitalEth] 0) = 0 3x + y [CapitalEth] z = 0 Therefore the equation of the plane is 3x + y [CapitalEth] z = 0. I was quite confident that it was correct. Except now I am attempting a question in which we are given a point and a plane, and asked to find the equation of the line perpendicular to the plane. I thought I would just work backwards using the above question, but I feel like my answers are all wrong. Any suggestions? Cassandra. === Subject: Re: Lines, points and planes > Aah, too tired to explain too much. I have answered the following > question but fear I am very incorrect. > Q) Find the equation of the plane that passes through the origin and is > perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) > A) > Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real > number. > v = 3i + j - k > Q1 = (0, 0, 0) > a(x ? x0) + b(y ? y0) + c(z ? z0) = 0 > 3(x ? 0) + 1(y ? 0) ? 1(z ? 0) = 0 > 3x + y ? z = 0 > Therefore the equation of the plane is 3x + y ? z = 0. > I was quite confident that it was correct. Except ... Omigod. Google turned some of your minus signs into ?s when I went into preview mode. Why, I don't know. (You didn't use some weird character like a dash instead of an ASCII hyphen did you?) Apart from the Google trauma, your answer looks fine to me. If you are unsure whether your answers are right, it's a great idea to check them with a numerical example (if you haven't already done so). In this example I would check the answer as follows... 1. Find the intersection of the line and the plane - call it point A. 2. Pick any point on the line (other than the point of intersection) - call it point B. 3. Pick any point on the plane (other than the point of intersection) - call it point C. Check that ABC is a right-angled triangle with right angle at A (use Pythagoras). If it is, then that gives you great confidence that you got the answer right. You have to be REALLY unlucky for the check to work but the answer to still be wrong. > ... now I am attempting a > question in which we are given a point and a plane, and asked to find > the equation of the line perpendicular to the plane. I thought I would > just work backwards using the above question, but I feel like my answers > are all wrong. > Any suggestions? > Cassandra. The equation of the plane will immediately give you the normal vector (i.e. the vector of the required perpendicular line). Specifically, if the equation of the plane is given in Cartesian form as ax + by + cz + d = 0, then the normal vector is (a,b,c) (or any non-zero scalar multiple thereof). But you knew that, right? Now that you have the line vector and the coordinates of one point on the line, it is straightforward to find the (parametric) equation of too, so I think you should be OK... You can check your answer using a similar right-angled triangle construction. === Subject: Re: Lines, points and planes >>Aah, too tired to explain too much. I have answered the following >>question but fear I am very incorrect. >>Q) Find the equation of the plane that passes through the origin and is >>perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) >>A) >>Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real >>number. >>v = 3i + j - k >>Q1 = (0, 0, 0) >>a(x ? x0) + b(y ? y0) + c(z ? z0) = 0 >>3(x ? 0) + 1(y ? 0) ? 1(z ? 0) = 0 >>3x + y ? z = 0 >>Therefore the equation of the plane is 3x + y ? z = 0. >>I was quite confident that it was correct. Except ... > Omigod. Google turned some of your minus signs into ?s when I went > into preview mode. Why, I don't know. (You didn't use some weird > character like a dash instead of an ASCII hyphen did you?) I am typing my assignment in MS Word, I copied the above workings straight from word. Yes, it would have been dashes not minus signs as I notice that MS Word does alot of auto-correcting with regard to minus-signs/dashes. Sorry about that. Cassandra. === Subject: Re: Lines, points and planes > Aah, too tired to explain too much. I have answered the following > question but fear I am very incorrect. > Q) Find the equation of the plane that passes through the origin and is > perpendicular to (x, y, z) = (1, 0, -2) + t(3, 1, -1) > A) > Line L is defined by (x, y, z ) = (1, 0, -2) + t(3, 1, -1), t any real > number. > v = 3i + j - k > Q1 = (0, 0, 0) > a(x [CapitalEth] x0) + b(y [CapitalEth] y0) + c(z [CapitalEth] z0) = 0 > 3(x [CapitalEth] 0) + 1(y [CapitalEth] 0) [CapitalEth] 1(z [CapitalEth] 0) = 0 > 3x + y [CapitalEth] z = 0 > Therefore the equation of the plane is 3x + y [CapitalEth] z = 0. Starting from the end... The normal vector of the plane Ax + By + Cz + D = 0 is (A, B, C), so the normal of your plane is (3, 1, -1). The direction vector of the line you were given is also (3, 1, -1), so the line is indeed perpendicular to the plane. 3 * 0 + 0 - 0 = 0, so your plane contains the origin. Thus, the plane you got is the right answer. > I was quite confident that it was correct. Except now I am attempting a > question in which we are given a point and a plane, and asked to find > the equation of the line perpendicular to the plane. I thought I would > just work backwards using the above question, but I feel like my answers > are all wrong. > Any suggestions? If you have point P = (Xp, Yp, Zp) and plane Ax + By + Cz + D = 0, then the normal to the plane is n = (A, B, C) and the line through P in the direction of n is simply l = (Xp, Yp, Zp) + t * (A, B, C). Based on the knowledge you demonstrated in the rest of your posts here, my hunch is that that's exactly what you did, that your answers are probably right, and that you just need to convince yourself of that :-) but if you really are wrong or unsure, post a specific example and we'll try to find specific errors. meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Proving planes are not parallel The definition that I have is as follows: Two planes are parallel if their normal vectors are parallel, that is, if the cross product of their normal vectors is zero. However my understanding is that a cross product yields another vector, not a scalar. So I am confused. My testbook has one so-called worked example, but it borders more along the lines of we'll leave the proof as an exercise, ie it leaves out the middle bits. *************************************** Two parallel planes. P_1: 2x + 3y - z = 3 P_2: -4x - 6y + 2z = 8 so n_1 = 2i + 3j - k, n_2 = -4i - 6j + 2k The planes are parallel because: n_2 = -2n_1 and n_1 x n_2 = 0 *************************************** I am trying to prove two planes are not parallel. Should I write them as vectors, then show that the vectors are not multiples of each other? ie show n_1 <> x(n_2) How does n_1 x n_2 = 0 in the above example? I thought n_1 x n_2 = -1i + 0j + 4k Cassandra === Subject: Re: Proving planes are not parallel > The definition that I have is as follows: > Two planes are parallel if their normal vectors are parallel, that is, > if the cross product of their normal vectors is zero. > However my understanding is that a cross product yields another vector, > not a scalar. So I am confused. There is a zero vector in every vector space, such that if 0 is the zero scalar and Z is the zero vector and . indicates product of scalar with vector then 0.Z = Z Since it is usually obvious from the context when a zero object is to be a scalar and when a vector, the same word is used ambiguously for both, as in this case. To say that the cross product of two vectors is zero clearly must mean the zero vector, not the zero scalar. Does this clear things up a bit? >... > Cassandra === Subject: Re: Proving planes are not parallel Cassandra Thompson a .8ecrit : > The definition that I have is as follows: > Two planes are parallel if their normal vectors are parallel, that is, > if the cross product of their normal vectors is zero. > However my understanding is that a cross product yields another vector, > not a scalar. So I am confused. In those formulations, zero stands for the number 0 or for the null vector > My testbook has one so-called worked example, but it borders more along > the lines of we'll leave the proof as an exercise, ie it leaves out > the middle bits. > *************************************** > Two parallel planes. > P_1: 2x + 3y - z = 3 > P_2: -4x - 6y + 2z = 8 > so > n_1 = 2i + 3j - k, > n_2 = -4i - 6j + 2k > The planes are parallel because: > n_2 = -2n_1 > and n_1 x n_2 = 0 > *************************************** > I am trying to prove two planes are not parallel. Should I write them as > vectors, then show that the vectors are not multiples of each other? > ie show > n_1 <> x(n_2) It is one possibility (better to write n_1 <> k(n_2) ) ; another one is to calculate the cross-product of n_1 and n_2 and to show it is not null > How does n_1 x n_2 = 0 in the above example? I thought > n_1 x n_2 = -1i + 0j + 4k You must have made a calculation mistake. First, n_1 is indeed equal to -2 n_2 (check it). Second, the i component of the cross-product n_1 x n_2 is y_1z_2-y_2z_1= 3*2-(-1)*(-6)=0 , and not -1 as you say... > Cassandra === Subject: Re: Proving planes are not parallel > Cassandra Thompson a .8ecrit : >> The definition that I have is as follows: >> Two planes are parallel if their normal vectors are parallel, that is, >> if the cross product of their normal vectors is zero. >> However my understanding is that a cross product yields another >> vector, not a scalar. So I am confused. > In those formulations, zero stands for the number 0 or for the null vector >> My testbook has one so-called worked example, but it borders more >> along the lines of we'll leave the proof as an exercise, ie it >> leaves out the middle bits. >> *************************************** >> Two parallel planes. >> P_1: 2x + 3y - z = 3 >> P_2: -4x - 6y + 2z = 8 >> so >> n_1 = 2i + 3j - k, >> n_2 = -4i - 6j + 2k >> The planes are parallel because: >> n_2 = -2n_1 >> and n_1 x n_2 = 0 >> *************************************** >> I am trying to prove two planes are not parallel. Should I write them >> as vectors, then show that the vectors are not multiples of each other? >> ie show >> n_1 <> x(n_2) > It is one possibility (better to write n_1 <> k(n_2) ) ; another one is > to calculate the cross-product of n_1 and n_2 and to show it is not null >> How does n_1 x n_2 = 0 in the above example? I thought >> n_1 x n_2 = -1i + 0j + 4k > You must have made a calculation mistake. First, n_1 is indeed equal to > -2 n_2 (check it). Second, the i component of the cross-product n_1 x > n_2 is y_1z_2-y_2z_1= 3*2-(-1)*(-6)=0 , and not -1 as you say... >> Cassandra simple algebra errors that I have posted to the newsgroup. Although I have to admit I find the nature of cross-products begs for mistakes. Perhaps I should learn to calculate the cross-product using a matrix, or better still get a graphing calculator! I have answered the assignment equation using the n_1 <> k(n_2) method, product and null vector. Cassie === Subject: Re: Proving planes are not parallel >simple algebra errors that I have posted to the newsgroup. Although I >have to admit I find the nature of cross-products begs for mistakes. >Perhaps I should learn to calculate the cross-product using a matrix, or >better still get a graphing calculator! >I have answered the assignment equation using the n_1 <> k(n_2) method, >product and null vector. >Cassie Don't feel ashamed Cassie. We all make mistakes, from inadvertent typo's to just flat being wrong. In my opinion it is always better to check whether v2 = c * v1 than v1 cross v2 = zero vector if testing for parallelness exactly because it is simpler and a less error - prone method. The cross product does have many uses, though, and you will need to be able to calculate them with a minimum of errors. There are several methods in use. My favorite, which requires a little more writing but, at least for me, keeps the errors to a minimum is the following. To cross with I first make a helper where I write the vectors starting and ending with the middle component: b c a b e f d e Now the components of the cross product are gotten by computing the left, middle, and right 2x2 determinants in the helper: In particular, this gets the sign of the middle component correct without giving it any additional thought. --Lynn === Subject: Re: Proving planes are not parallel >b c a b >e f d e >Now the components of the cross product are gotten by computing the >left, middle, and right 2x2 determinants in the helper: > Talk about typos...., of course the first one is bf - ec :-) --Lynn === Subject: Re: Proving planes are not parallel >>b c a b >>e f d e >>Now the components of the cross product are gotten by computing the >>left, middle, and right 2x2 determinants in the helper: >> Talk about typos...., of course the first one is bf - ec > :-) > --Lynn Yes, that is a much sinpler way of doing it. I have been writing the equations down, then writing down the general formula, then putting small labels on the equation (a_1, b_1 etc) then trying to look in three places at once to be sure I am putting the right number inthe right place. Of course this could easily be done on a calculator, scilab, or even using MS Excel, but I need to master it before I turn to technology. Cassandra === Subject: Re: Proving planes are not parallel >b c a b >e f d e >Now the components of the cross product are gotten by computing the >left, middle, and right 2x2 determinants in the helper: >> Talk about typos...., of course the first one is bf - ec >> :-) >> --Lynn >Yes, that is a much sinpler way of doing it. >I have been writing the equations down, then writing down the general >formula, then putting small labels on the equation (a_1, b_1 etc) then >trying to look in three places at once to be sure I am putting the right >number inthe right place. Ugh!!! Then I'm glad I posted that for you. I always told my classes to not even bother to memorize the formula. Just use the helper. --Lynn === Subject: Unit vectors... I am nearing the end of my bunch of vector questions (...hopefully...) I have found a unit vector in the direction of b = (3, 1, 0) below. I have also taken a crack at finding a vector four units long in the opposite direction. I am not sure about the math for the latter. |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) u (the unit vector) = (3/sqrt(10), 1/sqrt(10), 0) To find a vector four units long in the opposite direction, I imagine I would only need multiple the unit vector by the scalar -4? ie -4u = (-12/sqrt(10), -4/sqrt(10), 0). Is it this simple? cassandra === Subject: Re: Unit vectors... I have found a unit vector in the direction of b = (3, 1, 0) below. I > have also taken a crack at finding a vector four units long in the > opposite direction. I am not sure about the math for the latter. > |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) > u (the unit vector) = (3/sqrt(10), 1/sqrt(10), 0) Indeed, when |v| /= 0, |v/|v|| = (1/|v|)|v| = 1 > To find a vector four units long in the opposite direction, I imagine I > would only need multiple the unit vector by the scalar -4? > ie > -4u = (-12/sqrt(10), -4/sqrt(10), 0). > Is it this simple? Yup. Surprise, not everything is hard. Exercise: what's the length of (1,1,1,1)? Show (1/2, 1/2, 1/2, 1/2) is a unit vector. === Subject: Re: Unit vectors... >>I have found a unit vector in the direction of b = (3, 1, 0) below. I >>have also taken a crack at finding a vector four units long in the >>opposite direction. I am not sure about the math for the latter. >>|b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) >>u (the unit vector) = (3/sqrt(10), 1/sqrt(10), 0) > Indeed, when |v| /= 0, |v/|v|| = (1/|v|)|v| = 1 >>To find a vector four units long in the opposite direction, I imagine I >>would only need multiple the unit vector by the scalar -4? >>ie >>-4u = (-12/sqrt(10), -4/sqrt(10), 0). >>Is it this simple? > Yup. Surprise, not everything is hard. > Exercise: what's the length of (1,1,1,1)? > Show (1/2, 1/2, 1/2, 1/2) is a unit vector. Yes, when doing questions that take me hours, a 2 minute question makes me wonder what I am doing wrong. u = (1,1,1,1) = sqrt(4) = 2 v = (1/2, 1/2, 1/2, 1/2 = sqrt(1) = 1 Ahhh, very nice. This has been a great weekend, hard but satisfying, I really enjoy learning this stuff. I also really appreciated the help because there is something satisfying about understanding it at the end. Just lovely. === Subject: Re: Unit vectors... >I am nearing the end of my bunch of vector questions (...hopefully...) > I have found a unit vector in the direction of b = (3, 1, 0) below. I > have also taken a crack at finding a vector four units long in the > opposite direction. I am not sure about the math for the latter. > |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) > u (the unit vector) = (3/sqrt(10), 1/sqrt(10), 0) > To find a vector four units long in the opposite direction, I imagine I > would only need multiple the unit vector by the scalar -4? > ie > -4u = (-12/sqrt(10), -4/sqrt(10), 0). > Is it this simple? yes. === Subject: Finding the equation of a plane. Okay me again..I should explain that I am an external student. Which means I don't have any lectures, nor do I even have contact with any other students in my class. I hope this explains why I am asking so many questions. If I had study group I would be very involved in it. plane for vector a and b. I think I have done it correctly (hopefully..). a = (1, -2, 1) b = (3, 1, 0) w = a x b = (-1, 3, 7) Find the plane passing through the point A = (1, -2, 1) having the normal vector n = [CapitalEth]1i + 3j + 7k. a(x [CapitalEth] x0) + b(y [CapitalEth] y0) + c(z [CapitalEth] z0) = 0 -1(x [CapitalEth] 1) + 3(y + 2) + 7(z [CapitalEth] 1) = 0 [CapitalEth]x + 1 + 3y + 6 + 7z [CapitalEth] 7 = 0 [CapitalEth]x + 3y + 7z = 0 Confirm this by finding the plane passing through the point B = (3, 1, 0) having the normal vector n = [CapitalEth]1i + 3j + 7k. a(x [CapitalEth] x0) + b(y [CapitalEth] y0) + c(z [CapitalEth] z0) = 0 [CapitalEth]1(x [CapitalEth] 3) + 3(y [CapitalEth] 1) + 7(z [CapitalEth] 0) = 0 [CapitalEth]x + 3 + 3y [CapitalEth] 3 + 7z [CapitalEth] 0 = 0 [CapitalEth]x + 3y + 7z = 0 Therefore, the equation of the plain containing a and b is [CapitalEth]x + 3y + 7z = 0 Cassandra === Subject: Re: Finding the equation of a plane. >Okay me again.. Cassandra, many of us read several math newsgroups. I think you will find that you don't really need to post your questions to more than one newsgroup. When you cross-post lots of questions we find ourselves re-reading the same questions, not sure whether you are posting something new. So I would suggest you pick the newsgroup you like the best and try posting to just it, or at least don't post the same questions in different groups. --Lynn === Subject: Re: Finding the equation of a plane. > a(x - x0) + b(y - y0) + c(z - z0) = 0 > -1(x - 1) + 3(y + 2) + 7(z - 1) = 0 > -x + 1 + 3y + 6 + 7z - 7 = 0 > -x + 3y + 7z = 0 That's correct. In general, if you are given N, the normal to the plane, and P, the point the plane is to pass through, the equation of the plane is (where . is the dot product and r=(x, y, z)). N.(r - P) = 0 If you consider the geometrical meaning of the dot product of two vectors being zero, it should be obvious why that's the equation of the plane normal to N passing through P. Anyways, you can manipulate that into N.r = N.P or (-1, 3, 7).(x, y, z) = (-1, 3, 7).(1, -2, 1) -x + 3y + 7z = -1(1) + 3(-2) + 7(1) -x + 3y + 7z = 0 -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Finding the equation of a plane. > In general, if you are given N, the normal to the plane, and P, > the point the plane is to pass through, the equation of the > plane is (where . is the dot product and r=(x, y, z)). > N.(r - P) = 0 Absolutely. IMHO, this is by far the best way to remember both the geometry and vector algebra needed to represent a plane in R^{n}. Just as identifying a line in R^{n} requires a point x= (x1,x2,...,xn) and a direction v=(v1,v2,...,vn). Then the line is simply r(t) = x+vt, where r(t) is a function from the real numbers to R^{n}. DM === Subject: Magnitude and area of a parallelogram Firstly, I am not sure if it annoying that I am using the same thread to ask so many question, however I thought it might be equally annoying to be starting lots of new threads. At least if it is in the one thread you have a good idea of my background. This question is not part of the assignment, however working through the assignment has shed light on a couple a few equations, which in turn have confused me. I am hoping to ask some questions. We have two vectors: a = (1, -2, 1) b = (3, 1, 0) The cross-product of these two vectors is: a x b = (-1, 3, 7) let w = (-1, 3, 7) Okay all good so far I think. Now if I want to find the magnitude of w, I treat it exactly the same as I would if I wanted to find the magnitude of a and b. |w| = sqrt((-1)^2 + 3^2 + 7^2) = sqrt(17) However I also know that the following formulae is true: |a x b| = |a||b| sin& Is the |a x b| in this formulae the same as |w|? Does |w| = |a||b| sin & hold true? Does |a x b| = sqrt(17)? I guess I am just confused by the to different formulaes. ie |w| = sqrt(w_1^2 + w_2^2 + w_3^2) |a x b| = |a||b| sin& Cassandra. === Subject: Re: Magnitude and area of a parallelogram > Firstly, I am not sure if it annoying that I am using the same thread to > ask so many question, however I thought it might be equally annoying to > be starting lots of new threads. At least if it is in the one thread you > have a good idea of my background. > This question is not part of the assignment, however working through the > assignment has shed light on a couple a few equations, which in turn > have confused me. I am hoping to ask some questions. > We have two vectors: > a = (1, -2, 1) > b = (3, 1, 0) > The cross-product of these two vectors is: > a x b = (-1, 3, 7) > let w = (-1, 3, 7) > Okay all good so far I think. Now if I want to find the magnitude of w, > I treat it exactly the same as I would if I wanted to find the magnitude > of a and b. > |w| = sqrt((-1)^2 + 3^2 + 7^2) = sqrt(17) > However I also know that the following formulae is true: > |a x b| = |a||b| sin& > Is the |a x b| in this formulae the same as |w|? > Does |w| = |a||b| sin & hold true? > Does |a x b| = sqrt(17)? > I guess I am just confused by the to different formulaes. > ie > |w| = sqrt(w 1^2 + w 2^2 + w 3^2) > |a x b| = |a||b| sin& That's all correct. This is one way you can determine & (within -pi/2 to pi/2). Another way is via the dot products (within 0 and pi). And the parallelogram? Tomasso. === Subject: Re: Magnitude and area of a parallelogram - angle from cos and sin >>Firstly, I am not sure if it annoying that I am using the same thread to >>ask so many question, however I thought it might be equally annoying to >>be starting lots of new threads. At least if it is in the one thread you >>have a good idea of my background. >>This question is not part of the assignment, however working through the >>assignment has shed light on a couple a few equations, which in turn >>have confused me. I am hoping to ask some questions. >>We have two vectors: >>a = (1, -2, 1) >>b = (3, 1, 0) >>The cross-product of these two vectors is: >>a x b = (-1, 3, 7) >>let w = (-1, 3, 7) >>Okay all good so far I think. Now if I want to find the magnitude of w, >>I treat it exactly the same as I would if I wanted to find the magnitude >>of a and b. >>|w| = sqrt((-1)^2 + 3^2 + 7^2) = sqrt(17) >>However I also know that the following formulae is true: >>|a x b| = |a||b| sin& >>Is the |a x b| in this formulae the same as |w|? >>Does |w| = |a||b| sin & hold true? >>Does |a x b| = sqrt(17)? >>I guess I am just confused by the to different formulaes. >>ie >>|w| = sqrt(w_1^2 + w_2^2 + w_3^2) >>|a x b| = |a||b| sin& >> >That's all correct. This is one way you can determine & (within -pi/2 to pi/2). >Another way is via the dot products (within 0 and pi). >And the parallelogram? >Tomasso. The best way to determine & is from =both= its cosine and sine. If one uses only sin(&) to determine &, then one gets & with a severe loss of precision if sin(&) happens to be close to 1; in general, the absolute error in f(x) = |f '(x)| times the absolute error in x. And |d(arcsin x)/dx| = 1/sqrt (1 - xx). Whenever one knows both sin(&) and cos(&) (which is not always the case) one does not suffer loss of precision in &. Johan E. Mebius === Subject: Re: Magnitude and area of a parallelogram > That's all correct. This is one way you can determine & (within -pi/2 to pi/2). > Another way is via the dot products (within 0 and pi). > And the parallelogram? > Tomasso. I am very confused right now. Last night I went to bed thinking I understood this stuff, now I realise I have errors everywhere and non of it makes sense. If I were to give your two vectors, how would you go about finding the parallelogram formed by them? Last night I thought that that was what the cross-product formed. Now I have no idea why I came to that conclusion. The cross-products of the vectors a = (1, -2, 1) and b = (3, 1, 0) is w = (-1, 3, 7). Which clearly does not form a parallelogram. I have create a graph, this graph shows that the parallelogram formed by a and b has the fourth point x = (4, -1, 1). so how did I get this point? I got it straight off the graph. How do I get it mathematically? I could just add the points ie x = (3 + 1, -2 + 1, 1 + 0). But I don't recall seeing this formula. So is it the right thing to do. And what of the cross-product. Is it just some point in space that doesn't even lie on the parallelogram? If so is its sole relationship to the parallelogram the fact that its magnitude (length) is equal to the area of the parallelogram (in square units)? If this is all true, now I am going to attempt to use the cross-product of a and b (ie w) to find the equation of the plane containing a and b. I am assuming this plane doesn't contain w, but the vector w just has these wonderful properties that assist us in finding the plane? Time for breakfast I think. cassandra === Subject: Re: Magnitude and area of a parallelogram - >> That's all correct. This is one way you can determine & (within -pi/2 >> to pi/2). Another way is via the dot products (within 0 and pi). >> And the parallelogram? >> Tomasso. > I am very confused right now. Last night I went to bed thinking I > understood this stuff, now I realise I have errors everywhere and non > of it makes sense. > If I were to give your two vectors, how would you go about finding the > parallelogram formed by them? > Last night I thought that that was what the cross-product formed. Now > I have no idea why I came to that conclusion. > The cross-products of the vectors a = (1, -2, 1) and b = (3, 1, 0) is > w = (-1, 3, 7). Which clearly does not form a parallelogram. > I have create a graph, this graph shows that the parallelogram formed > by a and b has the fourth point x = (4, -1, 1). > so how did I get this point? I got it straight off the graph. How do I > get it mathematically? I could just add the points ie x = (3 + 1, -2 + > 1, 1 + 0). But I don't recall seeing this formula. So is it the right > thing to do. > And what of the cross-product. Is it just some point in space that > doesn't even lie on the parallelogram? If so is its sole relationship > to the parallelogram the fact that its magnitude (length) is equal to > the area of the parallelogram (in square units)? > If this is all true, now I am going to attempt to use the > cross-product of a and b (ie w) to find the equation of the plane > containing a and b. I am assuming this plane doesn't contain w, but > the vector w just has these wonderful properties that assist us in > finding the plane? > Time for breakfast I think. > cassandra Or perhaps time for a nightrest... I guess that you will learn about vectors and bivectors and about polar and axial vectors at a later time, but let me give you a foretaste. In 3D vector algebra and analysis it is common practice to identify a vector V of a certain length L with a pancake P of =area= L in a plane perpendicular to V. The exact positions of the vector and the pancake in space and the exact shape of the pancake do not matter. What does matter is the sense of orientation in the plane of the pancake; by convention the direction of V and the rotation sense in P fit to each other like the translation and rotation of a =right-handed= screw. Now look at the projections of V onto the axes of a Cartesian coordinate system (components of V), and also at the projections of P onto the coordinate planes of the same system (components of P). They are pairwise equal: Vx = Pyz, Vy = Pzx, Vz = Pxy. And now the main reason why V and P can be identified: their components change in identical ways when V and P are rotated together, maintaining their perpendicular relation. Finally the vector product: two vectors A and B with a common origin O span a parallelogram (O, A, A+B, B). Its area is |A|.|B|.|sin (A, B)|. This is also the length of the vector product A x B. It is a property of 3D space that all vectors that are perpendicalar to both A and B are scalar multiples of A x B. Try and find out for yourself what happens when one does not allow only rotations, but also reflections. Enjoy your breakfast! Johan E. Mebius === Subject: Re: Magnitude and area of a parallelogram Cassandra Thompson dixit: >> That's all correct. This is one way you can determine & (within -pi/2 to pi/2). >> Another way is via the dot products (within 0 and pi). >> And the parallelogram? >> Tomasso. >I am very confused right now. Last night I went to bed thinking I >understood this stuff, now I realise I have errors everywhere and non of >it makes sense. >If I were to give your two vectors, how would you go about finding the >parallelogram formed by them? >Last night I thought that that was what the cross-product formed. Now I >have no idea why I came to that conclusion. >The cross-products of the vectors a = (1, -2, 1) and b = (3, 1, 0) is w >= (-1, 3, 7). Which clearly does not form a parallelogram. >I have create a graph, this graph shows that the parallelogram formed by >a and b has the fourth point x = (4, -1, 1). >so how did I get this point? I got it straight off the graph. How do I >get it mathematically? I could just add the points ie x = (3 + 1, -2 + >1, 1 + 0). But I don't recall seeing this formula. So is it the right >thing to do. This is simple vector addition, if you consider vector 'a' starts at the origin then you have a+b = x, so yes this formula is correct to get the other point of the parallelogram. The vector (-1,3,7) which has magnitude sqrt(59) and not sqrt(17) as you said in your 1st message, is perpendicular to the plane that contains vectors a and b (and also x, if you consider it a vector). See this: http://members.tripod.com/~Paul_Kirby/vector/Vcrossproduct.html toward the bottom of the page there is a diagram explaining this. See the rest of the site too: http://members.tripod.com/~Paul_Kirby/vector/VectorLand.html >And what of the cross-product. Is it just some point in space that >doesn't even lie on the parallelogram? If so is its sole relationship to >the parallelogram the fact that its magnitude (length) is equal to the >area of the parallelogram (in square units)? >If this is all true, now I am going to attempt to use the cross-product >of a and b (ie w) to find the equation of the plane containing a and b. >I am assuming this plane doesn't contain w, but the vector w just has >these wonderful properties that assist us in finding the plane? For the plane equation see http://www.netcomuk.co.uk/~jenolive/vect11.html http://www.netcomuk.co.uk/~jenolive/vect12.html http://www.netcomuk.co.uk/~jenolive/vect15.html and the rest of the site also. >Time for breakfast I think. >cassandra === Subject: Re: Magnitude and area of a parallelogram > Cassandra Thompson dixit: >That's all correct. This is one way you can determine & (within -pi/2 to pi/2). >Another way is via the dot products (within 0 and pi). >And the parallelogram? >Tomasso. >>I am very confused right now. Last night I went to bed thinking I >>understood this stuff, now I realise I have errors everywhere and non of >>it makes sense. >>If I were to give your two vectors, how would you go about finding the >>parallelogram formed by them? >>Last night I thought that that was what the cross-product formed. Now I >>have no idea why I came to that conclusion. >>The cross-products of the vectors a = (1, -2, 1) and b = (3, 1, 0) is w >>= (-1, 3, 7). Which clearly does not form a parallelogram. >>I have create a graph, this graph shows that the parallelogram formed by >>a and b has the fourth point x = (4, -1, 1). >>so how did I get this point? I got it straight off the graph. How do I >>get it mathematically? I could just add the points ie x = (3 + 1, -2 + >>1, 1 + 0). But I don't recall seeing this formula. So is it the right >>thing to do. > This is simple vector addition, if you consider vector 'a' starts at > the origin then you have a+b = x, so yes this formula is correct to > get the other point of the parallelogram. The vector (-1,3,7) which > has magnitude sqrt(59) and not sqrt(17) as you said in your 1st > message, is perpendicular to the plane that contains vectors a and b > (and also x, if you consider it a vector). See this: > http://members.tripod.com/~Paul_Kirby/vector/Vcrossproduct.html > toward the bottom of the page there is a diagram explaining this. > See the rest of the site too: > http://members.tripod.com/~Paul_Kirby/vector/VectorLand.html >>And what of the cross-product. Is it just some point in space that >>doesn't even lie on the parallelogram? If so is its sole relationship to >>the parallelogram the fact that its magnitude (length) is equal to the >>area of the parallelogram (in square units)? >>If this is all true, now I am going to attempt to use the cross-product >>of a and b (ie w) to find the equation of the plane containing a and b. >>I am assuming this plane doesn't contain w, but the vector w just has >>these wonderful properties that assist us in finding the plane? > For the plane equation see > http://www.netcomuk.co.uk/~jenolive/vect11.html > http://www.netcomuk.co.uk/~jenolive/vect12.html > http://www.netcomuk.co.uk/~jenolive/vect15.html > and the rest of the site also. >>Time for breakfast I think. >>cassandra I found this link http://www.phy.syr.edu/courses/java-suite/crosspro.html off one of the above pages. The applet on that page, and the explanation below was extremely simple and did a wonderful job of explaining it all. I now have 2 vectors and the normal to those two vectors. I am now trying to find the equation of the plane. Not sure how to go about it. I see lots of equations for finding the equation of a plane, but none for the situation that I have two vectors and the normal. I am guessing I am going to need to do some manipulating? Cassandra === Subject: Re: Magnitude and area of a parallelogram > If I were to give your two vectors, how would you go about finding the > parallelogram formed by them? > Last night I thought that that was what the cross-product formed. Now I > have no idea why I came to that conclusion. >... > And what of the cross-product. Is it just some point in space that > doesn't even lie on the parallelogram? If so is its sole relationship to > the parallelogram the fact that its magnitude (length) is equal to the > area of the parallelogram (in square units)? Some questions to get your brain re-focussed: (1) What is the area of a triangle? Given the base and the height? (2) What is the area of a triangle? Given two sides and the included angle? (3) Can you construct a parallelogram from, say, two triangles? (4) What was your formula for ||w||? (5) Does the direction of w have anything to do with the area of the parallelogram? It's plane? etc. Tomasso. === Subject: Re: Magnitude and area of a parallelogram >>If I were to give your two vectors, how would you go about finding the >>parallelogram formed by them? >>Last night I thought that that was what the cross-product formed. Now I >>have no idea why I came to that conclusion. >>... >>And what of the cross-product. Is it just some point in space that >>doesn't even lie on the parallelogram? If so is its sole relationship to >>the parallelogram the fact that its magnitude (length) is equal to the >>area of the parallelogram (in square units)? > Some questions to get your brain re-focussed: > (1) What is the area of a triangle? Given the base and the height? > (2) What is the area of a triangle? Given two sides and the included angle? > (3) Can you construct a parallelogram from, say, two triangles? > (4) What was your formula for ||w||? > (5) Does the direction of w have anything to do with the area of the parallelogram? It's plane? etc. > Tomasso. Showing on a 3d graph that the formulae |u||v|sin& is the area of the paralleagram. But your questions 4 & 5 are where I am stuck. My formula for ||w|| is |w| = a_1*b_1i + a_2*b_2j + a_3*b_3k = -1j + 3j + 7k Does the direction of w have anything to do with the area of the parallelogram, and its plane? I have no idea. I am assuming by your question that it does. So I will go look. But all incite would be appreciated, because I am really clueless on this one. Cassandra === Subject: Re: Magnitude and area of a parallelogram > But your questions 4 & 5 are where I am stuck. > My formula for ||w|| is > |w| = a 1*b 1i + a 2*b 2j + a 3*b 3k > = -1j + 3j + 7k Nearly. That is w (vector). ||w|| = area of parallelogram = sqrt (1+9+49) = your answer. w is normal to the plan of the parallelogram. Tomasso. === Subject: Re: Magnitude and area of a parallelogram >>But your questions 4 & 5 are where I am stuck. >>My formula for ||w|| is >>|w| = a_1*b_1i + a_2*b_2j + a_3*b_3k >> = -1j + 3j + 7k > Nearly. That is w (vector). > ||w|| = area of parallelogram = sqrt (1+9+49) = your answer. > w is normal to the plan of the parallelogram. > Tomasso. had thought that area of the parallelogram to be sqrt(1 + 9 + 7^2). (But wasn't sure, and I appreciate the confirmation). doing the following: sqrt((-1)^2 + 3^2 + 7^2) = sqrt(1 + 9 + 7) = sqrt(17). I so much appreciated being able to show my results. Just to be able to check that I haven't done any silly typos like the above. Cassandra === Subject: Proving the cross product is orthogonal I hope I don't seem like one of those louts that only turn up at assignment time!! However I am struggling with a proof. My answer is not coming out as expected, and I think I am making a silly mistake somewhere. If anyone could have a glance over it, that would be appreciated. Q) a =(1, -2, 1) b =(3, 1, 0) Find a x b and prove that your cross-product vector is perpendicular to each of the vectors a and b. A) [I have worked out the cross-product, and I have proven that a is perpendicular to the cross-product, however I am struggling to prove that b is perpendicular to the cross-product] a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k = -3i + 2j + 7k Let w = -3i + 2j + 7k Prove vector a is perpendicular to w a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) cos&= (a.w)/(|a||w|) = 0 Prove vector a is perpendicular to w b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) I know that is wrong, can anyone see what my mistkae is?? Cassandra === Subject: Re: Proving the cross product is orthogonal > I hope I don't seem like one of those louts that only turn up at > assignment time!! However I am struggling with a proof. My answer is not > coming out as expected, and I think I am making a silly mistake > somewhere. If anyone could have a glance over it, that would be appreciated. > Q) a =(1, -2, 1) b =(3, 1, 0) > Find a x b and prove that your cross-product vector is perpendicular to > each of the vectors a and b. > A) [I have worked out the cross-product, and I have proven that a is > perpendicular to the cross-product, however I am struggling to prove > that b is perpendicular to the cross-product] Do you know about the dot product? === Subject: Re: Proving the cross product is orthogonal >>I hope I don't seem like one of those louts that only turn up at >>assignment time!! However I am struggling with a proof. My answer is not >>coming out as expected, and I think I am making a silly mistake >>somewhere. If anyone could have a glance over it, that would be appreciated. >>Q) a =(1, -2, 1) b =(3, 1, 0) >>Find a x b and prove that your cross-product vector is perpendicular to >>each of the vectors a and b. >>A) [I have worked out the cross-product, and I have proven that a is >>perpendicular to the cross-product, however I am struggling to prove >>that b is perpendicular to the cross-product] > Do you know about the dot product? Yes, I like the dot product...much easier then the cross product I think! It turned out my main mistake was silly mistakes in my workings (ie 2*0 = 2 etc) I did this two places, I think it was only fluke that my workings for a being perpendicular to a x b, after the silly errors where fixed both proofs worked. Cassandra === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE You simply overlooked the figures 0 in -2*0-1*1 and in 1*3-1*0. I guess that you were writing too small print too tightly in the left upper corner of your paper sheet. Check: (1, -2, 1) X (3, 1, 0) = (-1, 3, 7); etc., etc. IHTH - Johan E. Mebius > I hope I don't seem like one of those louts that only turn up at > assignment time!! However I am struggling with a proof. My answer is > not coming out as expected, and I think I am making a silly mistake > somewhere. If anyone could have a glance over it, that would be > appreciated. > Q) a =(1, -2, 1) b =(3, 1, 0) > Find a x b and prove that your cross-product vector is perpendicular > to each of the vectors a and b. > A) [I have worked out the cross-product, and I have proven that a is > perpendicular to the cross-product, however I am struggling to prove > that b is perpendicular to the cross-product] > a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k > = -3i + 2j + 7k > Let w = -3i + 2j + 7k > Prove vector a is perpendicular to w > a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 > |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) > |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) > cos&= (a.w)/(|a||w|) = 0 > Prove vector a is perpendicular to w > b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 > |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) > |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) > cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) > I know that is wrong, can anyone see what my mistkae is?? > Cassandra === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE > You simply overlooked the figures 0 in -2*0-1*1 and in 1*3-1*0. I guess > that you were writing too small print too tightly in the left upper > corner of your paper sheet. Check: (1, -2, 1) X (3, 1, 0) = (-1, 3, 7); > etc., etc. > IHTH - Johan E. Mebius >> I hope I don't seem like one of those louts that only turn up at >> assignment time!! However I am struggling with a proof. My answer is >> not coming out as expected, and I think I am making a silly mistake >> somewhere. If anyone could have a glance over it, that would be >> appreciated. >> Q) a =(1, -2, 1) b =(3, 1, 0) >> Find a x b and prove that your cross-product vector is perpendicular >> to each of the vectors a and b. >> A) [I have worked out the cross-product, and I have proven that a is >> perpendicular to the cross-product, however I am struggling to prove >> that b is perpendicular to the cross-product] >> a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k >> = -3i + 2j + 7k >> Let w = -3i + 2j + 7k >> Prove vector a is perpendicular to w >> a.w = (1)(-3)+(-2)(2)+(1)(7) = -3 -4 + 7 = 0 >> |a| = sqrt(1^2 + (-2)^2 + 1^2) = sqrt(6) >> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >> cos&= (a.w)/(|a||w|) = 0 >> Prove vector a is perpendicular to w >> b.w = (3)(-3)+(1)(2)+(0)(7) = -9 + 2 + 0 = -7 >> |b| = sqrt(3^2 + 1^2 + 0^2) = sqrt(10) >> |w| = sqrt((-3)^2 + 2^2 + 7^2) = sqrt(62) >> cos&= (b.w)/(|b||w|) = -7/(sqrt(10)*sqrt(62)) = -7/sqrt(620) >> I know that is wrong, can anyone see what my mistkae is?? >> Cassandra Much appreciated. I have been writing my equations in MS Word using the equation editor. Perhaps I should write them on paper 'big and large' first!! Cassie === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE > Much appreciated. I have been writing my equations in MS Word using the > equation editor. Perhaps I should write them on paper 'big and large' > first!! With a pencil. The trolls in this newsgroup post all kinds of nonsense, sometimes on purpose and sometimes because they don't check their work. === Subject: Re: Proving the cross product is orthogonal - WRITE WIDE AND LARGE >>Much appreciated. I have been writing my equations in MS Word using the >>equation editor. Perhaps I should write them on paper 'big and large' >>first!! > With a pencil. > The trolls in this newsgroup post all kinds of nonsense, sometimes on purpose > and sometimes because they don't check their work. Yes, one of my biggest downfalls in life is that I make lots of mistakes. I once had to do a speed & accuracy test as part of a pysch test. I scored full marks (not sure how they calculated it), but I remember the pysch saying something about me getting alot further through the test then most, but making more mistakes then most. I used to have a turtle on my desk to remind myself to slow down and double check things. I find it so difficult, but it is something I should really work on. I have made two very elementary mistakes in the course of this thread...I do need to improve. === Subject: Re: Proving the cross product is orthogonal >Q) a =(1, -2, 1) b =(3, 1, 0) >Find a x b and prove that your cross-product vector is perpendicular to >each of the vectors a and b. >A) [I have worked out the cross-product, and I have proven that a is >perpendicular to the cross-product, however I am struggling to prove >that b is perpendicular to the cross-product] >a x b = (-2*0-1*1)i+(1*3-1*0)j+(1*1-(-2)*3)k > = -3i + 2j + 7k Huh? Check your algebra. -- I'm not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Just checking line slope I hope this doesn't sound dumb, but is the following an acceptable formula for a straight line. I am sure it is, but just wanted to check. y = x/6 === Subject: Re: Just checking line slope > I hope this doesn't sound dumb, but is the following an acceptable > formula for a straight line. I am sure it is, but just wanted to check. > y = x/6 Add or subtract any number to your x,or y or 6 in this equation and it is still a straight line. I first learnt it by visual detection: If the plot or graph is straight then it is linear, but if bent it is not so.There is only one way keeping it straight, so many ways to bend it. === Subject: Re: Just checking line slope <3Mcsd.58178$K7.34752@news-server.bigpond.net.au> I hope this doesn't sound dumb, but is the following an acceptable > formula for a straight line. I am sure it is, but just wanted to check. > y = x/6 Yes. Checking your work is never dumb. In general the equation of a straight line in the plane is y = mx + b or x = a Are you familiar with this general equation? Your equation fits that general form for what values of m and b? === Subject: Re: Just checking line slope >>I hope this doesn't sound dumb, but is the following an acceptable >>formula for a straight line. I am sure it is, but just wanted to check. >>y = x/6 > Yes. Checking your work is never dumb. > In general the equation of a straight line in the plane is > y = mx + b or x = a > Are you familiar with this general equation? > Your equation fits that general form for what values of m and b? x/6 part. I have been struggling with some of those finer rules for what constitutes a linear equation. My textbook says this... The following *are* linear equations: 1) 3x + 2y = 7 2) x1-2x2 +10x3 + x4 = 0 3) 1/2x + y - (pi)z = sqrt(2) 4) (sin((pi)/2))x1 - 4x2 = e^2 The following *are not* linear equations: 5) xy + z = 2 6) e^x - 2y = 4 7) sin(x1) + 2x2 - 3x3 = 0 8) 1/x + 1/y = 4 Firstly my apologies for being unsure of ASCII syntax. Where I have written a number after the variable I am trying to indicate the number is a subscript, ie x1 is x subscript 1. What is the correct syntax? While I can understand why most of these are linear/not linear. Some have me confused. Why is 4 okay, but not 7? What is wrong with 8? Cassie === Subject: Re: Just checking line slope My textbook says this... > The following *are* linear equations: > 1) 3x + 2y = 7 > 2) x1-2x2 +10x3 + x4 = 0 > 3) 1/2x + y - (pi)z = sqrt(2) > 4) (sin((pi)/2))x1 - 4x2 = e^2 > The following *are not* linear equations: > 5) xy + z = 2 > 6) e^x - 2y = 4 > 7) sin(x1) + 2x2 - 3x3 = 0 > 8) 1/x + 1/y = 4 > Firstly my apologies for being unsure of ASCII syntax. Where I have > written a number after the variable I am trying to indicate the number > is a subscript, ie x1 is x subscript 1. What is the correct syntax? x_1 and x^n for n-th power of x. Best readable ascii style is to persist like you're doing for the most part with adequate spaces like x1 - 2x2 + 10x3 + x4 = 0 > While I can understand why most of these are linear/not linear. Some > have me confused. > Why is 4 okay, but not 7? sin pi/2 is a constant. sin x_1 isn't ax_1 for some a in R. > What is wrong with 8? It doesn't have the form ax + by = c, the most general equation for a line, hence the expression linear, line like. Indeed, y + x = 4xy Were you to graph the equations, what is linear and what is not would likely jump out at you for only lines are linear. === Subject: Re: Just checking line slope > Were you to graph the equations, what is linear and what is not > would likely jump out at you for only lines are linear. an equation looks linear on a graph, then it is linear. Aah, I enjoy maths so much...but it is also very hard at times...posibly why I like it so much! cassandra === Subject: Re: Just checking line slope === >Subject: Just checking line slope >Message-id: <3Mcsd.58178$K7.34752@news-server.bigpond.net.au>I hope this doesn't sound dumb, but is the following an acceptable >formula for a straight line. I am sure it is, but just wanted to check. >y = x/6 Yep. Slope is 1/6 and intercept is 0. y = x/6 is the same as saying y = (1/6)*x + 0 -- Mensanator Ace of Clubs === Subject: Re: Vectors question 2 > I am working through a question for a university assignment. I am not > wanting to be given the answer, but am after some pointers. > The question is: > A train is travelling in a straight line at 24km/h. A movie stunt-man > moves at 4km/h straight across its roof, at right angles to the > direction of its motion. Draw a clear sketch of his resultant movement > relative to the ground, and find the speed and direction of that movement. > My take on this is as follows (way below is a very bad ASCII > representation of the type of graph that is in my head...definetly not > to any sort of scale) > I first create a graph with x-axis distance travelled, and y-axis time. I would give up on attempting to depict time in this picture, unless you had some real attachment to drawing a 3-dimensional picture. Why? Well, the train's motion takes up one dimension, time a second one, and the stuntman's motion (which is not collinear with the train's, if I read the problem correctly) a third one. Rather, imagine the displacements traversed over one unit of time. That unit can be of your choosing: one second, one hour, one millenium, ... you get to name it (I prefer a unit that makes calculations trivial, and that would suggest one hour; the problem is that you would then need to imagine a train that is 4 km wide. While the mental image that emerges is difficult to believe in, the mental image is absolutely irrelevant; if the mental image is important, imagine the time interval to be 1 second or 1 millisecond). You can do this based on an assumption of constant velocities, and the simple (vector) relation s = vt, where s represents the (vector) displacment, v the (vector) velocity, and t the (scalar) time, which (by assumption) is a constant. If you have a fixed time interval, the equation s = vt takes a velocity v, multiplies it by a fixed time interval t, and obtains a vector that points from the starting position, to the position reached at the end, at time t. The problem is to deal with the vector addition (which is what is meant by the resultant) of velocities, and what you need to do is to represent that. So, let the train be travelling along the +x axis, and the stuntman travelling along the +y direction. The resultant, which is the sum of the two vectors v_train and v_stuntman, represents the velocity vector of the stuntman with respect to the ground. (It would alternatively represent the velocity of the train with respect to the ground, if it were the stuntman running along the ground carrying a moving train. I'll ignore that interpretation.) > I can then plot u = 24i + j to show the speed that the train is > travelling at. > The line perpendicular to to u is the speed and direction of the stunt > man, the length of that line will be 4. Note that you are devoting the vector i to displacement along a single (spatial) direction, and j to displacement along the time direction. This is incorrect. The vector u is fine. However, your vector perpendicular to u will be (proportional to) either -i + 24 j or i - 24 j, since those are along the (only) direction perpendicular to u in the plane containing the directions i and j. In the first case, the stuntman is moving along the -i direction. That can't be right, since he must be moving perpendicular to the train's direction. In the second case, time is moving backwards. That might be nice for a sci-fi flick, but it doesn't really happen that way in real life. You need to notice that the stuntman's motion is perpendicular in spatial terms: if the train moves east, he must be going north or south. It's the spatial content that you must model. Everyone and everything moves through time in exactly the same way (ignoring the subtleties that relativity forces us to deal with). Thus, the description of the motions of the train and the stuntman must show their motions to be perpendicular *in space*. Their behavior in time can be reflected in dealing with the magnitudes of their velocity vectors, and those can be made spatial in character by using the formula s = vt that I mentioned earlier. The problem is that the i: space and j: time coordinate system makes for a single spatial dimension, and that doesn't allow for two separate spatial directions. That's why I said earlier that this approach would require three dimensions to depict everything, and that it would be best to ignore devoting a dimension to the time variable. > The line that forms the hyptonuse (v), represents the speed (length), > and direction (x,y) that the stuntman has travelled. > Am I on the right track (no pun intended)? If not, could you offer some > suggestions. Sorry, you have missed the train entirely (that pun was irresistable). Another will be along shortly, provided you follow the instructions I gave earlier. > | 0 > | 0 0 > | 0 0 > | v 0 0 (24,1) > | 0 0 > | 0 0 | 0 0 u > | 0 0 > |0____________________________> Please note, this is an assignment, so I am not wanting anyone to solve > it for me. I do not want any reason to be accused of wrongdoings. > Cassandra I hope the information I gave was both sufficient to get you on your way, and yet not too much. Dale. === Subject: Re: Vectors question 2 >> I am working through a question for a university assignment. I am not >> wanting to be given the answer, but am after some pointers. >> The question is: >> A train is travelling in a straight line at 24km/h. A movie stunt-man >> moves at 4km/h straight across its roof, at right angles to the >> direction of its motion. Draw a clear sketch of his resultant movement >> relative to the ground, and find the speed and direction of that >> movement. > I hope the information I gave was both sufficient to get you on your > way, and yet not too much. > Dale. Dale, written a few more times I think before I understand it all. So after reading your response I began to think. Ignoring vectors and i & j etc. If I wanted to figure this out a few weeks ago (prior to learning this stuff) my method would have been as follows. Put the direction of the train on the x-axis, the direction of the stuntman on the y-axis. (this is I think what you are trying to say). So now for the first part of the question --Sketch the resultant movement of the stuntman relative to the ground-- The sketch would be the line y = x/6, for x >= 0. The second part of the question --find the speed--- is answered by finding the length of the line. speed = sqrt(4^2 + 24^2) = 4*sqrt(37) km/h The final part of the question then --find the...direction of that movement-- This would be where the vector math comes in. tan& = b / a = 4 / 24 = 1 / 6 & = tan^-1(1/6) = 0.165 Phew, this seems alot nicer. got me started at any rate. Any further comments? Now for question number three... Cassandra === Subject: Re: Vectors question 2 > So after reading your response I began to think. Ignoring vectors and i > & j etc. If I wanted to figure this out a few weeks ago (prior to > learning this stuff) my method would have been as follows. > Put the direction of the train on the x-axis, the direction of the > stuntman on the y-axis. (this is I think what you are trying to say). That method is correct, and produces the correct answers... so now my question is: what is it that you learned over the last few weeks that confused you into trying to plot time as well? meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Re: Vectors question 2 >>So after reading your response I began to think. Ignoring vectors and i >>& j etc. If I wanted to figure this out a few weeks ago (prior to >>learning this stuff) my method would have been as follows. >>Put the direction of the train on the x-axis, the direction of the >>stuntman on the y-axis. (this is I think what you are trying to say). > That method is correct, and produces the correct answers... so now my question > is: what is it that you learned over the last few weeks that confused you into > trying to plot time as well? > meeroh To answer your question, I overthought the question. We are studying vectors, including 3-dimensional vectors and projections. Perhaps it was because I had only just finished studying projections. Or because I thought that question must be more complex then it looked etc. I really don't know why, however I am sort of glad because it has allowed me to gain alot better understanding into using vectors, but then making mistakes does usually have that effect. cassandra === Subject: Re: Vectors question 2 > > >>So after reading your response I began to think. Ignoring vectors and i & j >>etc. If I wanted to figure this out a few weeks ago (prior to learning this >>stuff) my method would have been as follows. >Put the direction of the train on the x-axis, the direction of the stuntman >>on the y-axis. (this is I think what you are trying to say). > > > That method is correct, and produces the correct answers... so now my > question is: what is it that you learned over the last few weeks that > confused you into trying to plot time as well? > > To answer your question, I overthought the question. We are studying > vectors, including 3-dimensional vectors and projections. Perhaps it was > because I had only just finished studying projections. Or because I thought > that question must be more complex then it looked etc. > I really don't know why, however I am sort of glad because it has allowed me > to gain alot better understanding into using vectors, but then making > mistakes does usually have that effect. Well, if you want to consider this and include time as one of your dimensions, then you have to realize that you have to project the space-time coordinate system into a space-only coordinate system before trying to describe it in terms of visible effects. For example, consider a single point traveling along a straight line. Let the line of travel be the X axis and let time T be perpendicular to it. Then, in the XT plane, the point will be describing some curve (continuous, unless your universe includes teleportation). For example, if the point is oscillating about the origin, then the space-time curve describing its motion would be a sine wave on the T axis. But a description of that motion would probably not refer to a sine wave about the T axis, it would refer to an oscillatory motion about the origin. This seems like a trivial point until you start considering more dimensions: consider an XY plane and a time T axis perpendicular to it, and two points traveling in the plane -- one at a fixed velocity along the X axis, starting at the origin, and the other at the same velocity along the Y axis, also starting at the origin. If you look at them in the XY plane, then their trajectories (namely, the X and Y axes) are perpendicular. However, if you consider them to be points in XYT space, then the first one's trajectory is a straight line in the XT plane, and the second one's trajectory is a straight line in the YT plane, and those trajectories are not perpendicular! It is only their projections into the XY place (the space-only coordinate system) that are perpendicular. Which brings me to your question; your mistake in considering this with distance on the X axis and the time on the T axis was that when the description of physical motion of the stunt man was described as being perpendicular to the motion of the train, you drew the perpendicular in the XT plane, which doesn't work. What you need to do here is count the number of dimensions of space (clearly, you need at least two, as the man the the train are moving at right angles to each other. Then you add one for time, and you see that you need a three-dimensional system. In that system, the trajectory of the train will be a straight line; if the train travels along the X axis, as it does in your original graph, then its XYT trajectory will be a straight line in the XT plane, as it is in your original graph. However, the motion of the man is, according to the problem statement, perpendicular to that of the train -- but the problem statement refers to them in the space-only coordinate system, which means that you have to have the man moving perpendicular not to the XYT time-space trajectory of the train, but to the XY spatial trajectory of the train. If you do that, then everything comes out right, but it's somewhat complicated to draw that. Now, there is no really good reason to involve time in this, as the movement of the train (relative to the ground) and the man (relative to the train) are constant with time, so what you are asked for (movement of the man relative to the ground) is also constant and you can draw it without involving the T axis at all. However, I think you should know why your initial solution didn't work and how you could have made it work. meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: No Unique Initial Segment And No Characteristic Expansion. Infinite people each flip coins infinite times. Can you always find a different sequence of heads and tails? sci.math and sci.logic went quiet on this question for about a week, then against all logic, probability theory, and common sense they all agreed YES. Believers of hyperinfinities have no shame! How on Earth can you exhaust an infinite set? This is their solution, a standard application of Cantor's diagonalisation technique, the hailed method of proving infinity incomplete (infinite already means never ending or incomplete!) Take one of the people, whatever his 1st flip was, reverse it! If he flipped a head you select tail, if he flipped a tail, heads. That's your first outcome, cross him off and select someone else, whatever was their second flip, reverse it! Keep on going and you have an infinite sequence that is different to everyone's sequence in atleast one flip. THAT'S THE PROOF! That's hyperinfinity and cardinality theory 101. How many assumptions about infinity are they taking for granted here? What jumps straight out at most everyone is : Aren't all possible combinations of heads and tails for infinite flips already been done? This doesn't stir the Cantor supporter one bit. They think if the combination is on the list, it must be at some natural number position n, but the nth flip of person n is necessarily different! Voila a (somewhat contrived) contradiction. Hence the sequence does not appear anywhere on the list. Makes me reminiss to lectures on these types of problems. You decide for yourself no this is right. So the lecturer asks you for the further outcome of that conclusion, which goes around in a circle and proves what he wanted to. It's played out to everyone who studies theory, you can see the conversions taking place on sci.logic and sci.math each week. The other supporting argument Cantor followers have is demonstrated when you assume this infinite list: 0.222222.. 0.322222.. 0.332222.. 0.333222.. 0.333322.. 0.333332.. .. The diagonal is 0.22222.. If we modify every digit, we can get 0.33333.. Does 0.3.. occur on the list? 0.3333333 <--L--> There are unlimited 3s in sequence on the list! Although 0.333.. has No Unique Initial Segment, it does have a Characteristic Expansion. It ends in 333.., all members end in 222.. so no, 0.3.. does not occur on the list. Back to the infinite flippers list: httttt.. hhhttt.. hhhhhh.. tttttt.. hhhttt.. .. With probability 1, this list contains every initial segment possible of heads and tails sequences. Assume there is some initial segment that is not on the list. TTTHHH This sequence has a finite length L, there are 2^L possible sequences of length L, so with infinite amount of flippers that initial segment will be covered with probability 1. The diagonal sequence HHHTT.. inverted TTTHH.. has No Unique Initial Segment. That in itself does not prove it's on the list, remember 0.333..! But TTTHH.. has No Characteristic Expansion either. As far as can be determined the sequence appears on the list. So although we may not be able to disprove the hyperinfinity status of the diagonal yet, we can show that it is a N.U.I.S.A.N.C.E. Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. > Infinite people each flip coins infinite times. Infinite simply means not finite, which we know also means larger than any finite. But there may be lots of different sizes of larger-than-finite, so which particular one did you have in mind for the number of people and the number of coin flips per person? (That's two separate question you need to answer.) === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. sure. down further in the post I write out a hypothetical list Back to the infinite flippers list: httttt.. hhhttt.. hhhhhh.. tttttt.. hhhttt.. .. and the coin flips for each person are referred to as sequences. both list and sequence imply countable infinity can anyone confirm the claim here that the diag of such a random list has 1/ No Unique Initial Segment 2/ No Characteristic Expansion (as opposed to the standard examples used for Diag analysis) That's 2 separate questions. Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. H> Infinite people each flip coins infinite times. R>Infinite simply means not finite, which we know also means larger than R>any finite. But there may be lots of different sizes of R>larger-than-finite, so which particular one did you have in mind for R>the number of people and the number of coin flips per person? H> infinite flippers list ... and the coin flips for each person are H> referred to as sequences. both list and sequence imply countable H> infinity Are you saying your original statement was supposed to mean countably infinite number or people each flip coin(s) countable infinite number of times? (I guess it doesn't matter whether an individual person flips just one coin a countable infinite number of times, or switches coins every so often when the old coin wears out, so long as the total number of flips per person is countable infinite). H> Can you always find a different sequence of heads and tails? Do you mean if somehow you could collect all the h/t sequences from all the corresponding people, i.e. build a complete 2-d array, let's call it F, where both row index and column index run forever in one direction, can you find a single h/t sequence that is *not* exactly the same as any particular row in your array? Well of course! If the array is indexed from 1,2,... forever along each axis, you can define a single h/t sequence indexed along one axis as follows: For any m = 1,2,..., let A(m) be the opposite of F(m,m), that is A(M) is H if F(m,m) is T and vice versa. This is Cantor's diagonal method, which has been the topic of several threads you started recently. Why do you even need to ask the question unless you were sleeping through all your threads? > How on Earth can you exhaust an infinite set? You know as much as anyone else the Earth is finite in size, finite in you mean literally on Earth you cannot exhaust an infinite set. > Aren't all possible combinations of heads and tails for infinite > flips already been done? It depends on what sequences of h/t you are possible. If you have coins that aren't random, that for some perverse reason always go into a loop where they repeat the same sequence as before, so every sequence with a single coin generates a periodic sequence, then indeed there are only a countable number of possible sequences, and it's possible that the countable sequence of different coins might exactly cover all such repeatable sequences, and then Cantor's diagonal method produces a non-repeating sequence that is not a possible outcome of any single person flipping a single coin, so it does produce a sequence of h/t but such h/t sequence cannot be flipped exactly by any coin. Now it's actually pretty easy to make a list of all repeating patterns. Start with the two patterns of length 1: HHHHHHHH... TTTTTTTT... Then do the two additional patterns of length 2: HTHTHTHT... THTHTHTH... Then do the six additional patterns of length 3: HHTHHTHHT... HTTHTTHTT... HTHHTHHTH... THHTHHTHH... THTTHTTHT... TTHTTHTTH... etc. So using Cantor's diagonal on that sequence of sequences you can directly construct a non-repeating sequence. (Note it's even easier if you allow duplicates among the original sequences. Then you don't have to carefully omit any large pattern that is composed of smaller patterns, and it's easier to directly compute the mth generator and hence the mth digit of the mth generator hence the opposite of that mth-of-mth hence the mth of Cantor non-repeating sequence.) Back to the countable infinite sequence of people each flipping a coin a countable infinte sequence of times: > With probability 1, this list contains every initial segment possible > of heads and tails sequences. Assuming the coins are fair, that's an understatement. Any particular initial pattern is expected to occur an infinite number of times among the sequence of H/T sequences. > Characteristic Expansion I'm not familiar with that term, and a Google search turns up that term only in biochemistry/genetics, not in mathematics. Is it some term you invented yourself, or can you show me somewhere on the Web where it's defined as a mathematics term? > No Unique Initial Segment I'm not sure what you mean by that phrase, except in the trivial sense that there is more than one initial segment of the h/t sequences, i.e. not all h/t sequences start out exactly the same way. If you mean anything less trivial than that, please define what you mean. (Note typo: number or people should read number of people, sorry, didn't see until mid-upload.) === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. Right! My point is the diagonal you made, you call it unique yet it has No Unique Initial Segment. The length of all initial segments on an infinite list is unbounded. Therefore, the diagonal sequence is already present on the list to infinite number of flips. this is also true for 0.33333.. on the list 0.3 0.33 0.333 ... 0.3333... has No Unique Initial Segment. Why is 0.3333.. not on that list? Why is the diag HHHH.. not on the random list? Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. > My point is the diagonal you made, you call it unique yet it has No > Unique Initial Segment. Given any particular countably-infinite sequence, there is a unique initial segment of length 1, and a unique initial segment of length 2, etc. for any value of n you choose. Of course if you allow the length to vary, there's no overall unique initial segment, because a segment of length 1 is different from a segment of length 2, etc. So what does this trivial fact have to do with the topic being discussed?? The whole sequence may be unique in some context, and any initial segment is unique in the context of initial segments of the whole sequence of that particular length, but any initial segment is not unique in the context of all initial segments of all various lengths. > The length of all initial segments on an infinite list is unbounded. Obviously that's true, but who cares?? > Therefore, the diagonal sequence is already present on the list No, that's not true, and there's no valid reason why the previous statement has any bearing on this statement, so therefore makes no That's also irrelevant to the discussion of the anti-diagonal sequence. > 0.3 > 0.33 > 0.333 > ... (A countably infinite list of numbers, each of which is expressed by a finite bunch of digits. I.e. each element on the list is a terminating decimal fraction.) > 0.3333... (A single number, which is expressed by a countable-infinite sequence of digits. I.e. this number is *not* a terminating decimal fraction.) > Why is 0.3333.. not on that list? Because it is not a terminating decimal fraction, whereas every element on the list is a termianating decimal fraction. It's the same as if you made a list of eggs, each element is an egg, nothing else ever on your list except eggs, and you showed me a potato and asked why that potato is not on your list. Well, because it's a potato, whereas your list has only eggs on it. === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. >> Therefore, the diagonal sequence is already present on the list >No, don't misquote me. read the OP for a demonstration of Unique Initial Segment. try to work out what I mean, other people correspond but with you and Ullrich I have to go into maths text writing mode. A LIST xxxxxx xxxxx xxxxx A SEQUENCE yyyyyy yyyyyy has no unique initial segment if forall n, is an initial segment of some member *of the list.* The point was you can answer the 1st question but you can't give the same answer to the second question. To borrow David Ullrich's terminology its a version of the same question. Why is 0.3333.. not on that list? Why is the diag HHHH.. not on the random list? (why am I expecting a literal interpretation of HHHH.. in reply) Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. > yyyyyy By that, do you mean a finite sequence of length exactly six, or do you mean a finite sequence of some indeterminate length (in which case you should have written yyy...yyy instead), or do you mean an infinite sequence (in which case you should have written yyy... instead)? If you meant an infinite sequence, did you specifically mean the Cantor anti-diagonal sequence? I'll assume the answer is yes in my answer below. If that's not what you meant, you need to be more clear in expressing yourself. > has no unique initial segment if forall n, I have no idea what you mean to say there. is an initial segment of some member *of the list.* For any specific finite initial segment of the Cantor anti-diagonal sequence, it's possible for an infinite number of sequences in the proposed enumeration list to exactly match that initial segment, and yet not one of them exactly match the Cantor anti-diagonal all the way out forever. For example, consider this list of sequences: 987654321000000000... (each the rest of the digits '0') 876543210000000000... 765432100000000000... 654321000000000000... 543210000000000000... 432100000000000000... 321000000000000000... 210000000000000000... 100000000000000000... 554555510000000000... 554555501000000000... 554555500100000000... 554555500010000000... 554555500001000000... 554555500000100000... 554555500000010000... 554555500000001000... 554555500000000100... 554555500000000010... (continue that obvious pattern of first 7 digits same and '1' out one further each time, and each of the other digits '0') and let the Cantor anti-diagonal sequence be: 554555555555555555... (each of the rest of the digits '5') Now you see after the first 9 sequences in the list, each of the remaining sequences exactly matches the Cantor diagonal sequence for the first seven digits, yet not one of those sequences exactly matches the Cantor diagonal sequence all the way out, in fact not one of them matches even one digit after those first seven. > Why is 0.3333.. not on that list? You need to define what you mean by that notation, and what particular list you are talking about. If you mean the list of finite sequences of '3' digits followed by all the rest '0' digits, and if by the notation 0.3333.. you mean a sequence with *every* digit '3', not any '0' digit whatsoever after the decimal point, then the answer is so obvious you shouldn't need it explained to you again and again and yet again. (In case you are finally bothering to read: Every sequence in the list has only a finite number of '3' digits, whereas that one extra sequence has an infinite number of '3' digits, so of course none of those in the list can exactly match that one extra sequence all the way out, even if they match for a few digits at the start.) > Why is the diag HHHH.. not on the random list? I have no idea. It might by chance happen to be, but the probability of that occurring is zero unless the coins are *very* biassed! === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. For any specific finite initial segment of the Cantor anti-diagonal sequence, a number of sequences in the proposed enumeration list exactly match that initial segment That's what I mean by the anti-diagonal has No Unique Initial Segment. for this list 0.3 0.33 0.333 ... 0.3333... has No Unique Initial Segment. Why is 0.3333.. not on that list? BECAUSE IT'S A POTATO AND THEY'RE ALL TOMATOES The anti-diag of the infinite list of random coin tosses has No Unique Initial Segment. Why is the ant-diag of the list of coin tosses not on that list? Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. > Infinite people each flip coins infinite times. > Can you always find a different sequence of heads and tails? Wait for the data to fuly accumulate, then look. Idiot. Given n flips there will be 2^n states. That will nicely outrace any approximation to infinity you attempt. > sci.math and sci.logic went quiet on this question for about a week, > then against all logic, probability theory, and common sense they all > agreed YES. Believers of hyperinfinities have no shame! Georg Cantor. As for you, FOAD. [snip crap] -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. > > Infinite people each flip coins infinite times. > Can you always find a different sequence of heads and tails? > Wait for the data to fuly accumulate, then look. Idiot. Given n > flips there will be 2^n states. That will nicely outrace any > approximation to infinity you attempt. > sci.math and sci.logic went quiet on this question for about a week, > then against all logic, probability theory, and common sense they all > agreed YES. Believers of hyperinfinities have no shame! > Georg Cantor. As for you, FOAD. > [snip crap] Hi Al, Hey, for a length n, you get 2^n permutations, and they're all rational numbers. In base b, for p places, you get b^p permutations. You're a chemist, and respected for your knowledge of chemistry. Some people think you're great, others you're an over-the-top jerk, but it's generally accepted that you're a reliable scientist. I don't care about Herc, it's a free country. So anyways, I want to know if you ever use the uncountability of the reals or transfinite cardinality in any form. Georg Cantor's famous. He might be the most chronicled mathematician after Newton, and I never heard of him until the late 90's. He tackled the difficult task of trying to define the continuum in non-geometric terms. During the rush to reformalize mathematics in the 20th century, basically the Bourbaki school and perhaps Dresden with Russell and Whitehead being Englishmen readily used this notion of set theory, and along with it Cantor's powerset mapping result. As the century progressed, in terms of foundations there are basically the Zermelo and Fraenkel, and the Goedelian period and the 70's, and mathematics at large flourished outside of it, with tremendous gains in concrete and differential mathematics since the 50's and 60's, and of course the tremendous jumps in the tensorial and algebraic geometrical and geometric algebraic methods, and of course the probability density functions in your field, and the unfortunately ready application of numerical methods with digital computers. By and large those have little to do with foundations. Anyways, today there are theories that do not necessarily use ZF, for example, an excellent set theory with a somewhat restricted notion of a set, and calling everything else a class, except the class can not contain other classes, leading to no class of all classes, and no reolution of Burali-Forti. There are a variety of rigorous infinitesimals these days, some hundreds of years after the fact of their use for results. I retract that calling something a paradox is stupid, but stopping there is stupid. A paradox is a sign of insufficient knowledge or improper assumptions. About the semi-infinite binary sequences, they are that, but they might not adequately characterize the real numbers or the continuum. Today, there are a variety of modern approaches to foundations that attempt to revisit the assumptions and expand the knowledge about the behavior of fundamental mathematical objects. Some of these do not have the powerset result holding for variously many or all sets in the theory. One of them is mine, currently a theory with ubiquitous ordinals, and the ur-element, or that which is assumed, is either a) empty, or b) the universal set. Also it doesn't say. Basically its strength is to get over Burali-Forti, that the order type of all ordinals would itself be an ordinal, which is the same thing as Cantor, that the set of all sets would be its own powerset. So anyways, do you use transfinite cardinals in your day-to-day operations or ever? What's the use of integrating over the natural numbers and getting a meaningful result? Fondly, Frere Ross === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. > Infinite people each flip coins infinite times. > Can you always find a different sequence of heads and tails? > Wait for the data to fuly accumulate, then look. Idiot. Given n > flips there will be 2^n states. That will nicely outrace any > approximation to infinity you attempt. > sci.math and sci.logic went quiet on this question for about a week, > then against all logic, probability theory, and common sense they all > agreed YES. Believers of hyperinfinities have no shame! > Georg Cantor. As for you, FOAD. > [snip crap] > Hi Al, > Hey, for a length n, you get 2^n permutations, and they're all > rational numbers. In base b, for p places, you get b^p permutations. Are you flipping DnD dice instead of honest coins? > You're a chemist, and respected for your knowledge of chemistry. Some > people think you're great, others you're an over-the-top jerk, but > it's generally accepted that you're a reliable scientist. I don't > care about Herc, it's a free country. So anyways, I want to know if > you ever use the uncountability of the reals or transfinite > cardinality in any form. Our calculation of quantitative parity divergence of solid spheres of crystal lattice is not valid for an infinite number of atoms, but we only got to 4.44x10^19 atoms in alpha-quartz - so no problem in long_double_precision. My only use of uncountability is drawing a median line parallel to an equilateral triangle's base and inviting the viewer to show where lines drawn from apex to points in the base miss points in the half-length line. HERC is a chronic abusive boring idiot. He and his kind should not be allowed to proceed unmolested. What goes around comes around. Would you prefer the Department of Education and its policy of every child left behind? > Georg Cantor's famous. He might be the most chronicled mathematician > after Newton, and I never heard of him until the late 90's. He > tackled the difficult task of trying to define the continuum in > non-geometric terms. He never really got aleph_1 well identified, either. It's big stuff to juggle and there's lots of room for error. > During the rush to reformalize mathematics in the 20th century, > basically the Bourbaki school and perhaps Dresden with Russell and > Whitehead being Englishmen readily used this notion of set theory, > and along with it Cantor's powerset mapping result. As the century > progressed, in terms of foundations there are basically the Zermelo > and Fraenkel, and the Goedelian period and the 70's, and mathematics > at large flourished outside of it, with tremendous gains in concrete > and differential mathematics since the 50's and 60's, and of course > the tremendous jumps in the tensorial and algebraic geometrical and > geometric algebraic methods, and of course the probability density > functions in your field, and the unfortunately ready application of > numerical methods with digital computers. Bourbaki was condemned for negatively altering American math education (requiring intelligence and furnishing insight). The subsequently impressed solution was to diversely remove math education through high school. An advocate makes virtue of failure. The worse the cure the better the treatment - and the more that is required. > So anyways, do you use transfinite cardinals in your day-to-day > operations or ever? Not even on 15 April. > What's the use of integrating over the natural numbers and getting a > meaningful result? Suppose you only integrated over half of them... to save time. 8^>) -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. > Infinite people each flip coins infinite times. > Can you always find a different sequence of heads and tails? > Wait for the data to fuly accumulate, then look. Idiot. Given n > flips there will be 2^n states. That will nicely outrace any > approximation to infinity you attempt. > sci.math and sci.logic went quiet on this question for about a week, > then against all logic, probability theory, and common sense they all > agreed YES. Believers of hyperinfinities have no shame! > Georg Cantor. As for you, FOAD. > [snip crap] > Hi Al, > Hey, for a length n, you get 2^n permutations, and they're all > rational numbers. In base b, for p places, you get b^p permutations. > Are you flipping DnD dice instead of honest coins? Sure, why not. The dice only come in regular polyhedrons, though, the integer base can be any finite number. > You're a chemist, and respected for your knowledge of chemistry. Some > people think you're great, others you're an over-the-top jerk, but > it's generally accepted that you're a reliable scientist. I don't > care about Herc, it's a free country. So anyways, I want to know if > you ever use the uncountability of the reals or transfinite > cardinality in any form. > Our calculation of quantitative parity divergence of solid spheres of > crystal lattice is not valid for an infinite number of atoms, but we > only got to 4.44x10^19 atoms in alpha-quartz - so no problem in > long_double_precision. My only use of uncountability is drawing a > median line parallel to an equilateral triangle's base and inviting > the viewer to show where lines drawn from apex to points in the base > miss points in the half-length line. Basically you can only get continuity by passing in order through each element of the line segment. So, pencil drawing line on paper: continuous, darts covering board, not continuous. Then again, there are space-filling curves. > HERC is a chronic abusive boring idiot. He and his kind should not be > allowed to proceed unmolested. What goes around comes around. Would > you prefer the Department of Education and its policy of every child > left behind? blame the problems in the schools on the material culture. Also, I threaten terrorists, consider J. Lo's buttocks, consider my own vanity, and sanity, Miller's _Sexus _, sexual prowess, and too much porn, advocate the protection of freedoms, and express admiration for Virgil's intellect. I tend not to read Herc's posts, Herc I tend not to read your posts because I think you are a bonafide nutcase and I don't expect to learn something from you that is useful to promote my theories at large. That's OK, from what I understand other people might think I am a nutcase, but they're stupid. Where you are advocating revolutionary notions about Turing machines and uncountability, I am some years ahead of you in steadily promoting a similar viewpoint to the readers of sci.math, that being a vector into most academic math departments on the planet. So, I don't discourage you from your ruminations on the set theory, on the contrary, I encourage it, as I think it is a good outlet for creative and even divergent thinking, but you should realize that these forums are full of highly skilled individuals, and many are less tolerant than I am. About education, how about a two hour daily limit on television, and four hours a day on the Internet for kids under eighteen, two hours enforced reading at above grade level. It's been quite some time since I was in school myself, locales vary widely. > Georg Cantor's famous. He might be the most chronicled mathematician > after Newton, and I never heard of him until the late 90's. He > tackled the difficult task of trying to define the continuum in > non-geometric terms. > He never really got aleph_1 well identified, either. It's big stuff > to juggle and there's lots of room for error. Damn it, Al. There's no room for error, because otherwise if you multiply some finite thing by some infinite thing, and then divide it by some other infinite thing incorrectly, then it's just as bad as a sign error, or worse. Think how bad it would be if someone took the integral of a polynomial and multiplied it by negative one. Then it'd be both errors. > During the rush to reformalize mathematics in the 20th century, > basically the Bourbaki school and perhaps Dresden with Russell and > Whitehead being Englishmen readily used this notion of set theory, > and along with it Cantor's powerset mapping result. As the century > progressed, in terms of foundations there are basically the Zermelo > and Fraenkel, and the Goedelian period and the 70's, and mathematics > at large flourished outside of it, with tremendous gains in concrete > and differential mathematics since the 50's and 60's, and of course > the tremendous jumps in the tensorial and algebraic geometrical and > geometric algebraic methods, and of course the probability density > functions in your field, and the unfortunately ready application of > numerical methods with digital computers. > Bourbaki was condemned for negatively altering American math > education (requiring intelligence and furnishing insight). The > subsequently impressed solution was to diversely remove math education > through high school. An advocate makes virtue of failure. The worse > the cure the better the treatment - and the more that is required. Bourbaki was this panel of Frenchmen who published under the collective pseudonym N. Bourbaki. > So anyways, do you use transfinite cardinals in your day-to-day > operations or ever? > Not even on 15 April. Neither does anyone else. That's not correct. Some people manipulate them on paper. > What's the use of integrating over the natural numbers and getting a > meaningful result? > Suppose you only integrated over half of them... to save time. 8^>) > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf Well, that's a notion, because there are various ways to do that. It doesn't do me much good to start using EF, the natural/unit equivalency function, as a tool for further developments if I can't make any concrete, testable predictions using it. So, I am looking for something that has to do with a mapping between the analog and discrete that is either half or twice the expected experimental value. That has to do with that the points on the continuous line have only one side, distinct points two. I have some good news for you, Cantor isn't the only one with any ideas about how to address the continuum. There are others willing to help with that burden, some gladly. There are some strange behaviors of large numbers. Readers, you might not know that this post is basically to help indicate that Al has no opinion on transfinite cardinals, and as he doesn't use them he doesn't care if they are disproved, from the perspective that Al is a reliable scientist and extremely knowledgeable about the practice of industrial applied chemistry. Al, as much as you care, please put forth your opinion on the infinite and things infinite and so on. XOXO, Ross === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. <41B1812E.2C99A110@tiki-lounge.com> >Where you are advocating revolutionary notions about Turing machines and >uncountability, I am some years ahead of you in steadily promoting a similar viewpoint to the >readers of sci.math, that being a vector into most academic math departments on the planet. So, >I don't discourage you from your ruminations on the set theory, on the contrary, I encourage it, >as I think it is a good outlet for creative and even divergent thinking, but you should realize >that these forums are full of highly skilled individuals, and many are less tolerant than I am. Countability is central to some kind of global function system. God channels to me there is a proof that cardinality theory is bogus, its certainly my preferred view since the diagonal just ingores the assumption of all permutations to infinite length being possible. And a self referential 'non member' is to be expected from any formalism in any theory, cardinality is no exception, sorry but hyperinfinity is rubbish. How is UTM(n e N) possibly incomplete? There are paradoxes but the [CONTRADICTION] -> {conclusion} that current theory accept are fairy tales. A supreme ordered function list is probably on the cards for cyberspace in the next 5 years, I have to wip these computable-hating delinquents into shape. Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. if you love uncle al reply here === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. >Hey, for a length n, you get 2^n permutations, and they're all >rational numbers. In base b, for p places, you get b^p permutations. and....? are you going somewhere with this? is this what you get for putting maths in a physcis group? can you rename a spreadsheet aswell? The topic is: No Unique Initial Segment And No Characteristic Expansion??? I don't care if you don't care, put your love life with your Uncle in your own thread. THIS POST ADDRESSES THE COUNTABLE INFINITE POWER SET [HERC] What *is* the power set? A power set is a transform from subset space to element space. I run a banner exchange and to improve the targeting of the banner placements, everyone nominates the CATEGORY of their site. -------------------------------------------- SIGN UP SIGN UP FREE BANNER EXCHANGE WEBSITE : www.entrenchederrorsinmaths.com BANNER : www.entrenchederrorsinmaths.com/ban1.gif CATEGORY : --------------------------------------------- Categories are : sci, rec, alt, public_service_announcements So you can see the problem, I have 2 categories I want to use, sci and p_s_a. Now there's 2 ways to implement multiple categories. You maintain a list of several categories for each site, or... use a powerset of categories. CATS = {s, r, a, p} P(CATS) = {{s}, {r}, {a}, {p}, {s,r}, {s,a}, {s,p}, {r,a}, {r,p}, {a,p}, {s,r,a}, {s,r,p}, {r,a,p}, {s,r,a,p}} Now I can select the single category Sci & Public_Service_Announcement. We wanted a SUBSET, but instead used an ELEMENT of the power set. Is this process feasable for an infinite list? Mathematics texts will say no, as Cantor's proof is wrong it is possible to count an infinite power set. Say the set is the digits of pi, augmented with their digit place so each member is unique. SET = { <0,3>, <1,1>, <2,4>, <4,1>, <5,5> ... } Now we want to make a single selection of that set but containing multiple members! 1st Attempt: P_1(SET) = {{1000000000000.. AND SET}, {0100000000000.. AND SET}, {1100000000000.. AND SET}...} We use logical AND as a filter on the set with a reverse binary number and get: P_1(SET) = { {1 AND <0,3>, 0 AND <1,1>, 0 AND <2,4> ...}, {0 AND <0,3>, 1 AND <1,1>, 0 AND <2,4> ...}, {1 AND <0,3>, 1 AND <1,1>, 0 AND <2,4> ...}, ...} P_1(SET) = { {<0,3>}, {<1,1>}, {<0,3>, <1,1>}, ...} P_1(SET) = {{ <0,3> }, { <1,1> }, { <0,3>,<1,1> } ... } We get all the combinations of the SET but the members are all finite sets, we need something better than a binary counting mask. What can we use as a generator of *infinite* binary expansions? 2nd Attempt, (UTM is Universal Turing Machine): P_2(SET) = An, {UTM(neN, 0) AND SET} UTM(neN,0) will emulate every computer program from 1 to infinity and output a (potentially infinite) sequence of 1s and 0s for every natural. Every possible sequence and combination of 1s and 0s is present. Hence we can generate countable power sets of infinite sets. Herc [TINYURL..] > What *is* the power set? For a non-finite base set, how do you know it even has a power set? Even if there is a power set, how do you know it's unique, so that the word the can be used in that context? Until you answer both questions, you can't meaningfully use the phrase the power set. Hint: One of those two questions is easy to prove, while the other is impossible and requires an axiom to just assume it's so without proof. > emulate every computer program from 1 to infinity and output a > (potentially infinite) sequence of 1s and 0s for every natural. There will be an infinite number of duplicates of any particular output sequence. Do you allow duplicates, or do you have some computable method of eliminating the duplicates? Some programs generate only a finite sequence of output, and then just run forever never ever producing any more output. Do you completely discard such a machine's output, or append an infinite number of zeroes after the finite output the machine actually generated? In either case, how do you computably decide whether a given program will produce infinite output if you just wait long enough, or whether it'll stop producing output and you need to discard it or wait for that last output then append the zeroes? It's been proven that it's impossible to computably make that decision, so what what do you do when a program has run for a very long time and hasn't made any new output for a long time and you don't know whether it will ever again generate new output. [HERC] An injection is still a countable set, a bijection may be impossible since infinite sequences are impossible to compare. The complete output of all computer programs is theoretically achievable with multitasking, if a program infinite loops and there is no automated checker for that algorithm it will forever take a small portion of computing resources. Herc === Subject: Re: No Unique Initial Segment And No Characteristic Expansion. what is an approximation to infinity? big? Given f flippers, they cover all log(f) long combinations, as f->oo log(f)->oo. How are you planning on flipping MORE than infinite coins to make a new sequence? Herc === Subject: semi-infinite number There is a semi-infinite number of .... Anyone know how many that would be? === Subject: Re: semi-infinite number I'll take it to mean so many that no one can be bothered to count them. > There is a semi-infinite number of .... > Anyone know how many that would be? === Subject: Re: semi-infinite number > There is a semi-infinite number of .... > Anyone know how many that would be? A lot: it's used in phrases like there are a semi-infinite number of cranks on sci.math :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: semi-infinite number > There is a semi-infinite number of .... > Anyone know how many that would be? What's the context? Maybe it is something like cofinite, i.e. the complement is finite. -- Mitch Harris (remove q to reply) === Subject: Re: semi-infinite number > There is a semi-infinite number of .... > Anyone know how many that would be? > What's the context? Maybe it is something like cofinite, i.e. the > complement is finite. Conventional filters use the Z transform to describe the filter transfer characteristic, where Z-1 denoted a unit delay. There are a semi-infinite number of orthonormal functions for transform arithmetic. Also found on Google You can choose whether to use the Mandelbrot set, or generate a semi-infinite number of Julia sets of your own plus many more examples there; but no definition that I can fine. > Mitch Harris > (remove q to reply) === Subject: Re: semi-infinite number >There is a semi-infinite number of .... >Anyone know how many that would be? >>What's the context? Maybe it is something like cofinite, i.e. the >>complement is finite. > Conventional filters use the Z transform to describe the filter transfer > characteristic, where Z-1 denoted a unit delay. There are a semi-infinite > number of orthonormal functions for transform arithmetic. > Also found on Google > You can choose whether to use the Mandelbrot set, or generate a > semi-infinite number of Julia sets of your own > plus many more examples there; but no definition that I can fine. Oh. Duh. Google. Lots of examples. Anyway, it seems that semi-infinite or semiinfinite is a short way of saying infinite in one direction but not the other (e.g. along a ray or a halfplane). In such contexts it is a perfectly reasonable construction. As to its lack of inclusion in MathWorld, well, MathWorld has a lot of stuff in it, but it's not comprehensive. Then again, I'm not sure semiinfinite would be in a comprehensive source. -- Mitch Harris (remove q to reply) === Subject: Re: semi-infinite number > Anyway, it seems that semi-infinite or semiinfinite is a short way of > saying infinite in one direction but not the other (e.g. along a > ray or a halfplane). In such contexts it is a perfectly reasonable > construction. That might be one meaning, but I think it's more usual as an engineering term. The engineering meaning, as I recall it explained to me by my father is something like this: A magnitude is semi-infinite if it is so large that increasing it would make no practical difference. === Subject: Re: semi-infinite number >> There is a semi-infinite number of .... > Anyone know how many that would be? >> What's the context? Maybe it is something like cofinite, i.e. the >> complement is finite. > Conventional filters use the Z transform to describe the filter transfer > characteristic, where Z-1 denoted a unit delay. There are a semi-infinite > number of orthonormal functions for transform arithmetic. > Also found on Google > You can choose whether to use the Mandelbrot set, or generate a > semi-infinite number of Julia sets of your own > plus many more examples there; but no definition that I can fine. In the second of these examples, semi-infinite just means practically unlimitted - after all, there must be some limit to the number of diferent parameters you can specify in the program. I searched the online mathematical bible - http://mathworld.wolfram.com/search/ and semi-infinite doesn't appear - so it probably doesn't have any mathematical meaning. === Subject: Re: finite maze solving algorithm > I was wondering if anyone knows if all possible topologies of finite > 2d mazes can be solved by a finite algorithm. Yes, they can. > For example, we know > that all fully connected mazes can be solved by picking a wall and > exhaustively following it. Can a general solution work for all mazes > including the ones that are piecewise disconnected? If this is > possible, is the general solution a solved problem? Yes it is. Look for the Pledge Algorithm. It is well-described in the book Turtle Geometry, including a proof for how it solves every finite maze. The algorithm works by keeping track of the total amount of turning you do as you move throughout the maze. Using this purely local information (you don't need breadcrumbs, or to mark walls etc.) it is possible to escape. It is rather simple: 1. Initialize a counter to zero, and define an arbitrary direction to be 'north'. 2. Move straight north until an obstacle is met. 3. Turn left and follow the obstacle. Keep track of 'total turning' and increment the counter by +1 for every full 360 degree turn clockwise, and -1 for every full 360 turn anti-clockwise. 4. Leave the obstacle when it is possible to move straight north and the counter reads zero. Goto step 2. Steven === Subject: Re: finite maze solving algorithm > I was wondering if anyone knows if all possible topologies of finite > 2d mazes can be solved by a finite algorithm. For example, we know > that all fully connected mazes can be solved by picking a wall and > exhaustively following it. Can a general solution work for all mazes > including the ones that are piecewise disconnected? If this is > possible, is the general solution a solved problem? >>Interesting. How are you defining a maze, here? Usually, I think of a >>maze as just a graph, which you can solve using something simple like >>depth-first search, regardless of whether it is planar. It is probably >>better to think in terms of connectedness of rooms than connectedness of >>walls. > The basic problem with a depth-first search is the chance you'll get into a > loop going around and around a disconnected piece. The problem can be > avoided by making a rule to never cross your own path. > Or is there something more that I'm missing? If you keep going in circles, then you are not doing a depth-first search. Depth-first search will systematically explore *any* maze in its entirety. In depth-first-search, you keep traversing down previously untraversed paths until either you reach a place where all possible paths have already been traversed or you reach a dead end. In these situations, you backtrack along the path you've just traversed until reach a place where there is an unexplored corridor, and then you head down that corridor. To find your way through an actual maze (i.e. not one written on paper), you need some way of marking passages. A 19'th century algorithm for traversing a maze is as follows. Bring along a pile of pennies to mark passages as you enter and leave junctions. During the algorithm, passages without pennies haven't been traversed yet, those with one penny have been traversed exactly once, and those with two pennies have been traversed and then backtracked along in the opposite direction. Enter the maze. When you encounter a junction that you haven't seen before (no passageways with pennies), then choose a passage at random (remember to drop a penny in the passage you just left and in the passage you just entered). If you hit a dead end, turn around and go back. If you are walking down a passage for the first time and you encounter a junction you've seen before, then turn around and head back (remember to drop two pennies, one for entering the junction and one for leaving). If you are walking down a passage for the second time and encounter a junction, then take a new passage if there is one, otherwise take an old passage (marked with one penny). When you reach the solution, passages marked exactly once will indicate a direct path back to the start. If there is no solution, then you will end up back at the start with all passages marked twice. === Subject: FLTMA: c^n mod b - a^n mod b != (c^n - a^n) mod b (basic) Well, the title line says it all. I thought modular subtraction was, well, transitive? But with a,b,c = 3,4,5, 5 mod 4 - 3 mod 4 = 1 - 1 = 0 != 5-3 mod 4. So I've got a clear conceptual problem here. In the computatation of (a^n + b^n) mod c, (c^n - a^n) mod b, and (c^n - b^n) mod a, I use <-- receives the value of n<--AC<--BC<--AB<--CB<---BA<--CA<--1 for n = 1 to limit AC <-- AC*a mod c BC <-- BC*b mod c CB <-- CB*c mod b AB <-- AB*a mod b CA <-- CA*c mod a BA <-- BA*b mod a A.n <-- (CA - BA) mod a B.n <-- (CB - AB) mod b C.n <-- (AC + AB) mod c S.n <-- A.n + B.n + C.n if S.n == 0 then N = n and this inequality tells me I am doing this wrong, but to use conventional exponentiation is to invite overflow. How do I continue my explorations? I tolerance everything and tolerate everyone. I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri, alphabetically. I drive: A double-step Thunderbolt with 657% range. I fight terrorism by: Using less gasoline. === Subject: Re: FLTMA: c^n mod b - a^n mod b != (c^n - a^n) mod b (basic) > Well, the title line says it all. I thought modular subtraction was, well, > transitive? > But with a,b,c = 3,4,5, 5 mod 4 - 3 mod 4 = 1 - 1 = 0 != 5-3 mod 4. 3 mod 4 = 3 = -1 so... 5 mod 4 - 3 mod 4 = 1 - -1 = 2 = 5-3 mod 4 === Subject: Mechanical Turing Machine? I'm looking for information on any previous designs for a mechanical Turing Machine, i.e., a machine fabricated in the technology that Turing would have had available at the time of his thesis. I have conceived of a method for the (semi) infinite tape using techniques from clock-making, but I'm interested in seeing designs for a reprogrammable state-machine in mechanics. === Subject: Re: Mechanical Turing Machine? In 1936, Turing would not have been limited to mechanics. The relay had been known 100 years, and telephone people were building quite elaborate systems with them, as well as laying the foundations of formal switching theory. (Which was in turn, the basis for the wartime machines Turning worked on). The Eccles-Jordan flipflop (bistable) had been described in 1919, so vacuum-tube logic was also known. :I'm looking for information on any previous designs :for a mechanical Turing Machine, i.e., a machine :fabricated in the technology that Turing would have :had available at the time of his thesis. : :I have conceived of a method for the (semi) infinite :tape using techniques from clock-making, but I'm interested :in seeing designs for a reprogrammable state-machine :in mechanics. : === Subject: Re: Lyapunov Function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3E5lb31022; >x' = y - 2*x*(y^2) >y' = -(x^3) + (x^4)*y >I have tried V(x,y) of the form: >a(x^2) + b(y^2) >a(x^2) + b(y^2) + c(x^4) >a(x^2) + b(y^2) + c(y^4) >a(x^2) + c*x*y + b(y^2) -1with no luck. I even tried reverse engineering it, but still no luck. A hint or two would be better than an explicit answer, as I'm trying to learn here ;-) but I'm glad of any help you can give me. >Mike >>Try V(x,y) = x^(2m) + a*y^(2n) with m,n integers >=1 and a>0. >To be a little more specific: Use this Ansatz and compute >d/dt (V(x(t),y(t)). You will find that there is only one choice for m >and n such that the bad terms with odd exponents balance. Next you >can find an a which make the bad terms vanish *identically* so that >you are only left with even powers. >Fortunately this a is positive (otherwise we wouldn't have a >L.-function and d/dt (V(x(t),y(t)) <= 0 for all t ! >That might give you an idea. BTW: You have the solution in your list >(you don't need that parameters though) - somehow you skipped over it >or miscalculated. >Thomas >>Thomas Mike === Subject: Re: Lyapunov Function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3E5jk30913; >x' = y - 2*x*(y^2) >y' = -(x^3) + (x^4)*y >I have tried V(x,y) of the form: >a(x^2) + b(y^2) >a(x^2) + b(y^2) + c(x^4) >a(x^2) + b(y^2) + c(y^4) >a(x^2) + c*x*y + b(y^2) -1with no luck. I even tried reverse engineering it, but still no luck. A hint or two would be better than an explicit answer, as I'm trying to learn here ;-) but I'm glad of any help you can give me. >Mike >>Try V(x,y) = x^(2m) + a*y^(2n) with m,n integers >=1 and a>0. >To be a little more specific: Use this Ansatz and compute >d/dt (V(x(t),y(t)). You will find that there is only one choice for m >and n such that the bad terms with odd exponents balance. Next you >can find an a which make the bad terms vanish *identically* so that >you are only left with even powers. >Fortunately this a is positive (otherwise we wouldn't have a >L.-function and d/dt (V(x(t),y(t)) <= 0 for all t ! >That might give you an idea. BTW: You have the solution in your list >(you don't need that parameters though) - somehow you skipped over it >or miscalculated. >Thomas >>Thomas Ok got it now, V(x,y)= x^4 + 2*y^2 does it. This gives V'(x,y) = -(x^2)*(y^2)*(8 - 4*(x^2)) which is <= 0 for all (x,y) such that x <= sqrt(2) === Subject: Re: Gamma function/Mills ratio/Ineq by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3E5kQ30947; >> Let G be the Gamma function, and f(x)= G(x+0.5)/G(x+ 1) , (x>= 0). >> Question: to prove or disprove that for each pair (x,y), 0 =< x < y , >> there exists u(x,y) in ( 0, 1/2 ), such that >> f(y) >> ----- = sqrt{(x+u(x,y))/(y+u(x,y))} . >> f(x) >> Note:According to G.N.Watson [1] it is knownn hat for every x >= 0 >> there exists v(x) in ( 1/4, 1/pi ), such that >> f(x) =1/( sqrt(x+v(x)) ) . >> Reference: >> [1] G.N. Watson , ,,A note on gamma functions, Proc.Edinburgh >> Math.Soc., (2) 11 >> (1958/59), Edinburgh Math.Notes 42, (1959) 7-9. >I haven't seen the reference. But certainly if we intend to allow x to be >0, then the interval for v should actually be ( 1/4, 1/pi ] instead. And it >seems that that interval is as tight as possible. >Now to your question: I won't be giving a proof here, but surely what you >suspect is correct. However, the interval you've given for u, ( 0, 1/2 ), >is much looser than necessary, it seems to me. I propose that the >tightest interval for u is of the form ( 1/4, c ) where c is a constant >which is approximately 0.360674 . >David Cantrell May we use the fact that f(x) verifies this simple functional equation :f(x)*f(x+1/2)=1/(x+1/2)? Alain. === Subject: Re: Gamma function/Mills ratio/Ineq >> Let G be the Gamma function, and f(x)= G(x+0.5)/G(x+ 1) , (x>= 0). > Question: to prove or disprove that for each pair (x,y), 0 =< x < y >> , >> there exists u(x,y) in ( 0, 1/2 ), such that > f(y) >> ----- = sqrt{(x+u(x,y))/(y+u(x,y))} . >> f(x) > Note:According to G.N.Watson [1] it is knownn hat for every x >= 0 >> there exists v(x) in ( 1/4, 1/pi ), such that > f(x) =1/( sqrt(x+v(x)) ) . > Reference: >> [1] G.N. Watson , ,,A note on gamma functions, Proc.Edinburgh >> Math.Soc., (2) 11 >> (1958/59), Edinburgh Math.Notes 42, (1959) 7-9. >I haven't seen the reference. But certainly if we intend to allow x to >be 0, then the interval for v should actually be ( 1/4, 1/pi ] instead. >And it seems that that interval is as tight as possible. >Now to your question: I won't be giving a proof here, but surely what >you suspect is correct. However, the interval you've given for u, ( 0, >1/2 ), is much looser than necessary, it seems to me. I propose that >the tightest interval for u is of the form ( 1/4, c ) where c is a >constant which is approximately 0.360674 . > May we use the fact that f(x) verifies this > simple functional equation :f(x)*f(x+1/2)=1/(x+1/2)? Offhand, I don't know what using it will help us to do. (But perhaps I'm overlooking something obvious.) David === Subject: Re: Cantor's diagonal proof wrong? <41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> at 01:26 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >>Your ideas are perfectly reasonable, if you add the remark that an >>infinite sequence of digits (which would not form a natural number) >>could only emerge from a list with at least one line enumerated by >>omega or infinity. >> No. A sequence is a mapping from N to a set, and N has order type >> Omega, not Omega plus 1. > It is not a matter of order type, but a matter of finity. Each n is > finite. Each line-number of Cantors list is finite. Otherwise we could > not find and change the diagonal element a_nn. As long as we are > capable of doing so, the number of exchanged digits is finite. Wrong. There is an infinite number of finite numbers. > The infinite set of finite numbers is a contradicito in adjecto. Your saying so does not make it so. Not even when said in a fancy way. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor's diagonal proof wrong? >>Your ideas are perfectly reasonable, if you add the remark that an >>infinite sequence of digits (which would not form a natural number) >>could only emerge from a list with at least one line enumerated by >>omega or infinity. >> >> No. A sequence is a mapping from N to a set, and N has order type >> Omega, not Omega plus 1. > It is not a matter of order type, but a matter of finity. Each n is > finite. Each line-number of Cantors list is finite. Otherwise we could > not find and change the diagonal element a_nn. As long as we are > capable of doing so, the number of exchanged digits is finite. > Wrong. There is an infinite number of finite numbers. In terms of potential infinity, of course. But that does not change anything. Potential infinity means, that the sequence of natural numbers does never end, but that each one is finite. With each finite number n a finite sequence 1,2,3,...,n is connected (i.e. a bijection is possible), and the cardinal number of each of such sequences remains finite too. > The infinite set of finite numbers is a contradicito in adjecto. > Your saying so does not make it so. Not even when said in a fancy > way. If you insist on your opinion, then tell me what is wrong with my proof: For n of IN there are elements of {2, 4, 6, ..., 2n} surpassing its cardinal number, and, in particular with its inversion: If there are no elements of {2, 4, 6, ..., 2n} surpassing its cardinal number, than n is not a natural number. === Subject: Re: Cantor's diagonal proof wrong? <41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> It is not a matter of order type, but a matter of finity. Each n is >> finite. Each line-number of Cantors list is finite. Otherwise we could >> not find and change the diagonal element a_nn. As long as we are >> capable of doing so, the number of exchanged digits is finite. >> Wrong. There is an infinite number of finite numbers. > In terms of potential infinity, of course. No. The Peano axioms don't talk about processes or potential. They state properties that make a number belong to the set of naturals. An infinite set is one that can be placed into bijection with a proper subset of it, and the successor operation is a trivial bijection of the set of naturals onto a subset of it. > But that does not change anything. Potential infinity means, that > the sequence of natural numbers does never end, but that each one is > finite. Completely irrelevant. We are not interested in any fuzzy pseudophilosophical word plays. > With each finite number n a finite sequence 1,2,3,...,n is connected > (i.e. a bijection is possible), and the cardinal number of each of > such sequences remains finite too. Sure. But that tells you nothing about whether a union of an infinite number of sets with those properties still has the same properties. Because if you use induction for that proof, it is valid only for forming the union of a finite number of sets. >> The infinite set of finite numbers is a contradicito in adjecto. >> Your saying so does not make it so. Not even when said in a fancy >> way. > If you insist on your opinion, then tell me what is wrong with my > proof: For n of IN there are elements of {2, 4, 6, ..., 2n} > surpassing its cardinal number, No problem with that. > and, in particular with its inversion: If there are no elements of > {2, 4, 6, ..., 2n} surpassing its cardinal number, than n is not a > natural number. This is a heap of babbling hogwash, since it is not an inversion, but intended to be an equivalent statement. And the equivalent statement would be: There exists no n of IN such that no element of {2, 4, 6 ... 2n} surpasses the set's cardinal number. And indeed, that is true. But it tells us nothing about the set of all even numbers, since the set of all even numbers is not of the form {2, 4, 6 ... 2n} (with n in N). And since this set is not of the form for which you prove something by induction, your proof does not apply to it. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor's diagonal proof wrong? > As long as n is finite, the sequence {2, 4, 6, ..., 2n} contains > larger numbers than its cardinal number is. > If {2, 4, 6, ..., 2n} does not contain larger numbers than its > cardinal number is, then n cannot be finite. > The second sentence is meaningless as long as you no not specify what > the text > the sequence {2, 4, 6, ..., 2n} > means for some n which is not finite, i.e. the condition > {2, 4, 6, ..., 2n} does not contain larger numbers than its > cardinal number > cannot be assigned any of the truth values true or false for some > 2n which is not finite because you didn't explain what the text > {2,4,6,...2n} means for infinite n. I do not pretend to know anything about the properties of non-natural numbers. Contrary, I am convinced, that they are phantoms, at best. Bust my statement remains true. In case that there are no non-natural numbers, it simply means, that it is true in any case. Isn't ist obvious, to conclude from the statement If n is in IN, then n is positive that any n was positive, in case there were no other numbers than n? > If you use a logical statement using a variable n you have to > specify *before* which universe this variable comes from, and you > have to define the meaning fo any formula like {2,4,6,...,2n} for > *every* element of this universe. In our universe of natural numbers, my statement is correct. If one wants to circumvent it, he should define an extended universe (and prove the consistency of his definition). > A correct statement would be: > Every *finite* subset of the set of even natural numbers >=2 contains > an element which is larger than the cardinal number of this set. > If a subset of the set of even natural numbers >=2 does not contain an > element which is larger than the cardinal number of this subset then > this subset is not finite. Of course we could do that, unless finity and infinity had not been intermingles by statements about the actually infinite set of finite numbers. In order to circumvent wrong wording, I stated: In every senquence of even numbers 2, 4, 6, ..., 2n, where n is any natural number ... Here we see clearer than in your version, that there are no actually infinite sets. But I do not start with that obvious assumption, because I have set out to prove it. > Note that in this version the meaning of the term subset *is* > defined for *finite* and *not finite* because the assumed universe of > subset is the collection of all subsets of {2,4,6,.,,,,} > while you didn't even explain what 2n means for n not finite I repeat: I consider exclusively finite n. For those my proof is correct. It may become incorrect for other numbers, but I am not interested in those numbers. === Subject: Re: Cantor's diagonal proof wrong? > at 01:34 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: > >What does distinguish the limit from the diagonal? > > What distinguishes a banana from a townhouse? They're not remotely > similar. > > The diagonal is a sequence of digits, not a number. In the case of the > Cantor diagonal argument you're referring to, you take the sequence of > digits as coefficients in a series and it is trivial to prove that the > series converges. > > Ok, the abbreviation diagonal should expand to diagonal number. > >Do we need different words here? > > We need words that are applicable, and we need to ensure that we have > a common understanding of what they refer to. In particular, the term > limit has a precise meaning. > >However, how, then, can Cantor change all the digits of this >limit? > > He doesn't change anything. He defines a new number in terms of a > sequence of representations of numbers. > > He defines it by changing the digits a_nn of the diagonal number D to > a'_nn of the new number D', because he must make sure that in any case > a'_nn /= a_nn. > >And why can't we consider my proof in the limit? > > What proof? > > I have definined a Cantor-list, which always contains the diagonal > number D_n constructed up to line n in line Z(n+1) by construction. I > found this very same list also appearing in this thread: > > 0.000... > 0.100... > 0.11000... > 0.111000... > ... > > Changing the diagonal elements 0 -> 1, we have D_n = Z(n+1). > > We see that either of the two statements: > A) Cantors changed diagonal number differs from every real in a line > not A) Cantors diagonal number does not differ from every real in a > line > can be taken for granted. There is no logical priority in favour of A > or not A, as long as all lines are enumerated by natural, hence finite > numbers. > Why is > I) D_k =/= 0 for every k > II) for every Z_n: > not ( Z_n,k =/= 0 for every k ) > or, denoting E the property of a sequence to have *all* of its > elements =/= 0 > I) D has the property E > II) no Z_n has the property E > *not* a logical priority for ( D differs from every Z_n ) ? Just > *because* all lines are enumerated by natural numbers from which > follows that there is no other candidate for equality than *numbered* > lines, and because a sequence can only be equal to the sequence D if > it *has* got the property E. Your proof shows, that D does not completely consist of elements stemming of lines which are enumerated by natural numbers, because any D_n consisting of elements a_nn (with finite n) does contain only a finite number of elements equal 1 (and following zeros). But Cantor assumed his list to consist only of lines enumerated by finite numbers n. === Subject: Re: Cantor's diagonal proof wrong? >Your arguing is true only for finite sequences D_k of the diagonal. > No. >The number of digits of the complete diagonal D is infinite >(aleph_0), hence complete induction and your proof (or construction) >fail. > There's no induction to fail. >All three forms are equipollent, but wrong in case of the following >list: > No. >The diagonal squence D_n > The Cantor antidiagonal proof does not construct a sequence of real > num bers, it constructsw a single real number. Your D_n has nothing to > do with Cantor's proofs. >This fact does never change. Therefore it is true for the whole >diagonal D, > No, because the antidiagonal real is not any of your D_n. If you accept that the whole set of natural numbers is created by the Peano axioms, in particular the second, namely n has a successor n + 1, but if you simultaneously deny that induction is capable of reaching all natural numbers (all lines of the list), then we are talking about mathemagics, but not about mathematics. === Subject: Re: Cantor's diagonal proof wrong? <41ad4f31$14$fuzhry+tra$mr2ice@news.patriot.net> >If you accept that the whole set of natural numbers is created by the >Peano axioms, Axioms don't create, they describe. >but if you simultaneously deny that induction is capable of reaching >all natural numbers (all lines of the list), issue is not any imaginary claims that there is an unreachable natural number; the issue is Cantor's antidiagonal argument, which neither makes nor requires any such claim. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Cantor's diagonal proof wrong? >It is not a matter of order type, but a matter of finity. > Your at least one line enumerated by omega or infinity would seem to > rule out finitude. Adding a terminal element to an infinite ordered > set creates an ordered set with a different order type. It need not be a single element, it could be a set of infinitely many omegas. That, however, is irrelevant, because we are talking about the complete set IN of natural numbers n only. >It is not a matter of order type, but a matter of finity. Each n is >finite. Each line-number of Cantors list is finite. Otherwise we >could not find and change the diagonal element a_nn. > The list, however, is infinite. Every natural number is an ordinal number and is simultaneously the cardinal number of the sequence of all its successors including itself: 1,2,3,...,n. The sequence up to a natural number n can never have the cardinal number aleph_0. But IN does not contain other than natural numbers. >As long as we are capable of doing so, the number of exchanged > digits is finite. > The Cantor proof is not a construction or process. There is no > sequencing or temporality in it. It uses a purely functional > definition of the antidiagonal number in terms of the numbers on the > list. There need not be any sequencing or any temporality, but in order to identify the digit a_nn to be exchanged, both must be identified at first, n and a_nn. As long as n is natural, the sequence of lines has a finite cardinal number. But others than lines enumerated by natural numbers could not be identified. >The infinite set of finite numbers is a contradicito in adjecto. > No. There is no contradiction, because there are an infinit number of > finite numbers. I proved that for n of IN there are elements of {2, 4, 6, ..., 2n} surpassing its cardinal number. The true inversion of this statement is: If there are no elements of {2, 4, 6, ..., 2n} surpassing its cardinal number, than n is not a natural number. >Yes = No to No. > If there is a contradiction in finished infinity then produce it > with a logical argument rather than handwaving. Infinity is nothing else but the latin translation of unfinished or even unfinishable. Finished unfinished or finished unfinishable is a contradiction. Finished infinity means actual infinity like the infinite set of the finite natural numbers, which is self-contradictory (see above). === Subject: Re: Cantor's diagonal proof wrong? <41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net> <41ad5139$15$fuzhry+tra$mr2ice@news.patriot.net> >It need not be a single element, it could be a set of infinitely many >omegas. If you add more than one element at the end you have the same problem; it's no longer the order type of omega. >That, however, is irrelevant, because we are talking about the >complete set IN of natural numbers n only. Then there's nothing to add to the end. >Every natural number is an ordinal number and is simultaneously the >cardinal number of the sequence of all its successors including >itself: 1,2,3,...,n. The sequence up to a natural number n can never >have the cardinal number aleph_0. Cantor's proofs do not involve finite sequences. Omega is not a natural number. >There need not be any sequencing or any temporality, but in order to >identify the digit a_nn to be exchanged, both must be identified at >first, n and a_nn. As long as n is natural, the sequence of lines What sequence of lines? The sequence that is relevant is the entire list, which is not finite. Subsequences are irrelevant. >But others than lines enumerated by natural >numbers could not be identified. There are no others in a sequence. >I proved that for n of IN there are elements of {2, 4, 6, ..., 2n} >surpassing its cardinal number Which is irrelevant. >Infinity is nothing else but the latin translation of unfinished or >even unfinishable. Don't confuse definition with etymology. Infinity is no more unfinished than gauche is left. >Finished infinity means actual infinity like the infinite set of the >finite natural numbers, which is self-contradictory (see above). You showed no contradiction. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Cantor's diagonal proof wrong? > 0.000... > 0.100... > 0.11000... > 0.111000... > ... > Changing the diagonal elements 0 -> 1, we have D_n = Z(n+1). > We see that either of the two statements: > A) Cantors changed diagonal number differs from every real in a line > not A) Cantors diagonal number does not differ from every real in a > line > can be taken for granted. > If can be taken for granted is supposed to mean has to be true, > and line is supposed to mean the number indicated by a line. > There is no logical priority in favour of A or not A, as long as all > lines are enumerated by natural, hence finite numbers. > Uh, there is a proof for A. That's quite a logical priority. It does > not get better than that. This proof is necessarily wrong, if A is assumed. Hence, there is no logical priority. === Subject: Re: Cantor's diagonal proof wrong? > The diagonal squence D_n constructed by a'_nn = 1 + a_nn is always > contained in line Z(n+1). This is correct for every finite n by > construction of the matrix. > This fact does never change. Therefore it is true for the whole > diagonal D, > Wrong. Therefore it is true for any D_n. D itself is no such D_n. Of course it is. You must not intermingle my n with a fixed value. It stands for any conceivable natural number. Otherwise you could not construct IN by means of the Peano axioms. These axioms are the basis of the complete infinite set of all natural numbers. Should their application in complete induction be restricted to a smaller set? > unless it contains elements with non-natural numbers as indices. > No. The diagonal _entries_ contains only natural numbers as indices, > numbers that also form the indices of the elements of all the D_n. So > if you prove some property holding for all elements of every D_n, then > this property will also hold for all elements of D. That is my opinion too, and nothing more is required. But that does not > make D itself inherit any properties from the D_n. In particular, D > is not the image of any Z(n). Wrong (see above). Again: n is not a fixed value. n represents any natural number, and there is no natural number which is not represented by n. You can see this fact best, if you try to find a natural number which is in IN but is not covered by my proof. === Subject: Re: Cantor's diagonal proof wrong? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> The diagonal squence D_n constructed by a'_nn = 1 + a_nn is always >> contained in line Z(n+1). This is correct for every finite n by >> construction of the matrix. > This fact does never change. Therefore it is true for the whole >> diagonal D, >> Wrong. Therefore it is true for any D_n. D itself is no such D_n. > Of course it is. So which n are you talking about in particular? > You must not intermingle my n with a fixed value. In D_n it stands for that. > It stands for any conceivable natural number. Quite so. And for any conceivable natural number n, D is different from D_n. So D itself is no such D_n. > Otherwise you could not construct IN by means of the Peano > axioms. You are babbling. One has absolutely nothing to with the other. >> No. The diagonal _entries_ contains only natural numbers as indices, >> numbers that also form the indices of the elements of all the D_n. So >> if you prove some property holding for all elements of every D_n, then >> this property will also hold for all elements of D. > That is my opinion too, and nothing more is required. Wrong. For your proof, more is required. >> But that does not make D itself inherit any properties from the >> D_n. In particular, D is not the image of any Z(n). > Wrong (see above). Above was nothing but nonsense and babbling. > Again: n is not a fixed value. n represents any natural number, You know the difference between any and all, right? > and there is no natural number which is not represented by n. Rather which could not be assumed by n. That is right. And for no natural number n we have D_n = D. > You can see this fact best, if you try to find a natural number > which is in IN but is not covered by my proof. Your proof gives us a sequence D_n = (1-10^{-n})/9, and a value D=1/9. For any given value of n, we have D-D_n = 10^{-n}/9, so for any given value of n, D is different from D_n. The value of the difference, of course, depends on n, and there is no _fixed_ positive value epsilon such that _all_ D_n differ by more than that _fixed_ epsilon from D. But that is not required, unless you are suffering from operator dyslexia. It _does_ lead to the observation that D is the _limit_ of D_n. But it is never assumed by any of the D_n, nevertheless. It is my guess that you are also incapable of grasping the difference between continuous and uniformly continuous functions, right? Similar problem. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor's diagonal proof wrong? > > Your arguing is true only for finite sequences D_k of the diagonal. > I stated > for every k it is true that > the element #k of the sequence D is different from > the element #k of the sequence a_k > -> for every k it is true that > D is different from a_k > because element #k of D differs from element #k of a_k induces D > differs from k because two sequences differ whenever they differ by > at least one element. > -> There is no k such that a_k = D > Please point out _exactly_ which one of my two implications is > logically wrong in your system and why ? Can two sequences be > identical and still differ by at least one element? You seem to assume that k is a fixed number. It isn't. It can take any value of a natural number. There is no natural number which it could not represent. This meaning becomes clear if you consider the Peano axioms. In particulare the induction axiom (2) states that with n also its successor n+1 is a natural number. This axiom is the basis of the infinite set of all natural numbers. Should its application in complete induction be restricted to a smaller set? I proved that for n of IN there are elements of {2, 4, 6, ..., 2n} surpassing its cardinal number. The true inversion of this statement is: If there are no elements of {2, 4, 6, ..., 2n} surpassing its cardinal number, then n is not a natural number. But, of course, we are talking about natural numbers only. It is completely meaningless, whether there is an infinite set or no last element. The only assumption required for my proof is that IN consists of natural, hence finite, numbers. === Subject: Re: Cantor's diagonal proof wrong? > > Your arguing is true only for finite sequences D_k of the diagonal. > > I stated > > for every k it is true that > > the element #k of the sequence D is different from > the element #k of the sequence a_k > > -> > for every k it is true that > > D is different from a_k > > > because element #k of D differs from element #k of a_k induces D > differs from k because two sequences differ whenever they differ by > at least one element. > > > -> > There is no k such that a_k = D > > > Please point out _exactly_ which one of my two implications is > logically wrong in your system and why ? Can two sequences be > identical and still differ by at least one element? > You seem to assume that k is a fixed number. It isn't. You are pulling my leg, aren't you ? When I say for every k then I assume that k is a fixed number ? > It can take any > value of a natural number. There is no natural number which it could > not represent. That's what my sentence for every k it is true that the element #k of the sequence D is different from the element #k of the sequence a_k said. You don't have to explain it to me. Which part of this sentence don't you understand? -- Horst === Subject: Re: Cantor's diagonal proof wrong? > You seem to assume that k is a fixed number. It isn't. >> You are pulling my leg, aren't you ? When I say for every k then >> I assume that k is a fixed number ? >He does not understand quantifiers. He does not understand that the >quantifier variable does not mysteriously leak out of the quantified >expression in some manner. He does not understand that _inside_ of >the quantified expression, indeed k always stands for just one _fixed_ >number at a time, doesn't even understand the semantics of variables in a quantified expression. > and that _outside_ of the quantified expression, a >statement completely independent from k is made about _all_ _fixed_ >expressions resulting from assigning a possible fixed value to k. Horst === Subject: Re: Cantor's diagonal proof wrong? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Any line enumerated by a natural number has a finite number of lines > as predecessors. Hence, the diagonal of the line numbers 111...111 > does exist as a finite, hence natural number. Nonsense. The word Hence is used completely groundless. > But here is another puzzle: > As long as n is finite, the sequence {2, 4, 6, ..., 2n} contains > larger numbers than its cardinal number is. > If {2, 4, 6, ..., 2n} does not contain larger numbers than its > cardinal number is, then n cannot be finite. Where is the puzzle? Except in incredibly shoddy wording? First you make a claim about finite n, then you make some hypothetical claim without even specifying what values n is allowed to assume, then you wildly assume the existence of some set {2..2n} _still_ without specifying your base set of n, make a statement that refers to the previous set and find it contradictory given the original base set. This merely proves that you did something invalid. It does neither prove nor disprove the existence of some set {2..2n} with the desired property, in particular since you don't specify what base set n is allowed to be in in the first place! -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Cantor's diagonal proof wrong? <41aa5d5a$14$fuzhry+tra$mr2ice@news.patriot.net> at 10:24 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >I have definined a Cantor-list, which always contains the diagonal >number D_n constructed up to line n None of those is the number that Cantor defines. >We see that either of the two statements: >A) Cantors changed diagonal number differs from every real in a line >not A) Cantors diagonal number does not differ from every real in a >line can be taken for granted. What does that have to do with Cantor's proofs? The issue is not what appears in a line but rather what appears in a list. >There is no logical priority in favour of A or not A, as long as all >lines are enumerated by natural, hence finite numbers. Which is, of course, impossible, as Cantor proved. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Cantor's diagonal proof wrong? > at 10:24 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >I have definined a Cantor-list, which always contains the diagonal >number D_n constructed up to line n > None of those is the number that Cantor defines. Then he must include numbers which cannot be represented by n. That's not what he pretended to do. >We see that either of the two statements: >A) Cantors changed diagonal number differs from every real in a line >not A) Cantors diagonal number does not differ from every real in a >line can be taken for granted. > What does that have to do with Cantor's proofs? The issue is not what > appears in a line but rather what appears in a list enumerated by solely natural numbers. >There is no logical priority in favour of A or not A, as long as all >lines are enumerated by natural, hence finite numbers. > Which is, of course, impossible, as Cantor proved. Pardon, Cantor was talking about positive finite (today we say: natural) numers. If you think, that then his list would be impossible, we do agree. === Subject: Re: Cantor's diagonal proof wrong? <41aa5d5a$14$fuzhry+tra$mr2ice@news.patriot.net> <41ae4af6$6$fuzhry+tra$mr2ice@news.patriot.net> >Then he must include numbers which cannot be represented by n. That statement has no meaning. He defined a real number that is not in the list. >enumerated by solely natural numbers. Exactly, not, as you tried to do, enumerated by arbitrary real numbers. >Pardon, Cantor was talking about positive finite (today we say: >natural) numers. No. Cantor was talking about a list of real numbers. You are confusing the labels with the values labelled. >If you think, that then his list would be impossible, we do agree. No, we don't agree; he proved that no such list is possible. Or, to put it another way, he proved that for any list of real numbers there is a real number not on the list. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Cantor's diagonal proof wrong? <41aa5d5a$14$fuzhry+tra$mr2ice@news.patriot.net> <41ae4af6$6$fuzhry+tra$mr2ice@news.patriot.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> at 10:24 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >>I have definined a Cantor-list, which always contains the diagonal >>number D_n constructed up to line n >> None of those is the number that Cantor defines. > Then he must include numbers which cannot be represented by > n. That's not what he pretended to do. Wrong. D is not part of the list, it is merely derived from it. And it is not equal to any D_n. Because for any given n, you will find a non-zero difference |D-D_n|. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Lorenzini Can anyone comment on this http://www.uga.edu/news/artman/publish/041117lorenzini.shtml ? Their background presentation is the worst I've ever seen. Where can I read about the stated problem in greater details? Igor === Subject: Re: Lorenzini > Can anyone comment on this > http://www.uga.edu/news/artman/publish/041117lorenzini.shtml > Their background presentation is the worst I've ever seen. > Where can I read about the stated problem in greater details? > Igor Sounds like the Tate-Shafarevich and Selmer groups on an elliptic curve; however I haven't heard about this, so this is a TOTAL guess. In any case, I would like to know what the two 30-year-old 'results' were that were refuted. Jim Buddenhagen === Subject: Re: Lorenzini >Can anyone comment on this >http://www.uga.edu/news/artman/publish/041117lorenzini.shtml >Their background presentation is the worst I've ever seen. >Where can I read about the stated problem in greater details? At a wild guess, papers 20 and 19 at http://www.math.uga.edu/~lorenz/paper.html are the ones referenced. Lee Rudolph === Subject: A Question in differential geometry (or calculus) a friend has asked me a question, and although it looks simple, I can't find the answer: Let c(t) = (c1(t),c2(t)) be a curve in R2 given in unit speed. let's define: x(t) = c'(t) y(t) = c''(t) / k(t) (k(t) is the curvature of the curve, k(t) = ||a''(t)|| ) According the this friend of mine, there is a connection between x(t) and y(t). specifically, there is a constant linear transformation (which doesn't depend on c) which takes x(t) and returns y(t). Is this true? how can I prove it if it does? (sorry about the mistakes i probably have in english. it's not my native language - I'm from turkey) === Subject: Re: A Question in differential geometry (or calculus) > a friend has asked me a question, and although it looks simple, I > can't find the answer: > Let c(t) = (c1(t),c2(t)) be a curve in R2 given in unit speed. let's > define: > x(t) = c'(t) > y(t) = c''(t) / k(t) (k(t) is the curvature of the curve, k(t) = > ||a''(t)|| ) > According the this friend of mine, there is a connection between x(t) > and y(t). specifically, there is a constant linear transformation > (which doesn't depend on c) which takes x(t) and returns y(t). Is this > true? how can I prove it if it does? If I understand your question well, you're just asking for the Frenet formulas: http://mathworld.wolfram.com/FrenetFormulas.html Leave the torsion out if you are only interested in 2-D. > (sorry about the mistakes i probably have in english. it's not my > native language - I'm from turkey) No apologies needed. I think most of us find your English good enough. Han de Bruijn === Subject: Re: A Question in differential geometry (or calculus) >> Let c(t) = (c1(t),c2(t)) be a curve in R2 given in unit speed. let's >> define: >> x(t) = c'(t) >> y(t) = c''(t) / k(t) (k(t) is the curvature of the curve, k(t) = >> ||a''(t)|| ) ^^^ Should be c''(t) ?! >> According the this friend of mine, there is a connection between x(t) >> and y(t). specifically, there is a constant linear transformation >> (which doesn't depend on c) which takes x(t) and returns y(t). Is this >> true? how can I prove it if it does? > If I understand your question well, you're just asking for the Frenet > formulas: > http://mathworld.wolfram.com/FrenetFormulas.html > Leave the torsion out if you are only interested in 2-D. I think, he is asking for something more basic, namely for the fact that the vector of acceleration y(t) and the tangent vector x(t) form an orthonormal basis of R^2, so they can be transformed into each other by a rotation around c(t) with angle plus or minus 90 degrees. Problem is, that which of the two possible rotation angles must be used to turn x(t) into y(t) is not for all curves the same; but by continuity the angle remains the same along the curve as long as y(t) is defined, i.e. as long the curve does not contain a point of inflection (where k(t) becomes 0). For an example of this change of rotation angle happening look at a cubic parabola u |--> (u,u^3), re-parametrized to arc length, which contains an inflection point at time u=0. The only non-trivial step in the prove that (x(t),y(t)) forms an orthonormal system is the orthogonality of the vectors. This step follows by deriving < c'(t), c'(t) > = 1 with respect to t. === Subject: Re: JSH: Point of logic > A mathematical proof must be logical. What I aim to do here is > elaborate on the logic, which proves I'm correct, and let's see what > happens. > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the a's are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > That corresponds with the constant term of P(x) being 1078, which is > 7(7)(22). > Now I note that dividing both sides by 49 gives me > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, > which factors as > P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 > and I note that the constant term is now 22. > Now what do I know about the constant terms? > Well they are constant, and are specific numbers, specifically 7, 7 > and 22, for the factors of P(x). But for P(x)/49, the constant term > is 22. > So, logically, two of the constant terms were divided by 7. > That is the essential point on which everything hinges. > Notice how simple it is, the constant terms are actual numbers. > Numbers like 7 and 22 are NOT variables, and they remain the same > without regard to the value of x. > Dividing P(x) by 49 changes the constant terms. > They go from being 7, 7 and 22 to being 1, 1 and 22. > Therefore, exactly two of them were divided by 7. > There is no other way to get from 7 to 1, and you can write it out > algebraically. > 7/w = 1, giving w = 7. > By the distributive property for the constant terms of two of > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > to be divided by 7, the full factor has to be divided by 7, which > gives > P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) > indicating that two of the a's have 7 as a factor, and I've > arbitrarily selected a_1(x) and a_2(x). > What can be shown is that in the ring of algebraic integers, if > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > is irreducible over Q, with integer coefficients, then the a's cannot > have 7 as a factor in the ring of algebraic integers. > So is there a problem with the logic in my argument above? I thought it might be worthwhile revisiting this Harris argument, but with more of the detail filled in, as Harris provided in a post in another thread on November 14. This is quoted below: But P(x) is a multiple of 49 as *each* coefficient has 49 as a factor, >so I divide out the 49 to get >P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22. >Now assume each of the a's has some non-unit factor in common with 7, >for a given x, which in fact they do in the ring of algebraic integers >whenever all the a's are irrational, which is a point that has been >brought up repeatedly by people arguing with me. For a while I >resisted that fact, but now as I've said before I concede that they >are in fact correct--each of the a's, in the ring of algebraic >integers does in fact share a non-unit factor with 7 when all of the >a's are irrational. >So let w_1(x) w_2(x) w_3(x) = 49, where the w's are those factors, so >a_1(x) = w_1(x) b_1(x), >a_2(x) = w_2(x) b_2(x), >and >a_3(x) = w_3(x) b_3(x) >and where >w_1(x) v_1(x) = 7, w_2(x) v_2(x) = 7, and w_3(x) v_3(x) = 7, >and divide through by 49 to get >P(x)/49 = > (5 b_1(x) + v_1(x))(5 b_2(x) + v_2(x))(5 b_3(x) + v_3(x)) >and if you allow that the factors are each factors of the constant >term as before, then you have > v_1(0) v_2(0)(15 + v_3(0)) = 22 >at x=0, and when x does not equal 0, you have > v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 >where I introduce u_3(x) to handle any further weirdness with how >a_3(x) behaves, where u_3(0) = 3, to agree with previous results, and >remember >P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 >and the constant term doesn't change, as, well, it's constant, as it's >22. >So, notice, I now have that v_1(x) and v_2(x) are factors of 22. >Provably, they cannot be units in the ring of algebraic integers when >a_1(x), a_2(x), and a_3(x) are all irrational as that's what people >argued with me about for so long. >The full result was just so weird that it escaped even me for a while, >as following their arguments to their logical conclusion 22 and 7 must >share non-unit factors in the ring of algebraic integers. >But you can also appear to prove that 22 and 7 are coprime in the ring >of algebraic integers using various accepted definitions of >coprimeness, like you can find algebraic integers x, and y such that >22x + 7y = 1 >and claim to have proven that they are coprime. >So, in the ring of algebraic integers, using what's commonly accepted >you can prove that 22 and 7 are both coprime and that they share >non-unit algebraic integer factors. >That's the full result. Actually I thought this argument looked pretty good! Harris assumes that there exist algebraic integers w_1(x), w_2(x), and w_3(x) such that (1) w_1(x)*w_2(x)*w_3(x) = 49 (2) a_1(x) is divisible by w_1(x), etc. - that is, a_1(x)/w_1(x) = b_1(x), where b_1(x) is an algebraic integer, and similarly for a_2(x) and a_3(x). (3) 7 is divisible by w_1(x), w_2(x), and w_3(x),; that is, 7/w_1(x) = v_1(x), 7/w_2(x) = v_2(x), and 7/w_3(x) = v_3(x), where v_1(x), v_2(x), and v_3(x) are algebraic integers. (4) v_1(0) * v_2(0) * (5 b_3(0) + v_3(0)) = 22. All of this is correct! Harris then goes on to say, >... and when x does not equal 0, you have > v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 >where I introduce u_3(x) to handle any further weirdness with how >a_3(x) behaves, where u_3(0) = 3, to agree with previous results. And he then concludes that >So, notice, I now have that v_1(x) and v_2(x) are factors of 22. And this is the crucial statement. If Harris is right about this, then he has justified his claims. Is he? We need to go back to his assertion that v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 Is this correct, for x <> 0 ? v_1(x) * v_2(x) * v_3(x) = 7 * 7 * 7 / 49 = 7. Thus easily, v_1(x) v_2(x) (5 u_3(x) + v_3(x)) = 7 + 5 u_3(x) v_1(x) v_2(x). Now, this will equal 22 only if u_3(x) * v_1(x) * v_2(x) = 3. This is known to be true when x = 0, because u_3(0) = 3 and v_1(0) = 1 and v_2(0) = 1. But unless you assume what you want to prove - that is, that v_1(x) and v_2(x) are constant functions, and in addition that u_3(x) is a constant function, there is no reason to conclude that u_3(x) * v_1(x) * v_2(x) = 3 for values of x other than 0. Harris's statemtent that v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 is simply not justified. In fact it is false for x > 0. Again, this is where Harris abandons his own definition of constant term, which yields the correct statement v_1(0) v_2(0)(5u_3(0) + v_3(0)) = 22 and he assumes that the constant term is defined by its position in an expression rather than by a specific value. -------------------------------------------------------------------- Related to this: since w_1(x)*w_2(x)*w_3(x) = 49, it IS true that (7/w_1(x)) * (7/w_2(x)) * (22/w_3(x)) = 22, for any x. This looks a great deal like exactly what Harris wants! Better yet, unlike what he claimed above, it is actually correct !!! But does it imply what he wants, i.e., does it imply that v_1(x) and v_2(x) are divisors of 22 ? Noting that v_1(x) = 7/w_1(x), v_2(x) = 7/w_2(x), and v_3(x) = 7/w_3(x), the equation above implies that [***] v_1(x) * v_2(x) * v_3(x) * 22/7 = 22, which also follows trivially from the equation noted previously, v_1(x) * v_2(x) * v_3(x) = 7. Because 22/7 is not an algebraic integer, equation [***] does NOT tell you that v_1(x) and v_2(x) are divisors of 22 in the algebraic integers. Clearly from the equation just quoted, v_1(x), v_2(x), and v_3(x) are all divisors of 7. That is not surprising or informative, and it does not lead to the conclusion that Harris wants. ----------------------------------------------------------------------- v_1(x) v_2(x)(5u_3(x) + v_3(x)) = 22 is the following: If you are considering a product of functions of a variable x, then the product of the constant terms is the constant term of the product. And this is a true statement. However, it cannot be assumed, for example, that v_1(x) is the constant term of (5 b_1(x) + v_1(x)). By Harris's own definition, the constant term of this expression is 5 b_1(0) + v_1(0) = 5 * 0 + v_1(0) = v_1(0) = 1. Assuming that v_1(x) = 1 for x nonzero is just assuming what you (or Harris, actually) want to prove. The correct value for the constant term is simply v_1(0), i.e., neither more nor less than what Harris's definition of constant term specifies. --------------------------------------------------------------------- Another cut at the Harris reasoning. Harris notes correctly that P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22), where one may assume that a_1(x), a_2(x), and a_3(x) are functions that take their values in the ring of algebraic integers, and a_1(0) = 0, a_2(0) = 0, a_3(0) = 3, and b_3(0) = 0. [note that b_3(x) is defined to equal a_3(x) - 3 ]. So he has set up the factorization so that the constant terms are 7, 7, and 22, and of course P(0) = 49 * 22. Then Harris thinks: I must divide both sides by 49. The constant term of P(x)/49 is 22. The only way I can think of to divide 49 out of the three factors (which have constant terms 7, 7, and 22) is to divide 7 out of the first factor and second factor (giving constant terms of 1 in both cases), and divide the third term by 1 (giving constant term 22). The problem with that is, a_1(x) and a_2(x) are not divisible by 7. What you NEED to do is divide each of the three factors by numbers which DO divide a_1(x), a_2(x), and a_3(x), and also are divisors of 7. This *can* be done. There are three divisors which I will call w_1(x), w_2(x), and w_3(x), exactly as Harris describes above, and their product happens to be 49. They are functions of x. They HAVE to be, because when x = 0, a_1(x) and a_2(x) are divisible by 7, but when x > 0 , they are not. The result, however, is that 7/w_1(x) is NOT the constant term of (a_(x) + 7)/w_1(x). There is no reason it should be. The key thing here is that you divide both sides by 49 in such a way that all the numbers in sight are algebraic integers. Harris at this point has lost sight of what he wanted to do. He wanted to factor 49 out of both sides so that the factors are all algebraic integers. He has convinced himself that one of the objectives of the factoring was to preserve the constant terms. The key thing here is, you cannot have both. If you want to preserve the constant terms, then you will not end up with algebraic integers on both sides. If you want algebraic integers everywhere, then the constant terms cannot be preserved. Of these two goals, the essential one was the original one: to end up with algebraic integers on both sides after 49 is factored out. Preserving the constant terms amounts to nothing more than requiring that the functions w_1(x), etc., are constant, and there is no reason to do this. Nora B. === Subject: Re: JSH: Point of logic > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > > I factor the polynomial as > > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > > when the a's are the three roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > It's kind of strange to call f(0) the constant term of f when f > isn't a polynomial. It's called the constant term when f is a > polynomial, because f is a sum of finitely many monomial terms, > and the constant term is the one of degree 0. Your non-polynomial > factors of P(x) have no such expressions. Furthermore, the word > constant is really confusing you here. I suggest you stop using > it. The terms are constants. And they are 7, 7 and 22, factors of the constant term of the polynomial P(x), which is 1078. So, literally, the terms are constants, so they are constant terms, in that they do not vary. Understand? > By the distributive property for the constant terms of two of > > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > > to be divided by 7, the full factor has to be divided by 7, which > gives > > P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) > > indicating that two of the a's have 7 as a factor, and I've > arbitrarily selected a_1(x) and a_2(x). > Why should any of the a's have 7 as a factor? There's no reason > to assume this division stays inside the ring of algebraic > integers for all relevant values of x. The logic is trivial. The factors of P(x) MUST also contain factors of its constant term. You can determine what those factors are, and doing so shows that they are 7, 7 and 22. Now P(x) has 49 as a multiple, and dividing that multiple off divides 49 from the factors of P(x). Understand so far? Looking at the resultant factors of the constant term of P(x)/49, you find that they are 1, 1, and 22, indicating that 7 divides off from two of them. That means that two of the factors of P(x) must have been divided by 7. > What can be shown is that in the ring of algebraic integers, if > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > is irreducible over Q, with integer coefficients, then the a's cannot > have 7 as a factor in the ring of algebraic integers. > If this is true, then it just proves the above division leaves > the ring of algebraic integers for some x. There's nothing that > interesting about the fact that not everything in this ring is > divisible by 7. Well, the algebra says one thing, but the ring of algebraic integers says another. There is an apparent contradiction which has to be resolved. > I simply assert that given 7 to get to 1, you must > divide 7 by 7, and that by the distributive property, a factor that > has 7 as a constant term must itself by divided by 7. > Then you note that these factors have properties inconsistent with > divisibility by 7. No. They are inconsistent with it IN THE RING OF ALGEBRAIC INTEGERS. So the algebra says one thing, but in the ring of algebraic integers you get something contradictory if the cubic defining the a's is irreducible over Q. > When you see that a_1(x), for certain values of x, has a property > that rules out its being divisible by 7 in the ring of algebraic > integers, then you should conclude that a_1(x)/7 doesn't stay in > this ring. If the cubic defining the a's is irreducible then, yes, it's not in the ring, in apparent contradiction with what follows from the algebraic argument. It's neat. A one time thing in math history--apparent contradiction at the heart of what's believed to be mathematically true. Never been seen before like this in all of human history. > When I was first exposed to division in elementary school, we used > remainders when quotients didn't divide evenly. We said, 23 > divided by 7 equals 3, remainder 2. Later I learned about > fractions, and eventually became used to the idea that 23/7 is a > number. Now I'm used to being able to divide almost any number by > 7, but I still need to keep issues of divisibility in mind. It would > be a mistake to assume this sort of number shares all the properties > of integers. I can only apply the properties of integers to x/7 if I > know that x is divisible by 7. You have to follow the logic. If the logic escapes you then you'll never see the apparent contradiction. If you can't see the apparent contradiction, then you have no foothold for understanding the result. Logically, you must believe that the factors of the polynomial of P(x) multiply together to give P(x), including its constant term, 1078. Technically, g_1(x) g_2(x) g_3(x) = P(x) and then you must believe that those factors include factors of the constant term of P(x), and if you do believe then you can accept g_1(0) g_2(0) g_3(0) = P(0) and if those are in fact factors of the constant term of P(x), then you can accept that they themselves are NOT dependent on the value of x, and in fact since they are 7, 7 and 22, that is true. If you believe that constants do not change as x varies, then you can accept that the value of x has nothing to do with the value of 7, 7 and 22. But dividing the polynomial by 49--its multiple--gives constant terms that are 1, 1 and 22, so they changed. Since the polynomial has 49 as a multiple there's no RATIONAL reason to believe that you'll get some kind of fraction. If you go from 7, 7 and 22 to 1, 1 and 22, then you divided two by 7. It's simple. Now I suggest you try to attack the logic. For instance, do you disagree with the assertion that the factors of P(x) include factors of its constant term? If so, what is the mathematical basis for your disagreement? If not, then how can you not get it? James Harris === Subject: Re: JSH: Point of logic Discussion, linux) > The logic is trivial. The factors of P(x) MUST also contain factors > of its constant term. You can determine what those factors are, and > doing so shows that they are 7, 7 and 22. Now P(x) has 49 as a > multiple, and dividing that multiple off divides 49 from the factors > of P(x). Understand so far? I really don't get it. What is the difference between your situation and this one? Let f(x) = x g(x) = x h(x) = x^2 + 2x P(x) = (f(x) + 2)(g(x) + 2)(h(x) + 1) Since P(x) = (x + 2)^2 (x + 1)^2, clearly P(n) is divisible by 4 for every natural number n. 2 divides (f(0) + 2) and (g(0) + 2). Therefore, for every n, 2 divides (f(n) + 2) and (g(n) + 2). But this is clearly false, since 2 does not divide n + 2 for every n. Clearly, it's not the distributive law that's relevant, because the distributive law applies equally in my case. So what is it about your case that makes it different than mine? What additional facts (and additional arguments) justify your conclusion but not mine? -- Come on people!!! The US just blew up a lot of people in Iraq, don't you realize that a person with my exposure might just end up dead, by mysterious circumstances? --James Harris, on the dangers of proving Fermat's last theorem === Subject: Re: JSH: Point of logic >P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). >I factor the polynomial as >P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) >when the a's are the three roots of >a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). >I determine the constant terms of the three factors by setting x=0, >which reveals that for two of them the constant term is 7, while for >one it is 22. >>It's kind of strange to call f(0) the constant term of f when f >>isn't a polynomial. It's called the constant term when f is a >>polynomial, because f is a sum of finitely many monomial terms, >>and the constant term is the one of degree 0. Your non-polynomial >>factors of P(x) have no such expressions. Furthermore, the word >>constant is really confusing you here. I suggest you stop using >>it. > The terms are constants. And they are 7, 7 and 22, factors of the > constant term of the polynomial P(x), which is 1078. > So, literally, the terms are constants, so they are constant terms, in > that they do not vary. > Understand? Usually people start by defining what a term *is*, such as a section of an expression that is isolated from all other sections by addition (or subtraction). Based on that definition, 22 is not a term at all, simply the value of final factor to explicitly have 22 as a constant term was clearer about this. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Point of logic !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). >> >> I factor the polynomial as >> >> P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) >> >> when the a's are the three roots of >> >> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). >> >> I determine the constant terms of the three factors by setting x=0, >> which reveals that for two of them the constant term is 7, while for >> one it is 22. >> It's kind of strange to call f(0) the constant term of f when f >> isn't a polynomial. It's called the constant term when f is a >> polynomial, because f is a sum of finitely many monomial terms, >> and the constant term is the one of degree 0. Your non-polynomial >> factors of P(x) have no such expressions. Furthermore, the word >> constant is really confusing you here. I suggest you stop using >> it. > The terms are constants. And they are 7, 7 and 22, factors of the > constant term of the polynomial P(x), which is 1078. > So, literally, the terms are constants, so they are constant terms, in > that they do not vary. In that they don't vary exactly when _what_ changes? In short: constant with respect to what? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Point of logic > Notice that if you let x=0, you have that the factors give the > appropriate constants, as > 5 a_1(0)/7 + 1 = 1 > 5 a_2(0)/7 + 1 = 1 > 5 a_3(0) + 7 = 22. > The value '22' was arrived at by evaluating a_3(x) at x = 0. Therefore it is a >function of 'x' and Nope. 22 is NOT a function of x. It's the number 22. Here's how it works. You have the polynomial P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) and I factor it into three factors that here I'll just call g_1(x), g_2(x), and g_3(x), so you have g_1(x) g_2(x) g_3(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) and I assert that the factors of P(x) have factors of the constant term of P(x), and that constant term is 49(22). Well, I can *test* that assertion by setting x=0, and I find that g_1(x) contributes 7, g_2(x) contributes 7, and g_3(x) contributes 22. Technically, g_1(0) g_2(0) g_3(0) = 49(22) so the g's MUST have factors of the constant term. You see, 22 is not a function of x. It is a constant, and it is a factor of the constant term of P(x). > will change when 'x' changes. On the other hand the values 5 and 7 above are constants and will > *not* change with 'x'. Please clarify what you mean by 'constant'. If you mean a value that is > fixed and does not change with 'x', the value 22 is disqualified. > The algebra is basic. > Please apply it here. Locked inside the factors g_1(x), g_2(x), and g_3(x) are factors of the constant term. You see, the factors of the polynomial multiply together to give that polynomial, so they must have pieces of the polynomial to multiply together. The polynomial P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22) has one piece that is constant, as it is 49(22), and factors of the polynomial have to produce that piece. If it is not produced by the polynomial factors, then where does it come from? So they have constants within them, and you can find what those are by clearing out x, by setting it to 0. It's easy. Basic math. James Harris === Subject: Re: JSH: Point of logic > Notice that if you let x=0, you have that the factors give the > appropriate constants, as > 5 a_1(0)/7 + 1 = 1 > 5 a_2(0)/7 + 1 = 1 > 5 a_3(0) + 7 = 22. > The value '22' was arrived at by evaluating a_3(x) at x = 0. Therefore it is a >function of 'x' and > Nope. 22 is NOT a function of x. It's the number 22. Now your dissembling. You presented a factorization which contained the term(5*a_3(x) + 7) and demonstrated that (5*a_3(0) + 7) = (22). You also then claimed that the value (22) was *not* a function of 'x'. Of course it is a number, but let me be more precise: the number 22 appears between those parentheses when 'x = 0'. A different number will appear in that position when 'x' has a different value. That's what is meant by a number or value being a function of a variable. Stop kidding around. Everyone knows what's going on in this discussion. You have made a serious and unrecoverable blunder. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Point of logic it could already in a book of world records, or known as James' Last Recovered Fumble. > That's what is meant by > a number or value being a function of a variable. --Advice 0.05; free, if wrong! http://tarpley.net/bush22.htm === Subject: Re: JSH: Point of logic > Locked inside the factors g_1(x), g_2(x), and g_3(x) are factors of > the constant term. Locked inside the factors? Locked? What the hell does locked mean? Talk mathemtaics, idiot! > So they have constants within them, and you can find what those are by > clearing out x, by setting it to 0. But that only tells you about the situation when x=0. You cannot generalize what you find in that case with the general case where x=/=0. Can you see that? > It's easy. Basic math. It is not easy. It is meaningless. Do you wonder why so many people tell you are full of crap, if what you say is both easy and basic? Freaking idiot! === Subject: Re: JSH: Point of logic > Nope. 22 is NOT a function of x. It's the number 22. But it is 22 when x=0. It is other values when x =/= 0. So rather than being a constant, 22 is just a specific case of a variable that can take on values other than 22 as x varies. The problem is that you go on to genearlize from the case where the variable is 22 and x = 0 to other cases where the variable is not 22 for x =/= 0. That is your blunder... you make a generalization that is not valid for all values of x. === Subject: Re: how to measure entropy of music? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3F1ew04227; >I have vaguely heard about this method... and I am very interested in it... >could anybody give me some pointers? >After learning how to measure entropy of music... I can begin to measure >entropy of texts, etc.. that's going to be fun! Look at the work on Leonard Meyer for the musicologists point of view to entropy. Also read John PIerce's great Symbols, Signals and Noise, which has a short bit on musical entropy. In the audio domain, you can do various reductional analyses (such as the mentioned Eigenradio) or a simple gzip-type analysis. You can also compute 'classical' entropy by taking the log probabilities of the time domain samples, but this is a very noisy measure and won't give you the more informational 'structural entropy.' === Subject: Re: how to measure entropy of music? Has anybody applied Pincus's Approximate entropy to music? It seems to be what cardiac doctors are using for heart arrhythmia/heart attack symptoms. >>I have vaguely heard about this method... and I am very interested in it... >>could anybody give me some pointers? >>After learning how to measure entropy of music... I can begin to measure >>entropy of texts, etc.. that's going to be fun! >> >Look at the work on Leonard Meyer for the musicologists point of view to entropy. Also read John PIerce's great Symbols, Signals and Noise, which has a short bit on musical entropy. In the audio domain, you can do various reductional analyses (such as the mentioned Eigenradio) or a simple gzip-type analysis. You can also compute 'classical' entropy by taking the log probabilities of the time domain samples, but this is a very noisy measure and won't give you the more informational 'structural entropy.' -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn@netscape.net URL : http://home.earthlink.net/~tftn === Subject: Re: Lyapunov Function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3F1eG04235; >x' = y - 2*x*(y^2) >y' = -(x^3) + (x^4)*y >I have tried V(x,y) of the form: >a(x^2) + b(y^2) >a(x^2) + b(y^2) + c(x^4) >a(x^2) + b(y^2) + c(y^4) >a(x^2) + c*x*y + b(y^2) -1with no luck. I even tried reverse engineering it, but still no luck. A hint or two would be better than an explicit answer, as I'm trying to learn here ;-) but I'm glad of any help you can give me. >Mike >Try V(x,y) = x^(2m) + a*y^(2n) with m,n integers >=1 and a>0. >>To be a little more specific: Use this Ansatz and compute >>d/dt (V(x(t),y(t)). You will find that there is only one choice for m >>and n such that the bad terms with odd exponents balance. Next you >>can find an a which make the bad terms vanish *identically* so that >>you are only left with even powers. >>Fortunately this a is positive (otherwise we wouldn't have a >>L.-function and d/dt (V(x(t),y(t)) <= 0 for all t ! >>That might give you an idea. BTW: You have the solution in your list >>(you don't need that parameters though) - somehow you skipped over it >>or miscalculated. >>Thomas >Thomas >Ok got it now, V(x,y)= x^4 + 2*y^2 does it. This gives >V'(x,y) = -(x^2)*(y^2)*(8 - 4*(x^2)) which is <= 0 for all >(x,y) such that x <= sqrt(2) whoops miscalculation! V'(x,y)=-4(x^2)*(y^4) which means V'(x,y) is negative definite with respect to (0,0) which means v(x,y) is a STRICT Lyapunov function, which is confusing as the question implies you will only find a non-strict one. Does my result agree with what you got? === Subject: Re: Lyapunov Function >whoops miscalculation! V'(x,y)=-4(x^2)*(y^4) which means V'(x,y) is negative definite with respect to (0,0) which means v(x,y) is a STRICT Lyapunov function, which is confusing as the question implies you will only find a non-strict one. Does my result agree with what you got? You got V' right now. But it is not strict (at least not in the definition of Hirsch&Smale) since there is no strict inequality. V'(x,y) = 0 for x=0 *or* y=0. Not just for (x,y) =(0,0). Thomas === Subject: ladies blah blah blah ba ba ba mm m ohmm ohm ohh o o 0 i m supreme === Subject: Re: ladies reply-type=response > blah boring === Subject: Re: ladies > blah blah blah ba ba ba mm m ohmm ohm ohh o o 0 i m supreme The adult answer is yada yada. Save the drama for yer mama. Get down and push, maggot! -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: ladies shut up you twirt >> blah blah blah ba ba ba mm m ohmm ohm ohh o o 0 i m supreme > The adult answer is yada yada. Save the drama for yer mama. Get > down and push, maggot! > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: ladies In sci.math, Lord of Chaos(Suresh Devanathan) <31bp26F39llvmU1@individual.net>: > shut up you twirt He's much more coherent than you'll ever be. Answered my questions yet? :-) > blah blah blah ba ba ba mm m ohmm ohm ohh o o 0 i m supreme >> The adult answer is yada yada. Save the drama for yer mama. Get >> down and push, maggot! >> -- >> Uncle Al >> http://www.mazepath.com/uncleal/ >> (Toxic URL! Unsafe for children and most mammals) >> http://www.mazepath.com/uncleal/qz.pdf -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: two lines - finding point In R2 I have two points. Each point is the start point of a straight line with a specific length. At a certain point, the endings of these two lines meet each other. How can I compute this point? S. Nurbe === Subject: Re: two lines - finding point Mail-To-News-Contact: abuse@dizum.com >In R2 I have two points. Each point is the start point of a straight >line with a specific length. At a certain point, the endings of these >two lines meet each other. How can I compute this point? Let the two points have coordinates (x1,y1) and (x2,y2). Let the line segments starting from each have lengths l1 and l2. Use l1 as the radius of a circle centered on (x1,y1): (x-x1)^2 + (y-y1)^2 = l1^2 Similarly, use l2 as the radius of a circle centered on (x2,y2). Then, solve the system of two simultaneous equations in two unknowns. Unfortunately, they're not *linear* equations, so you might need to use numeric methods rather than being able to come up with a closed-form solution. Keep in mind that there could be zero, one, or two intersection points, depending upon whether (l1+l2)^2 is less than, equal to, or greater than (x1-x2)^2 + (y1-y2)^2. -- Michael F. Stemper #include A bad day sailing is better than a good day at the office. === Subject: Re: two lines - finding point Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >In R2 I have two points. Each point is the start point of a straight >line with a specific length. At a certain point, the endings of these >two lines meet each other. How can I compute this point? The meeting point is on two circles of know radius centred on the given points. Write down the the equations for the two circles and solve simultaneously. You will have to solve a quadratic to do this. -- Richard === Subject: 2-Adic Valuation in a Peculiar Continued Fraction Consider the constant: x=1.353871128429882374388894084016608124227333416812118556923672649787001842 ... which is defined by the continued fraction: [1;2,1,4,1,2,1,8,1,2,1,4,1,2,1,16,...,2^gde(n,2),...] where gde(n,2) = greatest dividing exponent of n base 2. This constant has the property that the continued fraction expansion of 2*x results in the continued fraction expansion of x interleaved with 2's: x = [1;2,1,4,1,2,1,8,1,2,1,4,1,2,1,16,...2^gde(n,2),... ] while 2*x = [2;1, 2,2, 2,1, 2,4, 2,1, 2,2, 2,1, 2,8,... 2,2^gde(n,2),...]. A surprising property of x is that the continued fraction of x^2 has large partial quotients that appear to be doubly exponential. The continued fraction of x^2 begins: [1,1,4,1,74,1,8457,1,186282390,1,1,1,2,1,430917181166219,11, 37,1,4,2,41151315877490090952542206046,11,5,3,12,2,34,2,9,8, 1,1,2,7,13991468824374967392702752173757116934238293984253807017,...] where x^2=1.8329670323960030544272195442104173240577165632272168977983897785571879 9.. The next large partial quotient has 106 digits. Does anyone see a pattern to the large PQs? The positions of these large PQs are: 1,3,5,7,9,15,21,35,71,143,291,635,1407,2979,6101,12339, 25019,50413,101339,202793,405745,811365,1624043,3249293, 6502711,13011309, ... (these terms were independently computed by Robert G. Wilson v and Hans Havermann). Hans Havermann has observed that the ratio of number-of-digits to position number is about 1.03, and seems to approach the reciprocal of Loch's constant: http://mathworld.wolfram.com/LochsTheorem.html FURTHER RESEARCH. In general, the following continued fraction has a similar property. Define y by: y = [1;m,1,m^2,1,m,1,m^3,1,m,1,m^2,1,m,1,m^4,...,m^gde(n,2),... ] then the CF of m*y equals the CF of y interleaved with m's: m*y = [m;1, m,m, m,1, m,m^2, m,1,..., m,m^gde(n,2),... ]. This raises the question: what is the function F(x) defined by: F(x) = [1;x,1,x^2,1,x,1,x^3,1,x,1,x^2,1,x,1,x^4,...,x^gde(n,2),... ] Perhaps this F(x), if known, would make it clear why F(2)^2 has such large partial quotients. With a slight change in variables, F(1-x) is a function with non-rational coefficients such that F(1-x)^2 has coefficients that oscillate near the value 2. If you wish, you may see these sequences at: http://www.research.att.com/projects/OEIS?Anum=A100338 http://www.research.att.com/projects/OEIS?Anum=A100863 http://www.research.att.com/projects/OEIS?Anum=A100864 http://www.research.att.com/projects/OEIS?Anum=A100865 http://www.research.att.com/projects/OEIS?Anum=A100866 Wanted to share this curiosity, and would welcome any comments. Paul === Subject: Re: 2-Adic Valuation in a Peculiar Continued Fraction > Consider the constant: > x=1.353871128429882374388894084016608124227333416812118556923672649787001842 . . > which is defined by the continued fraction: > [1;2,1,4,1,2,1,8,1,2,1,4,1,2,1,16,...,2^gde(n,2),...] > where gde(n,2) = greatest dividing exponent of n base 2. > This constant has the property that the continued fraction expansion of 2*x > results in the continued fraction expansion of x interleaved with 2's: > x = [1;2,1,4,1,2,1,8,1,2,1,4,1,2,1,16,...2^gde(n,2),... ] > while > 2*x = [2;1, 2,2, 2,1, 2,4, 2,1, 2,2, 2,1, 2,8,... 2,2^gde(n,2),...]. Alf van der Poorten reminds me that X[AY, BX, CY, DX, EY, FX, ...]=Y[AX, BY, CX, DY, EX, FY, ...] and in particular X[AY, BX, CY, DX, EY, FX, ...]=Y[AX, BY, CX, DY, EX, FY, ...] -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Surprising Pattern of Florida's Election Results <90b9p0dtd91jl0t2nht8gt6niom4tehuaj@4ax.com> <41ab9d84_4@news1.prserv.net> <41ac8f94_1@news1.prserv.net> <41ac9eef_3@news1.prserv.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Fingerprints? What are you talking about? All they need to do >> is look at the votes on the paper and then look at the face of >> the person who just handed them the paper. >> Where do you vote? Every place I've ever seen that uses paper >> ballots (including Russia) has a box with a slot where you insert >> your ballot. > Sure, but what happens to the box? It could easily be figured out, > if someone really wanted to. This application of the word easily in a new one to me. The people supervising the vote are community members from different parties, and the process is so transparent that tampering with it would be obvious, and have consequences just in a single voting district. Do you think you could design a ballot box that would tamper with votes, maybe by swallowing them and replacing them with prebuilt votes, getting the counts exactly right? Those boxes are checked by the voting officials from several parties. If some manufacturer would be caught with such stuff, he'd go to jail. In contrast, the people building vote machines are from a single source with vested interests, and there is no way that officials can check the results. If votes go missing, there is no way to take apart the machine and look for hidden vaults. There are no records, so you can't go to jail. In particular, if people swallow technical error as an explanation for stuff like a voting district having a negative count of Democratic votes. With wood boxes, any effective manipulation would convince any judge of premeditated malice. With computers, everybody accepts complete junk as plausible. Even when registered voting software gets replaced unofficially before an election (if that is not premeditated malice, I don't know what is), the responsible persons for that do not even get a slap on the wrist. It's just unbelievable. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Surprising Pattern of Florida's Election Results <90b9p0dtd91jl0t2nht8gt6niom4tehuaj@4ax.com> <41ab9d84_4@news1.prserv.net> <8cCdncH31tZwCTHcRVn-3w@giganews.com> <41ac9fd3_4@news1.prserv.net> <41acee3b_2@news1.prserv.net> <41ae0031_1@news1.prserv.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> If you're talking about ballot papers that are small relative to >> the size of the box, then I agree. Where I vote, the ballots are >> very large (11x14) and stack right up in the machine. >> That sounds completely lunatic. While there are some ballots with >> an even quite larger size in some regions here, you have to fold >> them small enough to make them fit the slot in the ballot box. >> They don't stack neatly from there. > Ours go into an optical reader - they don't get dropped in a slot > like back in the 1800s! In a lot of ways, the American democracy is not what is was back in the 1800s. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Surprising Pattern of Florida's Election Results <90b9p0dtd91jl0t2nht8gt6niom4tehuaj@4ax.com> <41ab9d84_4@news1.prserv.net> <8cCdncH31tZwCTHcRVn-3w@giganews.com> <41ac9fd3_4@news1.prserv.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> How are you going to link a paper ballot to an individual voter -- >> finger prints? >> I would rather trust my anonymity to a box full of paper ballots >> than to some encrypted ID. > I agree, sort of. But it would be relatively simple to figure out > who voted with the paper ballots. In fact, that's why I'd prefer > it. With the encryption, it's more sophisticated, so it's harder to > know who's spoofing who. But anyone with access to the ballot box > has access to who voted for who. You go there and check in with > your vote card or id, so they know who you are. Someone makes a > little note about what you look like. Or in sparsely populated or > unbusy precincts, they merely pay attention. Then when you put in > your ballot, they simply keep a list of the order that people put > their ballots in. At the end of the day, simply take out the > ballots and voila - a list in order with your list of names. Very > very easy. You have never seen a ballot box, right? No, the ballots are not neatly stacked. You have also never participated as a voting official, right? The idea that any single person has a chance to get a cleanly sorted stack of ballots out while the other officials are in the same room is a bit ridiculous. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Surprising Pattern of Florida's Election Results :> You have never seen a ballot box, right? No, the ballots are not :> neatly stacked. You have also never participated as a voting :> official, right? The idea that any single person has a chance to get :> a cleanly sorted stack of ballots out while the other officials are in :> the same room is a bit ridiculous. : 1) It's not clear whether or not I've seen a ballot box, but one thing is for : sure - you've never seen the ones we use around here And it seems on one else here has seen the sort of ballot box you are talking about. : 2) I didn't say it would take just one person You strongly implied it. Your first response in this thread was as follows: >> Doesn't that open the door to someone (either a hacker or insider) being >> able to determine how every person voted? > Well gosh, isn't that true with regular paper ballots anyway? It's true - > the only way to have fair elections is to have some sort of > verifiable paper trail (if not paper, then some other medium, but > definitely not bit bucket in computer terminology). The complaint about electronic voting was that *someone* could determine how every person voted. You responded that this was true of paper ballots. Now you are claiming that you did not mean 'someone' but a group of people with presumably unlimited and unsupervised access to the ballots. : 3) haven't you ever heard of people sneaking into files on their own with no one : looking? What files are you talking about? There are no files that contain the results of your vote when paper ballots are used. Stephen === Subject: Re: Surprising Pattern of Florida's Election Results > The complaint about electronic voting was that *someone* could > determine how every person voted. You responded that this was > true of paper ballots. Now you are claiming that you did not mean > 'someone' but a group of people with presumably unlimited and > unsupervised access to the ballots. What I said was that it didn't rule out getting someone else to turn a blind eye. And saying presumably unlimited is ridiculous. > : 3) haven't you ever heard of people sneaking into files on their own with no one > : looking? > What files are you talking about? There are no files that contain > the results of your vote when paper ballots are used. That's not the point. The point - which is not difficult to follow - is that people have many many many many times throughout history gotten access to things they weren't supposed to. === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3FX3208001; >> Zentralblatt f.9fr Mathematik< 9 publications >> of E. Escultura are listed. For 5 of them the >> Zentralblatt just mentions >not reviewed<, whatever >> that means. >> This is an indexing service like so many others. >No. It commissions reviews: not reviewed means not reviewed. >There are two reasons why a paper might not be reviewed: >(i) the editord did not commision a review since they considered >it without merit or otherwise unsuitable, >(ii) a review was commissioned but never written. >There is another reviewing service: Mathematical Reviews. >It lists ten items by Escultura. None of them have been reviewed. >-- >Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 >Francis Wheen, _How Mumbo-Jumbo Conquered the World_ But how can someone review something he does not know about because the ideas are new? Moreover, experts fear controversy. E. E. Escultura === Subject: Re: Escultura affair: publication scandal > Zentralblatt f?hematik< 9 publications > of E. Escultura are listed. For 5 of them the > Zentralblatt just mentions >not reviewed<, whatever > that means. > > This is an indexing service like so many others. >>No. It commissions reviews: not reviewed means not reviewed. >>There are two reasons why a paper might not be reviewed: >>(i) the editord did not commision a review since they considered >>it without merit or otherwise unsuitable, >>(ii) a review was commissioned but never written. >>There is another reviewing service: Mathematical Reviews. >>It lists ten items by Escultura. None of them have been reviewed. >>-- >>Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >>Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 >>Francis Wheen, _How Mumbo-Jumbo Conquered the World_ > But how can someone review something he does not know about > because the ideas are new? Moreover, experts fear controversy. ? The pleasure of reviewing lies in finding new ideas. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: (5+sqrt(26))^n gives n repeting decimals? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3FZCY08160; >Can someone help me solve this problem? I need theory that proves this is >correct... Each term c(n)=(5+sqrt(26))^n is fairly close to some integer b(n). Work out the first several values of b(n). They obey a recurrence relation something like that of the Fibonacci numbers. Use that recurrence relation to find an exact formula for b(n). Find a formula for b(n)-c(n). So how close is c(n) to an integer? === Subject: Re: (5+sqrt(26))^n gives n repeting decimals? >Can someone help me solve this problem? I need theory that proves this is >correct... > Each term c(n)=(5+sqrt(26))^n is fairly close to some integer b(n). > Work out the first several values of b(n). > They obey a recurrence relation something like that of > the Fibonacci numbers. > Use that recurrence relation to find an exact formula for b(n). > Find a formula for b(n)-c(n). > So how close is c(n) to an integer? The difference between (5+sqrt(26))^n and an integer is (5-sqrt(26))^n (see the minus sign?), so you should be able to tell roughly how fast this goes to zero. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3FbjS08490; >> Zentralblatt f.9fr Mathematik< 9 publications >> of E. Escultura are listed. For 5 of them the >> Zentralblatt just mentions >not reviewed<, whatever >> that means. >> >> This is an indexing service like so many others. >> No. It commissions reviews: >It was formerly, but haven't they changed lately? They publish >authors' summaries? Or did I just dream it? >-- >G. A. Edgar http://www.math.ohio-state.ed u/~edgar/ It is a challenge to review my work because the ideas are new. I know that experts fear controversy. E. E. Escultura === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3Fgqh09017; > > Zentralblatt fí.b9r Mathematik< 9 publications > of E. Escultura are listed. For 5 of them the > Zentralblatt just mentions >not reviewed<, whatever > that means. > > This is an indexing service like so many others. > > No. It commissions reviews: >> It was formerly, but haven't they changed lately? They publish >> authors' summaries? Or did I just dream it? >Quote It is not the fault of Zentralblatt that my work is not reviewed. It simply reflects the inadequacy of the experts. E. E. Escultura >Zentralblatt MATH is the world's most complete and longest running >abstracting and reviewing service in pure and applied mathematics. >They still publish reviews. >-- >Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 >Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3GBXA12128; >You've *proven* that you're not using the same language that other >professional mathematicians use. Anything else is simply a matter of >[probably correct] deduction in your axiom set which simply doesn't agree >with everyone else's. > Norm This is entirely new math and therefore new language. E. E. Escultura === Subject: Re: The Size of Graham's Number by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3GBXv12132; Coynil http://mathforum.org/discuss/sci.math/m/658618/658618 > Trying to explain the sheer hugeness of Graham's Number > in a popular way to someone, I kind of came up short. > I have read on the net somewhere that even if all matter > in the universe were converted to pen and paper, it > wouldn't be enough to write the number down. But that's > a pretty difficult concept to grasp as well. Is there > an easier way to a) estimate the number of digits in > Graham's Number and b) express this as something slighty > less unfathomable (i.e. it wouldrequier a hard drive > 40 times the size of our galaxy to store it or something) Graham's number is astronomically astronomically astronomically ... astronomically (astronomically many times) beyond this, and more. To store it, you'd need a hard drive that's a cube with edge lengths of a Graham's number of miles (or light-years, or observable universe diameters, etc.). In the same way that one plus a billion is essentially the same as a billion, or ten times the number one followed by one billion zeros is essentially the same as one followed by one billion zeros, or ten to the power of 'N', where 'N' is ten to the power of ten to the power of ten ... to the power of ten (one billion times) is essentially the same as ten to the power of N, Graham's number has little descriptive change if you multiply it by 100000, square it, raise it to its own power, raise it to its own power over and over again a Graham's number of times, and much more (e.g. Graham's number tetrated by Graham's number over and over again, a Graham's number of times). See the following posts: http://mathforum.org/discuss/sci.math/m/394644/394645 BIG NUMBERS #1 [April 8, 2002] http://mathforum.org/discuss/sci.math/m/403134/403134 BIG NUMBERS #2 [April 8, 2002] http://mathforum.org/discuss/sci.math/m/403126/403126 BIG NUMBERS #3 [April 8, 2002] http://mathforum.org/discuss/sci.math/m/403125/403125 Dave L. Renfro === Subject: Re: The Size of Graham's Number x*(y+1) = x + (x*y) multiplication x^(y+1) = x * (x^y) exponentiation x^^1 = x x^^(y+1) = x ^ (x^^y) tetration x^^^1 = x x^^^(y+1) = x ^^ (x^^^y) hyper-tetration 3^^3 = 3^3^3 = 3^27 = 7,625,597,484,987 3^^4 = 3^(3^^3) = 3^(3^27) = 3^(7,625,597,484,987) [Has about 8.4 x 10^12 digits.] Graham's number is between 2^^^66 and 3^^^131. Herc === Subject: Re: The Size of Graham's Number > x*(y+1) = x + (x*y) multiplication > x^(y+1) = x * (x^y) exponentiation > x^^1 = x > x^^(y+1) = x ^ (x^^y) tetration > x^^^1 = x > x^^^(y+1) = x ^^ (x^^^y) hyper-tetration > 3^^3 = 3^3^3 = 3^27 = 7,625,597,484,987 > 3^^4 = 3^(3^^3) = 3^(3^27) = 3^(7,625,597,484,987) > [Has about 8.4 x 10^12 digits.] > Graham's number is between 2^^^66 and 3^^^131. No, Graham's number is the 64th number in a rapidly growing sequence whose first element already far exceeds 3^^^131. In the hierarchy you've indicated above, let x{k} denote the operator with k ^'s , e.g. x^^^y = x{3}y. Then Graham's number is g_64 in the sequence defined by g_1 = 3{4}3 g_n = 3{g_(n-1)}3 (n>1). Thus, g_1 = 3{4}3 g_2 = 3{3{4}3}3 g_3 = 3{3{3{4}3}3}3 ... g_64 = 3{3{...3{4}3...}3}3 (64 pairs of {}) = Graham's number Note that the g-sequence *starts* with the number g_1 = 3{4}3 = 3^^^^3 = 3^^^(3^^^3) >> 3^^^131. --r.e.s. === Subject: Re: The Size of Graham's Number Completely different definition in Dave Renfro's reference. http://mathforum.org/discuss/sci.math/m/394644/394645 Here is how Smorynski defines Graham's number (p. 149) ---> Let N be the natural numbers. First, we define K: N^2 --> N. K(0,n) = n^n K(m+1, n) = K(m, K(m,n)) Next, we define G: N --> N using K. G(0) = K(3,3) G(n+1) = K(G(n), 3) Smorynski then defines Graham's number to be G(64). Will the real Graham's number please stand up! Herc === Subject: Re: The Size of Graham's Number > Completely different definition in Dave Renfro's reference. > http://mathforum.org/discuss/sci.math/m/394644/394645 > Here is how Smorynski defines Graham's number (p. 149) ---> Let N be the natural numbers. First, we define K: N^2 --> N. > K(0,n) = n^n > K(m+1, n) = K(m, K(m,n)) > Next, we define G: N --> N using K. > G(0) = K(3,3) > G(n+1) = K(G(n), 3) > Smorynski then defines Graham's number to be G(64). > Will the real Graham's number please stand up! Renfro's page discusses both definitions, and apparently none of his other references use Smorynski's version. (Regrettably, I didn't notice, in his section How Big is Graham's Number?, essentially the content of my posting.) It would be nice to know for sure which definition is given in the original paper (to which, unfortunately, I have no ready access): Graham, R. L. and Rothschild, B. L. Ramsey's Theorem for n-Parameter Sets. Trans. Amer. Math. Soc. 159, 257-292, 1971. --r.e.s. === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3GGKw12516; >>Zentralblatt f.9fr Mathematik< 9 publications >>of E. Escultura are listed. For 5 of them the >>Zentralblatt just mentions >not reviewed<, whatever >>that means. >>This is an indexing service like so many others. So it simply notes some contributions unless someone wants to try uncharted course. >I don't agree. The Zentralblatt and the Math Reviews are the >two major indexing services on this planet. Indeed together >they cover all countries. >>>wonder why persons like E. Escultura or J. Harris >>provoke such strong reactions. >>The most likely reason is fear of new ideas. >>E. E. Escultura >>University of the Philippines >This is one possibility. Why is it the most >likely one? Another is feeling of inadequacy. Which one is most likely, I don't know. === Subject: Elementary proof of R(3,7)=23? Is there an elementary proof of Ramsey Number R(3,7)=23? I founded the one with R(3,6)=18 and hope to construct a similar one, but probably need to get the complete classification of all (3,6,17)-good graphs... passed away... === Subject: Re: Elementary proof of R(3,7)=23? > Is there an elementary proof of Ramsey Number R(3,7)=23? I founded the one > with R(3,6)=18 and hope to construct a similar one, but probably need to get > the complete classification of all (3,6,17)-good graphs... I don't know what you call elementary, but perhaps you should look here: S. P. Radziszowski, D. L. Kreher, On R(3,k) Ramsey graphs: theoretical and computational results, Journal of Comb. Math. and Comb. Comp., 4 (1988), 37-52. Also a good reference for the bibliography is: S. Radziszewski, Small Ramsey numbers, Electron. J. Comb., 1 (1994), 1-30. Pawel Gladki === Subject: Re: Elementary proof of R(3,7)=23? > Is there an elementary proof of Ramsey Number R(3,7)=23? I founded the one > with R(3,6)=18 and hope to construct a similar one, but probably need to get > the complete classification of all (3,6,17)-good graphs... > passed away... This news deserves to be brought to the level of a named thread. Dale === Subject: Gamma Function/Mills ratio/Inequalities Let G be the Gamma function, and G(x+0.5) f(x)= -------- , x >= 0 . G(x+ 1) Question: to prove or disprove that for each pair (x,y), 0 =< x < y , there exists u(x,y) in ( 0, 1/2 ), such that f(y) ----- = sqrt{(x+u(x,y))/(y+u(x,y))} . f(x) Note:According to G.N.Watson [1] it is knownn hat for every x>= 0 there exists v(x) in ( 1/4, 1/pi ), such that 1 f(x) = ------------- . sqrt(x+v(x)) Reference: [1] G.N. Watson , ,,A note on gamma functions, Proc.Edinburgh Math.Soc., (2) 11 (1958/59), Edinburgh Math.Notes 42, (1959) 7-9. === Subject: Re: The Size of Graham's Number by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3GokK16022; === >>Subject: The Size of Graham's Number >>Message-id: >Trying to explain the sheer hugeness of Graham's Number in a popular >>way to someone, I kind of came up short. I have read on the net >>somewhere that even if all matter in the universe were converted to >>pen and paper, it wouldn't be enough to write the number down. But >>that's a pretty difficult concept to grasp as well. Is there an easier >>way to a) estimate the number of digits in Graham's Number and b) >>express this as something slighty less unfathomable (i.e. it >>wouldrequier a hard drive 40 times the size of our galaxy to store it >>or something) >Not likely. Even a tiny number like 10**600 is unfathonable. >See How big is a 2000-bit number? >http://members .aol.com/mensanator666/fun/2000_bit.htm-- >Mensanator >Ace of Clubs On the other hand, there are 1 googol (10^100) different ways to arrange 16 balls on a pool table, if the position of each is specified to within 1 mm. To me this is a pretty real world way to grasp a googol. Looked at in this way, a 2000 bit number (=10^602) can be grasped as far less than the numer of ways to arrange the total number of phil === Subject: fitting surface I am looking for routines (free and possibly in fortran) to fit functions except spline fitting. The data is smooth, can provide a sample plot if this would help. Would much appreciate response. Marl === Subject: Re: fitting surface >I am looking for routines (free and possibly in fortran) to fit functions >except spline fitting. The data is smooth, can provide a sample plot >if this would help. Would much appreciate response. >Marl Not really my field, but have you looked at Shepard's quadratic method and variations? Google shepard's quadratic fortran gives lots of hits, for example: http://ngwww.ucar.edu/ngdoc/ng/ngmath/shgrid/intro.html might get you started. --Lynn === Subject: Re: need help in understanding Torkel's ZFC comment > Summary of the discussion so far: > Frege> http://au.metamath.org/mpegif/2p2e4.html > Charlie-Boo> [6-line informal proof of 0'' + 0'' = 0''''] > Charlie-Boo> I beat him by 19,725 steps. > Ullrich> That's not a formal proof from the axioms of ZFC. > Neither is the MetaMath proof of 2+2=4. > Yes it is. If you believe otherwise then point out an axiom that is used > in the proof but is not part of ZFC. Very few of the terminal nodes of the proof tree are ZFC axioms, because ZFC has nothing to do with the fact that 2+2=4. (Only 1 set is involved, so we need not consider the question of what sets exist in general. 2+2=4 regardless of what ZFC says.) At these nodes you will find other axioms, rules that allow arbitrary wffs to be considered proven (unsound) and unverified definitions (also unsound) which must be set up to verify new theorems (a strict no-no in axiomatic systems.) There are also many nodes that have no links or justification, such as ECOPRASS at http://us.metamath.org/mpegif/ecoprass.html . What is the justification for lines 23 and 24 (and the other lines labeled ecoprass)? If you want to argue that certain rules don't count, that is not productive .9a the fact remains that each branch of Mathematics has its own axioms. The assertion that everything comes from ZFC is a lie and I would love to hear all sorts of professors claim that it is so. MetaMath derives almost nothing from any axioms. It uses a number of subterfuges to allow you to declare arbitrary expressions (even meaningless sequences of symbols) to be theorems. An axiomatic system is one that derives everything from a fixed set of axioms and rules (and perhaps definitions.) After it is set up, (1) you don't have to add anything to the system to develop a theorem, and (2) only actual theorems can be constructed. (Agree?) Otherwise you have no more than a word processor with a Mathematical font. C-B > Do you have any idea how the theorem was produced? How the proof was > produced? To what extent and how it was verified? > Do you? It's only been explained to you two or three times by now. > For the questions you ask, the web site really does > explain it in detail. Ok, then what are the possible justifications for the terminal nodes in a proof tree? === Subject: Re: need help in understanding Torkel's ZFC comment >>Summary of the discussion so far: >>Frege> http://au.metamath.org/mpegif/2p2e4.html >>Charlie-Boo> [6-line informal proof of 0'' + 0'' = 0''''] >>Charlie-Boo> I beat him by 19,725 steps. >>Ullrich> That's not a formal proof from the axioms of ZFC. >Neither is the MetaMath proof of 2+2=4. >>Yes it is. If you believe otherwise then point out an axiom that is used >>in the proof but is not part of ZFC. > Very few of the terminal nodes of the proof tree are ZFC axioms, > because ZFC has nothing to do with the fact that 2+2=4. (Only 1 set > is involved, so we need not consider the question of what sets exist > in general. 2+2=4 regardless of what ZFC says.) > At these nodes you will find other axioms, rules that allow > arbitrary wffs to be considered proven (unsound) and unverified > definitions (also unsound) which must be set up to verify new theorems > (a strict no-no in axiomatic systems.) There are also many nodes that > have no links or justification, such as ECOPRASS at > http://us.metamath.org/mpegif/ecoprass.html . What is the > justification for lines 23 and 24 (and the other lines labeled > ecoprass)? Looks like hypotheses of the theorem to me. > If you want to argue that certain rules don't count, that is not > productive [CapitalEth] the fact remains that each branch of Mathematics has its > own axioms. The assertion that everything comes from ZFC is a lie and > I would love to hear all sorts of professors claim that it is so. > MetaMath derives almost nothing from any axioms. It uses a number of > subterfuges to allow you to declare arbitrary expressions (even > meaningless sequences of symbols) to be theorems. > An axiomatic system is one that derives everything from a fixed set of > axioms and rules (and perhaps definitions.) After it is set up, (1) > you don't have to add anything to the system to develop a theorem, and > (2) only actual theorems can be constructed. (Agree?) > Otherwise you have no more than a word processor with a Mathematical > font. > C-B >Do you have any idea how the theorem was produced? How the proof was >produced? To what extent and how it was verified? >>Do you? It's only been explained to you two or three times by now. >>For the questions you ask, the web site really does >>explain it in detail. > Ok, then what are the possible justifications for the terminal nodes > in a proof tree? -- Replace Roman numerals with digits to reply by email === Subject: Re: need help in understanding Torkel's ZFC comment >>Ullrich> That's not a formal proof from the axioms of ZFC. >>Neither is the MetaMath proof of 2+2=4. >Yes it is. If you believe otherwise then point out an axiom that is used >>in the proof but is not part of ZFC. > > Very few of the terminal nodes of the proof tree are ZFC axioms, > because ZFC has nothing to do with the fact that 2+2=4. (Only 1 set > is involved, so we need not consider the question of what sets exist > in general. 2+2=4 regardless of what ZFC says.) > > At these nodes you will find other axioms, rules that allow > arbitrary wffs to be considered proven (unsound) and unverified > definitions (also unsound) which must be set up to verify new theorems > (a strict no-no in axiomatic systems.) There are also many nodes that > have no links or justification, such as ECOPRASS at > http://us.metamath.org/mpegif/ecoprass.html . What is the > justification for lines 23 and 24 (and the other lines labeled > ecoprass)? > Looks like hypotheses of the theorem to me. What are the hypotheses to 2+2=4? > An axiomatic system is one that derives everything from a fixed set of > axioms and rules (and perhaps definitions.) After it is set up, (1) > you don't have to add anything to the system to develop a theorem, and > (2) only actual theorems can be constructed. (Agree?) > > Otherwise you have no more than a word processor with a Mathematical > font. > > C-B You know, this is kind of like deja vue. All these people yelling and screaming and I'm the one who (as is often the case) puts out the formal definition to resolve technical issues - and they all go running for the hills until time for the next skirmish. What a life! === Subject: Re: need help in understanding Torkel's ZFC comment >>Ullrich> That's not a formal proof from the axioms of ZFC. >Neither is the MetaMath proof of 2+2=4. >Yes it is. If you believe otherwise then point out an axiom that is used >>in the proof but is not part of ZFC. >Very few of the terminal nodes of the proof tree are ZFC axioms, >because ZFC has nothing to do with the fact that 2+2=4. (Only 1 set >is involved, so we need not consider the question of what sets exist >in general. 2+2=4 regardless of what ZFC says.) >At these nodes you will find other axioms, rules that allow >arbitrary wffs to be considered proven (unsound) and unverified >definitions (also unsound) which must be set up to verify new theorems >(a strict no-no in axiomatic systems.) There are also many nodes that >have no links or justification, such as ECOPRASS at >http://us.metamath.org/mpegif/ecoprass.html . What is the >justification for lines 23 and 24 (and the other lines labeled >ecoprass)? >>Looks like hypotheses of the theorem to me. > What are the hypotheses to 2+2=4? What does that have to do with anything? The theorem in question isn't 2 + 2 = 4. Now that you mention it, why don't you show us a complete listing of a formal proof that 2 + 2 = 4 in the Charlie Boo system, so that we can get an idea of how your formal proof compares to a Metamath formal proof. >An axiomatic system is one that derives everything from a fixed set of >axioms and rules (and perhaps definitions.) After it is set up, (1) >you don't have to add anything to the system to develop a theorem, and >(2) only actual theorems can be constructed. (Agree?) >Otherwise you have no more than a word processor with a Mathematical >font. >C-B > You know, this is kind of like deja vue. All these people yelling and > screaming and I'm the one who (as is often the case) puts out the > formal definition to resolve technical issues - and they all go > running for the hills until time for the next skirmish. > What a life! -- Replace Roman numerals with digits to reply by email === Subject: How to measure randomness of a deck of cards? I read several papers, there are few approaches such as variation distances, birthday bound and markov chain. I need a method/algorithm to determine the randomness of my deck of cards (permutation) I would be appreciated if anybody can give me some ideas or good solutions. === Subject: Re: How to measure randomness of a deck of cards? Randomness would better be used to describe a specific shuffling method. If starting at any particular permutation the probability of getting each permutation is equal I think it's safe to say it's random. Partial-randomness could probably be measured with variance, but only for methods rather than particular combinations. Dan Rossul ??? > I read several papers, there are few approaches such as variation > distances, birthday bound and markov chain. I need a > method/algorithm to determine the randomness of my deck of cards > (permutation) > I would be appreciated if anybody can give me some ideas or good solutions. === Subject: Re: How to measure randomness of a deck of cards? > I read several papers, there are few approaches such as variation > distances, birthday bound and markov chain. I need a > method/algorithm to determine the randomness of my deck of cards > (permutation) Randomness is a negative property: It is the absence of pattern. Thus you can never prove that there is no pattern in a deck of cards. You can only prove that there _is_ a pattern and that additional shuffles have to be done. Thus you will have to rely on non-perfect methods like the ones you've mentioned above, submit the deck to each method and get it approved. And then hope that your deck is sufficently random for your purpose. Sad and simple :/ - Daniel Janzon === Subject: Re: How to measure randomness of a deck of cards? > I need a > method/algorithm to determine the randomness of my deck of cards Cannot be done. Any ordering is as random as any other ordering. Randomness is not a property of your deck of cards. Do you mean a random process of selecting a sequence of cards from your deck? === Subject: Re: How to measure randomness of a deck of cards? > I need a > method/algorithm to determine the randomness of my deck of cards > Cannot be done. Any ordering is as random as any other ordering. Randomness > is not a property of your deck of cards. Do you mean a random process of > selecting a sequence of cards from your deck? referred that cards are likely to be disturbuted randomly by 7 times of Perfect Shuffles. I wish to implement an algorithm to test the randomness after different kind of shuffling methods, not only Perfect Shuffle. Appreciated for help. Looking forward for everybody replies. === Subject: Re: How to measure randomness of a deck of cards? > I wish to implement an algorithm to test the > randomness after different kind of shuffling methods... Well, that is a much better question. Rather than asking if a deck is randomly shuffled, it makes much more sense to ask if a particular method (i.e. process) of shuffling is random, which can be measured statistically. Whether any process in physical reality is truly random is an open question in physics, and it may be true that nothing is truly random (like Einstein said... does god play dice)! If there are any truly random processes, it might require you to resort to quantum physical solutions, which is not practical in a casino, or a typical computer! Most if not all physical processes tend to generate patterns and are therefore to some extent predictable in their output, and thus, are not truly random. In particular, a computer algorithm can absolutely not be truly random! In any case, we can get close enough to a random process on a computer (for most purposes) using a PRNG (pseudo random number generator) algorithm. This is particularly important in cryptographic applications, so most of the PRNG research (including algorithm design and the measurement of randomness) has been done in the crypto community. A truly random process of shuffling would mean that every possible card ordering (i.e. all 52! of them) has an equal probability of being the result of shuffling the deck. How do you test it? It is actually not possible to measure, in physical reality, since you could test it a trillion times, and get exactly the same sequence every time! So what can you do? Well, you have to apply various statical calculations to see what the probability distribution looks like. There are several metrics that are used in crypto. The full answer to that would take many words, or at least many links to related web sites... and I am too lazy today! But I will make a suggestion. post this question on sci.crypt, and I guarantee that you will get lots of very good quality answers there! === Subject: Re: How to measure randomness of a deck of cards? > Well, you have to apply various statical calculations to see what the probability distribution looks like. May I know what are the various stat calculations for computing the probability distribution please? Secondly, what infomation can I obtain from the probability distribution graph? Is it something like here?: http://www.math.washington.edu/~chartier/Shuffle/simulation.html === Subject: Re: How to measure randomness of a deck of cards? OK... You will get more info on sci.crypt, but to get you going, you can learn about ent and diehard, which are stats tests for PRNG algos. ENT - see: http://www.fourmilab.ch/random/ DIEHARD - http://stat.fsu.edu/~geo/diehard.html === Subject: Re: How to measure randomness of a deck of cards? take the difference between consecutive cards, step 2 take the sum of squares. compare it to a shuffled deck. also do it for suits, 0 1 2 3 in mod 4. could also do it for every second card, every 3rd card. or a neural net could be trained to give you a yes or no if its shuffled. there's no absolute answer unless you're using a naive interpretation of random, say the deck is 314159, looks stacked to me! step 2: subtract that from the expected difference (I think) Herc === Subject: Re: How to measure randomness of a deck of cards? > take the difference between consecutive cards, > step 2 > take the sum of squares. > compare it to a shuffled deck. > also do it for suits, 0 1 2 3 in mod 4. > could also do it for every second card, every 3rd card. > or a neural net could be trained to give you a yes or no if its > shuffled. > there's no absolute answer unless you're using a naive interpretation > of random, > say the deck is 314159, looks stacked to me! > step 2: subtract that from the expected difference (I think) > Herc What is this method called? And what is its main purpose? === Subject: Re: How to measure randomness of a deck of cards? The sum of squares (of the differences) is standard for measuring total error of a sample. I'm not sure how it will go using the heuristic of an 'expected difference' between consecutive cards, but you should be able to detect a fresh deck from a shuffled one. Say you calculate the average difference between consecutive cards is 7. then a fresh deck will have differences of Cards = <1,2,3,4,5,6,7,8,9,10,11,12,13,1,2,3,4,5,6,7,8,9,10,11,12,13,1,2..> Delta = <1,1,1,1,1,1,1,1,1,1,1,1,1,1,13,1,1,1,1,1,1,1,1,1,1,1,1,1,13,1,1....> Errors = <6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,...> A random deck will have Delta = <2,4,1,10,12,4,2,6...> Errors = <5,1,3,5,1,5,1....> The standard deviation will pick it up (on Delta), that's a function of the sum of squares, so you don't need an expected difference or Errors. here's another technique : use GPS (General Problem Solver AKA Search algorithm) to sort the cards *into* order count the number of card swaps and you have a measure of the *distance* from an ordered list you could weight the cards swaps, *distance, *sqrt(distance), or just use adjacent swaps. Deck one took 5 card swaps to be ordered, deck two took 20, so deck two was more shuffled. Some problems... a reverse ordered deck will appear shuffled! Herc === Subject: Re: How to measure randomness of a deck of cards? > say the deck is 314159, looks stacked to me! Actually, the digits of pi look not too bad when you do PRNG stats tests... Not as good as SHA-1, but not too bad... === Subject: Re: How to measure randomness of a deck of cards? >> I need a >> method/algorithm to determine the randomness of my deck of cards > Cannot be done. Any ordering is as random as any other ordering. Randomness > is not a property of your deck of cards. Do you mean a random process of > selecting a sequence of cards from your deck? This is about as helpful as looking at the sequence: 1, 2, 3, 4, 5, ... and being unable to figure out the next number. Sure, if every arrangement is equiprobable, every arrangement is just as random as any other. But it's not always the case that all arrangements are equiprobable. If I get a deck of cards, break the shrink wrap and look at it, the sequence will be pretty predictable. What is true is that there is no one, true, absolute, objectively correct way of looking at the order of the deck and saying that deck is 50% random. Randomness is a property of the distribution of possible shuffles, not of the shuffle resulting from one particular trial. And you can't measure a complete distribution with a single sample. What is also true is that there are useful metrics that can produce figures of merit for how thoroughly the deck appears to have been shuffled. A single sample does provide some information about the underlying distribution. And it can provide a lot of information if you already know something about the set of distributions that are likely. If I see a card sharp riffle-shuffling a deck and see that the resulting deck order is consistent with a sequence of four perfect riffle shuffles starting from the store bought starting order, I wouldn't call that resulting order random. I might be wrong. Maybe the sharp was playing fair and it was just a one in fifty-two factorial freak occurrence. John Briggs === Subject: Re: How to measure randomness of a deck of cards? >If I see a card sharp riffle-shuffling a deck and see that the resulting >deck order is consistent with a sequence of four perfect riffle shuffles >starting from the store bought starting order, I wouldn't call that >resulting order random. >I might be wrong. Maybe the sharp was playing fair and it was just >a one in fifty-two factorial freak occurrence. > John Briggs If I recall correctly a programming exercise I worked out many years ago, in fact seven perfect shuffles leaves the deck as it started. --Lynn === Subject: Re: How to measure randomness of a deck of cards? >>I need a >>method/algorithm to determine the randomness of my deck of cards > Cannot be done. Any ordering is as random as any other ordering. Randomness > is not a property of your deck of cards. Do you mean a random process of > selecting a sequence of cards from your deck? I suspect the OP meant how far (in some sense) a particular arrangement of cards is from the canonical arrangement. There are lots of ways to do this. A good starting point is _The Art of Computer Programming_, vol. 2. Rick === Subject: invariant differential forms in a lie group I'm looking for proofs (of references) using the least possible lie theory of the following fact: if w is a left and right invariant differential form in a lie group G then w is closed. nojb. PS: is the following correct? Let G be a connected lie group, g in G and let L_g be the G -> G diffeomorphism given by left translation by g, then L_g s = s in homology for every singular chain. Proof: H(g,t) = f(t).g is a homotopy between id : G -> G and L_g where f is a path between e and g in G. === Subject: cone cone intersection hi, how can I compute the two intersection lines, when two cones intersect each other? Erich === Subject: Re: cone cone intersection >how can I compute the two intersection lines, when two cones intersect >each other? Write the equations for the two cones, and solve for two components in terms of the other. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Hilbert 16th Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3IkHx27203; Hello I Search new Interpretations For the Limit Cycle's Phenomena and unusuall formulation for the hilbert 16th problem: (any comments and new suggestions is appreciated), I begin with Following interpretations 1)The Number of attractores of a vector field X is equall to fredholm index of functional operator X(g)=X.g please review : http://www.arxiv.org/abs/math.DS/0408037 2)let g be a riamanian metrics compatible to a vecter field X:that is the trajectories of X are geodesics,so the limit cycles of X are closed geodesics,we can try to find various metrics g which curvature-function has an appropriate behaviour (please See Limit Cycles And complex Geometry,Romanovski St Ptrs Jour Of Math 1997) 3)Quantum Mechanics, however I have no background In Math. Aspects Of Quantum Mechanics,But I Have found The Following two Papers(with Intersting titles) in The Web,last year: i)IS QUANTUM MECHANICS A LIMIT CYCLE THEORY,By A.M.Cetto and de la Penna ii)Limit Cycles In Quantum Theories,Glazek Wilson Physical review letter 2002 Question :What is a Quantization Of Hilbert 16th Problem: Let we have A Vec. Field X on A surface M,it generate a flow and this flow naturally defines a flowb on C^inf(M),now let's Quantize M,that is to embedd C^inf(M) in Some B(H),now the flow in C^inf(M) would be converted to a flow in a subspace Of B(H)....(what process you suggest for continuation....) === Subject: bootstrap and Fisher's fiducial analysis Trying to understand the logic behind the bootstrap, it seems to me more and more that this is similar to Fisher's fiducial inference. Any opinion? Kjetil Halvorsen === Subject: Re: Takers and leavers: Synthesis > The way I see this, the Western society is torn between two lines of > thought. One is the unconditional growth of the civilization, whatever > the consequences for the planet and for the future generations. The > other is a reaction against it: a return-to-the-roots hippie > environmentalism that seeks to What old-fashioned hippie types (and their neoLuddite descendants) fail to take into account is the sheer self-serving strategic profligacy of Nature based on the policy of unconditional growth. Plants do not produce just enough seeds to ensure zero population growth for themselves, they produce enough for any given species to fill their local biome (in some cases, blanket the planet) in one generation. Similarly, animals do not produce just enough progeny to ensure ZPG for themselves. Fortunately, the Law Of Fang And Claw (for plants, chemical warfare and rapacious competition for water and sunlight) keeps any given species from taking over completely. So, who's actually acting in accordance with Nature, hippies or Capitalists? And before you bewail the sheer mass of human flesh that will occupy the planet after we've killed everything else off and have to resort to cannibalism because there's nothing else to eat, consider what happened to the poor, unsung anaerobes that formerly exclusively ruled the Earth. Now, us aerobes have to eat each other while trying to finish off the anaerobes. Nature is change in strobing neon caps, not any kind of imaginary idyllic steady state. Environmentalism, at its root, is narrow-minded arrogance. Mark L. Fergerson === Subject: Re: Takers and leavers: Synthesis I agree completely with you. You are an animal, and your law is the law of the jungle. === Subject: Re: Takers and leavers: Synthesis <41b24ea6$0$11754$8fcfb975@news.wanadoo.fr> Well ... _where_ do you live? He nailed it as far as I'm concerned. It's eons of evolution getting us to the point where the state can pass laws over-ruling God, Darwin, and anyone else who gets into the way :-) Mark ( ... ) === Subject: Re: Takers and leavers: Synthesis > I agree completely with you. > You are an animal, and your law is the law of the jungle. It's the New Barbarism. I hear it's all the rage in New York. === Subject: Re: What is a proof, exactly? >However, due to several technical difficulties formalising linear algebra I >have begun wondering what exactly it means to have a proof of a theorem. > I define a formal system as a finitistic function FS(Prf,Thm) that > distinguishes valid proofs from invalid proofs. I think you miss the point. He means the intuitive meaning of proof and how we formalize it. You are talking about particular formalizations (a lower level of abstraction.) There are plenty of formalizations around. Why is that one better than all the others? > A proof in a formal system FS is then a pair , such that > FS(Prf,Thm) = 1. Wouldn't putting it in terms of Recursion Theory be more general? > -Why could I not prove -say- that the set of natural numbers is as large > as the set of reals? > It's not necessarily impossible, but the existence of such a proof would > imply the inconsistency of the formal system in which it was checked, > given standard definitions of sets, the naturals, the reals, and the > predicate is as large as. Again, I think he means in the intitive sense that a proof means that it really is true and is also convincing, hopefully. Of course, I could be wrong. C-B === Subject: Re: What is a proof, exactly? > I define a formal system as a finitistic function FS(Prf,Thm) that > distinguishes valid proofs from invalid proofs. > I think you miss the point. He means the intuitive meaning of proof > and how we formalize it. You are talking about particular > formalizations (a lower level of abstraction.) I am talking about properties that should hold for *any* formal system. And I presume that we believe that an informal proof is valid just when we believe it could be formalized in a suitable formal system. > There are plenty of formalizations around. Why is that one better than > all the others? Representing a formal system as primitive-recursive proof predicate has a number of advantages. First of all, it is very general, but probably not overly so: the kinds of things that a primitive recursive function can do seem to neatly contain the functionality needed by a proof-checking algorithm. (If it is too general, then perhaps all that is really needed are the functions with polynomial time and space complexity.) Second, it is very expressive. It's possible to write a primitive recursive function that corresponds closely to our intuition of how a proof should be checked. Third, it allows us to calculate an upper bound on how long it will take to check a proof, since the running time of a FS(Prf,Thm) is bounded by f(length(Prf)+length(Thm)), where f is primitive recursive. > Wouldn't putting it in terms of Recursion Theory be more general? It already is - primitive recursive is part of the lexicon of recursion theory. === Subject: Re: What is a proof, exactly? > Generally speaking, this is true. But two important counter-examples > loom large: Megill's development of set theory in Metamath, and > Jutting's formalization of Landau's _ Foundations of Analysis _ in > Automath. Indeed. Jutting's work is the forefather of Automated Reasoning. Since then many many formalisations have been done, in various other systems. Notably Mizar (www.mizar.org) has a large library. Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? > Indeed. Jutting's work is the forefather of Automated Reasoning. Could you please elaborate a little more on this. The reason is that Jutting has been one of my mathematics instructors at the Eindhoven University of Technology. I'm just curious what happened next. Han de Bruijn === Subject: Re: What is a proof, exactly? >> - What is the status of = (equality) with respect to proofs? Is there >> a >> difference between meaning of and definition of? Is = defined or >> does it only have a (logical) meaning? Can we equate any two mathematical >> objects? > Well, first-order logic does give a preferential status to that symbol > over the other relational symbols, in that it appears in the axioms and is > not up for interpretation. This, in combination with other posts reminds me of something. I was raised on Dirk van Dalen's Logic and Structure, where he states that there are languages of 1st order logic where = doesn't appear, although AFAIR he gives no examples. But now discussion on set theory and specifically the axiom of extensionality makes me wonder about the status of = in ZF. Extensionality is (Ax Ay: x=y <=> (Az: zex <=> zey)). Does this define = in terms of e? That is: what is the language of set theory, does it have 2 relations (e and =) or 1 (only e)? In the latter case, is = a defined notion (by extensionality) or is it a logical thing which is only used in this axiom to axiomatize e? In that case, are the logical = and the relation symbol = two different things? (As to the question about your class' notion of truth - every model comes with an interpretation of the various constant, function and relation symbols of the language; a sentence of the language is true if it is true in all models (according to their respective interpretations)) > logically from the axioms or deduction. There is another way to insert > interpretations into logic, which is as a model of a set of logical > statements, namely a set in which each logical symbol corresponds to a > function or relation or element of the set, so that a logical sentence is > translated into a composition of these things concatenated with logical > connectives like -> or for all, and this sentence is either true or > false (or totally undecidable, of > course)...but again, this judgement is made by a person. Mathematical I don't quite get what you mean here. Could you expand, please? >> - Once we defined X as above, we can make additional definitions Y, Z, >> ... >> depending on X. Can we also make definitions depending on Thm? On prf? >> Can you comment on the definition of the reciprocal ($x to >> frac{1}{x}$), which seems to depend not only on $x$ but also on a >> 'theorem' stating that $x neq 0$? Does it also depend on a proof of such >> a theorem? How? > In what way does the definition of X allow us to make additional > definitions? Well, this is common, isn't it? We define what a ring is, and then we can prove that if there is a unit element (some definitions of ring don't require one) for its muliplication, then it's unique. Then one can define what a division ring (aka. skew field) is, and if we have that we can define what a field is (namely a commutative division ring). Similarly one can define x to be some real number and f a function on the reals, and then define y := f(x), depending on f and x. (By the way, do you consider these to be two different kinds of definitions?) > In the sense that the existence of X is a logical statement > with logical consequences, this is perhaps true (of course, that's not a > first-order > concept). Another way to look at it is that Y and Z depend on certain > logical non-axioms in addition to the axioms, and if X satisfies those > axioms then, having convinced ourselves that X is our universe of the day, > we will accept Y > and Z as well. That is a first order statement, and it really just says > that > the any model of Y's axioms is also a model of X's. I'm not sure I follow what you're saying here.What do you mean by X is our universe of the day? I meant X to be just some mathematical object: a real number, a graph, or a category perhaps. These are not universes in any way I'm aware of...? I also presume you mean to say that if the existence of object X requires axioms A1...Ai then we also have to accept objects Y and Z if their existence follows from these same A1...Ai? > However, the actual nature of Prf is irrelevant, since its existence is > the verification that Thm is provable, and conversely the provability of > Thm is (by > definition) the existence of Prf. And we can deal with Thm. Okay. So if Prf exists on the same level as X and Thm, could we define a new object Y:=f(prf)? Let's say Y:=1 if prf uses induction and Y:=0 otherwise? [about the 4colour theorem:] > reproduce the proof, and since as I've indicated that I belive, proofs are > eventually based on human ideas about how to determine truth. Okay, some systems already do that. They spew out a proof (encoded in a lambda term) of what they think is true. But these proofs are on the level of formal proofs: every reasoning step is minutely described, and so they become very big very soon. Even writing out the proof of n+k=k+n requires tens of lines - okay, it's a pretty-printed lambda term with lots of whitespace, but still it's a lot. So if you don't trust computers by themselves, and humans cannot check these proofs, what must we think of them? > If a tree falls in a forest, and no one witnesses it in any way, did it > fall? I > would say yes. But theorems aren't trees; if a logical statement is > consistent with the axioms then (by the completeness theorem) there exists ^^^^^^^^^^ Okay, I saw your other post :-) > a proof, and the > production of that proof does not change its logical status. All it does > is change our knowledge of it, and it's the part where we believe that the > RH is true not because we'd like to, but because it has been shown to be > the case by methods we also believe to be the perfect embodiment of our > intuition, that is > important to us. If aliens give us an incomprehensible proof it's no > good; and if the four-color theorem's proof is incomprehensible, that's no > good either. > Even if it's true. Of course, if I understand it and you don't, I can > feel free to use it (provided my understanding indicates it's true, of > course), whereas you would warily agree that my manipulation of the > logical symbols is correct, there is one spot in my proofs that needs > further verification. It is interesting to see that you appear to allow the human factor to play a role. I wonder how wide-spread this belief is. Responses, anyone? I know the official blurb that the human factor shouldn't play a role, but I don't believe it, and neither do you, it seems. I suppose that in some axiomatic systems there are theorems - well, statements really - whose minimum-sized proof is too long to even write out physically. What about these? >> I suppose that's enough questions for the time being. Feel free to >> include any pertinent thoughts and additional questions. > Let the tearing down of my amateurish responses begin! Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? Discussion, linux) > But now discussion on set theory and specifically the axiom of > extensionality makes me wonder about the status of = in > ZF. Extensionality is (Ax Ay: x=y <=> (Az: zex <=> zey)). Does this > define = in terms of e? That is: what is the language of set > theory, does it have 2 relations (e and =) or 1 (only e)? In the > latter case, is = a defined notion (by extensionality) or is it a > logical thing which is only used in this axiom to axiomatize e? In > that case, are the logical = and the relation symbol = two different > things? Howdy, Jasper. What's the difference, really? If one wanted, he could take equality to be defined by extensionality, so that there's really only one relation in ZF, but in this case, he has to introduce the axioms for equality in terms of elementhood. Thus, one would have to introduce a rule of inference something like: (A z)(z in x <-> z in y) P --------------------------- P[y/x] (where P[y/x] denotes a formula obtained by substituting some freely occurring x's in P by y's). Not particularly nice. Here's something worse than this cosmetic complaint. If we take this approach, we could have a model for ZF in which distinct elements satisfy all the same formulas. Nothing particularly bad happens, I guess, but it doesn't satisfy our intuitions. If instead we take equality as primitive and require that it is always interpreted as equality in the model (which seems the real meat of putting it into FOL), then we don't get these perverse models. When we have the freedom to say that equality must be interpreted as equal in the model, then indistinguishable elements are equal (in the model). This is obvious:if x and y are indistinguishable, then M |= x = y and thus x =_M y. Of course, as you know, it's been a while since I paid my first-order dues, so I could be just butt-wrong here. -- Jesse F. Hughes I thought it relevant to inform that I notified the FBI a couple of months ago about some of the math issues I've brought up here. -- James S. Harris gives Special Agent Fox a new assignment. === Subject: Re: What is a proof, exactly? > [about the 4colour theorem:] Not that it has so much to do with this thread, but if anybody wants to see some *real* map 4colouring, you could take a look at: http://hdebruijn.soo.dto.tudelft.nl/fcp/index.htm Han de Bruijn === Subject: Re: What is a proof, exactly? > - Suppose aliens gave us a pack of paper, full of english text and > mathematical symbols, ending in ... and hence the Riemann Hypothesis is > proved. QED. Suppose also that this text was so intricate and so involved > and so convoluted that no one would be able to understand it in full. Does > this pack of paper constitute a proof? The aliens have landed! Actually, there does _exist_ such a (would-be?) proof here on earth: http://www.math.purdue.edu/~branges/ Han de Bruijn === Subject: Re: What is a proof, exactly? >> [A proof] can't be required and empty at the same time. It can't at >> the same time be both necessarily present AND absent. > Good point. I was trying to say that x is a theorem just when > ProofPredicate(prf,x) is true, and ProofPredicate may be such that it is > true for those arguments even if prf is something that we might consider > empty, such as is the empty list or the empty tree. >> [...] FS(.,.) is never a proof of anything. That is NOT the sort of >> thing that a proof can be. > Right; that was sloppy of me. I should have written ``nil might be a > valid proof of 5=5'' (according to my (idiosyncratic?) definition of a > formal system as a ``finitistic function FS(Prf,x) that distinguishes > valid proofs of x from invalid proofs of x''). This is somewhat strange - does this mean that one (formal) proof can prove several things? At least I suppose that if <> proves 5=5 then it also proves e.g. 27=27. (It seems that I'm missing some of the posts to the newsgroup - I haven't seen George Greene's original post) Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? > This is somewhat strange - does this mean that one (formal) proof can prove > several things? Correct. Consider a Hilbert-style proof system, in which a proof is a sequence of statements, each of which is either an (instance of) an axiom, a premise, or follows from one or more previous statements in the sequence by a truth-preserving rule of inference. Thus, the proof can be considered to prove every sentence in the sequence. > (It seems that I'm missing some of the posts to the newsgroup - I haven't > seen George Greene's original post) Try Google Groups. === Subject: Re: What is a proof, exactly? Regarding the subject field contents: A proof is a complete justification for the truth of a statement. G C === Subject: Re: What is a proof, exactly? > Regarding the subject field contents: > A proof is a complete justification for the truth of a statement. > G C questions I raised? Please do. Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? > I wonder if Jasper Stein has benefitted from the 42 mails sent in > reply to his original post. How will they affect his formalization of > linear algebra; is he closer to being able to demonstrate to his > thesis committee that he has grasped the prologemena for automated > reasoning? is already finished (publicly available even), and although I intend to do some additional updating and polishing, this thread does not influence the contents of +that+. My questions were of a more general nature. I've been in this community for some while now, but I've started wondering if what we're doing is actually the same thing that we promise, namely computer-checked proofs. As I said I've encountered some strange problems and oddities in my formalisation, and I would know how the Average Mathematician thinks about some of these issues. Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? > My questions were of a more general nature. I've been in this community for > some while now, but I've started wondering if what we're doing is actually > the same thing that we promise, namely computer-checked proofs. As I said > I've encountered some strange problems and oddities in my formalisation, > and I would know how the Average Mathematician thinks about some of these > issues. I would say that, if computer proof-checking is ever going to take off, then that's a less important question than usability. I think everyone can live with the fact that a usable proof-checking system will have bugs. The real question is, is there a practical way for me to enter a proof I'm working on and have the system find my errors in reasoning, without requiring me to break my proof down into sand-grain size steps? I've often wished for that. I mean, when I'm writing a program, I just hit the compile button and it finds all sorts of errors; I don't have to personally pore over the document to find them. It would be very nice to have something similar when doing math, even if an occasional error is missed. But I'm somewhat skeptical about the notion, because it seems to me that one of the major goals of research in mathematics is to find new techniques, and the software would need constant updating to accomodate them. It's hard to see how it wouldn't always be years behind the state of the art. Only the less interesting new techniques can be chunked into theorems that say everything that needs to be said--most of them are more like very high-level, very complex, rules of inference. === Subject: Re: What is a proof, exactly? > it to what is inside the five in the blue set. If you don't have sets > all the way down, then evetually you get to something that doesn't > have equality defined and it causes equality to become undefined for > everything that contained it, and everything that contained that and > so on transfinitely. Do you mean set equality specifically, or is this more generally a denial of my question Can we equate any two mathematical objects?? > mean the usual things. The thing is that set equality itself is an > equivalence relation itself that is too big to be a set. So sets > can do relations on specific sets, but sets can't do relations on all > sets in general. So - would you say we'd need two different kinds of equalities, the set theoretic ones (equality between members of a set), and another one that spans all sets simultaneously? Would this be a logical equality? And since this equality is a proper class, what would this equality be if we consider Godel-Bernays instead of Zermelo-Frankel? Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? >> - Does the extremely simple theorem 5=5 need a proof? Why (not)? > Yes, even more you need a definition of 5 = and a standard of > proof. Why? Because that's what proving things is all about. If you Exactly. I believe that's what I'm after. You say we need a standard of proof - I'm trying to get to know what that standard looks like. As I've said, I come from the automated reasoning community, and I believe our own standard of proof is skewed. So I want to know what the 'standard' standard of proof is. People have answered that there is a difference between formal proof and informal proof, and that the informal proof should convince the reader that there is a formal proof, being a sequence of statements following from axioms or from rules of inference. But no one seems to bother actually carrying out this transition. Moreover, I expect that if I write a paper containing only formal proofs, it'd be (1)huge (2)too hard to follow even for relatively simple informal proofs (3)unpublishable in any regular math journal. So why do we care about formal proof? > define x=y as Az (zex <=> zey) then there IS a provable theorem > Ax x=x, Nice to see the defined equality boil down to a logical one in this case (after all, for all predicates P we have Az P(z)<=>P(z)) > so then as long as 5 is a set (usually > {{},{{}},{{{}}},{{{{}}}},{{{{{}}}}}}) I thought usually it something else, but never mind :-) { {}, {{}}, { {}, {{}} }, { {}, {{}}, { {}, {{}} } }, { {}. {{}}, { {}, {{}} } }, { {}, {{}}, { {}, {{}} } } } } > then 5=5 is a theorem too, the > proof of the more general theorem already covers when x happens to be > 5 assuming you defined 5 to be a set. So you assume everything to be a set. It is said that this can be done in principle, although personally I'd like to rephrase it as 'everything can be modeled using sets' since when I talk about numbers or graphs or vectorspaces, I don't think of these things as sets at all. They're more like urelements; a primitive notion that doesn't decompose into sets, although we can model them in pure set theory. I've been wondering: since math is all in the mind and on paper, wouldn't it be possible to actually have a foundation for mathematics based purely on non-sets? I'm thinking of a term model like kind of thing. >> - How is a proof different from just an explanation? Is there a >> difference >> at all? > The usual standard is that a proof (of T relative to A) is a > demonstration that if T were false then A would be false too. So if ??? That can't possibly be true. Deriving A->T from ~T -> ~A isn't even possible if you haven't got the law of excluded middle (ie. forall P, P or ~P) or equivalently double negation (forall P, ~~P -> P). But even if you do have them, this is only one means of inference, certainly not the only one. > the assumption is A, then A is a theorem, in the usual standard. > I'm assuming the is an ordinary first order logic system. This raises another interesting question: is informal maths formalisable using FOL only? I think we use higher order logic, actually. FOL might be enough if you say everything is a set, but if you don't think that's true then FOL seems rather restrictive. >> - Suppose we defined an object X. Then we state a theorem Thm. about X, >> which we subsequently prove by a proof prf. According to the main >> school(s) of mathematics, what is the ontological status of X, Thm, and >> prf? Does Thm 'exist' in the same sense as X 'exists'? Does prf 'exist' >> in the same sense as Thm? > I'm not sure you meant this as written, just defining something > doesn't entail proving that it exists, so the ontological status of X > is undetermined at this point. Okay, that's true. But if we can prove the existence of X, then what's your opinion? If you need an example, let's take this one: X := { {} } Thm := (forall a,b: (aeX and beX) -> a=b) (ie. X has 1 element) prf := (take any proof you like, there must be lots) >> - Once we defined X as above, we can make additional definitions Y, Z, >> ... >> depending on X. Can we also make definitions depending on Thm? On prf? >> Can you comment on the definition of the reciprocal ($x to >> frac{1}{x}$), which seems to depend not only on $x$ but also on a >> 'theorem' stating that $x neq 0$? Does it also depend on a proof of such >> a theorem? How? [...skipping definition == shorthand, with which I agree mostly - in type theory you ARE doing something: making a definition means extending the context, unfolding a definition is doing a delta reduction...] > the point is > that there is a function f that takes each non-zero real to it's > reciprocal, so 1/x can be thought of as f(x) and checking that ~x=0 is > just like checking that a number is in the domain of a function and > then finding the value of the function. My point in this question is exactly that: how do you check whether x is in the domain of the function? In general that's undecidable. If you're a classical mathematician (ie. believing in classical logic) then decidability isn't an issue in principle, but in practice it is. The prime example would be the number (let's call it a) defined by decimal expansion, where digit n is 1 if a sequence of 99 9's has occurred prior to position n in the expansion of pi, and 0 otherwise. So (classically speaking) a is either 0 or 0.000...01111... Now how does one check whether 1/a makes sense? For this we need a proof that actually a=/=0, ie. a proof that pi has a series of 99 9's somewhere in its expansion. (And as of today this is unproved.) > so why should I trust a proof by that is checked only by a computer? Good answer. The main automated-reasoning response is that (1) we have the de Bruijn criterion: the computer program code must be small enough that it can be checked by hand. (2) we can let the computer emit a 'proof object' that contains the reasoning it followed in proving the theorem. This proof object can then be checked independently. Does this influence your opinion? Why (not)? >> - This code calculated and checked a few thousands of 'configurations'. >> If >> these calculations and the checking were done by mathematicians, would >> the proof be still as controversial? Why (not)? > It's not one mathematician versus one computer, it's the fact that so > few independant minds would actually go through the bother of checking > it. It's just not as trustworthy, because it's been checked a smaller > number of both quality and independant times. Allow me to play the devil's advocate: in the middle ages, everyone thought the earth was flat. Today everyone thinks the theorem is true, that this-and-this reasoning is sound. You sound like you think maths is a sociological enterprise. Is that true? > If we could understand the symbols, then we could make a machine (from > scratch) to check the proof. More likely I would expect that > mathematicians would prefer to make a machine that attempted to > *simplify* the proof. Then you can just read the simplier proof, a > much better case. Now for 'pack of paper' substitute 'my computer's proof object', which is indeed a string of symbols that my computer has checked. How does this hold with your remark (see above) > so why should I trust a proof by that is checked only by a computer? questions. (And this goes for others who responded as well!) I hope you and others can find the time to answer my further questions too - they're quite important to me, because they bear on what I'm doing for my job every day, and I'm doubtful (as you may have noticed)... Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? More deconfusifying, again sorry for the inconvenience. Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? Test succeeded, further deconfusing my newsreader/news feed, sorry for the inconvenience... Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? Test, sorry... (My newsreader/news server combination is confused somehow, I hope this will deconfuse things) Test 1 2 3... Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? >-Supposealiensgaveusapackofpaper,fullofenglishtextand >mathematical symbols, ending in ... and hence the Riemann Hypothesis is >proved. QED. Suppose also that this text was so intricate and so >involved and so convoluted that no one would be able to understand it in >full. Does this pack of paper constitute a proof? > It would certainly not be a formal proof. If the stack of paper > could be read by a computer, if it was a formal proof, the > computer could go over it step by step, and verify it. If it > was a compressed formal proof, it gets much harder. The biggest > problem is likely to be the ambiguities of the English text. Okay, but let's forget about the english text. So there's only symbols. Still it is so convoluted that no one could follow it in full. Then the question becomes - for me - very interesting, because that's the kind of thing automated-reasoning people handle: we have typed lambda terms that are supposed to encode proofs. And indeed, the computer can check these terms step-by-step, and will tell us whether these terms are well-typed - ie. encode a formal proof. (These lambda terms don't look like lists of statements, though) But these terms become incredibly large. If you happen to have access to a paper copy of the Handbook of Logic in Computer Science, look up Henk before the references, it has one of these lambda terms. (Well, I hope the handbook has it. The preprint that I have at home has them, but no copy I find on the web includes them) It's some four pages long in tiny script, and only symbols. You won't understand it. Still the computer can check this kind of thing. Is it a proof? Shouldn't we be able to understand at least what it is a proof of? >>A proof must convince a mathematician of the truth of a theorem. And >>mathematicians are individuals. A proof must endure debate and >>examination. Voting isn't enough. > No; an informal proof is something which, in principle, can be > expanded to a formal proof. Then everything is an informal proof, isn't it? After any load of gibberish, just add now forget about all this, we'll restart and add a new fully formal proof. The informal proof needs to point the way to a formal proof, however vague that statement is. > We even have theorems, proved by > machines using known correct steps, whose proofs have not been > verified by humans. Do we? That's interesting. Or do you mean things like the four colour theorem? Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? >> Two sets are disjoint if they have no element in common. > (A is disjoint from B) iff (forall C (CeA iff (it is not the case that > CeB))). Is that more clear? It's certainly incorrect. A would be VB, a proper class...?! (V being the class of all sets) I propose: (A is disjoint from B) iff (forall C, (not (CeA and CeB))) or perhaps (more like your definition): (forall C in (A union B), (CeA iff (not CeB)) Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? >> Two sets are disjoint if they have no element in common. > > (A is disjoint from B) iff (forall C (CeA iff (it is not the case that > CeB))). Is that more clear? > It's certainly incorrect. A would be VB, a proper class...?! (V being the > class of all sets) > I propose: > (A is disjoint from B) iff (forall C, (not (CeA and CeB))) > or perhaps (more like your definition): > (forall C in (A union B), (CeA iff (not CeB)) > Jasper Yeah, I should remember not to post when I'm sleeping, obviously sticking to the easy stuff isn't good enough since there is no excuse for that error. === Subject: Re: What is a proof, exactly? >>The only thing you did is saying that distinct is synonymous >>with different. I don't find that a proper definition. > Come on Han, no word play now. You know as well as anyone else that > different means not equal. And the definition of equal is given by the > axiom of extensionality. OK. I will not pursue this argument any further. That is, until we would arrive at the final question I had in mind: is x different from {x} ? Which belongs to another thread: Re: Set inclusion and membership So I agree. We've gone far enough here. Han de Bruijn === Subject: Re: What is a proof, exactly? > OK. I will not pursue this argument any further. That is, until we would > arrive at the final question I had in mind: is x different from {x} ? Ah, but that's a very interesting question indeed! Although it has little bearing on this thread, as you say. > Which belongs to another thread: > Re: Set inclusion and membership I hope to have time to read this one As to your question, different set theories answer this differently. ZF says they're different, because you'd violate the axiom of foundation. Quine's NF (New Foundations) however +does+ equate them, or so I've been told. And then you have various forms of hyperset theory ZFA where there are some sets that obey x={x} and some don't. (In some forms there's only one such set, called Omega; in others you have plenty) Read Vicious circles if you're interested (by either Aczel or Barwise, I forget who) Jasper -- The problem with having an open mind is that people toss in garbage === Subject: Re: What is a proof, exactly? >Two sets are distinct if they are different. > Two sets A and B are different if there exists an element C such that > CeA but where it is NOT the case that CeB, OR if there exists an > element C such that CeB but where it is NOT the case the CeA. > (A is not different than B) iff (forall C (CeA iff CeB)). Is that > more clear? I will regard that as a technical definition. Yes. > I put some hard-core definitions above, I hope that helps. Another > way is to say that you understand disjoint, distinct means not > equal. Han de Bruijn === Subject: Re: How to measure randomness of a deck of cards? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3KH3I03194; >I read several papers, there are few approaches such as variation >distances, birthday bound and markov chain. I need a >method/algorithm to determine the randomness of my deck of cards >(permutation) >I would be appreciated if anybody can give me some ideas or good solutions. Estimate how much information is lost due to each shuffle. Ideally, it is as much as 1 bit. This would occur if you couldn't tell which half of the deck contains a given card (assuming you knew which half before the shuffle). Assuming each shuffle is independent, each destroys one bit of information. In practice about 8 or 9 shuffles provides pretty good randomness. Google for more including experiments phil === Subject: nearest common ancestor on Stern-Brocot tree? For example, the nearest common ancestor for 5/9 and 3/4 is 2/3. Indeed 5/9 = LRLLL 3/4 = LRR 2/3 = LR The straightforward way to calculate NCA is by converting the rational number into LR representation applying (extended) Euclidian algirithm. Is there a formula for NCA? === Subject: Circle-line intersection Hi there! Don't tell me to look at other posts in this forum. This is a special problem aimed to they who know math. The question how the get 't', look below. I thought I solved the problem but the outcome was a failure. I got a correct equation from mathworld but that is not what I want though. A line(L) intersects with a circle(C), tell me if something is wrong. All positions orients from circle position which is (0,0). t is what I want. L = A + (B-A)*t where D = B-A Lx = cos(v)*R Ly = sin(v)*R A combination of equations: cos(v)*R = Ax + (Dx)*t sin(v)*R = Ay + (Dy)*t I do this because the points of the line intersects on the circle-line which distance is R from origo. 'cos(v)*R' is the horizontal distance and sin(v)*R is the vertical distance. Hope you understand my thinking here. Only if could draw with a pen :). And no we do 'power of two' at both sides of equation. This I do because of a forumla which I've to merge with this equation, see below. cos^2(v)*R^2 = (Ax + Dx*t)^2 which also are ((Ax + Dx*t)^2) cos^2(v) = -------------------- R^2 sin^2(v)*R^2 = (Ay + Dy*t)^2 which also are ((Ay + Dy*t)^2) sin^2(v) = -------------------- R^2 Ok now to the hard part. This is a math rule: sin^2(v) + cos^2(v) = 1 and this gives cos^2(v) = 1 - sin^2(v) Now we combine the first combination with the formula above: ((Ax + Dx*t)^2) cos^2(v) = -------------------- R^2 will be ((Ax + Dx*t)^2) 1 - sin^2(v) = -------------------- R^2 And know we just take away the sin-function: ((Ay + Dy*t)^2) ((Ax + Dx*t)^2) 1 - -------------------- = -------------------- R^2 R^2 Now I can get 't'. The equation will be, after some flipping and flopping of the variables, look like this: t^2*(Dx^2 + Dy^2) + t*(2*Ax*Dx + 2*Ay*Dy) - R^2 + Ay^2 + Ax^2 = 0 If you don't see what this is I tell you this: ax^2 + bx + c = 0 Now if you know what math is you know what it is. Now the question: Why DOESN'T THIS WORK!!! :| === Subject: Re: Circle-line intersection days. My association with the Department is that of an alumnus. >A line(L) intersects with a circle(C), tell me if something is wrong. >All positions orients from circle position which is (0,0). You might want to say what it is you are trying to do... >t is what I want. >L = A + (B-A)*t >where D = B-A There was no D so far.... You mean: L is a line going through the points A and B, and we obtain it (by vector addition) as A + Dt, where D = (B-A) (A, B being vectors with initial point at the origin and terminal point at the point A and B, presumably, and t being a real number parameter...) Is that right? >Lx = cos(v)*R >Ly = sin(v)*R What are Lx and Ly supposed to mean? What is R? What is v? I assume Lx means the x coordinate of the point on the line, and Ly means the y coordinate (as below you have Ax, Dx, Ay, Dy, presumably being x coordinate of A, x coordinate of D, etc). Meaning... you are trying to figure out the intersection? (x,y) are the coordinates of the intersection? And then v is simply the angle, R the radius of the circle? Sounds reasonable to me... >A combination of equations: cos(v)*R = Ax + (Dx)*t > sin(v)*R = Ay + (Dy)*t >I do this because the points of the line intersects on the circle-line >which distance is R from origin. 'cos(v)*R' is the horizontal distance >and sin(v)*R is the vertical distance. Hope you understand my thinking >here. Only if could draw with a pen :). Let's see. We are working on the plane. You are describing the circle as being given by all points of the form (R*cos(v),R*sin(v)), as v ranges over, say, 0 to 2pi. You are describing the line as going through the points A=(a_1,a_2) and B=(b_1,b_2), so you let D = B-A = (b_1-a_1,b_2-a_2) = (d_1,d_2), and the line becomes (x,y) = (a_1,a_2) + t(d_1,d_2) where t ranges over all real numbers. So any intersection between the circle and the line would have to satisfy a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) for some value of t and some value of v. >And no we do 'power of two' at both sides of equation. You are squaring both equations... Okay... HOWEVER: remember that because you squared the original equations, you may have introduced new solutions that were not part of your original intersections. For example, if you find a point where R*(cos(v)) = - (a_1+td_1) R*(sin(v)) = (a_2 + td_2) then these values of v and t will ALSO satisfy R^2*cos^2(v) = (a_1+td_1)^2 R^2*sin^2(v) = (a_2+td_2)^2 even though they do not satisfy the original equation. In fact, the solutions to the following four systems: ORIGINAL: a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) Plus a_1 + td_1 = -R*cos(v) a_2 + td_2 = R*sin(v) a_1 + td_1 = R*cos(v) a_2 + td_2 = -R*sin(v) and a_1 + td_1 = -R*cos(v) a_2 + td_2 = -R*sin(v) will also satisfy (a_1 + td_1)^2 = R^2*cos^2(v) (a_2 + td_2)^2 = R^2*sin^2(v) So ->not every solution you find after squaring needs to be a solution of your original equation<-. Every solution to your original equation will also be a solution to the new one, but the new one may have more solutions. (It's like, if you start with x= 3, there is only one solution; but if you square it, you get x^2 = 9, and now both x=3 and x=-3 are solutions). >This I do >because >of a forumla which I've to merge with this equation, see below. > cos^2(v)*R^2 = (Ax + Dx*t)^2 > which also are > ((Ax + Dx*t)^2) > cos^2(v) = -------------------- > R^2 > > sin^2(v)*R^2 = (Ay + Dy*t)^2 > which also are > ((Ay + Dy*t)^2) > sin^2(v) = -------------------- > R^2 Fine. Since a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) then cos(v) = (a_1 + t*d_1)/R sin(v) = (a_2 + t*d_2)/R >Ok now to the hard part. >This is a math rule: > sin^2(v) + cos^2(v) = 1 Identity. Yes. >and this gives > cos^2(v) = 1 - sin^2(v) Yes. >Now we combine the first combination with the formula above: > ((Ax + Dx*t)^2) > cos^2(v) = -------------------- > R^2 > > will be > > ((Ax + Dx*t)^2) > 1 - sin^2(v) = -------------------- > R^2 >And know we just take away the sin-function: > ((Ay + Dy*t)^2) ((Ax + Dx*t)^2) > 1 - -------------------- = -------------------- > R^2 R^2 Easier to just add the two equations you had: a_1 + td_1 = R*cos(v) a_2 + td_2 = R*sin(v) we have R^2 = R^2(cos^2(v) + sin^2(v) = (a_1+td_1)^2 + (a_2 + td_2)^2. >Now I can get 't'. The equation will be, after some flipping and >flopping of the variables, look like this: >t^2*(Dx^2 + Dy^2) + t*(2*Ax*Dx + 2*Ay*Dy) - R^2 + Ay^2 + Ax^2 = 0 >If you don't see what this is I tell you this: ax^2 + bx + c = 0 >Now if you know what math is you know what it is. >Now the question: Why DOESN'T THIS WORK!!! :| Looks right. What do you mean by This does not work? Maybe you mean you sometimes get wrong answers? That's because of the observation above. You have to make sure that whatever answers you get actually satisfy the original equations. Plug them in. They should work fine. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Circle-line intersection t^2*(Dx^2 + Dy^2) + t*(2*Ax*Dx + 2*Ay*Dy) - R^2 + Ay^2 + Ax^2 = 0 but when I converted it to an equation of the second degree I forgot an expression (Dx^2 + Dy^2). ;D It will look like this (sqrt = square root of): Ax*Dx + Ay*Dy (Ax*Dx + Ay*Dy)^2 Ax^2 + Ay^2 - R^2 t = - ------------- +- sqrt(----------------- - ------------------) Dx^2 + Dy^2 (Dx^2 + Dy^2)^2 Dx^2 + Dy^2 I'm doing this as a hobby programmer. Maybe there are easier ways, but for me it's fun to solve this myself, and the purpose of this equation is to compare two vectors (which has the same angles but different origins). The vector which has the lowest 't' value above 0 is the vector which first intersects the circle. I'm planning a circle-circle collision-detection algorithm which tells the user where the first collision-point is, and the vectors in this equation has the same angle as the moving vector of the circle, but the origins is at the both sides of the circle, these two are needed because if I'd only one, the actual collision of the circles may miss. But this will not work if I take a bigger circle to collide with a smaller one because of the radius of the smaller one is lesser than the distance between the two vectors at the bigger circle. Therefore I made a rule to always check smaller circles against bigger ones (and not vice versa). Hope you understand my theory, and thank you for reading through and comment everything. /Johannes === Subject: Shiing-shen Chern (Chen Xingshen) passes away at 93 Just grabbed off Xinhua web site: Noted mathematician passes away in Tianjin TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a world-renowned overseas Chinese mathematician, 93, died of illness at his home at Nankai University in north China's Tianjin Municipality at around 7:15 p.m. Friday, the university announced. Chern, a US citizen, is best known for his achievements in the study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang Province, east China. He graduated from Nankai University in 1930 and received further education at Qinghua University and the University of Hamburg in Germany. He taught at several Chinese and US universities -- including Princeton University, the University of Chicago, and the University of California, Berkeley -- and is the only Chinese to win the Wolf Prize -- the most distinguished award in the international mathematics field. The International Astronomical Union officially named asteroid No. 1998CS2 after the noted mathematician in November for his outstanding contributions to human society. === Subject: Re: Shiing-shen Chern (Chen Xingshen) passes away at 93 > Just grabbed off Xinhua web site: > Noted mathematician passes away in Tianjin > TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a > world-renowned overseas Chinese mathematician, 93, died of illness at > his home at Nankai University in north China's Tianjin Municipality at > around 7:15 p.m. Friday, the university announced. > Chern, a US citizen, is best known for his achievements in the > study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang > Province, east China. He graduated from Nankai University in 1930 and > received further education at Qinghua University and the University of > Hamburg in Germany. > He taught at several Chinese and US universities -- including > Princeton University, the University of Chicago, and the University of > California, Berkeley -- and is the only Chinese to win the Wolf Prize -- > the most distinguished award in the international mathematics field. > The International Astronomical Union officially named asteroid No. > 1998CS2 after the noted mathematician in November for his outstanding > contributions to human society. I had the privilege of taking an upper division class (upper division = 3rd or 4th year undergrad math major) from Professor Chern in 1976 at U.C Berkeley. I remember him as a warm, down to earth man. Approachable to us students. He cared about his students and about people. I'm sad to hear of his passing. === Subject: Re: Shiing-shen Chern (Chen Xingshen) passes away at 93 Moment of Silence. W. Dale Hall ??? > Just grabbed off Xinhua web site: > Noted mathematician passes away in Tianjin > TIANJIN, Dec. 3 (Xinhuanet) -- Shiing-shen Chern (Chen Xingshen), a > world-renowned overseas Chinese mathematician, 93, died of illness at > his home at Nankai University in north China's Tianjin Municipality at > around 7:15 p.m. Friday, the university announced. > Chern, a US citizen, is best known for his achievements in the > study of differential geometry. He was born in 1911 in Jiaxing, Zhejiang > Province, east China. He graduated from Nankai University in 1930 and > received further education at Qinghua University and the University of > Hamburg in Germany. > He taught at several Chinese and US universities -- including > Princeton University, the University of Chicago, and the University of > California, Berkeley -- and is the only Chinese to win the Wolf Prize -- > the most distinguished award in the international mathematics field. > The International Astronomical Union officially named asteroid No. > 1998CS2 after the noted mathematician in November for his outstanding > contributions to human society. === Subject: Number of prime factors of an odd perfect number Content-Length: 215 Originator: rusin@vesuvius What is the known upper bound on the number of prime factors / distinct prime factors of an odd perfect number? Does anyone have a conjecture on what the tighest lower bound would be? Edmond === Subject: Re: Number of prime factors of an odd perfect number > What is the known upper bound on the number of prime factors / distinct > prime factors of an odd perfect number? Does anyone have a conjecture on > what the tighest lower bound would be? > Edmond You've already received some excellent replies, but it might be of interest to note that any constant upper bound here would imply there are only finitely many odd perfect numbers: Dickson showed that there are at most finitely many odd perfect numbers with a prescribed number of distinct prime factors. Dickson's result can be made effective. A striking result of Heath-Brown is that an odd perfect number with k distinct prime factors is bounded by 4^{4^k}. (The proof is entirely elementary.) This has been improved to 2^{4^k} by Nielsen in [1]. Hope this helps, Paul [1] Nielsen, Pace An Upper Bound for Odd Perfect Numbers http://www.emis.de/journals/INTEGERS/papers/d14/d14.pdf === Subject: Re: Number of prime factors of an odd perfect number > What is the known upper bound on the number of prime factors / distinct > prime factors of an odd perfect number? Does anyone have a conjecture on > what the tighest lower bound would be? I don't know much about recent advances, but there are a few classical results: J. J. Sylvester, L. E. Dickson and H. J. Kanold proved that there is no odd perfect number with 4 prime divisors. Next, I. S. Gradshtein, U. K.9fhnel and G. C. Weber disproved existence of numbers with 4 divisors and C. Pomerance and N. Robbins showed the same for 6 divisors. P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 divisors and I don't know about any newer results. Pawel Gladki === Subject: Re: Number of prime factors of an odd perfect number > P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous > theorem for 8 divisors and I don't know about any newer results. Of course we are talking about _distinct_ prime factors. For the bound of prime factors counted with repetitions it is known that there are at least 47 factors - see Kevin Hare's preprint: http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf Pawel Gladki === Subject: Re: Number of prime factors of an odd perfect number >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 >> divisors and I don't know about any newer results. > Of course we are talking about _distinct_ prime factors. For the bound of > prime factors counted with repetitions it is known that there are at least > 47 factors - see Kevin Hare's preprint: > http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf interested in known UPPER BOUNDS and a CONJECTURE on what would be the ACTUAL LOWER BOUND. Edmond === Subject: Re: Number of prime factors of an odd perfect number >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for 8 >> divisors and I don't know about any newer results. > Of course we are talking about _distinct_ prime factors. For the bound of > prime factors counted with repetitions it is known that there are at least > 47 factors - see Kevin Hare's preprint: > http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf > interested in known UPPER BOUNDS and a CONJECTURE on what would be the > ACTUAL LOWER BOUND. Known upper bounds? You mean a proof that if there is an odd perfect number it can't have more than so many prime factors? I don't think I've ever seen a result along those lines - have you? As for conjectured lower bounds, I think those would be infinite, as I think the conjecture is that there aren't any odd perfect numbers. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Number of prime factors of an odd perfect number > P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for > 8 > divisors and I don't know about any newer results. > Of course we are talking about _distinct_ prime factors. For the bound >> of >> prime factors counted with repetitions it is known that there are at >> least >> 47 factors - see Kevin Hare's preprint: > http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf >> interested in known UPPER BOUNDS and a CONJECTURE on what would be the >> ACTUAL LOWER BOUND. > Known upper bounds? You mean a proof that if there is an odd perfect > number it can't have more than so many prime factors? Exactly > I don't think > I've ever seen a result along those lines - have you? No, I haven't, but I'd like to know at least a reasonably tight upper bound. > As for conjectured lower bounds, I think those would be infinite, > as I think the conjecture is that there aren't any odd perfect numbers. Let us assume that someone does prove this conjecture. What would be the important corollaries thereof ? Edmond === Subject: Re: Number of prime factors of an odd perfect number days. My association with the Department is that of an alumnus. >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem for >> 8 >> divisors and I don't know about any newer results. > Of course we are talking about _distinct_ prime factors. For the bound > of > prime factors counted with repetitions it is known that there are at > least > 47 factors - see Kevin Hare's preprint: > http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf > interested in known UPPER BOUNDS and a CONJECTURE on what would be the > ACTUAL LOWER BOUND. >> Known upper bounds? You mean a proof that if there is an odd perfect >> number it can't have more than so many prime factors? > Exactly >> I don't think >> I've ever seen a result along those lines - have you? > No, I haven't, but I'd like to know at least a reasonably tight upper >bound. Myerson's point is that, as far as he is aware (and as far as I am aware), there are ->no<- results on upper bounds on the number of prime factors for an odd perfect number. I'm sure you would like to know one. But if there are none to be found, you'll have to prove one if you want to know one. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Number of prime factors of an odd perfect number >> P. Hagis Jr. and E. Z. Chein proved in 1975 the analogous theorem > for > 8 > divisors and I don't know about any newer results. > Of course we are talking about _distinct_ prime factors. For the >> bound >> of >> prime factors counted with repetitions it is known that there are at >> least >> 47 factors - see Kevin Hare's preprint: > http://www.math.uwaterloo.ca/~kghare/Preprints/PDF/P15_Odd.pdf > interested in known UPPER BOUNDS and a CONJECTURE on what would be the >> ACTUAL LOWER BOUND. > Known upper bounds? You mean a proof that if there is an odd perfect > number it can't have more than so many prime factors? >> Exactly > I don't think > I've ever seen a result along those lines - have you? >> No, I haven't, but I'd like to know at least a reasonably tight upper >>bound. > Myerson's point is that, as far as he is aware (and as far as I am > aware), there are ->no<- results on upper bounds on the number of > prime factors for an odd perfect number. > I'm sure you would like to know one. But if there are none to be > found, you'll have to prove one if you want to know one. I guess I have to set out to prove one myself. But I'm still curious as to how ( or maybe why ) so much work has been done on the lower bound, whereas we have no results regarding the upper bound. Also, is there a good example of a similar upper bound proof, i.e., an upper bound on some property of a hypothetical number ? Edmond === Subject: Re: Number of prime factors of an odd perfect number days. My association with the Department is that of an alumnus. >> Myerson's point is that, as far as he is aware (and as far as I am >> aware), there are ->no<- results on upper bounds on the number of >> prime factors for an odd perfect number. >> I'm sure you would like to know one. But if there are none to be >> found, you'll have to prove one if you want to know one. >I guess I have to set out to prove one myself. But I'm still curious as to >how ( or maybe why ) so much work has been done on the lower bound, whereas >we have no results regarding the upper bound. Because... how would you prove an upper bound on the number of distinct prime factors? The lower bounds are proven by using the fact that if m and n are relatively prime, then sigma(m*n) = sigma(m)*sigma(n), with sigma(k) = sum of all divisors of k, and showing that if q1,...,qs are distinct prime powers (for which it is easy to calculate sigma), then the product of the sigma(qi) simply does not add up to twice q1*...*qs, for small values of s. But how would you prove an upper bound? How would you even ->approach<- such a problem? The upper bound in the even case comes by showing that if you write the number as 2^n*q, with q odd, then the only way this works out is if q is a prime (and, more specifically, a prime of the form 2^{n}-1). It is not really a generalizable argument. > Also, is there a good example >of a similar upper bound proof, i.e., an upper bound on some property of a >hypothetical number ? Yes. Ramsey numbers have well known upper and lower bounds, but many of them are not exactly known. But here, you prove existence by proving an upper bound. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Number of prime factors of an odd perfect number > Myerson's point is that, as far as he is aware (and as far as I am > aware), there are ->no<- results on upper bounds on the number of > prime factors for an odd perfect number. > I'm sure you would like to know one. But if there are none to be > found, you'll have to prove one if you want to know one. >>I guess I have to set out to prove one myself. But I'm still curious as to >>how ( or maybe why ) so much work has been done on the lower bound, >>whereas >>we have no results regarding the upper bound. > Because... how would you prove an upper bound on the number of > distinct prime factors? The lower bounds are proven by using the fact > that if m and n are relatively prime, then sigma(m*n) = > sigma(m)*sigma(n), with sigma(k) = sum of all divisors of k, and > showing that if q1,...,qs are distinct prime powers (for which it is > easy to calculate sigma), then the product of the sigma(qi) simply > does not add up to twice q1*...*qs, for small values of s. > But how would you prove an upper bound? How would you even > ->approach<- such a problem? The upper bound in the even case comes > by showing that if you write the number as 2^n*q, with q odd, then the > only way this works out is if q is a prime (and, more specifically, a > prime of the form 2^{n}-1). It is not really a generalizable argument. participants. Edmond === Subject: Re: Smullyan's Quiz Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB3LRdl09441; >My own analysis is that the error is the insistance that the >students must have a rational basis for believing >that there is or is not going to be an exam. >In essence the professor is saying to the students: >you have a rational basis for believing there >will be an exam if and only if you do not have a rational >basis for believing there will be an exam. If the >students believe the professor, they must conclude >that they do not have a rational reason to either >to believe that there will be an exam or to believe that >there will not be an exam. The best resolution is that >the students should not believe the professor. >In real life, of course, the professor can only >say There is a very high probability that you >will have an unexpected exam next week. This statement >does not lead to a paradox, so the impression that >a professor can make this kind of statement and >communicate information is justified. >In other forms of the paradox, where there >is no prediction element and hence no probablity, >the statement (e.g. there is an unexpected egg) >cannot be believably made. perhaps one should think seriously what unexpected means. to do this one could try to formalise expectations. but this is readily done in elementary game theory, and from there one also gets the wiseness of formalising the situation as a game. with reasonable pay-off functions it is clear that there are no pure strategy equilibria but the professor uses a mixed strategy to determine the date of the exam. the students' expectations about the mixed strategy are, of course, correct in equilibrium but they do not know the realisation of the mixed strategy, i.e. the true date of the exam. in most cases then the exam will be a surprise exam in the sense that the studentd cannot predict the date with probability one. only if the realisation is friday they will be able to predict on thursday evening that the exam will be on friday with probability one. the point here is that this is not so much a problem of logic as a problem of formalising the situation correctly. and when one does! formalise it it turns out to be a game. === Subject: Re: Smullyan's Quiz Problem >My own analysis is that the error is the insistance that the >students must have a rational basis for believing >that there is or is not going to be an exam. >In essence the professor is saying to the students: >you have a rational basis for believing there >will be an exam if and only if you do not have a rational >basis for believing there will be an exam. If the >students believe the professor, they must conclude >that they do not have a rational reason to either >to believe that there will be an exam or to believe that >there will not be an exam. The best resolution is that >the students should not believe the professor. >In real life, of course, the professor can only >say There is a very high probability that you >will have an unexpected exam next week. This statement >does not lead to a paradox, so the impression that >a professor can make this kind of statement and >communicate information is justified. >In other forms of the paradox, where there >is no prediction element and hence no probablity, >the statement (e.g. there is an unexpected egg) >cannot be believably made. > perhaps one should think seriously what unexpected means. > to do this one could try to formalise expectations. > but this is readily done in elementary game theory, and from there > one also gets the wiseness of formalising the situation as a game. > with reasonable pay-off functions it is clear that there are no pure > strategy equilibria but the professor uses a mixed strategy to > determine the date of the exam. the students' expectations about the > mixed strategy are, of course, correct in equilibrium but they do not > know the realisation of the mixed strategy, i.e. the true date of the > exam. in most cases then the exam will be a surprise exam in the sense > that the studentd cannot predict the date with probability one. > only if the realisation is friday they will be able to predict on > thursday evening that the exam will be on friday with probability > one. the point here is that this is not so much a problem of logic > as a problem of formalising the situation correctly. and when > one does formalise it it turns out to be a game. Ok, but this does not change the basic nature of the paradox. Consider the following form of the paradox (I prefer this as it avoids discussion of prediction and states of knowledge, gives a more concrete meaning to trustworthy, and incorporates the idea of mixed strategy). We have two players A and B, and two cards, a king and an ace. A orders the cards and places them face down. The cards are turned over until the king is turned over. At this point the game ends. Before each card is turned over, B names an integer between 0 and 100. This is used as the % probability that B will place a $1 bet at even odds that the next card turned over will be the king. B's best strategy is to say 50 before the first card is turned over and 100 if the second card is to be turned over. No matter what strategy A uses, this ensures B an average return of $0.5 (So the game is worth $0.5 to B). A is trustworthy. That is if A says the first card is the king B will say 100 before the first card is turned over. A's problematic statement is (after he has placed the cards face down) If you act rationally, you will not say 100 during this play of the game B argues If the first card is the ace, then I will certainly say 100 before the second card is turned over, so the first card must be the king, so rationally I should say 100 before the first card is turned over. But then A knows his statement is always false, so I should not draw any inference from it and say 50 before the first card is turned over. But then A know that his statement is only true if the first card is the king, so I should treat his statement the same as if he had said the first card is the king and say 100 before the first card is turned over. But then A knows his statement is always false ... Surely it is not rational for B to refuse to play a game that has positive expectation. The unappealing, but seemingly inescapable conclusion is that B cannot behave rationally! -William Hughes === Subject: How Many M&Ms? Hello All. I am writing in reference to a contest a friend of mine is having. The rules are as follows. I bought a brand new clear plastic jar. It is filled with the standard size M & M candies. The jar is round and 8 inches tall. The diameter of the jar inside is 4 inches across. I bought over 4 pounds of M & Ms to fill it. I counted every M & M before filling the jar to the very top.. You can probably guess what I am seeking: How many M & Ms are in the jar as described? If anyone can help me on this it would be wonderful as I am not a good mathmatician. Joey === Subject: Re: How Many M&Ms? > Hello All. I am writing in reference to a contest a friend of mine is > having. The rules are as follows. > I bought a brand new clear plastic jar. It is filled > with the standard size M & M candies. > The jar is round and 8 inches tall. The diameter of > the jar inside is 4 inches across. > I bought over 4 pounds of M & Ms to fill it. I counted > every M & M before filling the jar to the very top.. > You can probably guess what I am seeking: > How many M & Ms are in the jar as described? > If anyone can help me on this it would be wonderful as I am not a good > mathmatician. > Joey The only precise method is to count them. Probably the next most accurate method is to esimate the number by weight. Count out as large a fraction of the total as you have patience for and weigh them to find an average weight per piece. Then weigh the total (less jar weight) and divide by the average weight per piece. Or maybe the manfacturer will give you the average weight per piece. === Subject: Re: How Many M&Ms? I don't get this post.... Joey counted the M&Ms but wants help knowing how many? oh she counted the 4 pounds of M&Ms then threw in a portion of them into the jar. If you're running the competition and people have to guess, they might expect you to count the exact number in the jar. Of course you could just take the median of the guesses and select them as winner, quoting the prize amount as a few different from what they guessed, but if the jar is the prize they might check, and be honest and hand it over to the actual best guesser, this could result in all sorts of bad publicity, you might find your next M&M competition you'll be eating them yourself. A few questions, the jar you bought was empty right? You and your friend are both running the competition? You counted the M&Ms of all the bags you bought? Can you count the remaining M&Ms after you put them in the jar and substract? Is this like that puzzle, if there are 5 apples and I take away 2 how many do I have? Are we eligible for the prize to calculate our guesses? 2601. Herc === Subject: Re: How Many M&Ms? Fortunately, there has been some research on M&M packing: The net is that the density of packing is around 71% (if packed in an irregular fashion). Further, it looks like the someone has actually measured an individual M&M: http://www.kleinbottle.com/Bernie_Tao.htm Which a result of 0.45239 cm. so, volume of jar is: PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 cm3 So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 Now the result is probably only good to around 2 significant figures, and there will probably be some rounding down due to the sides of the container, so I would say aound 2500 M&Ms ? Am I close? -Darren > Hello All. I am writing in reference to a contest a friend of mine is > having. The rules are as follows. > I bought a brand new clear plastic jar. It is filled > with the standard size M & M candies. > The jar is round and 8 inches tall. The diameter of > the jar inside is 4 inches across. > I bought over 4 pounds of M & Ms to fill it. I counted > every M & M before filling the jar to the very top.. > You can probably guess what I am seeking: > How many M & Ms are in the jar as described? > If anyone can help me on this it would be wonderful as I am not a good > mathmatician. > Joey === Subject: Re: How Many M&Ms? In sci.math, darrenn : > Fortunately, there has been some research on M&M packing: > The net is that the density of packing is around 71% (if packed in an > irregular fashion). > Further, it looks like the someone has actually measured an individual M&M: > http://www.kleinbottle.com/Bernie_Tao.htm > Which a result of 0.45239 cm. Pedant point. An M&M is generally a flattened round thing and probably should be measured in all three dimensions, or perhaps one can pour a number into a liquid (although water will melt off the coatings, methinks) and measure the displacement. > so, volume of jar is: > PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 > cm3 > So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 > Now the result is probably only good to around 2 significant figures, and > there will probably be some rounding down due to the sides of the container, > so I would say aound 2500 M&Ms ? > Am I close? > -Darren >> Hello All. I am writing in reference to a contest a friend of mine is >> having. The rules are as follows. >> I bought a brand new clear plastic jar. It is filled >> with the standard size M & M candies. >> The jar is round and 8 inches tall. The diameter of >> the jar inside is 4 inches across. >> I bought over 4 pounds of M & Ms to fill it. I counted >> every M & M before filling the jar to the very top.. >> You can probably guess what I am seeking: >> How many M & Ms are in the jar as described? >> If anyone can help me on this it would be wonderful as I am not a good >> mathmatician. >> Joey -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: How Many M&Ms? > In sci.math, darrenn > Fortunately, there has been some research on M&M packing: > The net is that the density of packing is around 71% (if packed in an > irregular fashion). > Further, it looks like the someone has actually measured an individual M&M: > http://www.kleinbottle.com/Bernie_Tao.htm > Which a result of 0.45239 cm. > Pedant point. An M&M is generally a flattened round thing and > probably should be measured in all three dimensions, or perhaps > one can pour a number into a liquid (although water will melt > off the coatings, methinks) and measure the displacement. > so, volume of jar is: > PI X 2 X 2 X 8 = 100.54 inches^3 = 100.54 X 2.54 X 2.54 X 2.54 = 1647.407 > cm3 > So, number of M&Ms is approx. = 1647.407 X 0.71 / 0.45239 ~ 2585 > Now the result is probably only good to around 2 significant figures, and > there will probably be some rounding down due to the sides of the container, > so I would say aound 2500 M&Ms ? > Am I close? > -Darren >> Hello All. I am writing in reference to a contest a friend of mine is >> having. The rules are as follows. > I bought a brand new clear plastic jar. It is filled >> with the standard size M & M candies. > The jar is round and 8 inches tall. The diameter of >> the jar inside is 4 inches across. > I bought over 4 pounds of M & Ms to fill it. I counted >> every M & M before filling the jar to the very top.. > You can probably guess what I am seeking: > How many M & Ms are in the jar as described? >> If anyone can help me on this it would be wonderful as I am not a good >> mathmatician. > Joey what does it matter if the water removes the coating? the coating will still contribute to the displacement. === Subject: Re: How Many M&Ms? In sci.math, David Bandel [snip for brevity -- discussing # of M&M's in a certain sized jar, and using water for one measurement of volume] > what does it matter if the water removes the coating? the coating will > still contribute to the displacement. There's a few issues regarding solubility that may complicate the analysis. I'd frankly have to look. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: How Many M&Ms? >[snip for brevity -- discussing # of M&M's in a certain sized jar, > and using water for one measurement of volume] >> what does it matter if the water removes the coating? the coating will >> still contribute to the displacement. >There's a few issues regarding solubility that may complicate >the analysis. I'd frankly have to look. M&M's require solving a polynomial of degree greater than 5 ? === Subject: Re: How Many M&Ms? > Which a result of 0.45239 cm. > Pedant point. An M&M is generally a flattened round thing and > probably should be measured in all three dimensions, or perhaps > one can pour a number into a liquid (although water will melt > off the coatings, methinks) and measure the displacement. True. This should say 0.45239 cm^3. -Darren === Subject: JSH: James, you be the judge... Here is how James Harris operates: 1. Start with some goofy polynomial that was a leftover from one of his many failed FLT proofs. No explanation of motivation or reasoning provided for choosing that particular polynomial or explanation of why it might be meaningful in his new context. 2. Avoid being specific about such important details as which ring he intends to work in. 3. Dream up a bunch of weird non-standard terminology like properly unit, dividing off, etc. Use terms incorrectly, such as referring to multiple when he should use the term factor. Also, make use of standard terminology like distributive property and constant term without actually understanding it or using it properly. Also, add lots of vague distractions such as the equation has no memory, you can see the 7's in there, can't you? to act as a smoke screen. 4. Apply a bunch of algebraic manipulations, some of which are just wrong. Make inappropriate generalizations from specific cases to general cases, etc. 5. Get a result that seems contradictory, and assume that the error was with the core of algebra rather than the much more likely explanation that he himself might have made an error. 6. Through out a bunch of paranoid ad hominem attacks on critics. Then hypocritically claim that all his critics resort to social crap and personal attacks rather than sticking to mathematics, logic, and reasoning. We are supposed to believe that his arbitrarily chosen goofy polynomial just happens to one that, when probed by the genius of James Harris, give results that shake mathematics to its core! James, you be the judge. Are you a pile of crap? === Subject: Pi in space I was having an argument with a friend here and he claimed that pi doesn't really have all this significance it's attached to it, since it is transcendental only in the theoretical space of Euclidian geometry and all know that this geometry does not represent true space-time. Is pi transcendental in ANY space except in the theoretical space of Euclidian geometry? For example, is it transcendental in Riemannian geometry? Lobatchevskian geometry? or any other geometry? What about in the relativistic universe? Is the ratio of the circumference of a circle to its diameter inside the Einsteinian universe rational or irrational? Anyone knows? -- I. N. G. --- http://users.forthnet.gr/ath/jgal/ === Subject: Re: Pi in space > I was having an argument with a friend here and he claimed that pi > doesn't really have all this significance it's attached to it, since it > is transcendental only in the theoretical space of Euclidian geometry > and all know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the > circumference of a circle to its diameter inside the Einsteinian > universe rational or irrational? Anyone knows? We don't know the true geometry of space. In particular, we don't know if it's continuous. Digital physicists think it is not: see www.digitalphysics.org The concept of Pi, however, should not be regarded transcendental or anything like that since it is represented perfectly with a small program. True, it is an idealization, but what is transcendental? In my opinion, nothing transcends physical reality, and if Pi is part of physical models, it is for a good reason: physics is mostly about ideal models that work in ideal conditions, etc. Note however, that, if the universe turns out to be discrete, then, obviously, the real number Pi does not describe any physical reality. The computation of Pi is still sensible, however, (e.g. computation of Pi up to the nth decimal place) and it would indeed describe physical reality in a discrete universe (if it's an actually infinite discrete universe, things get more complicated so just assume finite which seems to be the case for *our* universe;). Most mathematicians on this forum seem to be Platonists, or Platonists who lack the philosophical maturity to identify themselves as Platonists, so you could expect quite a lot of Platonist nonsense in response to your question. However, as one poster correctly pointed out, constructivists and formalists will most likely look at Pi as a useful mental construct of some sort, and no talk of transcending anything will be necessary. Pi is not in space: it's in your mind. [*] Indeed, constructivism is a more favorable position than Platonic Realism, in my opinion, and a point of view supported by great mathematicians, so it's an alternative you should think about. More recently, instrumentalism has been suggested as an alternative way to look at mathematics. In my opinion, that too is considerable. -- Eray Ozkural There is no perfect circle [*] A caveat, however, there are perfect discrete circles in the world. It just may be the case that there are no perfect continuous circles, except conceptually!!!!!! === Subject: Re: Pi in space |Most mathematicians on this forum seem to be Platonists, or Platonists |who lack the philosophical maturity to identify themselves as |Platonists, so you could expect quite a lot of Platonist nonsense in |response to your question. No, because such points of philosophy are thoroughly irrelevant to his question. |However, as one poster correctly pointed |out, constructivists and formalists will most likely look at Pi as a |useful mental construct of some sort, and no talk of transcending |anything will be necessary. Pi is not in space: it's in your mind. [*] I am disinclined to assume that you know that transcendental is a technical mathematical term meaning not a root of a nonzero polynomial with integer coefficients. |Indeed, constructivism is a more favorable position than Platonic |Realism, in my opinion, and a point of view supported by great |mathematicians, so it's an alternative you should think about. Kronecker, Brouwer, and Bishop were all outstanding mathematicians (whether we call them great sort of depends on how high a standard we set for greatness). Bishop supported constructivism; Brouwer supported some kind of constructivism; Kronecker appears to have supported something along those lines. I wouldn't say, however, that popularity among great mathematicians is a very good standard for judging such things. To the extent that it can be trusted at all, I see no obvious sign of the very best mathematicians differing very far in terms of the distribution of their opinions on Platonism (or realism), formalism, constructivism and so on from the run-of-the-mill. I also don't see sci.math as being much of a hotbed of Platonism or realism. Keith Ramsay === Subject: Re: Pi in space >The concept of Pi, however, should not be regarded transcendental or >anything like that since it is represented perfectly with a small >program. True, it is an idealization, but what is transcendental? In >my opinion, nothing transcends physical reality, and if Pi is part of >physical models, it is for a good reason: physics is mostly about >ideal models that work in ideal conditions, etc. I think you need to look up the definition of transcendental number. You seem to be confusing it with some philosophical notion. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Pi in space >The concept of Pi, however, should not be regarded transcendental or >anything like that since it is represented perfectly with a small >program. True, it is an idealization, but what is transcendental? In >my opinion, nothing transcends physical reality, and if Pi is part of >physical models, it is for a good reason: physics is mostly about >ideal models that work in ideal conditions, etc. > I think you need to look up the definition of transcendental number. > You seem to be confusing it with some philosophical notion. the concept of a transcendental number. AFAICT, the notion of a transcendental number is not relative to a give geometry, or the true geometry of our universe. Pi is a transcendental number, according to the definition of a transcendental number in mathematics. I do think his question ought to be philosophical. Why should he talk about relativity then? What difference is there between any continuous metric and a Riemann tensor regarding the fact that Pi is a transcendental number? Maybe, I think, his friend's intention was to give a better definition of what it means for a number to be transcendental than the ordinary usage. He might want to pick another term, though, or highlight the difference than the ordinary usage carefully. Otherwise, we become confused in argumentation. I thought so, and said that this philosophical sense is probably irrelevant to a computable real like Pi which captures a general geometric fact in a compact form! -- Eray Ozkural === Subject: Re: Pi in space > I was having an argument with a friend here and he claimed that pi > doesn't really have all this significance it's attached to it, since it > is transcendental only in the theoretical space of Euclidian geometry > and all know that this geometry does not represent true space-time. We know nothing whether space-time really exists and if it exists whether it has a given geometry of its geometry can be a matter of convention. You are asking metaphysical (ontological) questions that have no answers at this point. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? The geometries have relative consistency to Euclidean geometry. This, pi must be transcendental in all of them to maintain relative consistency. > What about in the relativistic universe? Is the ratio of the > circumference of a circle to its diameter inside the Einsteinian > universe rational or irrational? Anyone knows? You can draw a circle even on your pixelized TFT screen, consider it a perfect one and use pi to find its perimeter. I get the impression you are asking about the relation of abstract mathematical object to the actual universe. We do not know it. That's another metaphysical subject. Platonists would argue circles are really part of the ontology of space-time and Universe whereas Formalists and Constructivists will look at circles as useful geometrical models with no real existence apart from human mind. Mike === Subject: Re: Pi in space >I was having an argument with a friend here and he claimed that pi doesn't >really have all this significance it's attached to it, since it is >transcendental only in the theoretical space of Euclidian geometry and all >know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the circumference > of a circle to its diameter inside the Einsteinian universe rational or > irrational? Anyone knows? > -- > I. N. G. --- http://users.forthnet.gr/ath/jgal/ Defining pi to be the empirically measured ratio of a circle's circumference to its diameter, the only geometry in which this is ratio is a constant is Euclidean geometry. Pi doesn't exist as a constant that can be empirically measured in any other geometry, so its value in other geometries is a meaningless question. Of course, defining pi as the sum of a series, or the constant in a solution to a differential equation or proper integral, or any of the other billion ways of defining pi has absolutely nothing to do with the geometry of the Universe. So there is only one value of pi, and yes, it is transcendental. === Subject: Re: Pi in space |Defining pi to be the empirically measured ratio of a circle's circumference |to its diameter, the only geometry in which this is ratio is a constant is |Euclidean geometry. Pi doesn't exist as a constant that can be empirically |measured in any other geometry, so its value in other geometries is a |meaningless question. Obviously this is not a practical issue, but I thought as long as we were busy making points about what is true in principle, I would note that there often is a natural relationship between pi and other geometries. In some other geometries, pi is the limit of the ratio between the circumference and the diameter of a circle as the radius goes to 0, and could in principle (if you had such a space in your hands...) be measured that way. At a fixed radius, measuring the circumference and diameter would only be able to approximate pi up to a certain degree of precision, but measurements of lengths are always only up to a certain degree of precision anyway, so one is not any worse of in an essential way. On the hyperbolic plane, for example, you could find a length scale on which a triangle whose sides have lengths in a 3:4:5 ratio has an angle of between 90 and 91 degrees. One can then put an upper bound on the deviation of the ratio of the circumference to the diameter from pi, in terms of the ratio of the radius to the length of a side of this triangle. The circumference of a circle of radius r in the hyperbolic plane is 2*pi*a*sinh(r/a) where the length a depends on the curvature. The size of the 3:4:5 triangle with an angle of 91 degrees is likewise proportional to this length a. (I won't bother figuring out what the ratio is right here.) So to get n digits, we could make a circle whose radius is roughly 10^{-n/3} times the length of a side of this triangle, and measure it very precisely. Keith Ramsay === Subject: Re: Pi in space >I was having an argument with a friend here and he claimed that pi doesn't >really have all this significance it's attached to it, since it is >transcendental only in the theoretical space of Euclidian geometry and all >know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of > Euclidian geometry? > For example, is it transcendental in Riemannian geometry? Lobatchevskian > geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the circumference > of a circle to its diameter inside the Einsteinian universe rational or > irrational? Anyone knows? > -- > I. N. G. --- http://users.forthnet.gr/ath/jgal/ > Defining pi to be the empirically measured ratio of a circle's circumference > to its diameter, the only geometry in which this is ratio is a constant is > Euclidean geometry. It is not possible to empirically measure any irrational in any geometry, if by that you mean physical measurement. And if not, what do you mean? === Subject: Re: Pi in space >>I was having an argument with a friend here and he claimed that pi >>doesn't >>really have all this significance it's attached to it, since it is >>transcendental only in the theoretical space of Euclidian geometry and >>all >>know that this geometry does not represent true space-time. > Is pi transcendental in ANY space except in the theoretical space of >> Euclidian geometry? > For example, is it transcendental in Riemannian geometry? >> Lobatchevskian >> geometry? or any other geometry? > What about in the relativistic universe? Is the ratio of the >> circumference >> of a circle to its diameter inside the Einsteinian universe rational or >> irrational? Anyone knows? > -- >> I. N. G. --- http://users.forthnet.gr/ath/jgal/ >>